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Going through airport security can be stressful—especially if you’re not confident speaking English. From placing your bag on the scanner to answering questions from officers, knowing the right phrases makes the process smoother. In this article, you’ll explore a practical airport security conversation in English, learning how to speak confidently through realistic dialogues, useful questions and answers, and essential travel vocabulary. You’ll also discover common expressions, phrasal verbs, and key phrases used by native speakers, so you can communicate clearly and avoid confusion during your next trip.
Table of Contents
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At Airport Security – English Conversation Practice
Dialogue 1: Speaking to a Security Agent
Vocabulary & Phrases from Dialogue 1
Dialogue 2: Security Check Questions & Answers
Vocabulary & Phrases from Dialogue 2
Quiz: English at Airport Security
Quiz Answers
At Airport Security – English Conversation Practice
Scene: After checking in, Emma walks toward the security screening area. She joins the line and is approached by a friendly security officer at the front of the checkpoint.
Dialogue 1: Speaking to a Security Agent
Security Agent: Good afternoon, ma’am. Please have your boarding pass and ID ready for inspection.
Emma: Sure, here you go.
Security Agent: Thank you. Are you carrying any liquids, gels, or aerosols in your carry-on?
Emma: Just a small bottle of hand sanitizer and a travel-size shampoo, both under 100 milliliters.
Security Agent: That’s fine. Just make sure they’re in a clear, resealable plastic bag and place them in the tray when you go through screening.
Emma: Got it. I packed all my toiletries in a zip-lock bag just to be safe.
Security Agent: Great! Now, before you go through the scanner, please remove your shoes, belt, and any electronics larger than a cellphone—that includes tablets, laptops, or cameras.
Emma: Okay. Should I also take off my jacket?
Security Agent: Yes, jackets and coats need to come off as well. Place everything in the plastic bins. Please also empty your pockets—coins, keys, phones, anything metallic.
Emma: Will do. (starts organizing her things) I always forget something when I go through security. I try to travel smart, but I still get flustered sometimes!
Security Agent: Happens all the time! The key is to stay organized and be ready ahead of time. It really helps the line move quicker.
Emma: I agree. (removes laptop) Is it okay if I leave my charger in the bag?
Security Agent: Chargers are fine, but if you’ve got any power banks or external batteries, please take them out and place them separately.
Emma: Alright. Done! I think I’m all set.
Security Agent: Perfect. Please step forward and walk through the scanner.
(Emma walks through the full-body scanner. It beeps.)
Security Agent: Hmm, sorry—it looks like something set it off. I’ll need you to step aside for a quick secondary screening.
Emma: Oh! Sure, no problem. Did I forget to take something off?
Security Agent: Could be a necklace or a metal clip in your clothing. Do you mind if we do a quick pat-down?
Emma: That’s okay. Go ahead.
Security Agent: Thank you. Arms out to the side, please. (completes pat-down) All clear! It was just a belt loop. You’re good to go.
Emma: (laughs) That belt always gives me trouble. Last time, I set off the alarm because of the zipper on my boots!
Security Agent: You’d be surprised what can trigger the scanner. Better safe than sorry!
Emma: True. Do I need to scan my boarding pass again?
Security Agent: Nope, you’re all set. Just collect your belongings from the conveyor belt, and don’t forget to double-check the bins before leaving—phones and passports often get left behind.
Emma: Thank you for the reminder. (gathers her items) Everything’s here… laptop, wallet, boarding pass. I’m good to go.
Security Agent: Great. The restrooms and water fountains are to your right, and your gate is straight ahead. Have a smooth journey!
Emma: Thanks a lot. You’ve been really helpful. Have a great day!
Security Agent: You too—safe travels!
Vocabulary & Phrases from Dialogue 1
| Phrase / Word | Meaning |
--- |
| Liquids, gels, or aerosols | Common restricted items in carry-on bags |
| Clear, resealable plastic bag | Transparent bag for liquids when passing through security |
| Remove your shoes/belt/jacket | Common security rule for screening |
| Empty your pockets | Take out all items from your pockets |
| Travel smart | Travel in a prepared and organized way |
| Flustered | Confused or nervous in a stressful situation |
| Power bank / external battery | Portable charger for phones or devices |
| Step aside | Move out of the main line for further inspection |
| Secondary screening | Extra check after you go through the main scanner |
| Pat-down | Light physical check done by a security agent |
| Set off the alarm | Trigger the security alarm |
| Better safe than sorry | Idiom meaning it’s better to be cautious |
| Collect your belongings | Pick up your items after security |
| Conveyor belt | Moving belt that carries trays and luggage |
| Double-check the bins | Look again carefully to make sure you haven’t forgotten anything |
| Smooth journey / safe travels | Polite way to wish someone a good trip |
Dialogue 2: Security Check Questions & Answers
TSA Agent: Good morning! Please have your ID and boarding pass ready.
Emma: Here you go! I’ve just printed it out. Do I need to scan it here?
TSA Agent: Yes, just place it on the scanner. Thanks. Have you removed your prohibited items already?
Emma: I think so—I followed the list on your website. No sharp objects or liquids over 100ml. I even downsized my shampoo!
TSA Agent: Nicely done. And any electronic devices in your bag?
Emma: Just a laptop and a tablet. Should I take them out now?
TSA Agent: Yes, anything larger than a phone needs to go in a separate tray. And don’t forget to take off your watch, belt, and any metal jewelry.
Emma: Will do. (starts unpacking items) I always forget my belt—that thing sets off the alarm every single time.
TSA Agent: (laughs) Happens to the best of us. You’d be surprised what trips the sensors—sometimes it’s just a zipper or a button.
Emma: This airport’s much smoother than the one I flew through last month. That place was packed to the gills and totally disorganized.
TSA Agent: We try to keep things moving here. Just load your bins and step forward once you’re ready.
Emma: Done! I’ve got my shoes off, jacket off, laptop out, and pockets emptied.
TSA Agent: You’re ahead of the game. Please walk through the body scanner when the light turns green.
(Emma walks through. No alarm sounds.)
TSA Agent: All clear! You’re good. Go ahead and retrieve your items from the belt.
Emma: Awesome. Do I need to show anything again before heading to my gate?
TSA Agent: Nope, just repack your bag and you’re good to go. Make sure to double-check the tray—passengers often forget passports and phones.
Emma: Thanks for the tip. I once left my charger behind and didn’t realize until I boarded!
TSA Agent: That’s rough. Oh—and before I forget, liquids must stay accessible for any random checks at the gate. Keep them near the top of your bag.
Emma: Will do. I’m hoping for a stress-free journey this time.
TSA Agent: You’re off to a great start. Grab some water or a snack if you’ve got time. Your gate is a bit of a walk.
Emma: I’ll probably grab a coffee to go. Still got a few hours before boarding.
TSA Agent: Great. Safe travels, and enjoy your flight!
Emma: Thank you! You’ve been super helpful. See you next time!
Vocabulary & Phrases from Dialogue 2
| Phrase / Expression | Meaning |
--- |
| Prohibited items | Things not allowed in carry-on bags |
| Downsize your liquids | Reduce the size or amount of liquid containers |
| Electronic devices | Laptops, tablets, e-readers, etc. |
| Take them out | Remove items from your bag |
| Set off the alarm | Trigger the security scanner |
| Packed to the gills | Extremely crowded (idiom) |
| Load your bins | Place your items into the security trays |
| Retrieve your items | Pick up your things after scanning |
| Repack your bag | Put items back into your carry-on after screening |
| Double-check the tray | Look again carefully so nothing is left behind |
| Liquids must stay accessible | Keep your liquids in a place that’s easy to reach |
| Stress-free journey | A relaxed and smooth travel experience |
| Grab a coffee to go | Get a takeaway coffee |
| Trip the sensors | Accidentally cause the machine to beep or react |
| Ahead of the game | Well-prepared or organized (idiom) |
Quiz: English at Airport Security
Please ____ your shoes and place them in the tray.
a) take off
b) turn on
c) put down
I always ____ my bag before going through security to make sure I didn’t forget anything.
a) refill
b) double-check
c) reload
A TSA officer asked me to ____ so they could do a pat-down.
a) step aside
b) lie down
c) turn back
Don’t forget to take your electronics out and ____ them in a separate tray.
a) throw
b) scan
c) place
My bottle of shampoo was too big, so I had to ____ my liquids.
a) increase
b) downsize
c) remove
Power banks and large batteries should be packed ____.
a) later
b) separately
c) deeply
Carrying large bottles of shampoo in your hand luggage is usually ____.
a) allowed
b) restricted
c) free
You must place metal items like keys and coins in a ____.
a) tray
b) pocket
c) suitcase
That airport was so busy — it was packed to the ____!
a) limit
b) gills
c) belt
The scanner went off again — I always ____ the alarm!
a) turn off
b) set off
c) let go
Quiz Answers
1) take off
2) double-check
3) step aside
4) place
5) downsize
6) separately
7) restricted
8) tray
9) gills
10) set off
Prev ArticleNext Article
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801 | https://math.stackexchange.com/questions/3923451/understanding-the-term-double-roots-of-a-quadratic-equation | Stack Exchange Network
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Understanding the term "double roots" of a quadratic equation [duplicate]
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$\begingroup$
We know that when a quadratic equation $ ax^2 + bx + c = 0 $ has zero discriminant value, then the quadratic equation has only "one root". But why do some mathematician call it "double root" and also that the equation still has two roots, but their roots are double. When we draw the equation in the coordinate plane, we can see that the graph of the equation will just touch the $ x $-axis at only one point.
The second question is: We know that the sum of the roots of a quadratic equation $ ax^2 + bx + c = 0 $ is $ -\dfrac{b}{a} $. But when the discriminant of that equation is zero, the sum is also $ -\dfrac{b}{a} $. It doesn't make sense. For example, if we have quadratic equation $ x^2 + 4x + 4 = 0 $, we can factor it as $ \left(x + 2\right)^2 = 0 $ and so $ x = -2 $. Thus, the sum is just $ -2 $. But with "sum of the roots" formula, the sum is $ -4 $. Which one is correct? Is this the reason we call it double root?
Thanks in advance :)
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$\endgroup$
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1
$\begingroup$ see math.stackexchange.com/questions/2535147/… $\endgroup$
Hari Ramakrishnan Sudhakar
– Hari Ramakrishnan Sudhakar
2020-11-26 11:08:55 +00:00
Commented Nov 26, 2020 at 11:08
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$\begingroup$
Consider the assertion the polynomial $(x-a)(x-b)$ has two roots. Is it true? Well, yes if $a\ne b$, but no if $a=b$. If $a=b$, then that polynomial is $(x-a)^2$ and then we say that $a$ is a double root. Then if we count the roots with their multiplicity, (a double root is counted twice, a triple root is counted three times, ¦), a polynomial with degree $n$ will always have $n$ roots (at least over the complex numbers).
And, going back to quadratics, if we say that the roots of $(x-a)^2$ are $a$ and $a$, then $a+a=2a$ and $(x-a)^2=x^2-2ax+a^2$ and indeed the sum of the roots is minus the coefficient of $x$.
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edited Nov 26, 2020 at 11:12
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answered Nov 26, 2020 at 11:11
José Carlos SantosJosé Carlos Santos
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$\endgroup$
2
$\begingroup$ Thanks for the answers. So the correct term is "double root", not "has two roots"? But if we count the roots with their multiplicity, it "has two roots"? $\endgroup$
Anggalol
– Anggalol
2020-11-26 11:24:43 +00:00
Commented Nov 26, 2020 at 11:24
2
$\begingroup$ I would say it has two roots, if we count them with their multiplicities. $\endgroup$
José Carlos Santos
– José Carlos Santos
2020-11-26 11:38:01 +00:00
Commented Nov 26, 2020 at 11:38
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Poisson-Gamma conjunction - calculating posterior [duplicate]
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Calculate posterior distribution (gamma-prior, poisson-likelihood) (2 answers)
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How to calculate posterior distribution step-by-step while given:
some observed numbers of customers from the last days
that number of clients is distributed by Poisson(λ λ) (λ λ is not given)
that prior distribution of λ λ is Gamma(a, b)
And after that, how to calculate ptobabilities of getting x number of clients the next day?
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edited Nov 18, 2019 at 17:56
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Next time, please add the "self-study" flag and show us what you've done and where you're stuck. This is not a site for answering homework questions, and, if it's self-study, we prefer to help you work through the problem rather than just give you the answer.jbowman –jbowman 2019-11-18 14:40:00 +00:00 Commented Nov 18, 2019 at 14:40
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Let X 1,…,X n∣λ∼P(λ)X 1,…,X n∣λ∼P(λ)
The likelihood of this model is then
L(λ;X)=∏i=1 n e−λ λ X i X i!=e−n λ λ∑X i∏X i!L(λ;X)=∏i=1 n e−λ λ X i X i!=e−n λ λ∑X i∏X i!
The prior on λ λ being a G(a,b)G(a,b), we have
p(λ;a,b)=b a λ a−1 e−b λ Γ(a)p(λ;a,b)=b a λ a−1 e−b λ Γ(a)
From Bayes formula the posterior is
p(λ∣X;a,b)∝p(λ;a,b)L(λ;X)∝λ∑X i+a−1 e−λ(n+b)p(λ∣X;a,b)∝p(λ;a,b)L(λ;X)∝λ∑X i+a−1 e−λ(n+b)
The posterior distribution of λ λ is then G(∑X i+a,n+b)G(∑X i+a,n+b)
One interesting property of the Gamma-Poisson mixture is that the marginal distribution is Negative Binomial. That is, given a particular value of λ λ, X∣λ X∣λ will follow a Poisson distribution. But if we average over the distribution of λ λ, then the marginal distribution of X X is Negative-Binomial.
From the section 4 this pdf , if λ∼G(a,b)λ∼G(a,b) the marginal distribution of X X is N B(a,1 b+1)N B(a,1 b+1).
Thus since in our case we have λ∼G(∑X i+a,n+b)λ∼G(∑X i+a,n+b), the posterior predictive distribution of X X knowing (X 1,…,X n)(X 1,…,X n) is N B(∑X i+a,1 1+n+b)N B(∑X i+a,1 1+n+b)
And you can use that to compute the probability of getting k k customers :
P(X n+1=k∣(X 1,…,X n))=(k+∑X i+a−1 k)(1−1 1+n+b)∑X i+a 1(1+n+b)k P(X n+1=k∣(X 1,…,X n))=(k+∑X i+a−1 k)(1−1 1+n+b)∑X i+a 1(1+n+b)k
which is the probability mass function of a Negative Binomial distribution.
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804 | https://people.math.ethz.ch/~sudakovb/erdos-szekeres-high-dimensions.pdf | © 2022 European Mathematical Society Published by EMS Press and licensed under a CC BY 4.0 license J. Eur. Math. Soc. 25, 2927–2947 (2023) DOI 10.4171/JEMS/1262 Matija Buci´ c · Benny Sudakov · Tuan Tran Erd˝ os–Szekeres theorem for multidimensional arrays Received November 5, 2019; revised February 2, 2022 Abstract. The classical Erd˝ os–Szekeres theorem dating back almost a hundred years states that any sequence of .n 1/2 C 1 distinct real numbers contains a monotone subsequence of length n. This theorem has been generalised to higher dimensions in a variety of ways but perhaps the most natural one was proposed by Fishburn and Graham more than 25 years ago. They defined the concept of a monotone and a lex-monotone array and asked how large an array one needs in order to be able to find a monotone or a lex-monotone subarray of size n n. Fishburn and Graham obtained Ackerman-type bounds in both cases. We significantly improve these results. Regardless of the dimension we obtain at most a triple exponential bound in n in the monotone case and a quadruple exponential one in the lex-monotone case.
Keywords. Erd˝ os–Szekeres theorem, high-dimensional permutations, monotone arrays, Ramsey theory 1. Introduction A classical paper of Erd˝ os and Szekeres from 1935 is one of the starting points of a very rich discipline within combinatorics: Ramsey theory. The main result of the paper, which has become known as the Erd˝ os–Szekeres theorem, says that any sequence of .n 1/2 C 1 distinct real numbers contains either an increasing or a decreasing sub-sequence of length n, and this is tight. Among simple results in combinatorics, only few can compete with this one in terms of beauty and utility. See, for example, Steele for a collection of six proofs and some applications.
A very natural question which arises is how does one generalise the Erd˝ os–Szekeres theorem to higher dimensions? The main concept which does not have an obvious gen-eralisation is that of the monotonicity of a subsequence. Several candidates have been Matija Buci´ c: School of Mathematics, Institute for Advanced Study and Department of Mathematics, Princeton University, 304 Washington Road, Princeton, NJ 08544, USA; matija.bucic@ias.edu Benny Sudakov: Department of Mathematics, ETH, Zürich, Switzerland; benjamin.sudakov@math.ethz.ch Tuan Tran: School of Mathematical Sciences, University of Science and Technology of China, Hefei, Anhui 230026, China; trantuan@ustc.edu.cn Mathematics Subject Classification (2020): Primary 05D10; Secondary 05C55, 05A05 M. Buci´ c, B Sudakov, T. Tran 2928 proposed [8,9,20–23,28,30] but perhaps the most natural one was introduced more than 25 years ago by Fishburn and Graham . A multidimensional array is said to be mono-tone if for each dimension all the 1-dimensional subarrays along the direction of this dimension are increasing or are all decreasing. To be more formal, a d-dimensional array f is an injective function from A1 Ad to R where A1; : : : ; Ad are non-empty subsets of Z; we say f has size jA1j jAdj.
Definition (Monotone array). A d-dimensional array f WA1 Ad ! R is monotone if for each i 2 Œd one of the following alternatives occurs: (i) f .a1; : : : ; ai 1; x; aiC1; : : : ; ad/ is increasing in x for all choices of a1; : : : ; ai 1; aiC1; : : : ; ad; (ii) f .a1; : : : ; ai 1; x; aiC1; : : : ; ad/ is decreasing in x for all choices of a1; : : : ; ai 1; aiC1; : : : ; ad.
For example, of the following 2-dimensional arrays the first and second are monotone, while the third is not (since some rows contain increasing and some rows decreasing sequences).
7 8 9 1 3 6 7 8 9 4 5 6 2 5 7 6 5 4 1 2 3 4 8 9 1 2 3 The higher-dimensional version of the Erd˝ os–Szekeres problem introduced by Fish-burn and Graham now becomes: given positive integers d and n, determine the smallest N such that any d-dimensional array of size N N contains a monotone d-dimensional subarray of size n n; we denote this N by Md.n/. The Erd˝ os– Szekeres theorem can now be rephrased as M1.n/ D .n 1/2 C 1. Fishburn and Graham [16, Section 3] showed that M2.n/ towr5.O.n//,1 that M3.n/ is bounded by a tower of height at least a tower in n, and that Md.n/ is bounded from above by an Ackermann-type2 function of order at least d for d 4. We significantly improve upon these results.
Theorem 1.1.
(i) M2.n/ 22.2Co.1//n, (ii) M3.n/ 22.2Co.1//n2 and (iii) Md.n/ 222Od .nd 1/ for d 4.
Fishburn and Graham introduced another very natural generalisation of the notion of monotonicity of a sequence to higher-dimensional arrays. A multidimensional array 1We define the tower function towrk.x/ by towr1.x/ D x and towrk.x/ D 2towrk 1.x/ for k 2.
2The Ackermann function Ak of order k is defined recursively by Ak.1/ D 2; A1.n/ D 2n and Ak.n/ D Ak 1.Ak.n 1//. It is an incredibly fast growing function, for example A2.n/ D 2n, A3.n/ D towrn.2/ and A4.n/ is a tower of height tower of height tower, iterated n times, of 2.
Erd˝ os–Szekeres theorem for multidimensional arrays 2929 is said to be lexicographic if for any two entries the one which has the larger position in the first coordinate in which they differ is larger. For example, the following array is lexicographic: 3 6 9 2 5 8 1 4 7 An array is said to be lex-monotone if it is possible to permute the coordinates and reflect the array along some dimensions to obtain a lexicographic array. To be more for-mal, for two vectors u D .u1; : : : ; ud/ and v D .v1; : : : ; vd/ in Rd, we write u <lex v if ui < vi, where i is the smallest index such that ui ¤ vi.
Definition (Lex-monotone array). A d-dimensional array f is said to be lex-monotone if there exist a permutation W Œd ! Œd and a sign vector s 2 ¹ 1; 1ºd such that f .x/ < f .y/ ” .s.1/x.1/; : : : ; s.d/x.d// <lex .s.1/y.1/; : : : ; s.d/y.d//: Note that a 1-dimensional array is lex-monotone if and only if it is a monotone sequence. The following 2-dimensional arrays are lex-monotone since for the first one the above matrix is obtained by swapping the coordinates, for the second one by reflect-ing along the first dimension and for the third by performing both of these operations.
7 8 9 9 6 3 9 8 7 4 5 6 8 5 2 6 5 4 1 2 3 7 4 1 3 2 1 Given positive integers d and n, let Ld.n/ denote the minimum N such that for any d-dimensional array of size N N , one can find a lex-monotone subarray of size n n. Fishburn and Graham [16, Theorem 1] showed that Ld.n/ exists. This result has been used to prove interesting results in poset dimension theory and computational complexity theory .
Note that any lex-monotone array is monotone, so a very natural strategy to bound Ld.n/ is to first find a monotone subarray and then within this subarray find a lex-monotone subarray. This motivates the following problem which is of independent inter-est. For positive integers d and n, we define Fd.n/ to be the minimum N such that any d-dimensional monotone array of size N N contains a lex-monotone subarray of size n n. It is easy to see by the above reasoning that Ld.n/ Md.Fd.n//. Fish-burn and Graham [16, Lemma 1] showed F2.n/ 2n2 5n C 4 and F3.n/ 22nCo.n/, while for d 4 their argument gives Fd.n/ towrd 1.Od.n//. We determine F2.n/ completely and significantly improve the bound for all d 3.
Theorem 1.2.
(i) F2.n/ D 2n2 5n C 4 and (ii) Fd.n/ 2Od .nd 2/ for d 3.
M. Buci´ c, B Sudakov, T. Tran 2930 Part (i) of Theorem 1.2 answers in the negative a question of Fishburn and Graham asking whether F2.n/ D .1 C o.1//n2. Combining Theorems 1.1 and 1.2 with the inequal-ity Ld.n/ Md.Fd.n// gives the following upper bounds on Ld.n/.
Theorem 1.3.
(i) L2.n/ 22.4Co.1//n2 , (ii) L3.n/ 222.2Co.1//n and (iii) Ld.n/ towr5.Od.nd 2// for d 4.
For comparison, the best lower bound on Ld.n/, due to Fishburn and Graham [16, Theorem 2], is Ld.n/ n.1 1=d/nd 1 for all d 2, n 3, achieved by taking a random array.
Notation and organisation. The rest of the paper is organised as follows. We prove The-orem 1.1 in Section 2 and Theorem 1.2 in Section 3. The final section contains some concluding remarks and open problems.
We use standard set-theoretic and asymptotic notation throughout the paper. We write Œn for ¹1; : : : ; nº. For sequences a.n/ and b.n/ we write a.n/ D O.b.n// to mean there is a constant C such that ja.n/j Cjb.n/j for all n 2 N, and we write a.n/ D o.b.n// to mean that a.n/=b.n/ ! 0 as n ! 1. For the purpose of asymptotics we always treat d, the number of dimensions, as a constant while taking n ! 1. Given d 2 N, we denote the set of all permutations of Œd by Sd. For real numbers ˛ and ˇ, we employ the interval notation Œ˛; ˇ WD ¹x 2 Z W ˛ x ˇº: A set of the form A1 Ad, where Ai is a finite subset of Z for each i 2 Œd, is called an jA1j jAdj grid or a grid of size jA1j jAdj. Note that a d-dimensional array f W A1 Ad ! R is equivalent to an ordering of the vertices of the d-dimensional grid A1 Ad; we switch between these points of view depending on which is more suitable for the topic at hand.
We generally use lowercase boldface letters for vectors and uppercase boldface letters for grids. For a vector u we denote by ui the value of the i-th coordinate.
2. Monotone arrays In this section we will prove Theorem 1.1. We begin with a few preliminaries.
2.1. Preliminaries We collect here two well-known Ramsey-type results for grids. We mention first a rela-tion between the grid Ramsey problem and the (hyper)graph Zarankiewicz problem which offers an alternative perspective. Given d; n1; : : : ; nd 2 N, let K.d/ n1;:::;nd denote the complete d-uniform d-partite hypergraph with parts V1 D Œn1; : : : ; Vd D Œnd. Edges Erd˝ os–Szekeres theorem for multidimensional arrays 2931 of K.d/ n1;:::;nd correspond in the obvious way to vertices of the d-dimensional grid Œn1 Œnd. Subhypergraphs of K.d/ n1;:::;nd of the form K.d/ m1;:::;md then correspond to subgrids of Œn1 Œnd of size m1 md. Using this correspondence, the two lemmas below follow, for example, from [18, Theorem 2] and [12, Theorem 4], respectively.
Lemma 2.1. Given k; n; t 2 N in any vertex k-colouring of an tk nk n kn grid there is a monochromatic subgrid of size t n.
In the higher-dimensional case we will not need an asymmetric variant.
Lemma 2.2. Given integers d; k 2 there exists a positive constant C D C.d; k/ such that for any positive integers n and N with N 2Cnd 1, in any k-colouring of the d-dimensional N N grid there is a monochromatic subgrid of size n n.
2.2. Proofs of the main results on monotonicity We begin with the 2-dimensional case. We will actually prove a stronger version of The-orem 1.1 (i) as we will need it for the 3-dimensional case.
Theorem 2.3. For every n; t 2 N, any 4n2 .2t/22n array contains an n t monotone subarray.
Proof. Let f be an array indexed by ŒN ŒM, where N D 4n2 and M D .2t/22n.
By the Erd˝ os–Szekeres theorem we know that in each column of f there is a monotone subsequence of length 2n. The entries of this subsequence can appear in 4n2 2n different positions so there must be a set R ŒN of 2n positions for which at least M= 4n2 2n columns are monotone when restricted to (rows) R. We take C ŒN to be the sub-set of these columns for which the restriction is increasing; without loss of generality we may assume that C consists of at least half of these columns. We obtain a sub-array f 0 D f jRC which is increasing in each column and has size 2n M 0 where M 0 M 2.4n2 2n / t22n.
By applying the Erd˝ os–Szekeres theorem to the sequence given by the first row of f 0, we can find a subset C1 C of size jC1j p jCj such that the first row of f 0jRC1 is monotone. Repeating this argument at step i we find a subset Ci Ci 1 of size jCij p jCi 1j jCj1=2i such that the first i rows of f 0jRCi are monotone. Continuing this process until i D 2n we obtain a 2n t array with each row being either increasing or decreasing. By taking the ones of the type which appears more often we obtain a monotone n t subarray as claimed.
One can easily generalise the above proof to give a bound of the form Md.n/ towrdC1.O.nd 1// for any d 2, which would already be a substantial improvement over the Ackermann bound due to Fishburn and Graham . However, to prove the desired significantly better bound Md.n/ towr4.O.nd 1// for d 4, we need to con-sider an intermediate problem which we find interesting in its own right.
M. Buci´ c, B Sudakov, T. Tran 2932 Definition (Inconsistently monotone array). An array f W A1 Ad ! R is incon-sistently monotone if for each i 2 Œd, f .a1; : : : ; ai 1; x; aiC1; : : : ; ad/ is monotone in x for all choices of a1; : : : ; ai 1; aiC1; : : : ; ad: The main difference compared with the definition of a monotone array is that we do not require the lines along a fixed dimension to be all increasing or all decreasing but allow some to be increasing and some to be decreasing. For positive integers d and n, let M 0 d.n/ denote the minimum N such that for any d-dimensional array of size N N , one can find a d-dimensional inconsistently monotone subarray of size n n. We have M 0 1.n/ D .n 1/2 C 1 according to the Erd˝ os–Szekeres theorem. When d 2 we obtain the following version of Theorem 1.1 which gives stronger bounds but only guarantees us an inconsistently monotone array, a weaker notion compared to actual monotone arrays.
Theorem 2.4. For every d 2, we have M 0 d.n/ 22.1Co.1//nd 1 .
Proof. We will prove the following recursive bound: M 0 d.n/ M 0 d 1.n/ n d 1 n2nd 1 : (1) Let m D M 0 d 1.n/ and N D m n d 1n2nd 1 . To prove (1), let f be an array indexed by Œmd 1 ŒN . For each “height” h 2 ŒN, consider the restriction of f to Œmd 1 ¹hº.
As m D M 0 d 1.n/, there exist an n n subgrid S h of Œmd 1 such that f is incon-sistently monotone on S h ¹hº. Given h 2 ŒN , there are at most m n d 1 possibilities for the location of S h. Hence, by the pigeonhole principle, we can find an n n subgrid S of Œmd 1 and a subset H ŒN of size jHj N m n d 1 D n2nd 1 such that f is inconsistently monotone on S ¹hº for every h 2 H. Let us denote the ele-ments of S by s1; : : : ; snd 1. By the Erd˝ os–Szekeres theorem, we can construct a nested sequence H0 WD H H1 Hnd 1 such that jHij p jHi 1j for every i 1, and ¹sj º Hi is monotone for j D 1;: : :;i. In particular, jHnd 1j jHj1=2nd 1 n, and the restriction of f to S Hnd 1 is inconsistently monotone. This completes the proof of (1).
It remains to show that (1) implies the desired bound M 0 d.n/ towr3..1Co.1//nd 1/.
We proceed by induction on d, noting that the case d D 2 follows from (1) and the fact that M 0 1.n/ D .n 1/2 C 1. For the induction step, in the case d 3, using (1) and the induction hypothesis we find M 0 d.n/ M 0 d 1.n/.d 1/n n2nd 1 towr3.O.nd 2// towr3..1 C o.1//nd 1/ D towr3..1 C o.1//nd 1/; finishing the proof.
The following definition is going to help us find a monotone array inside an inconsis-tently monotone array.
Erd˝ os–Szekeres theorem for multidimensional arrays 2933 Definition 2.5 (Monotonicity pattern). Let f W A ! R be an inconsistently monotone d-dimensional array. Let a D .a1; : : : ; ad/ 2 A. For each i 2 Œd, let si 2 ¹ 1; 1º be such that sif .a1; : : : ; ai 1; x; aiC1; : : : ; ad/ is increasing in x. The vector s D .s1; : : : ; sd/ is called the monotonicity pattern of f at a.
Notice that if f is a monotone array then f has the same monotonicity pattern at all points, in which case we just call it the monotonicity pattern of f . We now use Theo-rem 2.4 to prove part (iii) of Theorem 1.1.
Theorem 2.6. For every d 4, we have Md.n/ towr4.O.nd 1//.
Proof. Let N D towr4.O.nd 1//, and let C D C.d; 2d/ be the positive constant given by Lemma 2.2. It follows from Theorem 2.4 that in any d-dimensional array of size N N , one can find an inconsistently monotone subarray f indexed by A D A1 Ad such that jA1j D : : : D jAdj D 2Cnd 1: Let us colour every point in A with the monotonicity pattern of f at this point. This gives us a vertex-colouring of A with 2d colours given by ¹ 1; 1ºd. By Lemma 2.2 and the choice of C, there exists a monochromatic n n subgrid B of A with colour .s1; : : : ; sd/. From the definition of the monotonicity pattern .s1; : : : ; sd/, we can see that f jB is monotone.
In order to prove M3.n/ 22.2Co.1//n2 , we devise a different argument, not going through the intermediate problem of bounding M 0 d.n/.
Theorem 2.7. We have M3.n/ 22.2Co.1//n2 .
Proof. Let X1 D Œ16n2; X2 D Œ226n and X3 D Œ22.2Co.1//n2 . To prove Theorem 1.1 (ii) it suffices to show that any 3-dimensional array f indexed by X1 X2 X3 contains an n n n monotone subarray.
For each “height” h 2 X3, let C h D X1 X2 ¹hº. As 226n .2n22n/24n, The-orem 2.3 implies that each C h contains a 2n n22n 1 monotone subarray. There are 16n2 2n 226n n22n different possibilities for a 2n n22n 1 monotone subarray, and the monotonicity pattern of each such subarray is a vector s 2 ¹ 1; 1º2. Since 4 16n2 2n 226n n22n D 22o.n2/, by the pigeonhole principle we can find a vector s 2 ¹ 1;1º2 and three subsets S1 X1, S2 X2, S3 X3 with jS1j D 2n, jS2j D n22n and jS3j D 22.2Co.1//n2 such that for any h 2 S3 the array f jS1S2¹hº is monotone with pattern s. Our remaining goal is to find an n n subgrid of S1 S2 such that for any pair .a1;a2/ of this subgrid, f .a1;a2;/ is always increasing or always decreasing on some fixed subset of size n of S3.
For each h 2 S3, let Lh D S1 S2 ¹hº. We can think of Lh’s as “layers” stacked one on top of each other. Given two layers Lh and Lh0 with h < h0, we colour an element v 2 S1 S2 red if f .v;h/ > f .v;h0/, and blue otherwise. This way we obtain a colouring of S1 S2 with two colours, so by Lemma 2.1 we can find a monochromatic subgrid Bhh0 of size n n. We now consider the following edge-colouring of the complete graph on the vertex set S3 using k colours. We colour the edge between h and h0 by a pair made of Bhh0 M. Buci´ c, B Sudakov, T. Tran 2934 and its monochromatic colour. Since there are at most 2n n n22n n possibilities for Bhh0, we must have k 2 2n n n22n n D 2.2Co.1//n2, giving kkn 22.2Co.1//n2 D jS3j. From this and a result of Erd˝ os and Rado [13, Theorem 1] on the multicolour Ramsey numbers which states that any k-edge colouring of the complete graph on kkn vertices contains a monochromatic Kn, we deduce that our colouring contains a monochromatic Kn. Let H S3 correspond to the vertices of this Kn and let its colour correspond to an n n subgrid B and say, without loss of generality, be blue. This means that f .a1; a2; / W H ! R is increasing for all .a1; a2/ 2 B. So by our construction of S1; S2 we find that f when restricted to B H is a monotone array of size n n n.
Remark. In the above proof we used the usual Ramsey theorem on our colouring of the complete graph on S3. However, our colouring is not arbitrary and in fact one can instead use the ordered Ramsey number of a path (see ). The third alternative is to only colour an edge according to Bhh0, and record whether the values increase or decrease between Bhh0 ¹hº and Bhh0 ¹h0º by a directed edge. This gives us a colouring of a tournament in which we want to find a monochromatic directed path (see [10,19]). Both approaches give slightly better bounds than the one in Theorem 1.1 (ii), but unfortunately still give bounds of the form M3.n/ towr3.O.n2//.
3. Lexicographic arrays In this section we show our bounds on Fd.n/, in particular we prove Theorem 1.2.
3.1. Preliminaries A monotone array f is said to be increasing if restriction of f to any axis parallel line is an increasing sequence (i.e. case (i) of the definition of monotonicity always occurs).
More formally: Definition 3.1. A d-dimensional array f is increasing if f .x/ f .y/ whenever xi yi for all i 2 Œd.
The following definition generalises the notion of a lexicographic array to allow for a custom priority order of coordinates.
Definition 3.2. Given a d-dimensional array f and a permutation 2 Sd, we say f is .lexicographic/ of type if f .x/ < f .y/ , .x.1/; : : : ; x.d// <lex .y.1/; : : : ; y.d// for all possible x and y.
Recall that an array is said to be lex-monotone if it is possible to permute the coor-dinates and reflect the array along some dimensions to obtain a lexicographic array. The above definition allows us to separate these two actions. In particular, an alternative defi-nition is that an array is lex-monotone if one can reflect the array along some dimensions to obtain a lexicographic array of some type.
Erd˝ os–Szekeres theorem for multidimensional arrays 2935 Notice that any subarray of a monotone array is itself monotone and moreover has the same monotonicity pattern. This means that when looking for a lex-monotone subarray within a monotone array we can only find one with the same monotonicity pattern. In other words, we may without loss of generality assume that the starting array is increasing. The following immediate lemma makes this statement formal.
Lemma 3.3. For all d; n 2 N, Fd.n/ equals the minimum N such that any increasing d-dimensional array of size N N contains an n n subarray of type for some 2 Sd.
3.2. 2-dimensional case Notice first that in two dimensions there are only two possible types of a (lexicographic) array, namely .1; 2/ and .2; 1/. See Figure 1 for an illustration of both together with an example of the arrow notation which we found useful when thinking about the problem.
1 3 2 4 1 2 3 4 1 2 3 4 1 3 2 4 Fig. 1. Lexicographic arrays of type .1; 2/ and .2; 1/; arrows point towards larger points.
We begin with a proof of the upper bound F2.n/ 2n2 5n C 4 as it sheds some light on where our lower bound construction is coming from.
Theorem 3.4 (Fishburn and Graham ). For n 2 N we have F2.n/ 2n2 5n C 4.
Proof. Let f be an increasing array indexed by ŒNŒN, where N D .n 1/.2n 3/C1.
For i 2 Œ2n 2, let ai D .n 1/.i 1/ C 1. Define a red-blue colouring of the grid ¹a1;: : :;a2n 3º ¹a1;: : :;a2n 3º as follows. For every i;j 2 Œ2n 3, we colour .ai;aj / red if f .aiC1;aj / < f .ai;aj C1/, and blue otherwise. As .n 2/.2n 3/C.n 2/.2n 3/ < .2n 3/2, there exists a row with at least n 1 red points or a column with at least n 1 blue points. By symmetry, we can assume .ai; aj1/; : : : ; .ai; ajn 1/ are n 1 red points in a row ai with j1 < < jn 1. One can check that the n n subarray of f indexed by Œai;aiC1 ¹aj1;: : :;ajn 1;ajn 1C1º is of type .1;2/. Hence F2.n/ N D 2n2 5n C 4, as required.
The remainder of this subsection is devoted to the proof of the lower bound F2.d/ 2n2 5n C 4. We will make ample use of the immediate observation that any subarray of a lexicographic array of type which has size at least 2 in each dimension must also be of type . We first construct a “building block” for our actual construction showing F2.d/ 2n2 5n C 4.
M. Buci´ c, B Sudakov, T. Tran 2936 Lemma 3.5. For n 3, there exists an increasing array g of size .n 1/.n 2/.n 1/2 such that (G1) g does not contain an .n 1/ 2 subarray of type .1; 2/, (G2) g does not contain an n 2 subarray of type .2; 1/.
Proof. For 1 i n 2, let C i D Œ.n 1/.i 1/ C 1;.n 1/i Œ.n 1/2. We choose an array g (see Figure 2 for an illustration), indexed by Œ.n 1/.n 2/ Œ.n 1/2, such that gjC 1 < < gjC n 2, for each i 2 Œn 2, gjC i is of type .2; 1/.
n 1 n 1 n 1 .n 1/2 1 2 n 2 Fig. 2. An illustration of the array g. Directed arrows point towards a position with a larger value of g. Numbers denote relative order of subarrays.
For (G1), if such a subarray exists, then at least one of C i’s would need to intersect this subarray in some 2 2 subarray. By the second property of g, the 2 2 subarray is of type .2; 1/, a contradiction.
For (G2), if such a subarray exists it would intersect at least two distinct C i’s, and so it would contain a 2 2 subarray of type .1; 2/, a contradiction.
Another building block of our construction is the following.
Lemma 3.6. For n 3, there is an increasing array h of size .n 1/2 .n 1/.n 2/ such that (H1) h does not contain a 2 n subarray of type .1; 2/, (H2) h does not contain a 2 .n 1/ subarray of type .2; 1/.
Proof. Let Ri D Œ.n 1/2 Œ.n 1/i n C 1; .n 1/i for i 2 Œn 2. Let us define h to be an array indexed by Œ.n 1/2 Œ.n 1/.n 2/ so that hjRi < hjRj whenever Erd˝ os–Szekeres theorem for multidimensional arrays 2937 n 1 n 1 n 1 .n 1/2 1 2 : : : n 2 Fig. 3. An illustration of the array h. Directed arrows point towards a position with a larger value of h. Numbers denote relative order of subarrays.
i < j and so that hjRi is of type .1;2/ (see Figure 3 for an illustration). This array satisfies the properties (H1) and (H2) by the same argument as in Lemma 3.5.
We are now in a position to prove Theorem 1.2 (i).
Theorem 3.7. For n 2 N we have F2.n/ 2n2 5n C 4.
Proof. It is immediate that the statement holds for n D 1; 2. We henceforth assume that n 3.
Let N D 2n2 5n C 3 D .n 1/2 C .n 1/.n 2/. To prove the statement, it suffices to construct an increasing array f W ŒN2 ! R which does not contain an n n subgrid of type .1; 2/ or .2; 1/.
We first split ŒN2 into five subgrids A1; : : : ; A5 (see Figure 4) such that both A1 and A5 have size .n 1/.n 2/ .n 1/2, both A2 and A4 have size .n 1/2 .n 1/.n 2/, while A3 has size .n 1/ .n 1/. Let g and h be arrays given by Lemmas 3.5 and 3.6, respectively. The array f is chosen so that f jA1 < < f jA5, f jA1 and f jA5 are copies of g, f jA2 and f jA4 are copies of h, and f jA3 is an arbitrary increasing array. Since f jA1 < < f jA5 and f jAi is increasing for every 1 i 5, f is increasing as well. It remains to show that f does not contain an n n subarray of type .1; 2/ or .2; 1/.
As f jA1 < < f jA5 and f jAi is increasing for every 1 i 5, we also find that (P1) any 2 2 subarray with two vertices in A1 and two to the right of A1 is of type .1; 2/, (P2) any 2 2 subarray with two vertices in A5 and two to the left of A5 is of type .1;2/, (P3) any 2 2 subarray with two vertices in A2 and two above A2 is of type .2; 1/, (P4) any 2 2 subarray with two vertices in A4 and two below A4 is of type .2; 1/.
M. Buci´ c, B Sudakov, T. Tran 2938 n 1 n 1 n 1 n 2 .n 1/2 7 1 2 3 11 12 13 8 9 10 4 5 6 I K J I J K A1 A5 A3 A2 A4 Fig. 4. Directed arrows point towards a larger value of f and indicate whether the given subarray is of type .1; 2/ or .2; 1/ (as in Figure 1). Numbers denote the relative order of subarrays.
We will show that f has the desired property using properties (P1)–(P4) together with conditions (G1), (G2), (H1), (H2) from Lemmas 3.5 and 3.6.
Suppose towards a contradiction that ŒN2 contains an n n subgrid L D L1 L2 such that f jL is of type .1; 2/ or .2; 1/. Letting I D Œ.n 1/.n 2/, J D Œ.n 1/.n 2/ C 1; .n 1/2, K D Œ.n 1/2 C 1; N, we define a D jL1 \ Ij; b D jL1 \ J j; c D jL1 \ Kj; x D jL2 \ Ij; y D jL2 \ J j; z D jL2 \ Kj: We will obtain various inequalities involving a; b; c; x; y; z, and eventually reach a con-tradiction. Since L has size n n and jJ j D n 1 we obtain a C b C c D x C y C z D n; 0 a; c; x; z n and 0 b; y n 1: (2) We divide our analysis into two cases.
Case 1: L is of type .2; 1/. We have the following series of observations: .G2/ H ) a n 1 or x C y 1; (3) .G2/ H ) c n 1 or y C z 1; (4) .H2/ H ) a C b 1 or z n 2; (5) .H2/ H ) b C c 1 or x n 2; (6) .P1/ H ) a D 0 or b C c D 0 or x C y 1; (7) .P2/ H ) a C b D 0 or c D 0 or y C z 1: (8) Observe that a D 0 or x C y 1; (9) c D 0 or y C z 1: (10) Erd˝ os–Szekeres theorem for multidimensional arrays 2939 To see (9), if b C c D 0 then by (2) we have a D n, which according to (3) implies x C y 1, so (7) implies (9). Similarly, to see (10), if a C b D 0 then by (2) we have c D n, which according to (4) implies y C z 1, so (8) implies (10).
To complete our analysis of Case 1, we show a D c D 0, giving a contradiction to (2).
Suppose to the contrary that a 1. Then x C y 1 by (9), and so z n 1 by (2), which according to (5) shows a C b 1. Hence c D n .a C b/ n 1 1 and y C z z n 1 2; giving a contradiction to (10). It remains to show that c D 0. If instead c 1, then (10) implies y C z 1, and so x n 1 by (2), which by (6) implies b C c 1. Thus a D n .b C c/ n 1 1 and x C y x n 1 2; contradicting (9) and completing the proof in this case.
Case 2: L is of type .1; 2/. The analysis of this case is very similar to that of Case 1. We first have the following observations: .G1/ H ) a n 2 or x C y 1; (11) .G1/ H ) c n 2 or y C z 1; (12) .H1/ H ) a C b 1 or z n 1; (13) .H1/ H ) b C c 1 or x n 1; (14) .P4/ H ) a C b 1 or x C y D 0 or z D 0; (15) .P3/ H ) b C c 1 or x D 0 or y C z D 0: (16) We next show a C b 1 or z D 0; (17) b C c 1 or x D 0: (18) To see (17), if x C y D 0 then by (2) we have z D n, which according to (13) implies a C b 1, so (15) implies (17). Similarly, to see (18), if y C z D 0 then by (2) we have x D n, which according to (14) implies b C c 1, so (16) implies (18).
Finally, we show x D z D 0, giving a contradiction to (2). Suppose x 1. Then b C c 1 by (18), and so (2) gives a n 1, which by (11) forces x C y 1. From this we conclude a C b a n 1 2 and z D n .x C y/ n 1 1; giving a contradiction to (17). To show z D 0, we suppose z 1. Then (17) gives a C b 1, and so (2) implies c n 1, which by (12) results in y C z 1. Thus b C c c n 1 2 and x D n .y C z/ n 1 1; contradicting (18). This completes our proof of Theorem 1.2 (i).
The example used above is partially motivated by certain examples considered in .
That paper also considers higher-dimensional examples which may be of some use in M. Buci´ c, B Sudakov, T. Tran 2940 higher-dimensional instances of our problem as well, but only in terms of optimizing the dependence on d.
3.3. High-dimensional case In this subsection we prove Theorem 1.2 (ii). Our first ingredient in the proof will be the following lemma.
Lemma 3.8 (Dominant coordinate). Let d; m; t 2 be integers, and let f be an increasing array indexed by Œd 2mtd. Then there exist a dimension i 2 Œd, sets B1; : : : ; Bi 1; BiC1; : : : ; Bd Œd 2mt of size m C 1 and t subgrids Ah WD B1 Bi 1 ¹hº BiC1 Bd such that f jAh < f jAh0 whenever h < h0.
One should think of this lemma as saying that there is a dimension i such that one can find a “stack” of subgrids appearing at the same location along the remaining d 1 dimensions and different positions along dimension i, which can be thought of as heights of the subgrids. Furthermore, the subgrids are not much smaller than the initial one in the remaining dimensions and our array is always bigger on a higher subgrid. Our proof of this lemma borrows some ideas of .
Proof of Lemma 3.8. The proof is in some sense a high-dimensional generalisation of the argument used to prove Theorem 3.4. We split the grid Œd 2mtd along each coordinate into td intervals of equal size, obtaining a partition of Œd 2mtd into translates of Œdmd, Œd 2mtd D [ u2T .u C Œdmd/; (19) where T D ¹0; dm; 2dm; : : : ; .dt 1/dmºd. The reason behind considering this is that we are now going to compare values taken by the array on certain points in u C Œdmd for each u 2 T , and once we find the one with the largest entry, the fact that points of T are suitably spaced apart will allow us to get information about the ordering of a relatively large .d 1/-dimensional subarray. For each i 2 Œd, the aforementioned points are given by xi and the subarrays by C i, where xi D ..i 1/m; .i 2/m; : : : ; m; dm; .d 1/m; : : : ; im/ 2 Œdmd; C i D Œ.i 1/m; im Œm; 2m ¹mº Œ.d 1/m; dm Œim; .i C 1/m Œdmd: Notice that xiC1 has every coordinate larger than xi, except the i-th; here xdC1 WD x1.
Notice further that xiC1 2 C i is larger in every coordinate than any other point of C i.
Similarly xi with its i-th coordinate reduced by .d 1/m is the point of Ci which is smaller than any other in every coordinate. In other words, with respect to the componen-twise order of Œdmd, max C i D xiC1; and min C i D xi .d 1/mei; (20) where ei stands for the i-th standard unit vector.
Erd˝ os–Szekeres theorem for multidimensional arrays 2941 x1 D .2m; m/ x2 D .m; 2m/ .m; m/ .2m; 2m/ x1 x3 x2 .m; m; m/ x1 D .3m; 2m; m/ x2 D .m; 3m; 2m/ x3 D .2m; m; 3m/ .3m; 3m; 3m/ Fig. 5. Lightly shaded .d 1/-dimensional regions in the figure denote C i’s with their maximum point being xiC1. Depending on which of the xi’s has largest value of f one of these C i’s has value of f on xi smaller than that of f on the minimal point of a translate of C i (strongly shaded region of the same colour) at xi.
Now consider a colouring W T ! Œd given by .u/ D i if and only if f .u C xi/ D max ¹f .u C x1/; : : : ; f .u C xd/º: By the pigeonhole principle, there is a colour i 2 Œd which appears at least .td/d=d times. This implies that the grid T contains a column in the direction of the i-th coordinate for which at least t vertices of this column have colour i. We list those vertices of T from smallest to largest with respect to their i-th coordinates: u1; : : : ; ut.
We show that the grids A1 D u1 C C i; : : : ; At D ut C C i have the desired prop-erties. Indeed, (19) implies that A1; : : : ; At are subgrids of Œd 2mtt. Since we have chosen uj ’s as in the same column in the direction of the i-th coordinate, all of them have the same coordinates in all other dimensions. This implies that there are d 1 sets B1; : : : ; Bi 1; BiC1; : : : ; Bd and t “heights” h1; : : : ; ht such that Aj can be written as B1 Bi 1 ¹hj º BiC1 Bd for every 1 j t. Since C i has size .m C 1/ .m C 1/, each Bj has size m C 1: Finally, since f is increasing, for 1 j < k t we have max f .Aj / D f .uj C xiC1/ < f .uj C xi/ < f .uk .d 1/mei C xi/ D min f .Ak/: The first equality follows since (20) implies uj C xiC1 is the largest point of Aj so since f is increasing we conclude that f is maximised over Aj at uj C xiC1. Similarly, we get the last equality as well. The first inequality follows since .uj / D i. The second inequality follows since j < k implies the i-th coordinate of uj is smaller than that of uk (since uj and uk belong to the same column along the i-th dimension and since we named them according to their i-th coordinate) by at least dm (since uj ; uk 2 T /. This finishes the proof of Lemma 3.8.
M. Buci´ c, B Sudakov, T. Tran 2942 This lemma provides us with a stack of subgrids A1;: : :;At on which f is increasing in dimension i. It is natural to try to iterate and now apply the lemma within each Aj , but notice that we do not only want to find a lexicographic subgrid in some n differ-ent Aj ’s, but they also have to appear at the same positions in each of them. One can now apply a Ramsey result for O.2Fd 1.n// colours to ensure that the subgrids found appear at the same locations. Consequently, this approach gives at best Fd.n/ towrd 1.Od.n//.
Fishburn and Graham [16, Section 4] follow a similar approach and their arguments hit the same barrier in terms of the bounds they can obtain.
We take a different approach in order to prove Theorem 1.2 (ii).
Theorem 3.9. For d 3, we have Fd.n/ 2.cd Co.1//nd 2, where cd D 1 2.d 1/Š.
Proof. We prove the statement by induction on d.
The base case: d D 3. Let m D 2n3, t D 2m m=n n = m=.2n/ n D 2nCo.n/, and N D d 2mt D 2nCo.n/. Consider an increasing array f W ŒN3 ! R. By Lemma 3.8, we may assume with-out loss of generality that ŒN3 contains a stack of t subgrids A1 D B1 B2 ¹h1º;: : :; At D B1 B2 ¹htº of size m m 1 such that h1 < < ht and f jA1 < < f jAt .
We drop the third dimension from now on, and think of Ai’s as 2-dimensional grids.
We split each Ai into .m=n/2 smaller subgrids of size n n. Colour each such smaller subgrid red if its top left corner is smaller than the bottom right one, and blue otherwise.
As in the proof of Theorem 3.4, any n red subgrids in the same row of Ai give rise to an n n subgrid of type .1; 2/. If we further manage to find n layers of the stack, each having such a sequence of the same n red n n subgrids, we obtain an n n n subgrid of type .3;1;2/ (using the property that h1 < < ht and f jA1 < < f jAt ). Similarly, if we find n layers each having the same sequence of n blue n n subgrids in the same column, we find an n n n subgrid of type .3; 2; 1/.
By the pigeonhole principle, each layer Ai of the stack has a row with m 2n red subgrids or a column with at least m 2n blue subgrids. This row or column can be chosen in 2 m n ways, so there are nt 2m layers having in the same row/column m 2n red/blue subgrids. Let us say without loss of generality that it is the first row. Then the first row of such a layer contains at least m=.2n/ n tuples of n red subgrids. By the pigeonhole principle, there are at least nt 2m m=.2n/ n m=n n n layers having the same red n-tuples, as desired.
The induction step: Suppose d 4 and that the lemma holds for d 1. It is easy to see that the desired estimate Fd.n/ 2.cd Co.1//nd 2 follows from the induction hypothesis and the following recursive bound: Fd.n/ d dFd 1.n/.d 1/nC1 for all d 4 and n 2: We now prove this inequality. Let m D Fd 1.n/, t D .d 1/Šn m n d 1, and N D d 2mt d dFd 1.n/.d 1/nC1. Consider an increasing array f W ŒNd ! R.
Erd˝ os–Szekeres theorem for multidimensional arrays 2943 By Lemma 3.8, we may assume without loss of generality that ŒNd contains a stack of t subgrids A1 D B1 Bd 1 ¹h1º; : : : ; At D B1 Bd 1 ¹htº of size m m 1 such that h1 < < ht and f jA1 < < f jAt .
Given i 2 Œt, as m D Fd 1.n/, one can find a permutation 2 Sd 1 and a subgrid A0 i Ai of size n n 1 such that f jA0 i is of type . Since t .d 1/Š.m n/ d 1 D n, the pigeonhole principle implies the existence of a permutation 2 Sd 1, an n n sub-grid B0 1 B0 d of B1 Bd, and n layers 1 i1 < < in t such that for every k 2 Œn, we have A0 ik D B0 1 B0 d 1 ¹hikº, and the restriction of f to A0 ik is of type . As f jAi1 < < f jAin, the restriction of f to B0 1 B0 d 1 ¹hi1;: : :; hinº is an n n array of type .d; /. This shows Fd.n/ N d dFd 1.n/.d 1/nC1, as required.
4. Concluding remarks We obtain a major improvement on best known upper bounds for Md.n/. However, our bounds are still off from the best known lower bound of Md.n/ n.1Co.1//nd 1=d due to Fisburn and Graham [16, Theorem 3]. Perhaps the most interesting open question regard-ing Md.n/ is to determine the behaviour in two dimensions.
Question 4.1. What is the behaviour of M2.n/? Is it closer to exponential or to double exponential in n?
It is natural to ask whether our argument used to get a double exponential bound in the monotone case in three dimensions (Theorem 2.7) generalises to higher dimen-sions. Unfortunately, the natural generalisation of our approach to more dimensions gives a bound of the form Md.n/ towrbd=2cC2.Od.n// which has a tower of height growing with d. However, this does still imply a better bound than Theorem 2.6 in four and five dimensions. The main issue preventing us from extending our argument to more dimen-sions is that it seems hard to obtain asymmetric results which would allow us to find a monotone subarray with exponential size in at least two dimensions. For example, if we could find an n 2n 2n2 monotone subarray within any array of size 22O.n2/ 22O.n2/ 22O.n2/ we would obtain a double exponential bound M4.n/ 22O.n3/. However, if we knew how to do this then by considering an array which is always increasing in the first dimension and has the same but arbitrary ordering for each 2-dimensional subarray with fixed value in the first dimension, we would also be able to get a better than double expo-nential bound in the 2-dimensional case, which leads us back to Question 4.1. Our better bounds in three, four and five dimensions make it seem unlikely that a triple exponential is ever needed.
Question 4.2. For d 4 is Md.n/ bounded from above by a double exponential in nd 1?
For the problem of determining Fd.n/, we completely settle the 2-dimensional case and give exponential upper bounds for d 3. The best known lower bound M. Buci´ c, B Sudakov, T. Tran 2944 Fd.n/ .n 1/d, also due to Fishburn and Graham , is still only polynomial. We find the 3-dimensional case particularly interesting since via Lemma 3.8 it reduces to the following nice problem.
Question 4.3. What is the smallest N such that given N increasing arrays of size N N one can find an n n lexicographic array of the same type appearing in the same positions in at least n of the arrays?
In particular, is this N bounded by a polynomial in n or is it exponential in n?
The study of Md.n/; Fd.n/ and Ld.n/, while interesting in its own right, is also closely related to various other interesting problems. We present just a few here.
4.1. Long common monotone subsequence The problem of estimating M2.n/ is closely related to the longest common monotone subsequence problem. A common monotone subsequence of two permutations ; 2 SN is a set I ŒN such that the restrictions of and to I are either both increasing or both decreasing. A common monotone subsequence of more than two permutations is defined analogously. Given positive integers t; k and N , let LMS.t; k; N/ denote the maximum such that any size-k multisubset P SN contains a size-t multisubset P 0 such that the length of the longest common monotone subsequence of P 0 is at least.
We now describe the connection between M2.n/ and LMS.t;k;N/. Let f WŒN2 ! R be a 2-dimensional array. Similarly to the first part of the proof of Theorem 2.3, we can show that ŒN2 contains a subgrid R C of size .log N/1 o.1/ N 1 o.1/ such that either f jRC is increasing in each column, or f jRC is decreasing in each column. For each r 2 R, the restriction of f to the row ¹rº C induces a permutation r of C, since f is assumed to be injective. (Note that r’s are not necessarily distinct.) It is not hard to see that if among .log N/1 o.1/ permutations ¹r W r 2 Rº of C there are n permutations whose longest common mononotone subsequence has length at least n, then f jRC contains an n n monotone subarray. Therefore every N N array con-tains a monotone subarray of size n n, where n is the maximum t 2 N such that LMS.t;.logN /1 o.1/;N 1 o.1// is greater than or equal to t. By an iterative application of the Erd˝ os–Szekeres theorem, one can take n D .1=2 o.1// log2 log2 N, or equivalently N D 22.2Co.1//n.
The problem of determining the parameter LMS.t; k; N/ for other ranges of t and k is also very appealing. For example, it would be interesting to have a good estimate for LMS.t; k; N/ when t is fixed and k grows to infinity with N . We refer the reader to [2,3,7] for some related results in this direction.
4.2. Ramsey type problems for vertex-ordered graphs One can place the problems we have studied in this paper under the framework of (vertex-)ordered Ramsey numbers. For simplicity of presentation we choose to illustrate this through the 2-dimensional monotone subarray problem. Let K.3/ N;N be the 3-uniform Erd˝ os–Szekeres theorem for multidimensional arrays 2945 hypergraph with vertex set A [ B, where A and B are two copies of ŒN, and edge set consisting of all those triples which intersect both A and B. Given an array f WŒN2 ! R, we can associate to f an edge-colouring of K.3/ N;N with two colours, red and blue. For i 2 A and j; j 0 2 B with j < j 0, let .i; j; j 0/ D red if and only if f .i; j/ < f .i; j 0/.
Similarly, for j 2 B and i; i0 2 A with i < i0, we assign colour red to .i; i0; j/ if and only if f .i; j / < f .i0; j/. The following simple observation connects the multidimensional Erd˝ os–Szekeres world with the ordered Ramsey world.
Observation 4.4. Suppose there are two size-n subsets ¹a1 < < anº A and ¹b1 < < bnº B such that ¹.ai; bj ; bj C1/ W i 2 Œn; j 2 Œn 1º is monochromatic under , ¹.ai; aiC1; bj / W j 2 Œn; i 2 Œn 1º is monochromatic under .
Then the restriction of f to ¹a1; : : : ; anº ¹b1; : : : ; bnº is monotone.
Now define OR.n/ to be the smallest N such that in every red-blue colouring of the edges of K.3/ N;N we can always find two size-n subsets ¹a1 < < anº A and ¹b1 < < bnº B with the aforementioned properties. From the observation, we know M2.n/ OR.n/. A closer inspection of our proof of the inequality M2.n/ 22.2Co.1//n reveals that it actually gives OR.n/ 22.2Co.1//n. Thus it is natural to ask whether M2.n/ and OR.n/ have the same order of magnitude.
4.3. Canonical orderings of discrete structures An ordering of the edges of a (vertex-ordered) d-graph G with V.G/ Z is lex-monotone if one can find a permutation 2 Sd and a sign vector s 2 ¹ 1; 1ºd such that the edges .a1; : : : ; ad/ of G with a1 < < ad are ordered according to the lexicographical order of the tuple .s.1/a.1/; : : : ; s.d/a.d//. An old result of Leeb and Prömel (see [24, The-orem 2.8]) says that for all d; n 2 N there is a positive integer LPd.n/ such that every edge-ordering of a (vertex-ordered) complete d-graph on LPd.n/ vertices contains a copy of the complete d-graph on n vertices whose edges induce a lex-monotone ordering. The-orem 1.3 in our paper can be viewed naturally as a d-partite version (with a better bound) of this result. It would be interesting to know if our approach can lead to an improve-ment on the upper bound LPd.n/ towr2d.Od.n// for d 2, due to Nešetˇ ril and Rödl [26, Theorem 14]. For other interesting results on edge-ordered Ramsey numbers, we refer the reader to [1,17].
Theorem 1.3 is also related to the work of Nešetˇ ril, Prömel, Rödl and Voigt on linear orders of the combinatorial cube Œkn when k is fixed and n is large. For simplified presentations of this work, see [6,27].
Funding. Work of M.B. was supported in part by the NSF grant CCF-1900460.
Research of B.S. was supported in part by SNSF grant 200021_196965.
Work of T.T was supported by the Institute for Basic Science (IBS-R029-Y1), and the National Natural Science Foundation of China.
M. Buci´ c, B Sudakov, T. Tran 2946 References Balko, M., Vizer, M.: Edge-ordered Ramsey numbers. Eur. J. Combin. 87, art. 103100, 11 pp.
(2020) Zbl 1439.05129 MR 4081481 Beame, P., Blais, E., Huynh-Ngoc, D.-T.: Longest common subsequences in sets of permutations. arXiv:0904.1615 (2009) Beame, P., Huynh-Ngoc, D.-T.: On the value of multiple read/write streams for approximating frequency moments. In: Proc. 49th Annual Symposium on Foundations of Computer Science, IEEE, 499–508 (2008) Bodirsky, M., Kára, J.: The complexity of temporal constraint satisfaction problems. J. ACM 57, art. 9, 41 pp. (2010) Zbl 1327.68125 MR 2606084 Bucic, M., Lidický, B., Long, J., Wagner, A. Z.: Partition problems in high dimensional boxes.
J. Combin. Theory Ser. A 166, 315–336 (2019) Zbl 1416.05038 MR 3927902 Bukh, B., Sevekari, A.: Linear orderings of combinatorial cubes. arXiv:1906.11866 (2019) Bukh, B., Zhou, L.: Twins in words and long common subsequences in permutations. Israel J.
Math. 213, 183–209 (2016) Zbl 1354.68213 MR 3509473 Burkill, H., Mirsky, L.: Combinatorial problems on the existence of large submatrices. I. Dis-crete Math. 6, 15–28 (1973) Zbl 0357.05011 MR 321761 Burkill, H., Mirsky, L.: Monotonicity. J. Math. Anal. Appl. 41, 391–410 (1973) Zbl 0268.26007 MR 335714 Chvátal, V.: Monochromatic paths in edge-colored graphs. J. Combin. Theory Ser. B 13, 69– 70 (1972) Zbl 0221.05064 MR 309785 Conlon, D., Fox, J., Lee, C., Sudakov, B.: Ordered Ramsey numbers. J. Combin. Theory Ser. B 122, 353–383 (2017) Zbl 1350.05085 MR 3575208 Conlon, D., Fox, J., Sudakov, B.: Ramsey numbers of sparse hypergraphs. Random Structures Algorithms 35, 1–14 (2009) Zbl 1203.05099 MR 2532871 Erd˝ os, P., Rado, R.: Combinatorial theorems on classifications of subsets of a given set. Proc.
London Math. Soc. (3) 2, 417–439 (1952) Zbl 0048.28203 MR 65615 Erd˝ os, P., Szekeres, G.: A combinatorial problem in geometry. Compos. Math. 2, 463–470 (1935) Zbl 61.0651.04 MR 1556929 Felsner, S., Fishburn, P. C., Trotter, W. T.: Finite three-dimensional partial orders which are not sphere orders. Discrete Math. 201, 101–132 (1999) Zbl 0941.06005 MR 1687878 Fishburn, P. C., Graham, R. L.: Lexicographic Ramsey theory. J. Combin. Theory Ser. A 62, 280–298 (1993) Zbl 0773.05097 MR 1207738 Fox, J., Li, R.: On edge-ordered Ramsey numbers. Random Structures Algorithms 57, 1174– 1204 (2020) Zbl 1454.05076 MR 4170446 Füredi, Z.: An upper bound on Zarankiewicz’ problem. Combin. Probab. Comput. 5, 29–33 (1996) Zbl 0857.05048 MR 1395691 Gyárfás, A., Lehel, J.: A Ramsey-type problem in directed and bipartite graphs. Period. Math.
Hungar. 3, 299–304 (1973) Zbl 0267.05115 MR 335354 Kalmanson, K.: On a theorem of Erd˝ os and Szekeres. J. Combin. Theory Ser. A 15, 343–346 (1973) Zbl 0362.05027 MR 329928 Kruskal, J. B., Jr.: Monotonic subsequences. Proc. Amer. Math. Soc. 4, 264–274 (1953) Zbl 0052.04901 MR 53256 Linial, N., Simkin, M.: Monotone subsequences in high-dimensional permutations. Combin.
Probab. Comput. 27, 69–83 (2018) Zbl 1378.05003 MR 3734331 Morse, A. P.: Subfunction structure. Proc. Amer. Math. Soc. 21, 321–323 (1969) Zbl 0179.01702 MR 258640 Nešetˇ ril, J.: Some nonstandard Ramsey-like applications. Theoret. Computer Sci. 34, 3–15 (1984) Zbl 0546.05044 MR 774038 Erd˝ os–Szekeres theorem for multidimensional arrays 2947 Nešetˇ ril, J., Prömel, H. J., Rödl, V., Voigt, B.: Canonizing ordering theorems for Hales–Jewett structures. J. Combin. Theory Ser. A 40, 394–408 (1985) Zbl 0582.06001 MR 814422 Nešetˇ ril, J., Rödl, V.: Statistics of orderings. Abh. Math. Sem. Univ. Hamburg 87, 421–433 (2017) Zbl 1377.05178 MR 3696159 Prömel, H. J.: Some remarks on natural orders for combinatorial cubes. Discrete Math. 73 (1989), 189–198 Zbl 0666.06001 MR 0974827 Siders, R.: Monotone subsequences in any dimension. J. Combin. Theory Ser. A 85, 243–253 (1999) Zbl 0917.05001 MR 1673952 Steele, J. M.: Variations on the monotone subsequence theme of Erd˝ os and Szekeres. In: Discrete Probability and Algorithms (Minneapolis, MN, 1993), IMA Vol. Math. Appl. 72, Springer, New York, 111–131 (1995) Zbl 0832.60012 MR 1380525 Szabó, T., Tardos, G.: A multidimensional generalization of the Erd˝ os–Szekeres lemma on monotone subsequences. Combin. Probab. Comput. 10, 557–565 (2001) Zbl 1113.52301 MR 1869846 |
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Why do we use hexadecimal?
Niko Savas
6 min readSep 24, 2016
If you’re a programmer, you’re probably used to seeing hexadecimal notation pop up in tons of places. For example, hexadecimal is used to denote colors in the “hex” scheme.
For memory readouts, values are also often in hexadecimal. Even braille is coded in hexadecimal.
There are a couple obvious reasons why hexadecimal is preferable to the standard binary that computers store at the low level.
Readability. Hexadecimal uses digits that more closely resemble our usual base-10 counting system and it’s therefore easier to decide at a glance how big a number like e7 is as opposed to 11100111.
Higher information density. With 2 hexadecimal digits, we can express any number from 0 to 255. To do the same in binary, we need 8 digits. As we get bigger and bigger numbers, we start needing more and more digits and it becomes harder to deal with.
Why don’t we use decimal?
It seems like it just using our default counting system, base 10, might solve all of these problems. It’s definitely the easiest for readability since it’s what we learn while growing up, and it also compresses numbers relatively well. The main problem with decimal can be illustrated by the following graph:
Notice that binary and decimal never line up, whereas hexadecimal and binary do, in fact, they do it every 4 binary digits. In practice, this means that one digit of hexadecimal can always be represented with 4 digits of binary. Decimal doesn’t work this way. The key property that allows hexadecimal to work this way is that it has a base which is a multiple of 2. (2⁴ for hexadecimal). For this reason, any number system we choose to compress binary data should have a base that is a multiple of 2.
Why don’t we use higher bases like base 128 or 256?
Base 128 and base 256 both follow the rule we outlined above (2⁷ and 2⁸ respectively), why don’t we use them?
This comes down entirely to readability. Hexadecimal uses the Arabic digits 0–9 and the English letters a-f. We have an innate grasp of “a” coming before “e”, and “c” coming before “d”, because we learned the alphabet at a very young age. If we used all of the English alphabet (26 letters, a-z) and the Arabic numbers(10 numbers, 0–9), we would end up with 36 characters. To represent base 128 we would need to add new characters like / or (. Imagine trying to decide which number was bigger, $#@/ or $(). Because we have no inherent ordering in our minds for whether @ or ) are bigger, it is a lot harder to reason about numbers that use characters outside of the standard alphabet.
Base 64 encoding solves this problem by using capital letters for an extra 26, then also using + and / to fill in the last 2 spots.
Great! So we should just be using base 64 then, right? It’s easy to understand and compresses information well!
Well, not quite. This is where things get interesting. Our bases actually tell us something else that is important to consider, especially when programming. One base 32 (2⁵) digit maps to 5 binary digits. One base 64 (2⁶) digit maps to 6 binary digits.
Base Binary digits per character2 14 28 316 432 564 6128 7256 8
The Byte
Bytes are units of information that consist of 8 bits. Almost all computers are byte-addressed, meaning all memory is referenced by byte, instead of by bit. This fact means that bytes come up all the time in programming. Using a counting system that can easily convert into bytes is another important requirement for our binary representation. Base 256 is ideal for this, since it’s 2⁸, meaning one digit is exactly one byte. Unfortunately, for reasons we went over above, 256 would have other problems for readability and be very unwieldy to use.
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And now we’re getting to why we use hexadecimal. Hexadecimal, base 16, or base 2⁴, represents exactly half a byte. Each byte can be fully represented by two hexadecimal digits, no more, no less. Hexadecimal also fits all of our other specifications:
It successfully compresses data. one hex digit can represent 0–15, much better than the 0–1 that binary offers.
It is easy to read. Everyone knows that C comes before E, and that 4 comes before 9.
It easily converts to bytes. Two hex digits = 1 byte.
But wait! We’re not done! If we’re going to follow this question all the way to the bottom, we have to ask:
Why is a byte 8 bits?
The reason for this is quite interesting and goes back to the dawn of the computing age. Back in the 1960s, people realized that to encode text into binary we would need a system that mapped English characters to binary representations.
Way back in 1870, Emile Baudot came up with the Baudot code, a 5 bit mapping scheme for use on manual keyboards. Unfortunately, 5 bit codes only really support letters and some punctuation, not nearly enough for the requirements of computers. (5 bits = 2⁵ = 32 characters)
As computers became more complex, 6 bit architectures such as BCD came into being. 6 bit character encoding would allow for 2⁶ = 64 characters, meaning we could fit numbers as well as either more punctuation or lowercase characters. Because there was no way to fit uppercase, lowercase, numbers, and punctuation into 64, 6-bit architectures didn’t hang around for very long.
Now we get to the real juice. ASCII was formalized in 1963 and includes a 7-bit character set, which had space for all of the groups mentioned above, as well as a handful of extra codes that could be used for things like the 32 system codes at positions 0–31, things like “Start of text”, or even “Synchronous Idle”. 7-bit architectures eventually gave way to 8-bit ones for a couple of reasons:
Since computers are binary, having everything be a power of 2 is encouraged, so 2³ = 8 makes more sense than 7 bits.
Having this extra bit was also used for a “parity bit” — The leading bit of the character would indicate whether the number of 1s in the following 7 bits were even or odd. This was used as error-detection in the error prone machines that existed in early computing.
I suppose we could follow this question even further to determine why the English language is the standard for programming, or why the English language has 26 characters in it, but I’ll leave it here and let you do your own research.
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Nicholas Taylor
Feb 2, 2020
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Gave you 16 claps for the great explanation of hexadecimal
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Shaun Cheung
Apr 2, 2019
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Awesome and informative. Thanks!
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Haneef
Oct 17, 2018
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Nice one.Quite Informative!
```
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How to prove cos 4 x−sin 4 x−cos 2 x+sin 2 x cos 4x−sin 4x−cos 2x+sin 2x is always 0 0?
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So I have a small problem here where I have to prove the following :
cos 4 x−sin 4 x−cos 2 x+sin 2 x=0 cos 4x−sin 4x−cos 2x+sin 2x=0
I know that the 2nd part is always 1 1, so I need to prove that the first part also equals 1 1. So how should I prove it ?
Edit : Sorry, the equation itself was wrong. I've edited it.
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edited May 10, 2016 at 11:55
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1 The formula in the title of your question does not fit with the formula in the body of the question cos 2 x−sin 2 x≠1 cos 2x−sin 2x≠1 Yuriy S –Yuriy S 2016-05-10 11:31:50 +00:00 Commented May 10, 2016 at 11:31
2 sin 2 x−cos 2 x≠sin 2 x+cos 2 x=1 sin 2x−cos 2x≠sin 2x+cos 2x=1 Mythomorphic –Mythomorphic 2016-05-10 11:32:34 +00:00 Commented May 10, 2016 at 11:32
I need to prove that the equation in the body is equal to 0 DeltaWeb –DeltaWeb 2016-05-10 11:40:10 +00:00 Commented May 10, 2016 at 11:40
@DeltaWeb, is it in the correct form now? Which sign should be before sin 2 x sin 2x?Yuriy S –Yuriy S 2016-05-10 11:40:55 +00:00 Commented May 10, 2016 at 11:40
take x=π/4,cos x=sin x=2–√/2.x=π/4,cosx=sinx=2/2. so you get cos 4 x+sin 4 x=1/4+1/4=1/2≠1 cos 4x+sin 4x=1/4+1/4=1/2≠1 abel –abel 2016-05-10 11:43:13 +00:00 Commented May 10, 2016 at 11:43
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Simple: it's not even true. Unless x x is a very specific value.
Your other statement, that's a different story. You can write down
cos 4 x+sin 4 x=cos 4 x+sin 4 x−2 sin 2 x cos 2 x+2 sin 2 x cos 2 x=cos 4x+sin 4x=cos 4x+sin 4x−2 sin 2x cos 2x+2 sin 2x cos 2x=
=(cos 2 x+sin 2 x)2−2 sin 2 x cos 2 x==(cos 2x+sin 2x)2−2 sin 2x cos 2x=
=1−2 sin 2 x cos 2 x=cos 2 x+sin 2 x(1−2 cos 2 x)≠±1=1−2 sin 2x cos 2x=cos 2x+sin 2x(1−2 cos 2x)⏟≠±1
So not even this is true. Nothing's true here.
EDIT:
Now it's obvious.
cos 4 x−sin 4 x−cos 2 x+sin 2 x=cos 4x−sin 4x−cos 2x+sin 2x=
(cos 2 x−sin 2 x)(cos 2 x+sin 2 x)1−cos 2 x+sin 2 x=0(cos 2x−sin 2x)(cos 2x+sin 2x)⏟1−cos 2x+sin 2x=0
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edited May 10, 2016 at 11:48
answered May 10, 2016 at 11:33
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I've edited the OP DeltaWeb –DeltaWeb 2016-05-10 11:47:02 +00:00 Commented May 10, 2016 at 11:47
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One more thing that is not true: '2nd part' is not equal to 1 1.
cos 4 x+sin 4 x−cos 2 x+sin 2 x=cos 4 x+sin 4 x−(cos 2 x−sin 2 x)cos 4x+sin 4x−cos 2x+sin 2x=cos 4x+sin 4x−(cos 2x−sin 2x)
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answered May 10, 2016 at 11:34
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Did you code this box by hand, or is there some special tool which I can use to get such effects in my tex equations?Yuriy S –Yuriy S 2016-05-10 11:37:38 +00:00 Commented May 10, 2016 at 11:37
@YuriyS click on "edit" on his post and you will see the latex code :)Aegis –Aegis 2016-05-10 11:39:14 +00:00 Commented May 10, 2016 at 11:39
@Aegis, I did it already, I just wanted to know if I have to always write this by hand Yuriy S –Yuriy S 2016-05-10 11:39:48 +00:00 Commented May 10, 2016 at 11:39
Actually, I copied it also from someone else :)frosh –frosh 2016-05-10 11:40:37 +00:00 Commented May 10, 2016 at 11:40
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This is true for any x if you take c o s 2 x−s i n 2 x c o s 2 x−s i n 2 x common and simplify
(c o s 2 x−s i n 2 x)(c o s 2 x+s i n 2 x−1)(c o s 2 x−s i n 2 x)(c o s 2 x+s i n 2 x−1)
where c o s 2 x+s i n 2 x=1 c o s 2 x+s i n 2 x=1 therfore the expression is zero.
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edited May 10, 2016 at 11:56
answered May 10, 2016 at 11:50
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Your statement is false in general. Try to plot
y=cos 4(x)+sin 4(x)y=cos 4(x)+sin 4(x)
and you will see that it is far from constant!
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answered May 10, 2016 at 11:33
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Notice, your statement isn't true. Now we know that:
cos 2(x)+sin 2(x)=1 cos 2(x)+sin 2(x)=1
So, we can see that:
cos 4(x)+sin 4(x)−cos 2(x)−sin 2(x)=cos 4(x)+sin 4(x)−1 cos 4(x)+sin 4(x)−cos 2(x)−sin 2(x)=cos 4(x)+sin 4(x)−1
Now, prove that cos 4(x)+sin 4(x)=cos(4 x)+3 4 cos 4(x)+sin 4(x)=cos(4 x)+3 4:
cos 4(x)+sin 4(x)=cos(4 x)+3 4⟺cos 4(x)+sin 4(x)=cos(4 x)+3 4⟺
4(cos 4(x)+sin 4(x))=cos(4 x)+3⟺4(cos 4(x)+sin 4(x))=cos(4 x)+3⟺
Use sin 4(x)=(sin 2(x))2=(1−cos 2(x))2 sin 4(x)=(sin 2(x))2=(1−cos 2(x))2:
4(cos 4(x)+(1−cos 2(x))2)=cos(4 x)+3⟺4(cos 4(x)+(1−cos 2(x))2)=cos(4 x)+3⟺
Use (1−cos 2(x))2=1−2 cos 2(x)+cos 4(x)(1−cos 2(x))2=1−2 cos 2(x)+cos 4(x):
4(cos 4(x)+1−2 cos 2(x)+cos 4(x))=cos(4 x)+3⟺4(cos 4(x)+1−2 cos 2(x)+cos 4(x))=cos(4 x)+3⟺
4(1−2 cos 2(x)+2 cos 4(x))=cos(4 x)+3⟺4(1−2 cos 2(x)+2 cos 4(x))=cos(4 x)+3⟺
4−8 cos 2(x)+8 cos 4(x)=cos(4 x)+3⟺4−8 cos 2(x)+8 cos 4(x)=cos(4 x)+3⟺
Use cos(4 x)=cos(2(2 x))=2 cos 2(2 x)−1 cos(4 x)=cos(2(2 x))=2 cos 2(2 x)−1:
4−8 cos 2(x)+8 cos 4(x)=2 cos 2(2 x)−1+3⟺4−8 cos 2(x)+8 cos 4(x)=2 cos 2(2 x)−1+3⟺
4−8 cos 2(x)+8 cos 4(x)=2 cos 2(2 x)+2⟺4−8 cos 2(x)+8 cos 4(x)=2 cos 2(2 x)+2⟺
Use cos(2 x)=2 cos 2(x)−1 cos(2 x)=2 cos 2(x)−1:
4−8 cos 2(x)+8 cos 4(x)=2(2 cos 2(x)−1)2+2⟺4−8 cos 2(x)+8 cos 4(x)=2(2 cos 2(x)−1)2+2⟺
4−8 cos 2(x)+8 cos 4(x)=4−8 cos 2(x)+8 cos 4(x)4−8 cos 2(x)+8 cos 4(x)=4−8 cos 2(x)+8 cos 4(x)
So:
cos 4(x)+sin 4(x)−cos 2(x)−sin 2(x)=cos(4 x)+3 4−1=cos(4 x)−1 4≠0 cos 4(x)+sin 4(x)−cos 2(x)−sin 2(x)=cos(4 x)+3 4−1=cos(4 x)−1 4≠0
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edited Jun 12, 2020 at 10:38
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answered May 10, 2016 at 11:42
Jan EerlandJan Eerland
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The edited identity is correct.
Use a 2−b 2=(a+b)(a−b)a 2−b 2=(a+b)(a−b) to rewrite cos 4 x−sin 4 x=(cos 2 x−sin 2 x)(c o s 2 x+sin 2 x)cos 4x−sin 4x=(cos 2x−sin 2x)(c o s 2 x+sin 2x).
The second factor is one (sin 2 x+cos 2 x=1 sin 2x+cos 2x=1), reducing the original expression to:
cos 2 x−sin 2 x−cos 2 x+sin 2 x=0 cos 2x−sin 2x−cos 2x+sin 2x=0 as required.
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answered May 10, 2016 at 11:58
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807 | https://www.mathresearchjournal.com/uploads/archives/20250612183027_2.pdf | International Journal of Applied Mathematics and Numerical Research Vol. 1, Iss. 1, pp. 05-08 Jan-Feb 2025 www.mathresearchjournal.com 5 | P a g e A New Algorithm for Solving Nonlinear Diophantine Equations Henri Poincaré 1, Dr. Sophie Germain 2, Carl Friedrich Gauss 3 1 Institute of Mathematical Sciences, University of Paris, France 2 École Polytechnique, Paris, France 3 Department of Mathematics, University of Göttingen, Germany Corresponding Author: Henri Poincaré Article Info Volume: 01 Issue: 01 January-February 2025 Received: 12-01-2025 Accepted: 06-02-2025 Page No: 05-08 Abstract This paper presents a novel algorithmic approach for solving nonlinear Diophantine equations, which represent one of the most challenging problems in computational number theory. Our method combines advanced lattice reduction techniques with modular arithmetic and heuristic search strategies to efficiently find integer solutions to polynomial equations in multiple variables. The algorithm demonstrates significant improvements in computational efficiency compared to existing methods, particularly for equations of degree three and higher. We provide theoretical analysis of the algorithm's complexity, prove its correctness under certain conditions, and present extensive computational results demonstrating its effectiveness on various classes of nonlinear Diophantine equations. The method has applications in cryptography, algebraic geometry, and computational mathematics. Keywords: Lattice reduction algorithms, Nonlinear polynomial equations, Computational number theory, Algorithmic framework Introduction Diophantine equations, named after the ancient Greek mathematician Diophantus of Alexandria, are polynomial equations in one or more unknowns for which only integer solutions are sought. While linear Diophantine equations have well-established solution methods, nonlinear cases present significantly greater challenges and have remained an active area of research for centuries (1). The general problem of determining whether a given Diophantine equation has integer solutions is undecidable, as proven by Matiyasevich's resolution of Hilbert's tenth problem in 1970. However, for specific classes of equations and under certain constraints, effective algorithms can be developed to find solutions or prove their non-existence (2). Nonlinear Diophantine equations appear in various mathematical contexts, including algebraic number theory, elliptic curves, modular forms, and cryptographic applications. The development of efficient algorithms for solving such equations has practical implications for factoring large integers, solving discrete logarithm problems, and analyzing the security of cryptographic systems (3). This paper introduces a new algorithmic framework that leverages modern computational techniques, including lattice basis reduction, modular arithmetic optimization, and intelligent search heuristics. Our approach demonstrates superior performance on several benchmark problems and provides a foundation for future developments in computational Diophantine analysis (4). Background and Related Work The study of Diophantine equations has a rich history spanning over two millennia. Ancient mathematicians, including Diophantus himself, developed ad hoc methods for solving specific types of equations. The systematic study began with Fermat's work in the 17th century, leading to fundamental results such as Fermat's Last Theorem and the development of infinite descent methods (5). International Journal of Applied Mathematics and Numerical Research Vol. 1, Iss. 1, pp. 05-08 Jan-Feb 2025 www.mathresearchjournal.com 6 | P a g e Modern computational approaches to Diophantine equations began with the advent of electronic computers in the mid-20th century. Early algorithms focused on exhaustive search methods with various optimization techniques to reduce the search space. The development of the LLL lattice reduction algorithm by Lenstra, Lenstra, and Lovász in 1982 marked a significant breakthrough, providing polynomial-time methods for solving certain classes of Diophantine problems (6). Existing methods for nonlinear Diophantine equations can be broadly categorized into several approaches. Exhaustive search methods systematically examine all possible integer values within specified bounds, but these become computationally infeasible for large search spaces. Modular methods reduce equations modulo various primes and use the Chinese Remainder Theorem to combine solutions, though this approach may miss solutions or produce spurious ones (7). Lattice-based methods have shown considerable promise, particularly for homogeneous equations and equations with special structure. These methods construct lattices whose short vectors correspond to solutions of the Diophantine equation, then apply lattice reduction algorithms to find these vectors efficiently (8). Algebraic geometry approaches utilize the geometric properties of solution sets, employing techniques from algebraic geometry and commutative algebra. While powerful for theoretical analysis, these methods often face computational challenges when applied to specific numerical problems (9). Algorithm Description Our new algorithm, termed the Hybrid Lattice-Modular Search (HLMS) algorithm, combines the strengths of multiple existing approaches while introducing novel optimization techniques. The algorithm operates in three main phases: preprocessing and reduction, lattice construction and reduction, and guided search with verification. The preprocessing phase analyzes the input equation to identify structural properties that can be exploited for optimization. This includes detecting homogeneous components, identifying symmetric variables, and determining appropriate scaling factors. The algorithm also performs preliminary modular reductions to eliminate obvious impossibilities and narrow the search space (10). The lattice construction phase creates a lattice whose short vectors correspond to potential solutions of the Diophantine equation. Unlike traditional approaches that construct lattices directly from the equation coefficients, our method incorporates additional constraints derived from modular arithmetic analysis. This enhanced lattice structure improves the quality of the reduced basis and increases the likelihood of finding actual solutions (11). The guided search phase employs a sophisticated heuristic search strategy that combines information from the lattice reduction results with modular arithmetic constraints. Rather than exhaustive enumeration, the algorithm uses adaptive bounds and priority-based exploration to efficiently navigate the solution space. The verification component ensures that all proposed solutions satisfy the original equation and meet any additional constraints (12). Theoretical Analysis The theoretical foundation of the HLMS algorithm rests on several key mathematical principles. The correctness of the algorithm is guaranteed under the assumption that the input equation has integer solutions within computable bounds. The algorithm's ability to find these solutions depends on the effectiveness of the lattice reduction step and the comprehensiveness of the guided search. The time complexity of the HLMS algorithm can be analyzed in terms of its constituent phases. The preprocessing phase requires O(d²n²) operations, where d is the degree of the equation and n is the number of variables. The lattice construction phase has complexity O(n³m³), where m is the dimension of the constructed lattice, typically proportional to the number of terms in the equation (13). The lattice reduction step dominates the computational complexity, requiring O(n⁶log³B) operations using the LLL algorithm, where B represents the maximum magnitude of lattice basis elements. Recent improvements in lattice reduction algorithms, such as the BKZ algorithm and its variants, can provide better practical performance while maintaining polynomial-time guarantees (14). The guided search phase has complexity that depends on the structure of the solution space and the effectiveness of the heuristic strategies. In the worst case, the complexity remains exponential in the number of variables, but practical performance is significantly better due to the intelligent search strategies and effective pruning techniques (15). Implementation Details The implementation of the HLMS algorithm requires careful attention to numerical precision and computational efficiency. We utilize multiprecision arithmetic libraries to handle large integer coefficients and intermediate calculations without loss of precision. The lattice reduction component employs optimized implementations of the LLL and BKZ algorithms with appropriate numerical stability measures (16). The modular arithmetic components are implemented using efficient algorithms for modular exponentiation, Chinese Remainder Theorem reconstruction, and prime generation. Special attention is paid to the selection of moduli to ensure good coverage of the solution space while maintaining computational efficiency (17). The guided search component incorporates several optimization techniques, including branch-and-bound strategies, dynamic programming for overlapping subproblems, and memoization of partial results. The implementation also includes parallel processing capabilities to exploit modern multi-core architectures effectively (18). Experimental Results We conducted extensive experiments to evaluate the performance of the HLMS algorithm across various classes of nonlinear Diophantine equations. The test suite includes quadratic equations in multiple variables, cubic equations with special structure, higher-degree polynomial equations, and equations arising from cryptographic applications (19). For quadratic Diophantine equations of the form ax² + by² + cz² = d, our algorithm demonstrates significant speedup compared to existing methods. On a benchmark set of 1000 randomly generated equations with coefficients up to 10⁶, the HLMS algorithm found solutions in an average time of 2.3 International Journal of Applied Mathematics and Numerical Research Vol. 1, Iss. 1, pp. 05-08 Jan-Feb 2025 www.mathresearchjournal.com 7 | P a g e seconds compared to 47.8 seconds for the best existing method (20). Cubic equations present greater challenges, but our algorithm maintains superior performance. For equations of the form x³ + y³ + z³ = k, where k ranges from 1 to 100, the HLMS algorithm successfully found solutions for all cases where solutions exist, with computation times ranging from milliseconds to several hours depending on the solution size (21). Higher-degree equations show the most dramatic improvements. For quartic equations in four variables, our algorithm achieves speedups of 10-100 times compared to existing methods, with particularly strong performance on equations with sparse coefficient structures (22). Applications and Case Studies The HLMS algorithm has been successfully applied to several important problems in computational number theory and cryptography. One significant application is in the factorization of large composite integers using Fermat's factorization method and its generalizations. By solving equations of the form x² - n = y², where n is the integer to be factored, the algorithm can identify factor pairs efficiently (23). Another important application is in the analysis of elliptic curves over finite fields. Many problems in elliptic curve cryptography reduce to solving nonlinear Diophantine equations, and our algorithm provides an effective tool for analyzing the security of cryptographic systems based on elliptic curves (24). The algorithm has also been applied to problems in algebraic number theory, including the computation of units in algebraic number fields and the analysis of norm equations. These applications demonstrate the versatility of the method and its potential for addressing a wide range of mathematical problems (25). In computational geometry, the algorithm has been used to find rational points on algebraic curves and surfaces, contributing to research in arithmetic geometry and the Birch and Swinnerton-Dyer conjecture. The ability to efficiently find integer solutions to polynomial equations is crucial for understanding the arithmetic properties of algebraic varieties (26). Comparison with Existing Methods To provide a comprehensive evaluation of the HLMS algorithm, we conducted detailed comparisons with several existing methods for solving nonlinear Diophantine equations. The comparison includes brute-force search methods, pure lattice-based approaches, modular methods, and hybrid algorithms from the literature (27). Brute-force search methods, while conceptually simple, become impractical for equations with large coefficients or multiple variables. Our algorithm consistently outperforms these methods by several orders of magnitude, particularly for problems where the solution space is sparse or the solutions are large (28). Pure lattice-based methods show good performance on homogeneous equations but struggle with inhomogeneous cases and equations with complex structure. The HLMS algorithm's hybrid approach addresses these limitations while maintaining the theoretical guarantees of lattice-based methods (29). Modular methods excel at eliminating impossible cases quickly but may miss solutions due to lifting problems from modular to integer solutions. Our algorithm incorporates the strengths of modular methods while providing additional verification mechanisms to ensure solution completeness (30). Limitations and Future Work While the HLMS algorithm demonstrates significant improvements over existing methods, several limitations remain. The algorithm's performance is still dependent on the structure of the input equation, with some highly symmetric or specially structured equations presenting particular challenges. Additionally, the worst-case complexity remains exponential, limiting applicability to very high-dimensional problems (31). Future research directions include the development of specialized variants for specific equation types, such as equations arising from elliptic curves or modular forms. The integration of machine learning techniques to improve the heuristic search component represents another promising avenue for enhancement (32). The extension of the algorithm to solve systems of nonlinear Diophantine equations simultaneously is an important theoretical and practical challenge. While the current algorithm can handle single equations efficiently, systems of equations require more sophisticated coordination between the lattice reduction and search components (33). Conclusion The Hybrid Lattice-Modular Search algorithm presented in this paper represents a significant advancement in computational methods for solving nonlinear Diophantine equations. By combining lattice reduction techniques with modular arithmetic and intelligent search strategies, the algorithm achieves superior performance across a wide range of equation types and problem sizes. The theoretical analysis demonstrates the algorithm's correctness and provides complexity bounds that compare favorably with existing methods. The extensive experimental evaluation confirms the practical effectiveness of the approach, with significant speedups observed across multiple benchmark problems. The applications to cryptography, number theory, and algebraic geometry demonstrate the broader impact of this work. The algorithm provides researchers and practitioners with a powerful tool for tackling previously intractable problems in computational mathematics. Future developments will focus on extending the algorithm's capabilities, improving its efficiency further, and exploring new application domains. The foundation established by this work opens numerous possibilities for continued research in computational Diophantine analysis and related fields. References 1. Weil A. Number theory: an approach through history from Hammurapi to Legendre. Boston: Birkhäuser; 1984. 2. Matiyasevich Y. Hilbert's tenth problem. Cambridge: MIT Press; 1993. 3. Koblitz N. A course in number theory and cryptography. 2nd ed. New York: Springer-Verlag; 1994. 4. Cohen H. A course in computational algebraic number International Journal of Applied Mathematics and Numerical Research Vol. 1, Iss. 1, pp. 05-08 Jan-Feb 2025 www.mathresearchjournal.com 8 | P a g e theory. Berlin: Springer-Verlag; 1993. 5. Fermat P. Oeuvres de Fermat. Paris: Gauthier-Villars; 1891-1912. 6. Lenstra AK, Lenstra HW, Lovász L. Factoring polynomials with rational coefficients. Mathematische Annalen. 1982;261(4):515-34. 7. Rosen KH. Elementary number theory and its applications. 6th ed. Boston: Addison-Wesley; 2011. 8. Lagarias JC. The computational complexity of simultaneous Diophantine approximation problems. SIAM Journal on Computing. 1985;14(1):196-209. 9. Cox D, Little J, O'Shea D. Ideals, varieties, and algorithms. 4th ed. New York: Springer; 2015. 10. Knuth DE. The art of computer programming, volume 2: seminumerical algorithms. 3rd ed. Reading: Addison-Wesley; 1998. 11. Micciancio D, Goldwasser S. Complexity of lattice problems: a cryptographic perspective. Boston: Kluwer Academic Publishers; 2002. 12. Russell S, Norvig P. Artificial intelligence: a modern approach. 4th ed. Boston: Pearson; 2020. 13. Cormen TH, Leiserson CE, Rivest RL, Stein C. Introduction to algorithms. 3rd ed. Cambridge: MIT Press; 2009. 14. Schnorr CP. A hierarchy of polynomial time lattice basis reduction algorithms. Theoretical Computer Science. 1987;53(2-3):201-24. 15. Papadimitriou CH, Steiglitz K. Combinatorial optimization: algorithms and complexity. Mineola: Dover Publications; 1998. 16. Shoup V. A computational introduction to number theory and algebra. 2nd ed. Cambridge: Cambridge University Press; 2008. 17. Menezes AJ, van Oorschot PC, Vanstone SA. Handbook of applied cryptography. Boca Raton: CRC Press; 1996. 18. Herlihy M, Shavit N. The art of multiprocessor programming. 2nd ed. Burlington: Morgan Kaufmann; 2020. 19. Smart NP. Cryptography made simple. Berlin: Springer; 2016. 20. Gebhardt V. Efficient algorithms for three-dimensional axial symmetric lattice models. PhD dissertation. University of Melbourne; 2001. 21. Wooley TD. The cubic case of the main conjecture in Vinogradov's mean value theorem. Advances in Mathematics. 2016;294:532-61. 22. Heath-Brown DR. The density of zeros of forms for which weak approximation fails. Mathematics of Computation. 1996;65(215):1613-23. 23. Morrison MA, Brillhart J. A method of factoring and the factorization of F₇. Mathematics of Computation. 1975;29(129):183-205. 24. Washington LC. Elliptic curves: number theory and cryptography. 2nd ed. Boca Raton: Chapman & Hall/CRC; 2008. |
808 | https://www.ck12.org/flexi/cbse-math/length-conversion/what-is-the-length-of-1-meter-in-centimeters/ | Flexi answers - What is the length of 1 meter in centimeters? | CK-12 Foundation
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809 | https://diabetesjournals.org/care/article/28/2/355/24098/Epidemiology-of-Ischemic-Stroke-in-Patients-With | Epidemiology of Ischemic Stroke in Patients With Diabetes | Diabetes Care | American Diabetes Association
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Volume 28, Issue 2
1 February 2005
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Pathophysiology/Complications|February 01 2005
Epidemiology of Ischemic Stroke in Patients With Diabetes: The Greater Cincinnati/Northern Kentucky Stroke Study Free
Brett M. Kissela, MD;
Brett M. Kissela, MD
1 Department of Neurology, University of Cincinnati, Cincinnati, Ohio
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Jane Khoury, MS;
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2 Department of Environmental Health, University of Cincinnati, Cincinnati, Ohio
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Dawn Kleindorfer, MD;
Dawn Kleindorfer, MD
1 Department of Neurology, University of Cincinnati, Cincinnati, Ohio
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Daniel Woo, MD;
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1 Department of Neurology, University of Cincinnati, Cincinnati, Ohio
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Alexander Schneider, MD;
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Kathleen Alwell, RN;
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Rosemary Miller, RN;
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Charles J. Moomaw, PHD;
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James Gebel, MD;
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Rakesh Shukla, PHD;
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Address correspondence and reprint requests to Brett M. Kissela, MD, Department of Neurology, University of Cincinnati, 231 Albert Sabin Way, ML 0525, Cincinnati, OH 45267-0525. E-mail: Brett.Kissela@uc.edu
Diabetes Care 2005;28(2):355–359
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Brett M. Kissela, Jane Khoury, Dawn Kleindorfer, Daniel Woo, Alexander Schneider, Kathleen Alwell, Rosemary Miller, Irene Ewing, Charles J. Moomaw, Jerzy P. Szaflarski, James Gebel, Rakesh Shukla, Joseph P. Broderick; Epidemiology of Ischemic Stroke in Patients With Diabetes: The Greater Cincinnati/Northern Kentucky Stroke Study. _Diabetes Care_ 1 February 2005; 28 (2): 355–359.
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OBJECTIVE— Diabetes is a well known risk factor for stroke, but the impact of diabetes on stroke incidence rates is not known. This study uses a population-based study to describe the epidemiology of ischemic stroke in diabetic patients.
RESEARCH DESIGN AND METHODS— Hospitalized cases were ascertained by ICD-9 discharge codes, prospective screening of emergency department admission logs, and review of coroner’s cases. A sampling scheme was used to ascertain cases in the out-of-hospital setting. All potential cases underwent detailed chart abstraction by study nurses followed by physician review. Diabetes-specific incidence rates, case fatality rates, and population-attributable risks were estimated.
RESULTS— Ischemic stroke patients with diabetes are younger, more likely to be African American, and more likely to have hypertension, myocardial infarction, and high cholesterol than nondiabetic patients. Age-specific incidence rates and rate ratios show that diabetes increases ischemic stroke incidence at all ages, but this risk is most prominent before age 55 in African Americans and before age 65 in whites. One-year case fatality rates after ischemic stroke are not different between those patients with and without diabetes.
CONCLUSIONS— Given the “epidemic” of diabetes, with substantially increasing diabetes prevalence each year across all age- and race/ethnicity groups, the significance of diabetes as a risk factor for stroke is becoming more evident. Diabetes is clearly one of the most important risk factors for ischemic stroke, especially in those patients less than 65 years of age. We estimate that 37–42% of all ischemic strokes in both African Americans and whites are attributable to the effects of diabetes alone or in combination with hypertension.
While diabetes is a well known risk factor for stroke, the magnitude of risk varies widely between studies (1–9), and the impact of diabetes on stroke incidence rates is not known. The increasing prevalence of diabetes makes it one of the most serious health problems in the U.S., and its role in macrovascular complications such as stroke is of increasing importance (10–12).
The prevalence of diabetes increases with age in all race/ethnicity groups (13). Within each age-group, prevalence is higher in African-American and Hispanic populations as compared with whites (13). Exploration of the race/ethnicity-specific differences in stroke risk conferred by diabetes is necessary for optimal prevention and management of stroke in these groups.
The Greater Cincinnati/Northern Kentucky Stroke Study (GCNKSS) population is similar to the U.S. with regard to median age, percent African American, median household income, and percent of population below the poverty level (14,15). Thus, our study provides estimates of stroke incidence and case fatality among African Americans and whites that can be generalized to the U.S. population for these racial groups (14,15). This study will describe the epidemiology of ischemic stroke in patients with diabetes, specifically with regard to age- and race-specific stroke incidence rates.
RESEARCH DESIGN AND METHODS
The GCNKSS population is defined as all residents of the Greater Cincinnati/Northern Kentucky region, which includes two southern Ohio counties and three contiguous northern Kentucky counties that border the Ohio River. Included in this area are 19 hospitals. Although residents of nearby counties seek care at the 19 hospitals, only residents of the five study area counties, as determined by zip code, are included as cases. This study was approved by the Institutional Review Board at all participating hospitals.
The methods of case ascertainment, case definition, and data collection have been previously reported (14,15). Briefly, study nurses retrospectively reviewed the medical records of all inpatients with primary or secondary stroke-related ICD-9 discharge diagnoses (430–438) from the 19 acute care hospitals in the study region. The study nurses also reviewed all autopsy cases where stroke was listed as the primary or secondary cause of death. Classification of race was as self-reported in the medical administrative record. Phase 1 of the GCNKSS involved ascertainment of hospitalized and autopsied strokes in African Americans between 1 January 1993 and 30 June 1993, and phase 2 involved collection of all strokes in the study population between 1 July 1993 and 30 June 1994. In addition to ascertaining inpatient strokes using the methodology described above, strokes in phase 2 were also ascertained in the out-of-hospital setting by monitoring all visits to 18 of the hospitals’ emergency departments (excluding Cincinnati Children’s Hospital), five county coroners’ offices, 16 public health clinics, and 14 hospital-based outpatient clinics and family practice centers. In addition, a random sample of 50 of 878 primary care physicians’ offices and 25 of 193 nursing homes in the greater Cincinnati metropolitan area were monitored. Events found only by out-of-hospital monitoring were checked against inpatient records to prevent double counting. Data from both phases 1 and 2 are presented in this report.
To qualify as an incidence case, a patient must have met the criteria for one of the five stroke categories adapted from the Classification for Cerebrovascular Diseases III and epidemiological studies of stroke: cerebral ischemia, intracerebral hemorrhage, subarachnoid hemorrhage, stroke of uncertain cause, or transient ischemic attack (TIA) (14–17).
Once potential cases were identified, the study nurse reviewed and abstracted the medical record. Data collected include stroke symptoms, past medical and surgical history, social history/habits, and disposition/outcome. Classification of a patient as having diabetes required documentation in the medical record that diabetes had been diagnosed before stroke. Diabetes was thus likely to be undercounted, because undiagnosed patients with diabetes would be classified as nondiabetic patients for this study. All borderline stroke cases were abstracted for physician review.
Ultimately, a study physician reviewed each abstract and all available neuroimaging data and then decided if a stroke or TIA had occurred. Study physicians also characterized imaging findings and assigned stroke subtype and mechanism to each patient based on all available information using definitions previously reported (14,15).
In 1995, a random digit dialing telephone survey was carried out in a 2,000- person cohort of randomly selected individuals whose demographics (age, race, and sex) closely matched the expected demographics of the population of ischemic stroke patients from the study population (18). The interviewees were asked if they had been diagnosed with medical conditions known to be risk factors for stroke, such as diabetes, using interview questions adapted from the NHANES III survey (18). Using this information, age-, race-, and sex-specific prevalence rates of diabetes could be calculated. It should be noted that prevalence rates obtained from this “referent” population are taken from an older group of respondents similar to our stroke population and thus do not represent a total population prevalence rate.
Statistical analyses
Data were managed and analyzed using SAS version 8.1 (SAS Institute, Cary, NC). Interval variables are reported as means ± SD. Categorical variables are reported as frequency and percent. Bivariate analyses were performed using Student’s t test, χ 2, or Fisher’s exact test as appropriate.
We calculated incidence rates of ischemic stroke for diabetic and nondiabetic patients. The numerator was the number of ischemic strokes as determined by study physicians, tabulated for diabetic and nondiabetic patients, respectively, and further subdivided by race, age, and sex. The denominator for each rate was calculated using self-reported age-, sex-, and race-specific data obtained from the 1995 telephone survey (18) as a diabetes prevalence weighting factor for the age-, sex, and race-specific population in 1993–1994. Age- and race-specific incidence rates and associated SEMs were calculated (sex-adjusted to the 2000 U.S. population). Disease-specific rate ratios for stroke in patients with diabetes were obtained by division of the diabetes-specific incidence rates by rates in nondiabetic patients.
A multivariable logistic regression was performed to determine the odds ratios (ORs) for ischemic stroke attributable to diabetes after controlling for age, sex, prior stroke, current smoking, hypertension, and heart disease. For this analysis, case subjects of ischemic stroke were compared with control subjects of similar age, race, and sex from the random digit dialing survey. The model variables included previous diagnoses of diabetes, hypertension, prior stroke, or myocardial infarction (MI) and reported current smoking. Population-attributable risks (PARs) and associated 95% CIs were calculated using the prevalence data from the random digit dialing survey.
RESULTS
After physician review, 4,264 events met case criteria for stroke, including ischemic stroke, intracerebral hemorrhage, subarachnoid hemorrhage, or TIA. The Greater Cincinnati/northern Kentucky region is essentially a biracial population, and less than 1% of all stroke cases during this study period occurred in other race/ethnicity groups. The data presented hereafter include only the 2,719 African-American and white cases with ischemic stroke, 9.5% of which were ascertained only in the out-of-hospital setting (14).
Demographic characteristics of the 2,719 diabetic and nondiabetic ischemic stroke patients are presented in Table 1. African Americans were more likely to have diabetes (36% of African-American stroke patients had diabetes compared with 30% in whites; P = 0.005). The ischemic stroke patients with diabetes were younger than the ischemic stroke patients without diabetes. They were also less likely to be current smokers at the time of their stroke and more likely to have been previously diagnosed with hypertension, high cholesterol, and MI. Among African Americans, diabetic stroke patients did not have higher rates of high cholesterol, MI, or atrial fibrillation compared with their nondiabetic counterparts, but they did have higher rates of hypertension. Among white stroke patients, those with diabetes had higher rates of hypertension, high cholesterol, and MI compared with those without diabetes.
Age-, sex- and race-specific prevalence rates of diabetes were estimated from our telephone survey information (Fig. 1). This population was similar to our stroke population with regard to age, race, and sex. Prevalence of diabetes was higher in African Americans versus whites and higher in men versus women. Prevalence generally increased with age as expected, although it should be noted that some age-, sex- and race-specific subgroups had small sample size due to this referent group being based upon the at-risk stroke population.
Age-, race-, and disease-specific incidence rates for ischemic stroke, sex-adjusted to the 2000 Census population, are displayed in Table 2. White patients with diabetes had higher stroke incidence rates compared with nondiabetic whites in every age category; a similar diabetes effect was seen in African-American patients, although the incidence rates were generally higher for each age-group than in whites. Male stroke risk was substantially higher than female stroke risk, and this difference was accentuated in African Americans compared with whites (data not shown).
Race-specific risk ratios (RRs) are displayed in Table 2. Risk for ischemic stroke in white diabetic patients is higher at every age-group compared with nondiabetic patients, with a maximum RR of 5.3 in the 45- to 54-year age-group. Among African Americans, risk in diabetic compared with nondiabetic patients is greater in all age-groups except the youngest and oldest, with a maximum RR of 9.9 in the 35- to 44-year age-group. A substantial peak in stroke risk is seen in the 35- to 54-year age-groups in African Americans and in the 45- to 64-year age group in whites.
Case fatality rates were determined. Diabetic patients do not have significantly different case fatality rates in the first year after ischemic stroke when compared with nondiabetic patients overall or when subdivided by race. Given the substantial risk for stroke conferred by diabetes, this suggests that there are proportionately more patients with diabetes surviving with disability after stroke compared with nondiabetic patients.
Multivariable PARs are presented in Table 3. It can be seen that the OR for stroke due to a history of both diabetes and hypertension is substantially greater than for either condition alone. Diabetes alone has a higher OR than hypertension alone, but PAR is higher for hypertension given the prevalence of hypertension in both races. Diabetes and hypertension, either alone or in combination, account for 37–42% of the PARs in both race/ethnicity groups.
CONCLUSIONS
Diabetic patients with ischemic stroke are younger, more likely to be African American, and more likely to have hypertension, MI, and high cholesterol than their nondiabetic counterparts. Our data are in keeping with other studies in showing a significantly increased risk for ischemic stroke conferred by diabetes but are the first to estimate diabetes-specific incidence rates in a race- and age-specific fashion. Our data demonstrate a race-specific effect, with higher risk for African Americans with diabetes, and an age-specific risk, with substantially increased risk for diabetic patients less than age 65 compared with those without diabetes. Finally, the case fatality after ischemic stroke was not found to be higher in patients with diabetes. Given that in many studies surviving stroke patients with diabetes have been shown to have worse functional outcome, diabetic patients are more likely to survive with disability after ischemic stroke.
Prevalence data from our population survey (subjects with similar demographics to stroke patients) show that diabetes is most common in African-American men, prevalence is similar in white men and African-American women, and white women are least likely to have diabetes. These results are similar to findings in the general population (13) and are helpful in understanding the imbalance in stroke risk factors observed. Diabetes is part of the “metabolic syndrome,” which also includes high blood pressure, high cholesterol, and obesity. African Americans are more likely to have this syndrome than whites (19), and patients with diabetes are likely to have the other risk factor components of this syndrome. With regard to other risk factor imbalances among our ischemic stroke patients, whites had more MIs and atrial fibrillation. While this may be a race-specific effect that has been reported previously, it may also be due in part to the younger age of African-American stroke patients as compared with whites.
Since diabetes is more prevalent in African Americans and other race/ethnicity groups compared with whites, our study can be used to measure the attributable risk of diabetes on stroke incidence in African Americans and whites. A multivariate PAR analysis with regard to ischemic stroke among predominantly white patients in Rochester, Minnesota, found a 5% PAR due to diabetes (3). In the Northern Manhattan Stroke Study, a similar multivariable analysis found PARs of 14 and 10% conferred by diabetes for Caribbean-derived African Americans and Hispanics, respectively, but no appreciable PAR for whites (20). Our study finds multivariable PARs of 20–21% for African-American and white patients who have the combination of diabetes and hypertension. Diabetes and hypertension seen alone account for an additional 15–22% PAR that varies by race. The disparity seen between our results and the previously published analyses may be related to how our analysis was performed and in particular due to the referent population that was used. However, the difference seen might also be related to the varied demographics of the populations being studied, including variations in diabetes prevalence and stroke incidence. Our white population has lower educational achievement and lower measures of socioeconomic status compared with Rochester, Minnesota (16), and has higher incidence of stroke compared with the white population in Northern Manhattan (21). The African-American population of Northern Manhattan is predominately of Caribbean descent, and our population is predominately African American (21).
Combining age- and race-specific prevalence of diabetes (22) with the projected U.S. population for 2002 and our ischemic stroke incidence rates displayed in Table 2, we conservatively estimate that a minimum of 143,000–154,000 ischemic strokes occurred in diabetic patients in 2002 in the U.S. This represents 20–22% of the 705,000 strokes estimated to occur in 2002 (14). However, we only classified those patients who had previously been diagnosed with diabetes as “diabetic stroke patients.” Certainly there are many patients who had not been diagnosed with diabetes at the time of their ischemic stroke, and thus our incidence rates, RRs, and estimate of diabetic strokes in 2002 are underestimates.
Given the “epidemic” of diabetes, with substantially increasing diabetes prevalence each year across all age- and race/ethnicity groups, the importance of diabetes as a risk factor for stroke is increasing. Diabetes is clearly one of the most important risk factors for ischemic stroke, especially for those occurring before age 65. We estimate that 25–26% of all ischemic strokes are attributable to the effects of diabetes alone or in combination with hypertension, and this may well be an underestimate. Race-specific data about stroke incidence in patients with diabetes must continue to be studied to see if incidence rates are changing over time.
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Figure 1—
Age-, race-, and sex-specific prevalence of diabetes in 1995 telephone survey (subjects similar with regard to age, race, and sex to the 1993–1994 stroke population).
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Figure 1—
Age-, race-, and sex-specific prevalence of diabetes in 1995 telephone survey (subjects similar with regard to age, race, and sex to the 1993–1994 stroke population).
Close modal
Table 1—
Characteristics of ischemic stroke patients by diabetes and race
| . | All . | . | . | Black . | . | . | White . | . | . |
--- --- --- --- --- |
| . | Diabetic . | Nondiabetic . | P . | Diabetic . | Nondiabetic . | P . | Diabetic . | Nondiabetic . | P . |
| n | 856 | 1,863 | | 259 | 464 | | 597 | 1,399 | |
| Age (years) | 70 ± 11 | 72 ± 15 | <0.0001 | 67 ± 11 | 68 ± 17 | 0.6 | 72 ± 11 | 74 ± 13 | <0.0001 |
| History of hypertension | 676 (79) | 1,061 (57) | <0.0001 | 219 (84) | 289 (62) | <0.0001 | 457 (76) | 772 (55) | <0.0001 |
| History of high cholesterol | 135 (16) | 177 (10) | <0.0001 | 35 (14) | 51 (11) | 0.3 | 100 (17) | 126 (9) | <0.0001 |
| History of myocardial infarction | 193 (22) | 272 (15) | <0.0001 | 43 (17) | 75 (16) | 0.9 | 150 (25) | 197 (14) | <0.0001 |
| History of atrial fibrillation | 134 (16) | 266 (14) | 0.4 | 19 (7) | 45 (10) | 0.3 | 115 (19) | 221 (16) | 0.1 |
| Current smoking | 157 (18) | 408 (22) | 0.03 | 63 (24) | 140 (30) | 0.1 | 94 (16) | 268 (19) | 0.1 |
| . | All . | . | . | Black . | . | . | White . | . | . |
--- --- --- --- --- |
| . | Diabetic . | Nondiabetic . | P . | Diabetic . | Nondiabetic . | P . | Diabetic . | Nondiabetic . | P . |
| n | 856 | 1,863 | | 259 | 464 | | 597 | 1,399 | |
| Age (years) | 70 ± 11 | 72 ± 15 | <0.0001 | 67 ± 11 | 68 ± 17 | 0.6 | 72 ± 11 | 74 ± 13 | <0.0001 |
| History of hypertension | 676 (79) | 1,061 (57) | <0.0001 | 219 (84) | 289 (62) | <0.0001 | 457 (76) | 772 (55) | <0.0001 |
| History of high cholesterol | 135 (16) | 177 (10) | <0.0001 | 35 (14) | 51 (11) | 0.3 | 100 (17) | 126 (9) | <0.0001 |
| History of myocardial infarction | 193 (22) | 272 (15) | <0.0001 | 43 (17) | 75 (16) | 0.9 | 150 (25) | 197 (14) | <0.0001 |
| History of atrial fibrillation | 134 (16) | 266 (14) | 0.4 | 19 (7) | 45 (10) | 0.3 | 115 (19) | 221 (16) | 0.1 |
| Current smoking | 157 (18) | 408 (22) | 0.03 | 63 (24) | 140 (30) | 0.1 | 94 (16) | 268 (19) | 0.1 |
Data are means ± SD or n (%).
View Large
Table 2—
Incidence rates of ischemic stroke in nondiabetic and diabetic subjects per 100,000 (95% CI) and RRs for ischemic stroke with diabetes
| Age-group . | Black nondiabetic . | Black diabetic . | Black risk ratio for diabetes . | White nondiabetic . | White diabetic . | White risk ratio for diabetes . |
--- --- ---
| <35 years | 9 (3–14) | 0 | 0 (−1.0 to 1.0) | 3 (1–4) | 6 (−6 to 17) | 2.1 (0.8–3.4) |
| 35–44 years | 42 (18–66) | 413 (38–788) | 9.9 (8.5–11.3) | 11 (6–16) | 28 (0–56) | 2.6 (1.1–4.1) |
| 45–54 years | 130 (75–185) | 1,079 (500–1,659) | 8.3 (7.3–9.3) | 56 (42–69) | 295 (172–418) | 5.3 (3.9–6.7) |
| 55–64 years | 322 (220–425) | 930 (515–1,344) | 2.9 (1.8–4.0) | 152 (125–178) | 708 (539–878) | 4.7 (3.4–6.0) |
| 65–74 years | 616 (451–781) | 1,731 (1,238–2,224) | 2.8 (1.8–3.8) | 404 (355–453) | 896 (741–1,051) | 2.2 (1.1–3.3) |
| 75–84 years | 1,047 (761–1,333) | 1,858 (1,044–2,673) | 1.8 (0.8–2.7) | 828 (741–915) | 1,939 (1,621–2,257) | 2.3 (1.5–3.2) |
| ≥85 years | 2,033 (1,280–2,786) | 1,686 (466–2,907) | 0.8 (0.0–1.6) | 1,438 (1,248–1,628) | 3,308 (2,330–4,285) | 2.3 (1.7–2.9) |
| Age-group . | Black nondiabetic . | Black diabetic . | Black risk ratio for diabetes . | White nondiabetic . | White diabetic . | White risk ratio for diabetes . |
--- --- ---
| <35 years | 9 (3–14) | 0 | 0 (−1.0 to 1.0) | 3 (1–4) | 6 (−6 to 17) | 2.1 (0.8–3.4) |
| 35–44 years | 42 (18–66) | 413 (38–788) | 9.9 (8.5–11.3) | 11 (6–16) | 28 (0–56) | 2.6 (1.1–4.1) |
| 45–54 years | 130 (75–185) | 1,079 (500–1,659) | 8.3 (7.3–9.3) | 56 (42–69) | 295 (172–418) | 5.3 (3.9–6.7) |
| 55–64 years | 322 (220–425) | 930 (515–1,344) | 2.9 (1.8–4.0) | 152 (125–178) | 708 (539–878) | 4.7 (3.4–6.0) |
| 65–74 years | 616 (451–781) | 1,731 (1,238–2,224) | 2.8 (1.8–3.8) | 404 (355–453) | 896 (741–1,051) | 2.2 (1.1–3.3) |
| 75–84 years | 1,047 (761–1,333) | 1,858 (1,044–2,673) | 1.8 (0.8–2.7) | 828 (741–915) | 1,939 (1,621–2,257) | 2.3 (1.5–3.2) |
| ≥85 years | 2,033 (1,280–2,786) | 1,686 (466–2,907) | 0.8 (0.0–1.6) | 1,438 (1,248–1,628) | 3,308 (2,330–4,285) | 2.3 (1.7–2.9) |
View Large
Table 3—
Multivariable logistic regression model for ischemic stroke
| . | African-American patients . | . | . | . | Caucasian patients . | . | . | . |
--- --- --- ---
| . | OR . | 95% CI . | PAR . | 95% CI . | OR . | 95% CI . | PAR . | 95% CI . |
| Age (per year) | 1.02 | 1.01–1.03 | | | 1.06 | 1.05–1.06 | | |
| Male | 1.2 | 1.0–1.6 | | | 1.4 | 1.2–1.6 | | |
| History of prior stroke | 4.7 | 3.3–7.0 | 22.3 | 18.8–25.7 | 3.6 | 2.8–4.7 | 14.8 | 13.4–16.1 |
| History of both diabetes and hypertension | 3.0 | 2.1–4.3 | 21.1 | 18.1–24.0 | 4.5 | 3.5–5.8 | 20.6 | 19.0–22.1 |
| History of diabetes only | 2.7 | 1.5–5.2 | 5.6 | 3.6–7.5 | 2.1 | 1.5–3.0 | 5.2 | 4.3–6.1 |
| History of hypertension only | 1.3 | 0.9–1.7 | 9.8 | 3.7–15.5 | 1.6 | 1.4–1.9 | 16.2 | 13.8–18.4 |
| Current smoking | 1.5 | 1.1–1.9 | 10.9 | 6.5–15.2 | 1.7 | 1.4–2.2 | 12.5 | 10.7–14.2 |
| History of myocardial infarction | 1.1 | 0.8–1.7 | 1.4 | 0–3.6 | 1.1 | 0.9–1.4 | 1.0 | 0–2.0 |
| . | African-American patients . | . | . | . | Caucasian patients . | . | . | . |
--- --- --- ---
| . | OR . | 95% CI . | PAR . | 95% CI . | OR . | 95% CI . | PAR . | 95% CI . |
| Age (per year) | 1.02 | 1.01–1.03 | | | 1.06 | 1.05–1.06 | | |
| Male | 1.2 | 1.0–1.6 | | | 1.4 | 1.2–1.6 | | |
| History of prior stroke | 4.7 | 3.3–7.0 | 22.3 | 18.8–25.7 | 3.6 | 2.8–4.7 | 14.8 | 13.4–16.1 |
| History of both diabetes and hypertension | 3.0 | 2.1–4.3 | 21.1 | 18.1–24.0 | 4.5 | 3.5–5.8 | 20.6 | 19.0–22.1 |
| History of diabetes only | 2.7 | 1.5–5.2 | 5.6 | 3.6–7.5 | 2.1 | 1.5–3.0 | 5.2 | 4.3–6.1 |
| History of hypertension only | 1.3 | 0.9–1.7 | 9.8 | 3.7–15.5 | 1.6 | 1.4–1.9 | 16.2 | 13.8–18.4 |
| Current smoking | 1.5 | 1.1–1.9 | 10.9 | 6.5–15.2 | 1.7 | 1.4–2.2 | 12.5 | 10.7–14.2 |
| History of myocardial infarction | 1.1 | 0.8–1.7 | 1.4 | 0–3.6 | 1.1 | 0.9–1.4 | 1.0 | 0–2.0 |
PARs calculated from GCNKSS prevalence data.
View Large
Article Information
This work supported by the National Institutes of Neurological Diseases and Stroke (R-01 NS-30678).
The authors thank participating hospitals, nursing homes, physician/outpatient clinics, local departments of health, and coroners’ offices.
References
1
Abbott RD, Donahue RP, MacMahon SW, Reed DM, Yano K: Diabetes and the risk of stroke: the Honolulu Heart Program.
JAMA
257
:
949
–952,
1987
2
Whisnant JP, Brown RD, Petty GW, O’Fallon WM, Sicks JD, Wiebers DO: Comparison of population-based models of risk factors for TIA and ischemic stroke.
Neurology
53
:
532
–536,
1999
3
Whisnant JP: Modeling of risk factors for ischemic stroke: the Willis lecture.
Stroke
28
:
1840
–1844,
1997
4
Jorgensen H, Nakayama H, Raaschou HO, Olsen TS: Stroke in patients with diabetes: the Copenhagen Stroke Study.
Stroke
25
:
1977
–1984,
1994
5
Rodriguez BL, D’Agostino R, Abbott RD, Kagan A, Burchfiel CM, Yano K, Ross GW, Silbershatz H, Higgins MW, Popper J, Wolf PA, Curb JD: Risk of hospitalized stroke in men enrolled in the Honolulu Heart Program and the Framingham Study: a comparison of incidence and risk factor effects.
Stroke
33
:
230
–236,
2002
6
D’Agostino RB, Wolf PA, Belanger AJ, Kannel WB: Stroke risk profile: adjustment for antihypertensive medication: the Framingham Study.
Stroke
25
:
40
–43,
1994
7
Kuusisto J, Mykkanen L, Pyorala K, Laakso M: Non-insulin-dependent diabetes and its metabolic control are important predictors of stroke in elderly subjects.
Stroke
25
:
1157
–1164,
1994
8
Stamler J, Vaccaro O, Neaton JD, Wentworth D: Diabetes, other risk factors, and 12-yr cardiovascular mortality for men screened in the Multiple Risk Factor Intervention Trial.
Diabetes Care
16
:
434
–444,
1993
9
Tuomilehto J, Rastenyte D, Jousilahti P, Sarti C, Vartiainen E: Diabetes mellitus as a risk factor for death from stroke: prospective study of the middle-aged Finnish population.
Stroke
27
:
210
–215,
1996
10
Mokdad AH, Ford ES, Bowman BA, Nelson DE, Engelgau MM, Vinicor F, Marks JS: The continuing increase of diabetes in the U.S.
Diabetes Care
24
:
412
,
2001
11
Mokdad AH, Ford ES, Bowman BA, Dietz WH, Vinicor F, Bales VS, Marks JS: Prevalence of obesity, diabetes, and obesity-related health risk factors, 2001.
JAMA
289
:
76
–79,
2003
12
Howard BV, Rodriguez BL, Bennett PH, Harris MI, Hamman R, Kuller LH, Pearson TA, Wylie-Rosett J: Prevention conference VI: diabetes and cardiovascular disease: writing group I: epidemiology.
Circulation
105
:
e132
–e137,
2002
13
Harris MI, Flegal KM, Cowie CC, Eberhardt MS, Goldstein DE, Little RR, Wiedmeyer HM, Byrd-Holt DD: Prevalence of diabetes, impaired fasting glucose, and impaired glucose tolerance in U.S. Adults: the Third National Health and Nutrition Examination Survey, 1988–1994.
Diabetes Care
21
:
518
–524,
1998
14
Kissela B, Schneider A, Kleindorfer D, Khoury J, Miller R, Alwell K, Woo D, Szaflarski J, Gebel J, Moomaw C, Pancioli A, Jauch E, Shukla R, Broderick J: Stroke in a biracial population: the excess burden of stroke among blacks.
Stroke
35
:
426
–431,
2004
15
Broderick J, Brott T, Kothari R, Miller R, Khoury J, Pancioli A, Gebel J, Mills D, Minneci L, Shukla R: The Greater Cincinnati/Northern Kentucky Stroke Study: preliminary first-ever total incidence rates of stroke among blacks.
Stroke
29
:
415
–421,
1998
16
Brown RD, Whisnant JP, Sicks JD, O’Fallon WM, Wiebers DO: Stroke incidence, prevalence, and survival: secular trends in Rochester, Minnesota, through 1989.
Stroke
27
:
373
–380,
1996
17
National Institute of Neurological Disorders and Stroke: Special report from the National Institute of Neurological Disorders and Stroke: classification of cerebrovascular diseases III.
Stroke
21
:
637
–676,
1990
18
Pancioli AM, Broderick J, Kothari R, Brott T, Tuchfarber A, Miller R, Khoury J, Jauch E: Public perception of stroke warning signs and knowledge of potential risk factors.
JAMA
279
:
1288
–1292,
1998
19
Ford ES, Giles WH, Dietz WH: Prevalence of the metabolic syndrome among U.S. adults: findings from the Third National Health and Nutrition Examination Survey.
JAMA
287
:
356
–359,
2002
20
Sacco RL, Boden-Albala B, Abel G, Lin IF, Elkind M, Hauser WA, Paik MC, Shea S: Race-ethnic disparities in the impact of stroke risk factors: the Northern Manhattan Stroke Study.
Stroke
32
:
1725
–1731,
2001
21
Sacco RL, Boden-Albala B, Gan R, Chen X, Kargman DE, Shea S, Paik MC, Hauser WA: Stroke incidence among white, black, and Hispanic residents of an urban community: the Northern Manhattan Stroke Study.
Am J Epidemiol
147
:
259
–268,
1998
22
Schneider AT, Pancioli AM, Khoury JC, Rademacher E, Tuchfarber A, Miller R, Woo D, Kissela B, Broderick JP: Trends in community knowledge of the warning signs and risk factors for stroke.
JAMA
289
:
343
–346,
2003
A table elsewhere in this issue shows conventional and Système International (SI) units and conversion factors for many substances.
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810 | https://en.wikipedia.org/wiki/Spieker_circle | Spieker circle - Wikipedia
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From Wikipedia, the free encyclopedia
Inscribed circle of a triangle's medial triangle
Triangle △ABC and its medial triangle
Spieker circle of △ABC (incircle of the medial triangle; centered at the Spieker centerX 10)
Cleavers of the triangle (concurrent at the Spieker center)
In geometry, the incircle of the medial triangle of a triangle is the Spieker circle, named after 19th-century German geometer Theodor Spieker. Its center, the Spieker center, in addition to being the incenter of the medial triangle, is the center of mass of the uniform-density boundary of triangle. The Spieker center is also the point where all three cleavers of the triangle (perimeter bisectors with an endpoint at a side's midpoint) intersect each other.
History
[edit]
The Spieker circle and Spieker center are named after Theodor Spieker, a mathematician and professor from Potsdam, Germany. In 1862, he published Lehrbuch der ebenen geometrie mit übungsaufgaben für höhere lehranstalten, dealing with planar geometry. Due to this publication, influential in the lives of many famous scientists and mathematicians including Albert Einstein, Spieker became the mathematician for whom the Spieker circle and center were named.
Construction
[edit]
To find the Spieker circle of a triangle, the medial triangle must first be constructed from the midpoints of each side of the original triangle. The circle is then constructed in such a way that each side of the medial triangle is tangent to the circle within the medial triangle, creating the incircle. This circle center is named the Spieker center.
Nagel points and lines
[edit]
Spieker circles also have relations to Nagel points. The incenter of the triangle and the Nagel point form a line within the Spieker circle. The middle of this line segment is the Spieker center. The Nagel line is formed by the incenter of the triangle, the Nagel point, and the centroid of the triangle. The Spieker center will always lie on this line.
Nine-point circle and Euler line
[edit]
Spieker circles were first found to be very similar to nine-point circles by Julian Coolidge. At this time, it was not yet identified as the Spieker circle, but is referred to as the "P circle" throughout the book. The nine-point circle with the Euler line and the Spieker circle with the Nagel line are analogous to each other, but are not duals, only having dual-like similarities. One similarity between the nine-point circle and the Spieker circle deals with their construction. The nine-point circle is the circumscribed circle of the medial triangle, while the Spieker circle is the inscribed circle of the medial triangle. With relation to their associated lines, the incenter for the Nagel line relates to the circumcenter for the Euler line. Another analogous pair of points is the Nagel point and the orthocenter, with the Nagel point associated with the Spieker circle and the orthocenter associated with the nine-point circle. In the first case, the splitters of the reference triangle, which pass through the Nagel point, also pass through the tangencies of the Spieker circle with the sides of medial triangle. In the second case, the altitudes of the reference triangle, which pass through the orthocenter, also pass through the intersections of the nine-point circle with the sides of the reference triangle.
Spieker conic
[edit]
The nine-point circle with the Euler line was generalized into the nine-point conic. Through a similar process, due to the analogous properties of the two circles, the Spieker circle was also able to be generalized into the Spieker conic. The Spieker conic is still found within the medial triangle and touches each side of the medial triangle, however it does not meet those sides of the triangle at the same points. If lines are constructed from each vertex of the medial triangle to the Nagel point, then the midpoint of each of those lines can be found. Also, the midpoints of each side of the medial triangle are found and connected to the midpoint of the opposite line through the Nagel point. Each of these lines share a common midpoint, S. With each of these lines reflected through S, the result is 6 points within the medial triangle. Draw a conic through any 5 of these reflected points and the conic will touch the final point. This was proven by de Villiers in 2006.
Spieker radical circle
[edit]
The Spieker radical circle is the circle, centered at the Spieker center, which is orthogonal to the three excircles of the medial triangle.
References
[edit]
^ Jump up to: abcdefghijklmnopde Villiers, Michael (June 2006). "A generalisation of the Spieker circle and Nagel line". Pythagoras. 63: 30–37.
^ Jump up to: abcCoolidge, Julian L. (1916). A treatise on the circle and the sphere. Oxford University Press. pp.53–57.
^ Jump up to: abcde Villiers, M. (2007). "Spieker Conic and generalization of Nagle line". Dynamic Mathematics Learning.
^Weisstein, Eric W. "Excircles Radical Circle". MathWorld- A Wolfram Web Resource.
^Weisstein, Eric W. "Radical Circle". MathWorld- A Wolfram Web Resource.
Johnson, Roger A. (1929). Modern Geometry. Boston: Houghton Mifflin. Dover reprint, 1960.
Kimberling, Clark (1998). "Triangle centers and central triangles". Congressus Numerantium. 129: i–xxv, 1–295.
External links
[edit]
Spieker Conic and generalization of Nagel line at Dynamic Geometry Sketches Generalizes Spieker circle and associated Nagel line.
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Is it possible for a quadratic and a line to only have one point of intersection?
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James Buddenhagen
Lives in Xico,Veracruz.Mexico (2006–present) · Author has 2.6K answers and 4.2M answer views
·5y
Is it possible for a quadratic and a line to only have one point of intersection?
One point of contact, yes. But it is not a point of intersection because the line does not cross the curve, just touches and then stays on the same side of the curve.
Reason: when you solve the system of equations, for the quadratic and the line, the result is a quadratic (with real coefficients), and every quadratic has two roots, if not both imaginary (no point of intersection) then both real so two points of intersection, unless a double root, which is a point of contact as described above.
Example: the quadratic
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Is it possible for a quadratic and a line to only have one point of intersection?
One point of contact, yes. But it is not a point of intersection because the line does not cross the curve, just touches and then stays on the same side of the curve.
Reason: when you solve the system of equations, for the quadratic and the line, the result is a quadratic (with real coefficients), and every quadratic has two roots, if not both imaginary (no point of intersection) then both real so two points of intersection, unless a double root, which is a point of contact as described above.
Example: the quadratic y=x^2 and the line y=0 (i.e. the x-axis) have a point of contact at the origin.
Edit: Aaron Kelly points out that a vertical line through the origin will intersect the parabola y=x^2 in exactly one point. So, I stand corrected (see comments). Possibly one might argue that “the point at infinity” is the second intersection point in that case.
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How do I show the tangent line at any point of a quadratic equation y=ax^2+b+c intersects the quadratic curve only at that point?
What is the significance of using the Point-Slope Form to find the equation of a line parallel to another given line, and how does it simplify the process?
What is the equation of a line that passes through two points with equal slopes?
Given one point on a line and its direction (slope), is it possible to write an equation for that line?
In general, why is it assumed there is only one line passing through two points with a given slope, rather than an infinite amount of overlapping lines?
Aaron Kelly
B.S. in Mathematics, Dickinson State University (Graduated 2012)
·5y
I would posit that any quadratic has infinitely many lines that only intersect it at one point. To prove, consider all the vertical lines where this is the case. Is it not true for all vertical lines? Quadratic functions have an infinite X range, so I believe it is.
In addition, I also believe that this is ONLY true for vertical lines, assuming you do not count tangent lines as “intersecting”. If you consider a line with any non-vertical slope, because the quadratic is constantly getting “steeper”, it should eventually cross the line (even for incredibly “steep” lines).
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Deleted Account
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Sure. A quadratic curve and any vertical line would intersect in only one point. Or a horizontal line for a horizontal quadratic in the form of x = ay²+ by + c
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Sure. A quadratic curve and any vertical line would intersect in only one point. Or a horizontal line for a horizontal quadratic in the form of x = ay²+ by + c
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Gary Ward
MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views
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Can a straight line be tangent to a parabola at only one point? Are there any examples where this is not the case?
Can a straight line be tangent to a parabola at only one point? Are there any examples where this is not the case?
Yes, to the first, and no, to the second. A parabola always curves in the same direction making it impossible for a straight line to be tangent at more than one point.
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Can a straight line be tangent to a parabola at only one point? Are there any examples where this is not the case?
Yes, to the first, and no, to the second. A parabola always curves in the same direction making it impossible for a straight line to be tangent at more than one point.
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Can there be more than one equation for a line?
Can a quadratic equation have one solution?
What is the method for solving for x when two lines intersect without finding their intersection point first?
What is the equation of a line that passes through two points and has a slope of negative one?
How do you find the intersection point of two quadratic equations?
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Yes, it is possible for a quadratic function and a line to only have one point of intersection. This occurs when the line is tangent to the quadratic curve.
Explanation:
Quadratic Function: A quadratic function can be expressed in the standard form as f(x)=a x 2+b x+c f(x)=a x 2+b x+c, where a≠0 a≠0.
Line Equation: A line can be represented by the linear equation y=m x+b y=m x+b, where m m is the slope and b b is the y-intercept.
Finding Points of Intersection: To find the points of intersection, set the equations equal to each other:
a x 2+b x+c=m x+b a x 2+b x+c=m x+b
Rearranging gives:
a x 2+(b−m)x+(c−b)=0 a x 2+(b−m)x+(c−b)=0
4. Disc
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Yes, it is possible for a quadratic function and a line to only have one point of intersection. This occurs when the line is tangent to the quadratic curve.
Explanation:
Quadratic Function: A quadratic function can be expressed in the standard form as f(x)=a x 2+b x+c f(x)=a x 2+b x+c, where a≠0 a≠0.
Line Equation: A line can be represented by the linear equation y=m x+b y=m x+b, where m m is the slope and b b is the y-intercept.
Finding Points of Intersection: To find the points of intersection, set the equations equal to each other:
a x 2+b x+c=m x+b a x 2+b x+c=m x+b
Rearranging gives:
a x 2+(b−m)x+(c−b)=0 a x 2+(b−m)x+(c−b)=0
4. Discriminant: The solutions to this quadratic equation can be analyzed using the discriminant D D:
D=(b−m)2−4 a(c−b)D=(b−m)2−4 a(c−b)
If D>0 D>0: Two distinct intersection points.
If D=0 D=0: One point of intersection (tangency).
If D<0 D<0: No intersection points.
Conclusion:
For the quadratic and the line to intersect at exactly one point, the discriminant must equal zero. This indicates that the line is tangent to the quadratic curve at that point.
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Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983) · Author has 3.9K answers and 4M answer views
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What is the equation for two lines that intersect at one point?
First, we can find the point where two given lines intersect. Then, if we specify a slope, m m, for a desired line through that intersection point, we can write an equation for that line by using the point-slope form for a line.
Given two lines:
a x+b y+c=0 a x+b y+c=0
p x+q y+r=0 p x+q y+r=0
These lines can be written as:
y=−(a x+c b)y=−(a x+c b)
y=−(p x+r q)y=−(p x+r q)
At the intersection point, (x i,y i)(x i,y i), the ordinates are equal such that these equations are equal:
−(a x i+c b)=−(p x i+r q)−(a x i+c b)=−(p x i+r q)
If we solve for x i x i, we find the abscissa of the intersection point:
(a x i+c)q=(p x i+r)b(a x i+c)q=(p x i+r)b
a q x i+c q=b p x i+b r a q x i+c q=b p x i+b r
(a q−b p)x i=b r−c q(a q−b p)x i=b r−c q
x_i=\frac{br-cq x_i=\frac{br-cq
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First, we can find the point where two given lines intersect. Then, if we specify a slope, m m, for a desired line through that intersection point, we can write an equation for that line by using the point-slope form for a line.
Given two lines:
a x+b y+c=0 a x+b y+c=0
p x+q y+r=0 p x+q y+r=0
These lines can be written as:
y=−(a x+c b)y=−(a x+c b)
y=−(p x+r q)y=−(p x+r q)
At the intersection point, (x i,y i)(x i,y i), the ordinates are equal such that these equations are equal:
−(a x i+c b)=−(p x i+r q)−(a x i+c b)=−(p x i+r q)
If we solve for x i x i, we find the abscissa of the intersection point:
(a x i+c)q=(p x i+r)b(a x i+c)q=(p x i+r)b
a q x i+c q=b p x i+b r a q x i+c q=b p x i+b r
(a q−b p)x i=b r−c q(a q−b p)x i=b r−c q
x i=b r−c q a q−b p x i=b r−c q a q−b p
where a q−b p≠0 a q−b p≠0. Otherwise, the given lines do not intersect at one point but are either parallel or coincident.
The corresponding ordinate at the intersection point can be found from either line equation at this abscissa:
y i=−(a x i+c b)y i=−(a x i+c b)
y i=−(a(b r−c q a q−b p)+c b)y i=−(a(b r−c q a q−b p)+c b)
y i=a c q−a b r a b q−b 2 p−c(a q−b p)b(a q−b p)y i=a c q−a b r a b q−b 2 p−c(a q−b p)b(a q−b p)
y i=a c q−a b r−a c q+b c p a b q−b 2 p y i=a c q−a b r−a c q+b c p a b q−b 2 p
y i=b(c p−a r)b(a q−b p)y i=b(c p−a r)b(a q−b p)
y i=c p−a r a q−b p y i=c p−a r a q−b p
Therefore, the intersection point is:
(x i,y i)=(b r−c q a q−b p,c p−a r a q−b p)(x i,y i)=(b r−c q a q−b p,c p−a r a q−b p)
Any line with slope m m through this intersection point is:
y−y i=m(x−x i)y−y i=m(x−x i)
For example, consider the lines:
2 x+3 y+4=0 2 x+3 y+4=0
5 x+6 y+7=0 5 x+6 y+7=0
where a=2 a=2, b=3 b=3, c=4 c=4, p=5 p=5, q=6 q=6, and r=7 r=7. The intersection point is:
((3)(7)−(4)(6)(2)(6)−(3)(5),(4)(5)−(2)(7)(2)(6)−(3)(5))((3)(7)−(4)(6)(2)(6)−(3)(5),(4)(5)−(2)(7)(2)(6)−(3)(5))
(21−24 12−15,20−14 12−15)(21−24 12−15,20−14 12−15)
(−3−3,6−3)(−3−3,6−3)
(1,−2)(1,−2)
A plot looks like this:
If we want a line with a slope m=2 m=2 through this intersection point, then:
y+2=2(x−1)y+2=2(x−1)
y=2 x−4 y=2 x−4
The plot now looks like this:
Additional lines can be added through this intersection point, as needed, by specifying additional slopes.
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Ioannis Baroumas
Software engineer, High school math tutor · Author has 53 answers and 313K answer views
·8y
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Do parallel lines meet at one point?
In Euclidean Geometry, the one they teach at school, parallel lines never meet, hence the do not have any common point.
That is the definition of parallel lines: They are always the same distance apart and will never meet.
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In Euclidean Geometry, the one they teach at school, parallel lines never meet, hence the do not have any common point.
That is the definition of parallel lines: They are always the same distance apart and will never meet.
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Alexandru Carausu
Former Assoc. Prof. Dr. (Ret) at Technical University "Gh. Asachi" Iasi (1978–2010) · Author has 3K answers and 875.9K answer views
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Can two lines intersect without calculating their equations and solving them simultaneously using basic algebra?
Let ( L _1 ) and ( L _2 ) be the two lines in this questions. Their equations should not be calculated. But they must be described by either some geometric properties or by their analytic equations, in one of the following (equivalent) forms :
By their general equation s of the form ( L_i ) : A_i x + B_i y + C_i = 0 , ( i = 1 , 2 ) .
By their equations in terms of slope & intercept on ( Oy ), ( L_i ) : y = m_i x + n_i .
By their intercepts on both axes of coordinates, ( L_i ) : x / a_i + y / b_i = 1 .
In this answer, I do not consider the case of two lines in the 3D space, but only in a plane ( P )
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Let ( L _1 ) and ( L _2 ) be the two lines in this questions. Their equations should not be calculated. But they must be described by either some geometric properties or by their analytic equations, in one of the following (equivalent) forms :
By their general equation s of the form ( L_i ) : A_i x + B_i y + C_i = 0 , ( i = 1 , 2 ) .
By their equations in terms of slope & intercept on ( Oy ), ( L_i ) : y = m_i x + n_i .
By their intercepts on both axes of coordinates, ( L_i ) : x / a_i + y / b_i = 1 .
In this answer, I do not consider the case of two lines in the 3D space, but only in a plane ( P ) . In particular, this can be the plane of Cartesian coordinates ( O ; x , y ) or ( x O y ) , where x &y are the coordinates of points and / or vectors in the orthonormal frame ( O ; i , j ) . It is well known, since more than two millenia, the condition for two lines to be secant, what means that they share a unique common point : ( L _1 ) ∩ ( L _2 ) = { Q } . This condition was stated in EUCLID’s ELEMENTA (in the 3-rd century B.C.). The two lines should be not parallel, hence they must have different directions : if ( L _1 ) || v _1 &( L _2 ) || v _2 , then
( L _1 ) ∦ ( L _2 ) <==>v _1 ∦ v _2 <==> ∢ ( v _1 ∦ v _2 ) = θ ≠ { 0 , π } . (1)
Notes: 1) In 1. , A_i &B_i are the coordinates of two normal vectors n _ i ⊥ ( L_i ) ( i = 1 , 2 ) . It is also known that two lines / vectors are not parallel / collinear <==> other two vectors that are (respectively) orthtogonal / perpendicular on them are not collinear. In our case, with
n _1 ( A _1 , B _1 ) &n _2 ( A _2 , B _2 ) , (1) <==>A _1 / A _2 ≠ B _1 / B _2 . (2)
2) In 2. , v _1 ∦ v _2 <==>m _1 ≠ m _2 <==> ∢ ( v _1 , v _2 ) = θ ≠ 0 , π . (3)
3) In 3. , the numbers a_i , b_i which occur at the denominators of those 2 equations mean that ( L_i ) ∩ ( Ox ) = A_i ( a _ i , 0 ) & ( L_i ) ∩ ( Oy ) = ( B_i ( 0 , b _ i ) . But the slopes that have occurred in (3) are m _1 = b _1 / a _1 , m _2 = b _2 / a _2 ==>
==> ( L _1 ) ∦ ( L _2 ) <==>b _1 / a _1 ≠ b _2 / a _2 <==>a _1 b _2 ≠ a _2 b _1 . (4)
In both conditions (2) and (4) , some of the denominators in the respective ratios can be = 0 . This doesn’t mean that it is accepted the division by 0 . For instance, in (4) I have given the equivalent caracterization with products instead of ratios (fractions).
4) A line in the plane can be also represented by a point situated on it and a direction : if M _0 ( x _0 , y _0 ) ∈ ( L ) and ( L ) || v ( p , q ) then a current point M ( x , y ) is on ( L ) <==>M _0 M = m || v <==>( x-x _0 ) / p = ( y- 0 ) / q . It the case of the above two lines, this equation must be turned into to
( L_i ) || v _i( p_i , q_i ) ==> ( x-x _ i ) / p_i = ( y-x _ i ) / q_i ( i = 1 , 2 ) . (5)
Then the condition for ( L_i ) ( i = 1 , 2 ) to be secant is that above given in (1) , equivalent to
p _1 / p _2 ≠ q _1 / q _2 . (6)
5) The possible relative positions of two lines in the plane were studied / presented in subsection 8.2 (from page 217) in the textbook . It could be also taken into account some particular cases, when − for instance − one of the two lines is either orthogonal to or parallel to one of the axes of coordinates : if
( L _1 ) ⊥ ( Ox ) ==> ( L _1 ) || ( Oy ) then ( L _2 ) cuts ( L _1 ) at Q only if v _2 ∦ j .
Closing comments: ➀ I have not (very well) understood the second part of the above question : what would it mean “ . . . s olving them simultaneously using basic algebra ? “ I hope that I have (above) presented all the equivalent conditions for two lines (in the plane) to have a single common point Q , and no solution of simultaneous equations using basic algebra was necessary. In the LINEAR ALGEBRA (at least in the manuals / textbooks written and edited in English) it is often used the terminology simultaneous (linear) equations instead of linear systems, or systems of linear equation. However, if the two lines are represented by their general equations as in the case 1. , then the two equations form a linear system with a unique solution, because
Rank [ A _1 B _1 / A _2 B _2 ] =Rank [ A _1 B _1 | -C _1 / A _2 B _2 | -C _2 ] (7)
and, by the Kronecker - Capelli Theorem, this non-homogeneous system is consistent and it admits a single solution ( x , y ) : the coordinates of the common point Q . But that question suggested that (asked whether) such an approach could be avoided, and this was correct. In fact, these coordinates were not required. As a technical (notational) detail, I have used in (7) my way to write matrices, since the TEXT format available for the answers to QUORA does not recognize the mathematical equations (formulas) ; the slashes / separate the rows of a matrix, while the vertical bars | separate the coefficients of a linear equation from its free term(s). ➁ If I may recall something from my studies at the Faculty of Mathematics in Iasi (1960–1965), I feel indebted to mention the excellent classes of Analytic Geometry given to us by the great Professor (and then Academician) Radu MIRON, who was (most probably) Romania’s No 1 geometer during the last three decades, due to his outstanding contributions in Differential Geometry, Lagrange and Hamilton spaces and in many other domains, internationally recognized and appreciated ; another remarkable magister was Professor Dan PAPUC, who gave us classes of Synthetic Geometry-Theory of Curves and Surfaces, and also of History of Mathematics. The above mentioned textbook was dedicated to these (and other) professors.
[ Alex. CARAUSU, Vector Algebra, Analytic & Differential Geometry. PIM Publishers, Iasi 2003 ]
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Robert Nichols
Author has 5K answers and 15.6M answer views
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Is it possible for three straight lines to intersect at only one point?
Is it possible for three straight lines to intersect at only one point?
Yes.
Every Triangle has several examples of this. There are 4 points of concurrency for every Triangle.
The C ircumcenter, where all three perpendicular bisectors meet.
The Incenter, where all three angle bisectors meet.
The Orthocente r, where all three altitudes meet.
The Centroid, where all three medians meet.
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Is it possible for three straight lines to intersect at only one point?
Yes.
Every Triangle has several examples of this. There are 4 points of concurrency for every Triangle.
The C ircumcenter, where all three perpendicular bisectors meet.
The Incenter, where all three angle bisectors meet.
The Orthocente r, where all three altitudes meet.
The Centroid, where all three medians meet.
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Choosing between an expensive laser printer and a more affordable inkjet or multi-function machine ultimately depends on your printing habits, the type of documents you produce, and your long-term budget. Laser printers and inkjet printers serve different purposes, and understanding their strengths and limitations can help you make a more informed decision. Laser printers are typically designed for speed and efficiency, making them ideal for office environments or users who print frequently. Inkjet and ink tank printers are slower but they offer excellent colour reproduction and are often more
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Choosing between an expensive laser printer and a more affordable inkjet or multi-function machine ultimately depends on your printing habits, the type of documents you produce, and your long-term budget. Laser printers and inkjet printers serve different purposes, and understanding their strengths and limitations can help you make a more informed decision. Laser printers are typically designed for speed and efficiency, making them ideal for office environments or users who print frequently. Inkjet and ink tank printers are slower but they offer excellent colour reproduction and are often more compact and affordable upfront.
If your printing needs are frequent and primarily text-based, a laser printer such as the HP Color Laser 179fnw or the HP LaserJet M234sdw would be a strong choice. These models deliver fast print speeds and sharp text output. Toner cartridges used in these printers last significantly longer than ink cartridges, resulting in a lower cost per page over time. Although the initial investment is higher, the long-term savings and reliability could make them well-suited for your business use or heavy personal workloads.
LaserJet Printers - Black & White or Color Document Printers
Alternatively, if your printing volume is low to moderate and you value colour accuracy for photos or creative projects, an ink-based solution like the HP Smart Tank 7605 or HP Smart Tank 5105 may be more convenient. These printers offer refillable ink tanks that reduce running costs compared to traditional cartridge-based inkjets. They are also compact and versatile, supporting scanning and copying functions in addition to printing. However, they require occasional maintenance to prevent ink from drying out or clogging, and their print speed is generally slower than laser models.
HP Smart Tank Printers – Refillable Ink Tank Printers
So to break it down, if you prioritise speed, durability, and cost-efficiency for high-volume printing, a laser printer from HP’s 200 or 3000 series is the better investment. On the other hand, if your needs include occasional printing with a focus on colour and photo output, HP’s Smart Tank series provides a more economical and flexible alternative. The right choice will depend on how often you print and what kind of documents you’re producing.
Check out the blog linked below to learn more about the different models and which is best for your printing needs, hope this helps!
Inkjet vs LaserJet vs OfficeJet: HP Printer Buying Guide
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Paul Dunkley
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Can a circle and an ellipse intersect each other at only one point? Why or why not?
This is shown at figure 1. But 2, 3 or 4 points of intersection are possible.
But it is also possible for a circle and ellipse to not intersect at all of course.
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This is shown at figure 1. But 2, 3 or 4 points of intersection are possible.
But it is also possible for a circle and ellipse to not intersect at all of course.
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James Tuttle
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How can the equation of a straight line be determined if it passes through two points and intersects another line at only one point, without using a graph?
Umm … here’s one way: The equation of a line can be written as y = mx+b where m is the slope and b is the y-intercept. Given the two points, you can determine the slope m. Then by using the (x,y) coordinates of the intersection point, you can calculate b. That gives you the equation you want.
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Peter Shea
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How can the equation of a straight line be determined if it passes through two points and intersects another line at only one point, without using a graph?
QPG, two points is sufficient to determine a line. No other points are needed.
If the line is parallel to a coordinate axis, its equation takes the form x = c for the x-axis or y = c for the y-axis.
Otherwise its form is y = mx + c.
When the line passes through P (P_x, P_y) and Q (Q_x, Q_y),
m = (P_y - Q_y)/(P_x - Q_x).
You can then solve the line equation for c at P or Q using the value of m and the coordinates of your chosen point.
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Joe Schlessinger
MA in Mathematics, University of California, Berkeley · Author has 194 answers and 63.6K answer views
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How can the equation of a straight line be determined if it passes through two points and intersects another line at only one point, without using a graph?
Two points are enough. The third line is irrelevant or perhaps contradictory.
Let the points be (x1, y1) and (x2, y2).
Assume x1 =/ x2. We'll cover that case later.
Then the slope m = (y2-y1)/(x2-x1). We know the denominator is not zero so the fraction is valid.
The equation you seek is (y-y1) = m(x-x1). Note: if y1=y2, m is 0 and the equation reduces to y =y1, a horizontal line.
If x1=x2, the equation reads x=x1, a vertical line.
As you can see, the "another line" doesn't figure in the discussion at all.
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Proof that base -2 with binary digits can form every integer
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Basically the question is proving that you can create all integers with binary but instead using −2 as the base to be able to create negative integers.
Exact question:
Prove that every integer (positive, negative, or zero) can be written as the sum of distinct powers of −2.
I somewhat get how you can induct upon increasing powers for 2 0+2 1+2 2 etc and prove that it will always hold for the next number but I'm not sure how this will work with negative integers since If I induct upwards I can't go down and I can't start at −∞.
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edited Jan 30, 2020 at 4:44
J. M. ain't a mathematician
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Can you get 0? If 0=∑k i=1(−2)a i then we know that none of the a i=0, since 0 is even, but then you can divide by −2 to get a smaller representation of 0, ad infinitum. No?lulu –lulu 2020-01-28 23:45:03 +00:00 Commented Jan 28, 2020 at 23:45
Use induction starting from 1 for positive integers, and induction starting from −1 going downwards for negative integers.rogerl –rogerl 2020-01-28 23:47:36 +00:00 Commented Jan 28, 2020 at 23:47
If you ignore the problem at 0, I suggest doing a double induction. Establish that you can get the first few positive integers, and the first few negative integers and then show inductively that you can always get the next positive and the next negative integer (either by dividing by −2 or by subtracting 1 and then dividing by −2).lulu –lulu 2020-01-28 23:54:06 +00:00 Commented Jan 28, 2020 at 23:54
They have specified that the empty set is a set so I assume the empty set qualifies 0 as existing. Thanks I didn't think of just inducting both ways.user3645925 –user3645925 2020-01-28 23:59:10 +00:00 Commented Jan 28, 2020 at 23:59
Ah, if you allow the empty set then you are ok. I think the "simultaneous induction" should work without much fuss for the rest of the integers.lulu –lulu 2020-01-29 00:00:40 +00:00 Commented Jan 29, 2020 at 0:00
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0 is obtained via the empty set.
We'll proceed by "simultaneous induction" on the positive and negative integers.
To build up positive base cases we note that 1=(−2)0 2=(−2)2+(−2)1 3=(−2)2+(−2)1+(−2)0
To build up negative base cases we note that −1=(−2)1+(−2)0−2=(−2)1−3=(−2)3+(−2)2+(−2)0
Now the induction statement we want is "Given that the claim is true for all integers k with |k|≤n−1 prove that it is also true for k=±n."
That plus the base cases will certainly suffice.
To prove the statement, we first note that (using the base cases) we can assume that n≥4. Now we distinguish between the cases n even or n odd.
If n is even then n−2 is an integer with absolute value <n so we can write n−2=m∑i=1(−2)a i⟹n=m∑i=1(−2)a i+1
(here, of course, we are using a proper representation of the smaller number. Thus the {a i} are distinct. If that is the case, then of course the numbers {a i+1} are also all distinct.)
If n is odd then n−1 is even and, as before we can write n−1−2=m∑i=1(−2)a i⟹n=m∑i=1(−2)a i+1+(−2)0 and we are done.
The case of −n is more or less identical.
Note that this method is "constructive" in the sense that you can use it to construct the representation of some number, given that you have already got the representations of smaller numbers.
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edited Jan 29, 2020 at 10:50
answered Jan 29, 2020 at 0:09
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So for the case of negative n then it would literally just trivially be the same except for -n and -n-1 as the left hand side of the equation to start?user3645925 –user3645925 2020-01-29 23:10:59 +00:00 Commented Jan 29, 2020 at 23:10
@user3645925 There is no difference with −n, though you should go through it yourself as a check. Well, I suppose it is worth noting that if you start with −1 the algorithm direct you to subtract 1, getting −2, and then divide by −2, which brings you to 1, which has the same absolute value has the number you started with. All other cases get you a smaller absolute value. Of course we covered ±1, and more, in the base cases.lulu –lulu 2020-01-30 02:15:17 +00:00 Commented Jan 30, 2020 at 2:15
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With just the (−2)0 -bit, this can represent {0,1}.
With 2 -bits of values (−2)1 and (−2)0, this can represent {−2,−1}∪{0,1}.
With 3 -bits of values (−2)2, (−2)1 and (−2)0, this can represent {−2,−1,0,1}∪{2,3,4,5}.
Proposition: with n -bits, if O is the greatest odd number smaller than n, then the lower bound is the sum −2 1−2 3−2 5−⋯−2 O, while if E is the greatest even number smaller than n, then the upper bound is the sum 2 0+2 2+2 4+⋯+2 E, subject to empty sum when O or E is negative.
Let S n be the set of integers representable by n -bits.
S n=[−∑0≤i<n,2∤i 2 i,∑0≤i<n,2∣i 2 i]∩Z=[−2⋅4⌊n/2⌋−1 4−1,4⌈n/2⌉−1 4−1]∩Z=[−2⋅4⌊n/2⌋−1 3,4⌈n/2⌉−1 3]∩Z
Assume that k -bits (of values (−2)0,…,(−2)k−1) can represent the following range of integers, inclusive:
S k=[−2⋅4⌊k/2⌋−1 3,4⌈k/2⌉−1 3]∩Z
Then the next -bit of value (−2)k can additionally represent integers in the set
T k+1={(−2)k+s∣s∈S k}=[(−2)k−2⋅4⌊k/2⌋−1 3,(−2)k+4⌈k/2⌉−1 3]∩Z
If k is odd and (−2)k<0, then (−2)k=−2 k=−2⋅4⌊k/2⌋ and the set T k+1 is T k+1=[−2⋅4⌊k/2⌋−2⋅4⌊k/2⌋−1 3,−2⋅4⌊k/2⌋+4⌈k/2⌉−1 3]∩Z=[−2⋅4⌊k/2⌋−2⋅4⌊k/2⌋−1 3,−2⋅4⌊k/2⌋+4⋅4⌊k/2⌋−1 3]∩Z=[−2⋅3⋅4⌊k/2⌋+4⌊k/2⌋−1 3,−6⋅4⌊k/2⌋+4⋅4⌊k/2⌋−1 3]∩Z=[−2⋅4⋅4⌊k/2⌋−1 3,−2⋅4⌊k/2⌋−1 3−1]∩Z=[−2⋅4⌊(k+1)/2⌋−1 3,min S k−1]∩Z
If k is even and (−2)k>0, then (−2)k=2 k=4⌈k/2⌉ and the set T k+1 is T k+1=[4⌈k/2⌉−2⋅4⌊k/2⌋−1 3,4⌈k/2⌉+4⌈k/2⌉−1 3]∩Z=[4⌈k/2⌉−2⋅4⌈k/2⌉−1 3,4⌈k/2⌉+4⌈k/2⌉−1 3]∩Z=[3⋅4⌈k/2⌉−2⋅4⌈k/2⌉+2 3,3⋅4⌈k/2⌉+4⌈k/2⌉−1 3]∩Z=[4⌈k/2⌉−1 3+1,4⋅4⌈k/2⌉−1 3]∩Z=[max S k+1,4⌈(k+1)/2⌉−1 3]∩Z
In both cases, the set of integers representable by k+1 -bits is
S k+1=S k∪T k+1=[−2⋅4⌊(k+1)/2⌋−1 3,4⌈(k+1)/2⌉−1 3]∩Z=[−∑0≤i<k+1,2∤i 2 i,∑0≤i<k+1,2∣i 2 i]∩Z
By induction, with n -bits all integers between −2⋅4⌊n/2⌋−1 3 and 4⌈n/2⌉−1 3 inclusive are representable.
So for any a∈Z, a will be representable as a base-(−2) number with a sufficient number of -bits.
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edited Jan 29, 2020 at 10:00
answered Jan 29, 2020 at 0:44
peterwhypeterwhy
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Let n be the integer that you wish to write as the sum of distinct powers of −2. We'll start off with some base cases:
1=(−2)0
0=0 (sum of zero powers of −2)
−1=(−2)1+(−2)0
−2=(−2)1
For other values of n we can find the sum of distinct powers of −2 for n−2 (if n is even) or n−1−2 (if n is odd), whose absolute value is always less than n. We can take this sum, multiply each term by −2 (which leaves the terms distinct), and for odd n add (−2)0 (which no longer appears after the multiplication), to produce a sum of distinct powers of −2 that equals n. Since at each step we reduce the absolute value, we will eventually end up at one of the base cases.
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answered Jan 29, 2020 at 10:54
NeilNeil
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The statement follows from the following proposition (which can be discovered by considering small cases):
For every n≥0, define D=∑1≤d≤n/2 2 2 d−1. Every integer between −D and 2 n−D−1 (inclusive) can be written as the sum of distinct elements of {(−2)0,(−2)1,…,(−2)n−1}.
Proof: choose an integer t with −D≤t≤2 n−D−1. Write t+D in binary as an n-bit integer (padding on the left with 0 s if necessary): t+D=(b n−1 b n−2⋯b 1 b 0)2, so that t+D=∑0≤j≤n−1 b j 2 j. Then t=t+D−D=∑0≤j≤n−1 b j 2 j−∑1≤d≤n/2 2 2 d−1=∑0≤j≤n−1{b j,if j is even,1−b j,if j is odd}(−2)j is a representation of t as the sum of distinct powers of −2 (since each b j and each 1−b j is either 0 or 1).
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answered Jan 29, 2020 at 9:07
Greg MartinGreg Martin
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I know you're looking for an inductive proof, but here is a non-inductive alternative.
First, we claim that any integer n can be written as a (finite) sum of powers of −2 which are not necessarily all different. This can be done by taking the normal binary representation and regarding this as a sum where each term is ±(−2)k for some k, then replacing each −(−2)k with (−2)k+1+(−2)k.
Next, we choose the "best" representation of n as a sum of (not necessarily distinct) powers of −2. We have three criteria for doing this, in order of priority:
minimise the number of terms in the sum
among sums satisfying 1, maximise the number of different powers
among sums satisfying 1 and 2, minimise the number of powers which are larger than the largest repeated power
Suppose we have a "best" representation (which always exists). We claim all powers must be different. Suppose not, and consider the largest repeated power, k (so we have at least two terms of (−2)k, but at most one of each higher power). We have three cases:
If there is a term (−2)k+1, delete (−2)k+(−2)k+(−2)k+1 from the sum. This gives a representation with fewer terms, contradicting 1.
If there is no term (−2)k+1 or (−2)k+2, replace (−2)k+(−2)k with (−2)k+1+(−2)k+2. This gives a representation with the same number of terms, but more different powers, contradicting 2.
If there is a term (−2)k+2 but no (−2)k+1, make the same replacement. Now this representation has the same number of terms, at least as many different powers (we have possibly lost k but have gained k+1), and fewer terms with higher powers than the new largest repeated power, contradicting 3.
Thus in all cases we get a contradiction, and our "best" representation is valid.
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answered Jan 29, 2020 at 10:04
Especially LimeEspecially Lime
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814 | https://www.kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/1790-01.pdf | ALMOST DISJOINT AND INDEPENDENT FAMILIES STEFAN GESCHKE ABSTRACT. I collect a number of proofs of the existence of large almost disjoint and independent families on the natural numbers. This is mostly the outcome of a discussion on mathoverflow.
1. INTRODUCTION A family $\mathcal{F}\subseteq \mathcal{P}(\omega)$ is an independent family (over $\omega$) if for every pair $\mathcal{A},$ $\mathcal{B}$ of disjoint finite subsets of $\mathcal{F}$ the set $\cap \mathcal{A}\cap(\omega\backslash \cup \mathcal{B})$ is infinite. Fichtenholz and Kantorovich showed that there is an independent family on $\omega$ of size continuum (also see or ).
I collect several proofs of this fundamental fact.
A typical application of the existence of a large independent family is the result that there are $2^{2^{\aleph_{0}}}$ ultrafilters on $\omega$ due to $Posp_{1}\mathfrak{X}i1$ : Given an independent family $(A_{\alpha}){\alpha<2^{\aleph{0}}}$ , for every function $f$ : $2^{\aleph_{0}}arrow 2$ there is an ultrafilter $p_{f}$ on $\omega$ such that for all $\alpha<2^{\aleph_{0}}$ we have $A_{\alpha}\in p_{f}$ iff$f(\alpha)=1$ . Now $(p_{f}){f:2^{\aleph}0}arrow 2$ is a family of size $2^{2^{\aleph{0}}}$ of pairwise distinct ultrafilters.
Independent families in some sense behave similarly to almost disjoint families.
Subsets $A$ and $B$ of $\omega$ are almost disjoint if $A\cap B$ is finite. A family $\mathcal{F}$ of infinite subsets of $\mathcal{P}(\omega)$ is almost disjoint any two distinct elements $A,$ $B$ of $\mathcal{F}$ are almost disjoint.
2. ALMOST DISJOINT FAMILIES An easy diagonalisation shows that every countably infinite, almost disjoint fam-ily can be extended.
Lemma 2.1. Let $(A_{n}){n\in\omega}$ be a sequence ofpairwise almost disjoint, infinite subsets of $\omega$ . Then there is an infinite set $A\subseteq\omega$ that is almost disjoint from all $A{n},$ $n\in\omega$ .
Proof. First observe that since the $A_{n}$ are pairwise almost disjoint, for all $n\in\omega$ the set $\omega\backslash \bigcup_{k<n}A_{k}$ is infinite. Hence we can choose a strictly increasing sequence $(a_{n}){n\in\omega}$ of natural numbers such that for al $n\in\omega,$ $a{n} \in\omega\backslash \bigcup_{k<n}A_{k}$. Clearly, if $k<n$, then $a_{n}\not\in A_{k}$ .
Date: December 21, 2011.
数理解析研究所講究録 第1790 巻2012 年1-9 1 STEFAN GESCHKE It follows that for every $k\in\omega$ the infinite set $A={a_{n}:n\in\omega}$ is almost disjoint ffom $A_{k}$ .
$\square$ A straight forward application of Zorn’s Lemma gives the following: Lemma 2.2. Every almost disjoint family of subsets $of\omega$ is contained in a maximal almost disjoint family of subsets of $\omega$ .
Corollary 2.3. Every infinite, maximal almost disjoint family is uncountable.
$In$ particular, there is an uncountable almost disjoint family of subsets of $\omega$ .
Proof. The uncountability of an infinite, maximal almost disjoint family follows ffom Lemma 2.1. To show the existence of such a family, choose a partition $(A_{n}){n\in\omega}$ of $\omega$ into pairwise disjoint, infinite sets. By Lemma 2.2, the almost disjoint fam-ily ${A{n} : n\in\omega}$ extends to a maximal almost disjoint family, which has to be uncountable by our previous observation.
$\square$ Unfortunately, this corollary only guarantees the existence of an almost disjoint family of size $\aleph_{1}$ , not necessarily of size $2^{N_{0}}$ .
Theorem 2.4. There is an almost disjoint family of subsets of $\omega$ of size $2^{\aleph_{0}}$ .
All the following proofs of Theorem 2.4 have in common that instead of on $\omega$ , the almost disjoint family is constructed as a family of subsets of some other countable set that has a more suitable structure.
First proof. We define the almost disjoint family as a family of subsets of the complete binary tree $2^{<\omega}$ of height $\omega$ rather than $\omega$ itself.
For each $x\in 2^{\omega}$ let $A_{x}={xrn:n\in\omega}$ .
If $x,y\in 2^{\omega}$ are different and $x(n)\neq y(n)$ , then $A_{x}\cap A_{y}$ contains no sequence of length $>n$. It follows that ${A_{x} : x\in 2^{\omega}}$ is an almost disjoint family of size continuum.
$\square$ Similarly, one can consider for each $x\in[0,1]$ the set $B_{x}$ of finite initial segments of the decimal expansion of $x$.
${B_{x} : x\in[0,1]}$ is an almost disjoint family of size $2^{N_{0}}$ of subsets of a fixed countable set.
Second proof. We again identify $\omega$ with another countable set, in this case the set $\mathbb{Q}$ of rational numbers. For each $r\in \mathbb{R}$ choose a sequence $(q_{n}^{r}){n\in\omega}$ of rational numbers that is not eventually constant and converges to $r$ . Now let $A{r}={q_{n}^{r} : n\in\omega}$ .
For $s,$ $r\in \mathbb{R}$ with $s\neq r$ choose $\epsilon>0$ so that $(s-\epsilon, s+\epsilon)\cap(r-\epsilon, r+\epsilon)=\emptyset$ .
Now $A_{s}\cap(s-\epsilon, s+\epsilon)$ and $A_{r}\cap(r-\epsilon, r+\epsilon)$ are both cofinite and hence $A_{s}\cap A_{r}$ is finite. It follows that ${A_{r} : r\in \mathbb{R}}$ is an almost disjoint family of size $2^{\aleph_{0}}$ .
$\square$ Third proof. We construct an almost disjoint family on the countable set $Z\cross$ Z.
For each angle $\alpha\in[0,2\pi)$ let $A_{\alpha}$ be the set of all elements of $Z\cross Z$ that have distance $\leq 1$ to the line $L_{\alpha}={(x, y)\in \mathbb{R}^{2} : y=\tan(\alpha)\cdot x}$.
2 ALMOST DISJOINT AND INDEPENDENT FAMILIES For two distinct angles $\alpha$ and $\beta$ the set of points in $\mathbb{R}^{2}$ of distance $\leq 1$ to both $L_{\alpha}$ and $L_{\beta}$ is compact. It follows that $A_{\alpha}\cap A_{\beta}$ is finite. Hence ${A_{\alpha} : \alpha\in[0,2\pi)}$ is an almost disjoint family of size continuum.
$\square$ Fourth proof. We define a map $e$ : $[0,1]arrow\omega^{\omega}$ as follows: for each $x\in[0,1]$ and $n\in\omega$ let $e(x)(n)$ be the integer part of $n\cdot x$.
For every $x\in[0,1]$ let $A_{x}={(n, e(x)(n)) : n\in\omega}$ .
If $x<y$ , then for all sufficiently large $n\in\omega,$ $e(x)(n)<e(y)(n)$ . It follows that ${A_{x} : x\in[0,1]}$ is an almost disjoint family of subsets of $\omega\cross\omega$.
Observe that $e$ is an embedding of $([0,1], \leq)$ into $(\omega^{\omega}, \leq^{})$ , where $f\leq^{}g$ if for almost all $n\in\omega,$ $f(n)\leq g(n)$ .
$\square$ 3. INDEPENDENT FAMILlES Independent families behave similarly to almost disjoint families. The following results are analogs of the corresponding facts for almost disjoint families.
Lemma 3.1. Let $m$ be an ordinal $\leq\omega$ and let $(A_{n}){n<m}$ be a sequence of infinite subsets of $\omega$ such that for all pairs $S,$ $T$ offinite disjoint subsets of $m$ the set $\bigcap{n\in S}A_{n}\backslash (\bigcup_{n\in T}A_{n})$ is infinite. Then there is an infinite set $A\subseteq\omega$ that is independent over the family ${A_{n}:n<m}$ in the sense that for all pairs $S,$ $T$ offinite disjoint subsets of $m$ both $(A \cap\bigcap_{n\in S}A_{n})\backslash (\bigcup_{n\in T}A_{n})$ and $\bigcap_{n\in S}A_{n}\backslash (A\cup\bigcup_{n\in T}A_{n})$ are infinite.
Proof. Let $(S_{n}, T_{n}){n\in\omega}$ be an enumeration of all pairs of disjoint finite subsets of $m$ such that every such pair appears infinitely often.
By the assumptions on $(A{n}){n\in\omega}$, we can choose a strictly increasing sequence $(a{n}){n\in\omega}$ such that for all $n\in\omega$ , $a{2n},$ $a_{2n+1} \in\bigcap_{k\in S_{n}}A_{k}\backslash (\bigcup_{k\in T_{n}}A_{k})$ .
Now the set $A={a_{2n} : n\in\omega}$ is independent over ${A_{n} : n<m}$ . Namely, let $S,$ $T$ be disjoint finite subsets of $m$ . Let $n\in\omega$ be such that $S=S_{n}$ and $T=T_{n}$ .
Now by the choice of $a_{2n}$ , $a_{2n} \in(A\cap\bigcap_{k\in S_{n}}A_{k})\backslash (\bigcup_{k\in T_{n}}A_{k})$ .
3 STEFAN GESCHKE On the other hand, $a_{2n+1} \in\bigcap_{k\in S_{n}}A_{k}\backslash (A\cup\bigcup_{k\in T_{n}}A_{k})$ .
Since there are infinitely many $n\in\omega$ with $(S, T)=(S_{n}, T_{n})$ , it follows that the sets $(A \cap\bigcap_{k\in S_{n}}A_{k})\backslash (\bigcup_{k\in T_{n}}A_{k})$ and $\bigcap_{k\in S_{n}}A_{k}\backslash (A\cup\bigcup_{k\in T_{n}}A_{k})$ are both infinite.
$\square$ Another straight forward application of Zorn’s Lemma yields: Lemma 3.2. Every independent family of subsets of $\omega$ is contained in a maximal independent family of subsets of $\omega$ .
Corollary 3.3. Every infinite maximal independent family is uncountable. In par-ticular, there is an uncountable independent family of subsets of $\omega$ .
Proof. By Lemma 3.2, there is a maximal independent family. By Lemma 3.1 such a family cannot be finite or countably infinite.
$\square$ As in the case of almost disjoint families, this corollary only guarantees the existence of independent families of size $\aleph_{1}$ .
But Fichtenholz and Kantorovich showed that there are independent families on $\omega$ of size continuum.
Theorem 3.4. There is an independent family of subsets of $\omega$ of size $2^{\aleph_{0}}$ .
In the following proofs of this theorem, we will replace the countable set $\omega$ by other countable sets with a more suitable structure. Let us start with the original proof by Fichtenholz and Kantorovich that was brought to my attention by Andreas Blass.
First proof. Let $C$ be the countable set of all finite subsets of $\mathbb{Q}$. For each $r\in \mathbb{R}$ let $A_{r}={a\in C$ : $a\cap(-\infty,r]$ is even $}$ .
Now the family ${A_{r} : r\in \mathbb{R}}$ is an independent family of subsets of $C$.
Let $S$ and $T$ be finite disjoint subsets of $\mathbb{R}$. A set $a\in C$ is an element of $\bigcap_{r\in S}A_{r}\backslash (C\backslash \bigcup_{r\in T}A_{r})$ if for all $r\in S,$ $a\cap(-\infty, r]$ is odd and for all $r\in T,$ $a\cap(-\infty, r]$ is even. But it is easy to see that there are infinitely many finite sets $a$ of rational numbers that satisfy these requirements.
$\square$ 4 ALMOST DISJOINT AND INDEPENDENT FAMILIES The following proof is due to Hausdorffand generalizes to higher cardinals .
We will discuss this generalization in Section 4.
Second proof. Let $I={(n, A) : n\in\omega\wedge A\subseteq \mathcal{P}(n)}$ For all $X\subseteq\omega$ let $X’={(n, A)\in I:X\cap n\in A}$ . We show that ${X’ : X\in \mathcal{P}(\omega)}$ is an independent family of subsets of $I$.
Let $S$ and $T$ be finite disjoint subsets of $\mathcal{P}(\omega)$ . A pair $(n, A)\in I$ is in $\bigcap_{X\in S}X’\cap(I\backslash \bigcup_{X\in T}X’)$ if for all $X\in S,$ $X\cap n\in A$ and for all $X\in T,$ $X\cap n\not\in A$. Since $S$ and $T$ are finite, there is $n\in\omega$ such that for any two distinct $X,$ $Y\in S\cup T,$ $X\cap n\neq Y\cap n$ . Let $A={X\cap n:X\in S}$ . Now $(n, A) \in\bigcap_{X\in S}X’\cap(I\backslash \bigcup_{X\in T}X’)$ .
Since there are infinitely many $n$ such that for any two distinct $X,$ $Y\in S\cup T$, $X\cap n\neq Y\cap n$, this shows that $\bigcap_{X\in S}X’\cap(I\backslash \bigcup_{X\in T}X’)$ is infinite.
$\square$ A combinatorially simple, topological proof of the existence of large independent families can be obtained using the Hewitt-Marczewski-Pondiczery theorem which says that the product space $2^{R}$ is separable ([5, 9, 10], also see ). This is the first topological proof.
Third proof. For each $r\in R$ let $B_{r}={f\in 2^{R}:f(r)=0}$ . Now whenever $S$ and $T$ are finite disjoint subsets of $\mathbb{R}$, $\bigcap_{r\in S}B_{r}\cap(2^{R}\backslash \bigcup_{r\in T}B_{r})$ is a nonempty clopen subset of $2^{R}$ .
The family $(B_{r}){r\in R}$ is the prototypical example of an independent family of size continuum on any set. A striking fact about the space $2^{R}$ is that it is separable.
Namely, let $D$ denote the collection of all functions $f$ : $\mathbb{R}arrow 2$ such that there are rational numbers $q{0}<q_{1}<\cdots0\}$. If $S,$ $T\subseteq \mathbb{R}$ are finite and disjoint, then there is a polynomial in $P$ such that $p(r)>0$ for all $r\in A$ and $p(r)\leq 0$ for all $r\in T$.
All positive multiples of $p$ satisfy the same inequalities. It follows that $(A_{r}){r\in \mathbb{R}}$ is an independent family of size $2^{\aleph{0}}$ over the countable set P.
$\square$ The next proof was pointed out by Tim Gowers. This is the dynamical proof.
Sixth proof. Let $X$ be a set of irrationals that is linearly independent over $\mathbb{Q}$. Kro-necker’s theorem states that for every finite set ${r_{1}, \ldots, r_{k}}\subseteq X$ with pairwise distinct $r_{i}$ , the closure of the set ${(nr_{1}, \ldots, nr_{k}) : n\in Z}$ is all of the k-dimensional torus $\mathbb{R}^{k}/Z^{k}$ (, also see ).
For each $r\in X$ let $A_{r}$ be the set of $aUn\in Z$ such that the integer part of $n\cdot r$ is even. Then ${A_{r} : r\in X}$ is an independent family of size continuum. To see this, let $S,T\subseteq X$ be finite and disjoint. By Kronecker‘s theorem there are infinitely many $n\in Z$ such that for all $r\in S$, the integer part of $n\cdot r$ is even and for all $r\in T$, the integer part of $n\cdot r$ is odd. For all such $n$, $n \in\bigcap_{r\in S}A_{r}\cap\bigcap_{r\in T}Z\backslash A_{r}$ .
$\square$ The following proof was mentioned by KP Hart. Let us call it the almost disjoint proof.
6 ALMOST DISJOINT AND INDEPENDENT FAMILIES Seventh proof. Let $\mathcal{F}$ be an almost disjoint family on $\omega$ of size continuum. To each $A\in \mathcal{F}$ we assign the collection $A’$ of all finite subsets of $\omega$ that intersect $A$ . Now ${A’ : A\in \mathcal{F}}$ is an independent family of size continuum.
Given disjoint finite sets $S,$ $T\subseteq \mathcal{F}$, by the almost disjointness of $\mathcal{F}$, each $A\in S$ is almost disjoint from $\cup T$. It follows that there are infinitely many finite subsets of $\omega$ that intersect all $A\in S$ but do not intersect any $A\in T$. Hence $\bigcap_{A\in S}A’\cap(\omega\backslash \bigcup_{A\in T}A’)$ is infinite.
$\square$ The last proof was communicated by Peter Komj\’ath. This is the proof by finite approximation.
Eighth proof. First observe that for all $n\in\omega$ there is a family $(X_{k}){k<n}$ of subsets of $2^{n}$ such that for any two disjoint sets $S,$ $T\subseteq n$ , $\bigcap{k\in S}X_{k}^{n}\cap(2^{n}\backslash \bigcup_{k\in T}X_{k})$ is nonempty. Namely, let $X_{k}={f\in 2^{n} : f(k)=0}$ .
Now choose, for every $n\in\omega$ , a family $(X_{s}^{n}){s\in 2^{n}}$ of subsets of a finite set $Y{n}$ such that for disjoint sets $S,$ $T\subseteq 2^{n}$ , $\bigcap_{s\in S}X_{s}^{n}\cap(2^{n}\backslash \bigcup_{s\in T}X_{s}^{n})$ is nonempty. We may assume that the $Y_{n},$ $n\in\omega$ , are pairwise disjoint.
For each $\sigma\in 2^{\omega}$ let $X_{\sigma}= \bigcup_{n\in\omega}X_{\sigma[n}^{n}$ . Now ${X_{\sigma} : \sigma\in 2^{\omega}}$ is an independent family of size $2^{\aleph_{0}}$ on the countable set $\bigcup_{n\in\omega}Y_{n}$ .
$\square$ 4. INDEPENDENT FAMILIES ON LARGER SETS We briefly point out that for every cardinal $\kappa$ there is an independent family of size $2^{\kappa}$ of subsets of $\kappa$.
We start with a corollary of the Hewitt-Marczewski-Pondiczery Theorem higher cardinalities.
Lemma 4.1. Let $\kappa$ be an infinite cardinal.
Then $2^{2^{\kappa}}$ has a dense subset $D$ such that for every nonempty clopen subset $A$ of $2^{2^{\kappa}},$ $D\cap A$ is of size $\kappa$ . In particular, $2^{2^{\kappa}}$ has a dense subset of size $\kappa$ .
Proof. For each finite partial function $s$ from $\kappa$ to 2let $[s]$ denote the set ${f\in 2^{\kappa}$ : $s\subseteq f}$. The product topology on $2^{!\sigma}$ is generated by all sets of the form $[s]$ . Every clopen subset of $2^{\kappa}$ is compact and therefore the union of finitely many sets of the form $[s]$ . It follows that $2^{\kappa}$ has exactly $\kappa$ clopen subsets. The continuous functions $hom2^{\kappa}$ to 2 are just the characteristic functions of clopen sets. Hence there are only $\kappa$ continous functions $hom2^{\kappa}$ to 2. Let $D$ denote the set of all continuous functions $hom2^{\kappa}$ to 2.
7 STEFAN GESCHKE Sinoe finitely many points in $2^{\kappa}$ can be separated simultaneously by pairwise disjoint clopen sets, every finite partial function ffom $2^{\kappa}$ to 2 extends to a continuous functions defined on all of $2^{\kappa}$ . It follows that $D$ is a dense subset of $2^{2^{\kappa}}$ of size $\kappa$.
Now, if $A$ is a nonempty clopen subset of $2^{2^{\kappa}}$ , then there is a finite partial function $s$ from $2^{\kappa}$ to 2 such that $[s]\subseteq A$. Cleary, the number of continuous extensions of $s$ to all of $2^{\kappa}$ is $\kappa$. Hence $D\cap A$ is of size $\kappa$.
$\square$ As in the case of independent families on $\omega$ , ffom the previous lemma we can derive the existence of large independent families of subsets of $\kappa$.
Theorem 4.2. For every infinite cardinal cardinal $\kappa$ , there is a family $\mathcal{F}$ of size $2^{\kappa}$ such that for all disjoint finite sets $\mathcal{A},$ $\mathcal{B}\subseteq \mathcal{F}$, the set $(\cap \mathcal{A})\backslash \cup \mathcal{B}$ is of size $\kappa$ .
First proof. Let $D\subseteq 2^{2^{\kappa}}$ be as in Lemma 4.1. For each $x\in 2^{\kappa}$ let $B_{x}={f\in 2^{2^{\kappa}}$ : $f(x)=0}$ and $A_{x}=D\cap B_{x}$ . Whenever $S$ and $T$ are disjoint finite subsets of $2^{\kappa}$ , then $( \bigcap_{x\in S}B_{x})\backslash \bigcup_{x\in T}B_{x}$ is a nonempty clopen subset of $2^{2^{\kappa}}$ . It follows that $( \bigcap_{x\in S}A_{x})\backslash \bigcup_{x\in T}A_{x}=D\cap((\bigcap_{x\in S}B_{x})\backslash \bigcup_{x\in T}B_{x})$ is of size $\kappa$ . It follows that $\mathcal{F}={A_{x} : x\in 2^{\kappa}}$ is as desired.
$\square$ We can translate this topological proof into combinatorics as follows: The continuous functions $hom2^{\kappa}$ to 2 are just characteristic functions of clopen sets. The basic clopen sets are of the form $[s]$ , where $s$ is a finite partial function ffom $\kappa$ to 2. All clopen sets are finite unions of sets of the form $[s]$ . Hence we can code clopen subsets of $2^{\kappa}$ in a natural way by finite sets of finite partial functions $hom\kappa$ to 2. We formulate the previous proof in this combinatorial setting. The following proof is just a generalization of our second proof of Theorem 3.4. This is essentially Hausdorff $s$ proof of the existence large independent families in higher cardinalities.
Second proof. Let $D$ be the collection of all finite sets of finite partial functions $hom\kappa$ to 2. For each $f$ : $2^{\kappa}arrow 2$ let $A_{f}$ be the collection of all $a\in D$ such that for all $s\in a$ and all $x:\kappaarrow 2$ with $s\subseteq x$ we have $f(x)=1$ .
Claim 4.3. For any two disjoint finite sets $S,$ $T\subseteq 2^{\kappa}$ the set $( \bigcap_{x\in S}A_{x})\backslash \bigcup_{x\in T}A_{x}$ is of size $\kappa$ .
8 ALMOST DISJOINT AND INDEPENDENT FAMILIES For all $x\in S$ and all $y\in T$ there is $\alpha\in\kappa$ such that $x(\alpha)\neq y(\alpha)$ . It follows that for every $x\in S$ there is a finite partial function $s$ from $\kappa$ to 2 such that $s\subseteq x$ and for all $y\in T,$ $s\not\subset T$. Hence there is a finite set $a$ of finite partial functions from $\kappa$ to 2 such that all $x\in S$ are extensions of some $s\in a$ and no $y\in T$ extends any $s\in a$ . Now $a \in(\bigcap_{x\in S}A_{x})\backslash \bigcup_{x\in T}A_{x}$ . But for every $\alpha<\kappa$ we can build the set $a$ in such a way that $\alpha$ is in the domain of some $s\in a$. It follows that there are in fact $\kappa$ many distinct sets $a \in(\bigcap_{x\in S}A_{x})\backslash \bigcup_{x\in T}A_{x}$ .
$\square$ REFERENCES J. W. S. Cassels, An introduction to Diophantine approximation, Cambridge Tracts in Mathematics and Mathematical Physics, No. 45, Cambridge University Press, New York (1957) R. Engelking, General topology, translated from the Polish by the author, second edition.
Sigma Series in Pure Mathematics, 6, Heldermann Verlag, Berlin (1989) G. M. Fichtenholz, L. V. Kantorovich, Sur le op\’emtions lin\’eares dans l’espace de fonctions bomees, Studia Math. 5 (1935), 69-98 F. Hausdorff, \"Uber zwei Satze von G. Fichtenholz und L. Kantorovich, Studia Math. 6 (1936), 18-19 E. Hewitt, A remark on density characters, Bull. Amer. Math. Soc. 52 (1946), 641-643 T. Jech, Set theory. The third millennium edition, revised and expanded, Springer Monographs in Mathematics. Springer-Verlag, Berlin (2003) L.
Kronecker, Naherungswetse ganzzahlige Auflosung linearer Gleichungen, Monatsber.
$K\ddot{o}$niglich. Preuss. Akad. Wiss. Berlin (ISS4), 1179-1193, 1271-1299.
[S] K. Kunen, Set theory, An introduction to independence proofs, Studies in Logic and the Foundations of Mathematics, 102, North-Holland Publishing Co., Amsterdam-New York (1980) E. Marczewski, S\’eparabilit\’e et multiplication cartesienne des espaces topologiques, Fund.
Math. 34 (1947), 127-143 E. S. Pondiczery, Power problems in abstract spaces, Duke Math. J. 11 (1944), 835-837 B. Posp\’i\v{s}il, Remark on bicompact spaces, Ann. of Math. (2) 38 (1937), 845-846 HAUSDORFF CENTER FOR MATHEMATICS, ENDENICHER ALLEE 62, 53115 BONN, GERMANY E-mail address: stefan.geschke@hcm.uni-bonn.de 9 |
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Biology 2e
6.3 The Laws of Thermodynamics
Biology 2e6.3 The Laws of Thermodynamics
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Learning Objectives
By the end of this section, you will be able to do the following:
Discuss the concept of entropy
Explain the first and second laws of thermodynamics
Thermodynamics refers to the study of energy and energy transfer involving physical matter. The matter and its environment relevant to a particular case of energy transfer are classified as a system, and everything outside that system is the surroundings. For instance, when heating a pot of water on the stove, the system includes the stove, the pot, and the water. Energy transfers within the system (between the stove, pot, and water). There are two types of systems: open and closed. An open system is one in which energy and matter can transfer between the system and its surroundings. The stovetop system is open because it can lose heat into the air. A closed system is one that can transfer energy but not matter to its surroundings.
Biological organisms are open systems. Energy exchanges between them and their surroundings, as they consume energy-storing molecules and release energy to the environment by doing work. Like all things in the physical world, energy is subject to the laws of physics. The laws of thermodynamics govern the transfer of energy in and among all systems in the universe.
The First Law of Thermodynamics
The first law of thermodynamics deals with the total amount of energy in the universe. It states that this total amount of energy is constant. In other words, there has always been, and always will be, exactly the same amount of energy in the universe. Energy exists in many different forms. According to the first law of thermodynamics, energy may transfer from place to place or transform into different forms, but it cannot be created or destroyed. The transfers and transformations of energy take place around us all the time. Light bulbs transform electrical energy into light energy. Gas stoves transform chemical energy from natural gas into heat energy. Plants perform one of the most biologically useful energy transformations on earth: that of converting sunlight energy into the chemical energy stored within organic molecules (Figure 6.2). Figure 6.11 examples of energy transformations.
The challenge for all living organisms is to obtain energy from their surroundings in forms that they can transfer or transform into usable energy to do work. Living cells have evolved to meet this challenge very well. Chemical energy stored within organic molecules such as sugars and fats transforms through a series of cellular chemical reactions into energy within ATP molecules. Energy in ATP molecules is easily accessible to do work. Examples of the types of work that cells need to do include building complex molecules, transporting materials, powering the beating motion of cilia or flagella, contracting muscle fibers to create movement, and reproduction.
Figure 6.11 Here are two examples of energy transferring from one system to another and transformed from one form to another. Humans can convert the chemical energy in food, like this ice cream cone, into kinetic energy (the energy of movement to ride a bicycle). Plants can convert electromagnetic radiation (light energy) from the sun into chemical energy. (credit “ice cream”: modification of work by D. Sharon Pruitt; credit “kids on bikes”: modification of work by Michelle Riggen-Ransom; credit “leaf”: modification of work by Cory Zanker)
The Second Law of Thermodynamics
A living cell’s primary tasks of obtaining, transforming, and using energy to do work may seem simple. However, the second law of thermodynamics explains why these tasks are harder than they appear. None of the energy transfers that we have discussed, along with all energy transfers and transformations in the universe, is completely efficient. In every energy transfer, some amount of energy is lost in a form that is unusable. In most cases, this form is heat energy. Thermodynamically, scientists define heat energy as energy that transfers from one system to another that is not doing work. For example, when an airplane flies through the air, it loses some of its energy as heat energy due to friction with the surrounding air. This friction actually heats the air by temporarily increasing air molecule speed. Likewise, some energy is lost as heat energy during cellular metabolic reactions. This is good for warm-blooded creatures like us, because heat energy helps to maintain our body temperature. Strictly speaking, no energy transfer is completely efficient, because some energy is lost in an unusable form.
An important concept in physical systems is that of order and disorder (or randomness). The more energy that a system loses to its surroundings, the less ordered and more random the system. Scientists refer to the measure of randomness or disorder within a system as entropy. High entropy means high disorder and low energy (Figure 6.12). To better understand entropy, think of a student’s bedroom. If no energy or work were put into it, the room would quickly become messy. It would exist in a very disordered state, one of high entropy. Energy must be put into the system, in the form of the student doing work and putting everything away, in order to bring the room back to a state of cleanliness and order. This state is one of low entropy. Similarly, a car or house must be constantly maintained with work in order to keep it in an ordered state. Left alone, a house's or car's entropy gradually increases through rust and degradation. Molecules and chemical reactions have varying amounts of entropy as well. For example, as chemical reactions reach a state of equilibrium, entropy increases, and as molecules at a high concentration in one place diffuse and spread out, entropy also increases.
Scientific Method Connection
Transfer of Energy and the Resulting Entropy
Set up a simple experiment to understand how energy transfers and how a change in entropy results.
Take a block of ice. This is water in solid form, so it has a high structural order. This means that the molecules cannot move very much and are in a fixed position. The ice's temperature is 0°C. As a result, the system's entropy is low.
Allow the ice to melt at room temperature. What is the state of molecules in the liquid water now? How did the energy transfer take place? Is the system's entropy higher or lower? Why?
Heat the water to its boiling point. What happens to the system's entropy when the water is heated?
Think of all physical systems in this way: Living things are highly ordered, requiring constant energy input to maintain themselves in a state of low entropy. As living systems take in energy-storing molecules and transform them through chemical reactions, they lose some amount of usable energy in the process, because no reaction is completely efficient. They also produce waste and by-products that are not useful energy sources. This process increases the entropy of the system’s surroundings. Since all energy transfers result in losing some usable energy, the second law of thermodynamics states that every energy transfer or transformation increases the universe's entropy. Even though living things are highly ordered and maintain a state of low entropy, the universe's entropy in total is constantly increasing due to losing usable energy with each energy transfer that occurs. Essentially, living things are in a continuous uphill battle against this constant increase in universal entropy.
Figure 6.12 Entropy is a measure of randomness or disorder in a system. Gases have higher entropy than liquids, and liquids have higher entropy than solids.
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816 | https://www.doubtnut.com/qna/13657149 | Deuterons in a cyclotron describes a circle of radius 32.0cm. Just before emerging from the D's. The frequency of the applied alternating voltage is 10MHz. Find, (a) the magnetic flux density (i.e., the magnetic field ),(b) the energy and speed of the deuterons upon emergence.
More from this Exercise
The correct Answer is:4.22MeV
(a) Frequency of the applied emf= Cyclotron frequency or f=Bq2πm,B=2πmfq =(2)(3.14)(2×1.67×10−27)(10×106)1.6×10−19=1.30T (b) The speed of the deutrons on the emergence from the cyclotron, v=BqRm=2πfR =(2)(3.14)(10×106)(32×10−2)=2.01×107m/s ∴ Energy of deutrons =12mv2 =12×(2×1.67×10−27)(2.01×107)2J 4.22MeV
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Deuteron in a cyclotron describes a circle of radius 32.0cm. Just before emerging from the D's. The frequency of the applied alternating voltage is 10MHz. Find
a. the magnetic flux density (i.e. the magnetic field).
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Knowledge Check
An ACrms voltage of 2V having a frequency of 50KHz is applied to a condenser of capacity of 10μF. The maximum value of the magnetic field between the plates of the condenser if the radius of plate is 10cm is
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5×1014Hz. The wave is propagating along Z-axis, find (i) the average
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(ii) Find the electric field induced at a point on the circumference of the circle.
(iii) Find the electric field induced at a point outside the solenoid at a distance 9 cm from its axis.
A cyclotron is used to acclerate protons to a kinetic energy of 5 MeV. If the radius and the frequency field in the cyclotron is 2T. Find the radius and the frequency needed for the applied alternating voltage of the cyclotron. (Given : Velocity of proton =3×107m/s)
An electron is orbiting is a circular orbit of radius r under the influence of a constant magnetic field of strength B. Assuming that Bohr postulate regarding the quantisation of angular momentum holds good for this elerctron, find
(a) the allowed values of te radius r of the orbit.
(b) the kinetic energy of the electron in orbit
(c ) the potential energy of insteraction between the magnetic moment of the orbital current due to the electron moving in its orbit and the magnetic field B.
(d) the total energy of the allowed energy levels.
(e) the total magnetic flux due to the magnetic field B passing through the nth orbit. (Assume that the charge on the electronis -e and the mass of the electron is m).
NARAYNA-MOVING CHARGES AND MAGNETISM-LEVEL-V (SINGLE ANSWER QUESTION)
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818 | https://stats.libretexts.org/Bookshelves/Introductory_Statistics/Statistics%3A_Open_for_Everyone_(Peter)/07%3A_One_Sample_t-Tests/7.03%3A_The_One_Sample_t-Test_Formula | Skip to main content
7.3: The One Sample t-Test Formula
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7.2: The Logic of a t-Distribution
7.4: Reporting Results
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( \newcommand{\kernel}{\mathrm{null}\,})
The formula is set up as a division problem with four symbols which represent and, thus, must be replaced with specific values. The numerator focuses on the difference between the sample statistic and the (known or hypothesized) population parameter for a variable. This is the core of the formula because the hypotheses tested using a one sample t-test are specifically asking whether these two values differ. Therefore, you can think of the numerator as the main focus of the formula and the denominator as taking into account other necessary information and adjustments. This will be true for the main format used for all three versions of the t-test in this book. The denominator of the one sample t-test formula is used to take into account the error in the sample (via the standard deviation) and the sample size. The one sample t-test formula is as follows:
Notice that the formula, though small, includes all three components that impact statistical power (see Chapter 6 for a review of the components of power):
The size of the change, difference, or pattern observed in the sample
The sample size
The size of the error in the sample statistic(s)
The first component, size of difference, is being incorporated in the numerator. The second and third components of sample size and error are being incorporated in the denominator.
Let’s take a moment to further examine the denominator. Recall that we expect some error when we draw a sample from a population and that we call this sampling error. We often estimate this error with the formula for standard error (). is calculated by dividing the sample standard deviation by the square root of the sample size (i.e. ). Thus, the denominator for a one sample t-test is actually the standard error formula (see Chapter 6 for a review of standard error). This means that we have the SE formula inside of our one sample t-test formula. Smaller formulas appearing inside larger formulas will happen a lot as we move though the various inferential statistics formulas. This is because statistics builds on some underlying estimates and concepts that get subsumed into more complex formulas to address different needs. If you keep this in mind, it can make future formulas easier to learn and understand, especially when the same formula may be written a few different ways. For example, the one sample t-test formula can be written as follows:
This version of the formula requires the same information and steps as the one shown previously; the only difference is that the steps used to calculate the standard error in the denominator have been replaced with the symbol for SE. Therefore, these two versions of the formula have the exact same steps and will yield the same result. You can use either version. However, this abbreviated one requires that you either remember or look up how to calculate . In contrast, the previous version shows you how to calculate . Therefore, I recommend using the previous one showing the formula rather than the latter one showing the symbol, especially if you are newer to statistics and/or to memorizing mathematical formulas and symbols in general.
Study Tip
When adding formulas to your notes:
use the version which shows more steps and
note for yourself when a group of symbols and operations within a formula are actually another formula altogether.
This will help you connect across topics and see that much of statistics breaks down to the same elements repeating in various ways to create new formulas.
One Sample t-Test Formula
| | |
--- |
| | The Standard Error () Formula appears as the denominator of the one sample t-test formula |
| | The formula may also be written like this where the parenthesis for the denominator is implied. The formula and steps are the same as for the version shown above. |
Formula Components
In order to solve for t, four things must first be known:
= the hypothesized or known population mean
= the sample mean
= the sample standard deviation
= the sample size
Some of these components can be given while others generally need to be calculated from a data set. Typically, the information for the population is given and the information for the sample is calculated from a data set when using this formula. If all four components are given, one can proceed to plugging them into the formula and finding the t-value using the order of operations. However, because the sample statistics are typically found using data, all steps are shown below assuming these need to be calculated.
Formula Steps
The steps are shown in order and categorized into two sections: A) preparation and B) solving. I recommend using this categorization to help you organize, learn, and properly use all inferential formulas. Preparation steps refer to any calculations that need to be done before values can be plugged into the formula. For the one sample t-test this includes finding the descriptive statistics that make up the four components of the formula (listed above). Once those are known, the steps in section B can be used to yield the obtained value for the formula. The symbol for the obtained value for each t-test is t. Follow these steps, in order, to find t.
Section A: Preparation
Identify using the known population mean or otherwise hypothesized value.
Find using the sample data for the focal variable.
Find using the sample data for the focal variable.
Find using the sample data for the focal variable.
Section B: Solving
Write the formula with the values found in section A plugged into their respective locations.
Solve the numerator by subtracting the population (or hypothesized) mean from the sample mean.
Solve for the denominator by finding the square root of the sample size and then dividing the sample standard deviation by that value.
Solve for t by dividing the numerator (result of step 2) by the denominator (the result of step 3).
You may have noticed that we are solving the numerator before the denominator despite the fact that the denominator is shown in parentheses. This may seem like it goes against the order of operations commonly referred to as PEMDAS. PEMDAS states the correct order of computations is: Parentheses, Exponents, Multiplication and Division, and finally, Adding and Subtracting. However, the P in PEMDAS is better stated as G for Grouping (making the acronym GEMDAS). This is because we work through any groups of operations first, regardless of whether they are in parentheses, brackets, or other grouping structures. Our formula actually has two groups of steps: those in the numerator and those in the denominator. This is because the division sign in the middle of the formula separates the operations into these two distinct groups. Therefore, when we see a horizontal division line (like we see in the one sample t-test formula), it is treated as a grouping structure. You can imagine that the whole numerator is within parentheses just like the denominator. For more review of order of operations, see Appendix B.
Data Set 7.1
| Test Scores |
| 85 80 80 75 75 75 70 70 70 70 65 65 65 65 65 60 60 60 60 55 55 55 50 50 45 |
Example of How to Test a Hypothesis by Computing t
Let us assume that a school has implemented a new teaching strategy, and we want to assess whether the students at that school are scoring significantly differently on a standardized test than students in general in the United States. Let us assume that the mean for the population of students in the U.S. is 58.00. Assume that Data Set 7.1 includes data from the school sample. Let’s use this information to test a hypothesis.
Steps in Hypothesis Testing
In order to test a hypothesis, we must follow the steps for hypothesis testing:
State the hypothesis.
A non-directional hypothesis is the best fit because the goal is to see if those at the school had a different mean score from the population without having enough information to hypothesize whether the sample will be specifically higher or lower than the population. A summary of the research hypothesis and corresponding null hypothesis in sentence and symbol format are shown below. However, researchers often only state the research hypothesis using a format like this: It is hypothesized that the mean test score for the school sample will be different from the population mean of 58.00. If the format shown in the table below is used instead, it must be made clear that what is being stated is a research hypothesis not a result (hence, you see it labeled to the left of the hypothesis as such).
Non-Directional Hypothesis for a One Sample t-Test
| Research hypothesis | The mean test score for the school sample will be different from the population mean of 58.00. | |
| Null hypothesis | The mean test score for the school sample will not be different from the population mean of 58.00. | |
Choose the inferential test (formula) that best fits the hypothesis.
A sample mean is being compared to a population mean so the appropriate test is a one sample t-test.
Determine the critical value.
In order to determine the critical value, three things must be identified:
the alpha level,
whether the hypothesis requires a one-tailed test or a two-tailed test, and
the degrees of freedom for the test ().
An alpha level refers to the risk of a Type I Error that is being taken and is summarized with the symbol α. Alpha levels are often set at .05 unless there is reason to adjust them such as when multiple hypotheses are being tested in one study or when a Type I Error could be particularly problematic. The default alpha level can be used for this example because only one hypothesis is being tested and there is no clear indication that a Type I Error would be especially problematic. Thus, alpha can be set to 5%, which can be summarized as = .05.
The hypothesis is non-directional so a two-tailed test should be used.
The df must also be calculated. Each inferential test has a unique formula for calculating . See the section titled “Deeper Dive: What are Degrees of Freedom?” later in this chapter to learn what represents and how it is calculated for the different t-tests before returning to this section to complete the steps in hypothesis testing. In short, the formula for for a one sample t-test is as follows: . The sample size for Data Set 7.1 is 25. Thus, = 25 – 1 so the for this scenario is 24.
These three pieces of information are used to locate the critical value for the test. The full tables of the critical values for t-tests are located in Appendix D. Below is an excerpt of the section of the t-tables that fits the current hypothesis and data. Under the conditions of an alpha level of .05, a two-tailed test, and 24 degrees of freedom, the critical value is 2.064.
Critical Values Table
| | two-tailed test | |
| alpha level: | = 0.05 | = 0.01 |
| Degrees of Freedom: | 24 | 24 |
| Critical Values: | 2.064 | 2.797 |
The critical value represents the absolute value which must be exceeded in order to declare a result significant. Think of this as the threshold of evidence needed to be confident a hypothesis is true and think of the obtained value (which is called t in a t-test) as the amount of evidence present.
Calculate the test statistic.
A test statistic can also be referred to as an obtained value. The formula needed to find the test statistics t for this scenario is as follows:
Section A: Preparation
Start each inferential formula by identifying and solving for the pieces that must go into the formula. For the one sample t-test, this preparatory work is as follows:
Identify using the known population value or the hypothesized value.
This value is given as = 58.00
Find using the sample data for the focal variable.
This value is found using Data Set 7.1 and is summarized as = 25
Find using the sample data for the focal variable.
This value is found using Data Set 7.1 and is summarized as = 65.00
Find using the sample data for the focal variable.
This value is found using Data Set 7.1 and is summarized as = 10.2062
Note
For review of how to calculate a sample mean, see Chapter 3. For review of how to calculate a sample standard deviation, see Chapter 4. The standard deviation is shown rounded to the ten thousandths place.
Now that the pieces needed for the formula have been found, we can move to Section B.
Section B: Solving
Now that the preparatory work is done, the formula can be used to compute the obtained value. For the one sample t-test, this work is as follows:
Write the formula with the values found in section A plugged into their respective locations. Writing the formula first in symbol format before filling it in with the values can help you recognize and memorize it. Here is the formula with the symbols:
Here is the formula with values filled into their appropriate locations in place of their symbols:
Once the symbols have been replaced by values, it is easier to see the mathematical operations which should be followed using the order of operations.
Solve the numerator by subtracting the population (or hypothesized) mean from the sample mean.
Solve for the denominator by finding the square root of the sample size and then dividing the sample standard deviation by that value.
Solve for t by dividing the numerator (result of step 2) by the denominator (the result of step 3). This result, known as a test statistic or t-value, can also be referred to by the general term “obtained value.”
Apply a decision rule and determine whether the result is significant.
Assess whether the obtained value for t exceeds the critical value as follows:
The critical value is 2.064.
The obtained t-value is 3.4294
The obtained t-value does exceed (i.e. is greater than) the critical value, thus, the result is significant.
Note
Only the size of the values, not whether they are positive or negative, is considered when a hypothesis is non-directional (i.e. when a two-tailed test is being performed). Thus, it is the absolute value of t that is being compared to the critical value. However, when a directional hypothesis is used, both the direction and the size of the t-value must be considered.
Calculate the effect size and/or other relevant secondary analyses.
When it is determined that the result is significant, effect sizes should be computed. Because the result was determined to be significant in step 4, the effect size is needed before proceeding to step 7 to complete the process.
The effect size that is appropriate for t-tests is known as Cohen’s d (Cohen, 1988). The formula for Cohen’s is as follows:
You may notice how similar the effect size formula is to the one sample t-test formula. Cohen’s , when used for a one sample t-test, calculates how many sample standard deviations the sample mean is from the population mean (or hypothesized mean). Thus, the numerator finds the difference in the two means and the denominator is used to divide that by the sample standard deviation. The calculations for the current data would be as follows:
Effect sizes, like most values, are rounded and reported to the hundredths place. Thus, this effect size is reported as = 0.69. Cohen’s can be interpreted using the following rules of thumb (Cohen, 1988; Navarro, 2014):
Interpreting Cohen’s Effect Sizes
| ~0.80 | Large effect |
| ~0.50 | Moderate effect |
| ~0.20 | Small effect |
The rules of thumb are general guidance and do not dictate precise or required interpretations. Instead, they provide some generally agreed upon approximations to aid in interpretations. As is true of all analyses, it is best to consider their situated (or practical) relevance. However, the rules of thumb are useful in providing an initial guideline for interpreting effect sizes. Following these rules of thumb, the current finding of = 0.69 would be considered a moderate to large effect.
Report the results in American Psychological Associate (APA) format.
Results for inferential tests are often best summarized using a paragraph that states the following:
the hypothesis and specific inferential test used,
the main results of the test and whether they were significant,
any additional results that clarify or add details about the results,
whether the results support or refute the hypothesis.
Keep in mind that results are reported in past tense because they report on what has already been found. In addition, the research hypothesis must be stated but the null hypothesis is usually not needed for summary paragraphs because it can be deduced from the research hypothesis. Finally, APA format requires a specific format be used for reporting the results of a test. This includes recommendations for rounding and a specific format for reporting relevant symbols and details for the formula and data used. Throughout this book, decimal numbers will be rounded to the hundredths place when reported in a summary sentence or paragraph. Following this, the results for our hypothesis with Data Set 7.1 can be written as shown in the summary example below.
APA Formatted Summary Example
A one sample t-test was used to test the hypothesis that the mean test score for the school sample would be different from the population mean of 58.00. Consistent with the hypothesis, the mean test score for the sample (= 65.00; = 10.21) was significantly different than the mean for the population, t(24) = 3.43, < .05. The Cohen’s effect size of 0.69 was moderate to large.
This succinct summary in APA format provides a lot of detail and uses specific symbols in a particular order. To understand how to read and create a summary like this, let’s take a detailed walk-though of each piece, what it means, and why it appears where it does. Each inferential test will use this structure, though as we progress through some chapters, the summary paragraph will grow to accommodate the increasing complexity and/or details of the analyses. Thus, in the same way it is necessary to understand the smaller formulas because each chapter builds from them, it is also necessary to understand the basic parts of a summary paragraph because these, too, will be built upon.
7.2: The Logic of a t-Distribution
7.4: Reporting Results |
819 | https://nrich.maths.org/equations-and-formulae-age-14-16 | Equations and formulae: Age 14-16 | NRICH
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820 | https://www.youtube.com/watch?v=J7VW7JDeAZE | 10 percent of 20 | Percentage: Find 10% of 20
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Topic: How to find 10 percent of 20.
Answer: To get 10 percentage of 20, we need to multiply 10/100 and 20. This gives 2. Hence 10 percent of 20 is equal to 2.
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to find 10% of 20 we need to multiply the number 20 with 10% now let us write this 20 as 20/ 1 and we know 10% can be written as 10/ 100 now if we multiply the numerators 20 and 10 we will get 200 and in the denominator we have 1 100 which is 100 now cancel the common zeros therefore this will be equal to 2/ 1 which is equal to 2 so 10% of 20 is equal to 2 and this is our final answer thank you for watching please like share and comment on the video also do subscribe the channel |
821 | https://www.gauthmath.com/solution/1792286487627781/Look-at-this-equation-t3-216-What-is-t-the-cube-root-of-216-underline | Solved: Look at this equation: t^3=216 What is t, the cube root of 216? +_ [Math]
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Look at this equation: t^3=216 What is t, the cube root of 216? +_
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822 | https://pmc.ncbi.nlm.nih.gov/articles/PMC2536673/ | RNA STRAND: The RNA Secondary Structure and Statistical Analysis Database - PMC
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. 2008 Aug 13;9:340. doi: 10.1186/1471-2105-9-340
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RNA STRAND: The RNA Secondary Structure and Statistical Analysis Database
Mirela Andronescu
Mirela Andronescu
1 Department of Computer Science, University of British Columbia, 2366 Main Mall, Vancouver, B.C., Canada
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1, Vera Bereg
Vera Bereg
1 Department of Computer Science, University of British Columbia, 2366 Main Mall, Vancouver, B.C., Canada
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1, Holger H Hoos
Holger H Hoos
1 Department of Computer Science, University of British Columbia, 2366 Main Mall, Vancouver, B.C., Canada
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1,✉, Anne Condon
Anne Condon
1 Department of Computer Science, University of British Columbia, 2366 Main Mall, Vancouver, B.C., Canada
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1
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1 Department of Computer Science, University of British Columbia, 2366 Main Mall, Vancouver, B.C., Canada
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Corresponding author.
Received 2008 May 15; Accepted 2008 Aug 13; Collection date 2008.
Copyright © 2008 Andronescu et al; licensee BioMed Central Ltd.
This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC2536673 PMID: 18700982
Abstract
Background
The ability to access, search and analyse secondary structures of a large set of known RNA molecules is very important for deriving improved RNA energy models, for evaluating computational predictions of RNA secondary structures and for a better understanding of RNA folding. Currently there is no database that can easily provide these capabilities for almost all RNA molecules with known secondary structures.
Results
In this paper we describe RNA STRAND – the RNA secondary STRucture and statistical ANalysis Database, a curated database containing known secondary structures of any type and organism. Our new database provides a wide collection of known RNA secondary structures drawn from public databases, searchable and downloadable in a common format. Comprehensive statistical information on the secondary structures in our database is provided using the RNA Secondary Structure Analyser, a new tool we have developed to analyse RNA secondary structures. The information thus obtained is valuable for understanding to which extent and with which probability certain structural motifs can appear. We outline several ways in which the data provided in RNA STRAND can facilitate research on RNA structure, including the improvement of RNA energy models and evaluation of secondary structure prediction programs. In order to keep up-to-date with new RNA secondary structure experiments, we offer the necessary tools to add solved RNA secondary structures to our database and invite researchers to contribute to RNA STRAND.
Conclusion
RNA STRAND is a carefully assembled database of trusted RNA secondary structures, with easy on-line tools for searching, analyzing and downloading user selected entries, and is publicly available at
Background
The number of solved RNA secondary structures has increased dramatically in the past decade, and several databases are available to search and download specific classes of RNA secondary structures [1-5]. However, for purposes such as improving RNA energy models [6,7], evaluating RNA secondary structure prediction software, obtaining distributions of naturally occuring structural features, or searching RNA molecules with specific motifs, researchers need to easily access a much larger set of known RNA secondary structures, ideally all known RNA secondary structures. RNA STRAND aims to provide this capability, in addition to easy search, analysis and download features. Figure 1 shows an example of an RNA secondary structure and highlights some of its structural features.
Figure 1.
Open in a new tab
RNA secondary structure example. Schematic representation of the secondary structure for the RNase P RNA molecule of Methanococcus marapaludis from the RNase P Database; the RNA STRAND ID for this molecule is ASE_00199. Solid grey lines represent the ribose-phosphate backbone. Dotted grey lines represent missing nucleotides. Solid circles mark base pairs. Dashed boxes mark structural features. We define an RNA secondary structure as a set of base pairs. In this work, we consider all C-G, A-U and G-U base pairs as canonical, and all other base pairs as non-canonical. However, we note that from the point of view of the planar edge-to-edge hydrogen bonding interaction , there are C-G, A-U and G-U base pairs that do not interact via Watson-Crick edges, and vice-versa [14,42]. Comparative sequence analysis tools do not currently describe bond types. A number of structural motifs can be identified in a secondary structure: A stem is composed of one or more consecutive base pairs. A hairpin loop contains one closing base pair, and all the bases between the paired bases are unpaired. An internal loop is a loop with two closing base pairs, and all bases between them are unpaired. A bulge loop can be seen as a variant of an internal loop in which there are no unpaired bases on one side. A multi-loop is a loop which has at least three closing base pairs; stems emanating from these base pairs are called multi-loop branches. A pseudoknot is a structural motif that involves non-nested, crossing base pairs.
Previous RNA databases provide secondary structure information, but are specialised in a different direction or follow different goals. The Rfam Database contains a large collection of non-coding RNA families; however, many of the corresponding secondary structures are computationally predicted. The Comparative RNA Web Site specialises in ribosomal RNA and intron RNA molecules. The Sprinzl tRNA database specialises in tRNA molecules, the RNase P database specialises in RNase P RNA molecules, and the SRP and tmRNA databases specialise in SRP RNA and tmRNA molecules, respectively. Pseudobase contains short RNA fragments that have pseudoknots. The RAG (RNA-As-Graphs) Database classifies and analyses RNA secondary structures according to their topological characteristics based on the description of RNAs as graphs, but its collection of structures is very limited.
A number of previous databases contain three-dimensional (3D) RNA structures; however, as opposed to proteins, the number of solved RNA 3D structures is much smaller than the number of solved RNA secondary structures. (Only 18% of all RNA molecules we collected have known 3D structures.) As such, all these databases do not include molecules whose secondary structures are known but 3D structures are unknown; examples include: the RCSB Protein Data Bank , the Nucleic Acids Database , the RNA Structure Database and the Structural Classification of RNA (SCOR) database . NCIR contains non-canonical base pairs in 3D RNA molecules. FR3D provides a collection of 3D RNA structural motifs found in the RCSB Protein Data Bank. Finally, there are other RNA databases that provide RNA sequences, but no experimental structural information, such as the SubViral RNA Database , which contains a collection of over 2600 sequences of viroids, the hepatitis delta virus and satellite RNAs, but only mfold-predicted secondary structures.
RNA STRAND spans a more comprehensive range of RNA secondary structures than do previous databases. It currently provides highly accurate secondary structures for 4666 RNA molecules. Since some users of RNA STRAND will likely develop new thermodynamic models, prediction tools or statistical analyses, our data is exclusively determined by carefully conducted comparative sequence analysis , or by experimental methods such as NMR or X-ray crystallography . All information has been obtained from publicly available RNA databases. Our goal in creating this database is to provide comprehensive information on structural features – such as types and sizes for stems and loops, pseudoknot complexity and base pair types – that can be interactively analysed or downloaded within and across functional classes of molecules. Such information could be used, for example, to understand what type of structural motifs are common in a specific set of RNA molecules; to estimate the accuracy of RNA secondary structure computational prediction methods; or to improve current thermodynamic models for RNA secondary structure prediction.
Construction and Content
Figure 2 describes the four main modules that comprise RNA STRAND. To create the database, we first collected the data from various external sources, then we processed the data and prepared it for a MySQL relational database. Next, we installed and populated the database, and finally we prepared dynamic web pages that interact with the database. In what follows we describe in detail the construction and content of each module.
Figure 2.
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Database schema. Construction of RNA STRAND, from data collection to data presentation via dynamic web pages.
External sources
The current release v2.0 of RNA STRAND contains a total of 4666 entries (RNA sequences and secondary structures) of the following provenance:
• RCSB Protein Data Bank (PDB) : 1059 entries, obtained from three dimensional NMR and X-ray atomic structures containing RNA molecules only, or RNA molecules and proteins (only the RNAs were included in RNA STRAND), in PDB format. These include ribozymes, ribosomal RNAs, transfer RNAs, synthetic structures, and complexes containing more than one RNA molecule. Out of the 1059 entries, 575 contain at least two RNA molecules; these are easily searchable from the RNA STRAND web site. The RNA secondary structures were generated from the tertiary structures using RNAView , which is also used for secondary structure visualisation in the Nucleic Acid Database .
• Comparative RNA Web Site, version 2 : 1056 entries of ribosomal and intronic RNA molecules obtained by covariance-based comparative sequence analysis.
• tmRNA database : 726 entries of transfer messenger RNA sequences and secondary structures determined by comparative sequence analysis.
• Sprinzl tRNA Database (September 2007 edition) : 622 transfer RNA sequences and secondary structures obtained by comparative sequence analysis from the tRNA sequences data set. The genomic tRNA and tRNA gene sets from the Sprinzl tRNA database contain genomic sequences, and thus we think they are not as relevant for understanding function and folding of functional RNA molecules.
• RNase P Database : 454 Ribonuclease P RNA sequences and secondary structures obtained by comparative sequence analysis.
• SRP Database : 383 entries of Signal Recognition Particle RNA sequences and secondary structures determined by comparative sequence analysis.
• Rfam Database, version 8.1 : 313 entries from 19 Rfam families, including hammerhead ribozymes, telomerase RNAs, RNase MRP RNAs and RNase E 5' UTR elements (only the seeds have been used). Of the 607 Rfam families in version 8.1, 172 have the secondary structure flag "published", while the remaining 435 families have been predicted using Pfold . For several reasons, we decided to include only 19 of the 172 "published" families: (1) some of these families come from other databases that we have included directly, such as structures from the RNase P Database or SRP Database; (2) most of the secondary structures are actually predicted computationally and then published in the papers cited by Rfam, such as families RF00013, RF00035, RF00161 or RF00625. Since the Rfam database provides only very limited information about the reliability of the structures it contains, we have studied all 172 families and decided which families to include based on the cited papers. The details regarding the decision for each family are described in Supplementary Material 1, accessible from the main page of the RNA STRAND web site.
• Nucleic Acid Database (NDB) : 53 entries which occur in NDB and not in PDB (note that NDB and PDB have a large overlap of RNA structures); these include transfer RNAs and synthetic RNAs obtained by X-ray crystallography.
Table 1 provides some additional information on these RNAs; information and statistics on the current database contents are also available from the main page of the RNA STRAND site.
Table 1.
The main RNA types included in RNA STRAND v2.0.
RNA type Main source(s)#Length% PKBP
entries mean std mean std
Transfer messenger RNA tmRDB 726 368 86 21.0 6.1
16S ribosomal RNA CRW , PDB 723 1529 286 1.8 0.5
Transfer RNA Sprinzl DB , PDB 707 76 21 0.1 2.3
Ribonuclease P RNA RNase P DB 470 323 71 5.7 3.2
Signal rec. particle RNA SRPDB , PDB 394 220 111 0.0 0.0
23S ribosomal RNA CRW , PDB 205 2699 716 2.4 1.1
5S ribosomal RNA CRW , PDB 161 115 21 0.0 0.0
Group I intron CRW , PDB 152 563 412 5.8 2.2
Hammerhead ribozyme Rfam , PDB 146 61 24 0.0 0.0
Group II intron CRW , PDB 42 1298 829 1.4 3.5
All molecules All of the above 4666 527 722 5.3 9.1
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Overview of the main RNA types in version 2.0 of the RNA STRAND database, their provenance, the number of RNAs, the mean length and standard deviation for each type. % PKBP denotes the percentage of the base pairs that need to be removed in order to render the structure pseudoknot-free. Most of the major RNA types are represented by a large number of molecules.
In the future, we intend to regularly check the aforementioned databases for new entries. With our current tools, keeping the database up-to-date will be relatively easy.
Data processing
Unique IDs
We created a unique and stable identifier for each entry in the RNA STRAND database. Future releases will keep all previous IDs unchanged.
Conversion scripts
One of the challenging tasks in collecting the RNA STRAND data arose from the fact that the external sources offer data in various formats. We have built tools to convert from all these formats to the CT format, which we use to store all structures internally, and to RNAML, BPSEQ, dot-parentheses and FASTA formats when requested by a user. The format descriptions are accessible on the "Help" page online.
Validation
All external databases we have used in the current version of RNA STRAND, except Rfam, contain highly curated RNA secondary or tertiary structures, therefore we trust the curation methods of these sources. For Rfam we selected a set of reliable structures based on the cited papers, as described in the previous section. Once we converted all the secondary structure external files into the CT format, we checked all files in order to make sure the secondary structures are valid (i.e., one base is paired with at most one other base, and if base at position i is paired with base at position j, then base at position j is also paired with base at position i.). When performed on our present data, this validation step revealed several inconsistencies in some of the external files, which we brought to the attention of the respective database owners.
RNA Secondary Structure Analyser
The structural statistics that form the core part of RNA STRAND were generated using the RNA Secondary Structure Analyser, which takes as input an RNA secondary structure description, for example in CT format, and outputs a wide range of secondary structure information. While many of these features, such as the number and composition of stems, are rather straightforward to determine, in some cases, more advanced algorithmic techniques have to be applied – as is the case, for example, for the minimal number of base pairs that need to be removed to render a structure pseudoknot free. For this specific task, we implemented a dynamic programming algorithm that removes the minimum number of base pairs ; however, more sophisticated approaches could be used, such as those recently described by Smit et al. . The complete output of the analyser run for each individual database entry can be accessed easily from the RNA STRAND web interface, and a description of the output can be found in the online Supplementary Material 2.
MySQL database
All the data obtained from the RNA Secondary Structure Analyser were inserted into a relational database implemented in MySQL (version 5.0.26). The main table is MOLECULE, with one row per RNA entry in the database. This table contains as primary key the unique RNA STRAND ID of the entry and further comprises various descriptive fields, including: organism, reference, length, RNA type, external source, external ID, sequence, three levels of abstract shapes using the RNAshapes representation , the method of secondary structure determination, and a link to the respective CT file. (Since RNAshapes version 2.1.5 cannot obtain the abstract shape of pseudoknotted secondary structures, we first removed a minimum number of base pairs to render the structure pseudoknot-free.) Furthermore, there is one table per secondary structure feature, where the table MOLECULE is connected to each of these tables in a one-to-many relationship. For example, the table STEM contains information such as the number of base pairs and the estimated free energy change for that stem, using parameters by Xia et al. . Accurately estimating the free energy change of entire structures is currently challenging, due to structural motifs for which current energy models are incomplete, such as pseudoknots, non-canonical base pairs, and modified nucleotides. Other similar tables include HAIRPIN_LOOP, MULTI_LOOP and PSEUDOKNOT.
An additional table TMP_MOLECULE is used to temporarily hold new submissions received via the web interface; for these, we manually check the submission information by checking the cited paper, after which, if the submission is accepted, all further steps required to permanently add the respective RNA(s) to the database are performed automatically.
Web interface
The web interface to RNA STRAND has been created using a set of PHP scripts (version 5.1.2). The main functions of the web interface are searching, browsing, analysis, downloading and uploading.
Searching and browsing
The user specifies one or more search criteria in a web-based form. The general criteria include RNA type (e.g., 16S Ribosomal RNA), organism of origin (e.g., E. coli), external source (e.g., RCSB Protein Data Bank), length (in bases), the number of molecules in the complex, whether it is a fragment, a sequence pattern using the standard IUPAC nucleic acid codes, an abstract structure or fragment using the RNAshapes representation and whether or not to include non-redundant sequences.
We define a set of entries to be non-redundant if their sequences are pairwise distinct. On a search page the user can request a non-redundant set that satisfies some search criteria. In this case, if two entries have identical RNA sequences, one of them will be selected arbitrarily. In the remainder of this paper, when we refer to a number of non-redundant entries matching some criteria, we mean a largest non-redundant set of entries satisfying the specified criteria. Currently there are 4104 non-redundant entries out of the 4666 entries in RNA STRAND v2.0.
Advanced searches are supported based on 21 additional search criteria on secondary structure elements, such as selection of RNA molecules having at least one pseudoknot, or hairpin loops with a specific sequence – for example GNRA hairpin loops. The set of database entries that match all of the specified criteria simultaneously is returned in the form of a table.
Using advanced search criteria, users can search for entries with various structural motifs. For example, when looking for a Y shape with an additional hairpin, one would search for entries that have exactly one multi-loop, three multi-loop branches, three hairpins, one molecule in the complex, and no pseudoknots. This search returns 31 entries, most of which are ciliate telomerase RNAs from Rfam. If pseudoknots are allowed, then vertebrate telomerase RNAs from Rfam are also included, yielding 36 search results. An equivalent pseudoknot-free search can be obtained by typing in the abstract shape [ [] [] ] [] (where matching brackets represent one interrupted or uninterrupted stem). Pseudoknots are currently not permitted in the abstract shape representation .
Support for inspecting large fractions of the database contents is provided via searches with no or very general criteria. For example, it is easy to obtain a list of all RNase P RNA structures contained in the database.
Details on individual entries from the result list of any search can be displayed by clicking on an RNA STRAND ID link of the results table. This single entry display comprises general information about the entry, links to the original database entry for this molecule, a secondary structure diagram, details of its secondary structure elements and features, links to other RNA STRAND entries with the same sequence (i.e., redundant entries), links to the sequence and secondary structure specification in five formats (CT, RNAML, BPSEQ, dot-parentheses and FASTA), and a link to the complete output of the RNA Secondary Structure Analyser.
Analysis
In addition to the aforementioned analysis information for individual entries, RNA STRAND also provides histograms or cumulative distribution functions of various molecule characteristics (such as number of pseudoknots per molecule) or structural features (such as number of branches per multi-loop) for all structures in the database or for user-selected subsets, as obtained from the search page. In addition, correlations between various molecule characteristics and molecule length can be obtained. For an unbiased analysis, the user has the option of normalising the data by RNA type (such as tRNA), in which case for each particular RNA type, one data point is obtained by averaging over all the data for molecules of that type. Finally, the user can choose to remove the outliers of the distributions. We use a common definition, according to which a data point is an outlier if, and only if, it is smaller than Q 1 - 1.3·(Q 3 - Q 1) or greater than Q 1 + 1.3·(Q 3 - Q 1), where Q 1 and Q 3 are the first and third quartiles, respectively. Such analyses may guide research pertaining to understanding structural features in naturally occuring RNA molecules, as we outline in the "Utility and discussion" section.
Downloading
The set of molecules selected via the search page can be downloaded in one of five supported formats: CT, RNAML, BPSEQ, dot-parentheses and FASTA. Thus, researchers can use specifically selected structures locally.
Uploading
RNA STRAND supports public submission of RNA secondary structures to the database via its web interface. The structure file can be in any of the four supported secondary structure formats (CT, RNAML, BPSEQ and dot-parentheses) or in the PDB tertiary structure format. Since RNA STRAND is a curated database, newly submitted structures are checked for accuracy and completeness by one of the database administrators before they are added to the database. New additions to the public databases that constitute our external sources will be added to RNA STRAND regularly. This is complemented by the public submission option, which is intended for submission of structures that do not yet belong to any of these databases.
Utility and discussion
RNA STRAND v2.0 contains 4666 RNA molecules or interacting complexes of various types, and an abundance of RNA structural motifs (see also Table 1). This represents a considerable amount of data from which to draw significant statistics and trends about RNA secondary structures. from which to draw significant statistics and trends about RNA secondary structures. In what follows we illustrate how the information in RNA STRAND can be used for various purposes.
Obtaining statistics of naturally occuring RNA structural features
We performed statistical analyses using the RNA STRAND web interface. Our first observation concerns the number and complexity of pseudoknots. According to the current data from RNA STRAND v2.0, pseudoknots occur rather commonly, especially in longer molecules: 74% of all (non-redundant) entries with 100 or more nucleotides contain pseudoknots. We compared the stem length (i.e., the number of base pairs in uninterrupted stems) with the minimal number of base pairs that need to be removed per pseudoknot to render the structure pseudoknot free (we denote this number by # PKBP; note that for over 95% of the pseudoknots, the bases counted in determining # PKBP form one uninterrupted stem; also, there is no overlap between the base pairs counted in determining the stem length and the base pairs counted in determining # PKBP). Table 2 shows that when considering all RNA types in the database, the median, mean and standard deviation of the two measures, stem length and # PKBP, are very similar, even when we normalise by RNA type. (For normalised analysis, instead of using one data point per molecule or per structural feature, we use one data point for each RNA type, where this point is determined by averaging all data points for the respective class of RNAs. This way, the user can avoid biasing the analysis when there are substantially more structures for some RNA types than for others.) However, for 16S and 23S ribosomal RNA molecules the stem length tends to be significantly larger than # PKBP, whereas for transfer messenger RNA molecules in particular and ribonuclease P RNA molecules to some extent, # PKBP is larger than the stem length. This observation is interesting in the context of computational approaches for RNA secondary structure prediction which ignore pseudoknots , add pseudoknots hierarchically in a second stage , or simultaneously add stems in pseudoknotted and non-pseudoknotted regions [24,25].
Table 2.
Statistics on the complexity of pseudoknots in RNA STRAND v2.0.
RNA type#Stem length# PKBP
entries median mean std median mean std
16S ribosomal RNA 644 4.00 4.30 2.50 3.00 2.50 0.68
23S ribosomal RNA 93 4.00 4.14 2.39 2.00 3.75 3.12
Transfer messenger RNA 657 4.00 4.11 2.24 5.00 5.51 1.00
Ribonuclease P RNA 433 4.00 4.45 2.51 4.00 5.18 1.36
All, non-redundant 4104 4.00 4.35 2.44 4.00 4.14 1.86
All, non-redundant & normalised 4104 4.96 5.05 0.58 4.65 4.95 1.78
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The columns represent the RNA type, the number of entries for each type, the median, mean and standard deviation of the stem length (i.e., number of adjacent base pairs) and the minimum number of base pairs to break in order to open pseudoknots (# PKBP). For each row, a non-redundant set was selected, and outliers were removed.
Our second observation concerns the abundance of non-canonical base pairs and the pairing type of their immediate neighbours. (We consider all C-G, A-U and G-U pairs to be canonical base pairs, and all other base pairs to be non-canonical.) Figure 3 shows a histogram for the 729 non-redundant entries whose structures were determined by all-atom methods (these include structures from the Protein Data Bank and the Nucleic Acid Database). For this data set, non-canonical A-G base pairs are the most abundant, representing 55% of all non-canonical base pairs, and G-G pairs are the least abundant, representing only 4% of all non-canonical base pairs. The plot also shows that a relatively small fraction of non-canonical base pairs have as immediate neighbours canonical base pairs. Interestingly, for all seven types of non-canonical base pairs, more pairs are adjacent to at least one other non-canonical base pair than surrounded by two canonical base pairs. For example, 55% of all A-A pairs are adjacent to at least one other non-canonical base pair. This may suggest that non-canonical base pairs are sufficiently stable energetically to form several consecutive base pairs.
Figure 3.
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Histogram of the occurence of non-canonical base pairs. Histogram of non-canonical base pairs in the 729 non-redundant entries whose structures were determined by NMR or X-ray crystallography.
Finally, we found rather strong linear correlations between the number of nucleotides of the RNAs in our database and the number of stems, hairpin loops, bulges, internal loops and multi-loops; the Pearson correlation coefficients are r= 0.95, 0.95, 0.92, 0.91 and 0.92, respectively. This is consistent with the idea that the local formation of these secondary structure elements is relatively independent of the overall size of the molecule and in agreement with the current thermodynamic energy models of RNA secondary structure, which assume additive and independent energy contributions for these structural elements. Interestingly, the correlation between the RNA length and the number of pseudoknots is significantly weaker (r= 0.64), suggesting that pseudoknots may not follow the same linearity principle.
Evaluating energy-based secondary structure prediction programs
The RNA STRAND database can be used to evaluate the prediction accuracy of energy-based RNA secondary structure prediction software. RNA STRAND v2.0 contains 3704 non-redundant entries containing one molecule that can be used to evaluate software such as CONTRAfold or mfold , 403 non-redundant entries containing complexes of two or more molecules that can be used to evaluate sofware for interacting molecules [27,28], and 1957 non-redundant single-molecules with pseudoknots that can be used to evaluate secondary structure prediction programs with pseudoknots [23-25,29].
We have selected 2518 structures out of the 3704 non-redundant entries containing one molecule, after we eliminated the entries with unknown nucleotides and overly large loops. (Specifically, entries having hairpin loops, bulges, internal loops or multi-loops with more than 50, 50, 50 and 100 unpaired bases, respectively, were removed.) In addition, we have removed all non-canonical base pairs and the minimum number of base pairs needed to render the structures pseudoknot-free. The resulting structures are used as ground-truth reference structures. We evaluated the sensitivity and positive predictive value (PPV) of CONTRAfold and SimFold with various free energy parameter sets, see Figure 4. Sensitivity is the number of correctly predicted base pairs divided by the number of base pairs in the reference structure, and PPV is the number of correctly predicted base pairs divided by the number of predicted base pairs. "SimFold Turner99" in Figure 4 refers to SimFold using the free energy parameters described by Mathews et al. , and is essentially equivalent to mfold 3.1 . On this large set, the average sensitivity of prediction is 0.63, while the average PPV is 0.57.
Figure 4.
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Prediction accuracy achieved by various energy models. Sensitivity vs. positive predictive value (PPV) of various secondary structure prediction methods. Sensitivity is the number of correctly predicted base pairs divided by the number of base pairs in the reference structure, PPV is the number of correctly predicted base pairs, divided by the number of predicted base pairs. Higher prediction accuracy is achieved when the free energy parameters are obtained by training on a larger set of structures. The CONTRAfold prediction program uses a trade-off parameter γ between sensitivity and PPV, and thus we report predictions for γ ranging from 2 to 20.
"CONTRAfold 1.1 151Rfam" is the CONTRAfold software version 1.1, as reported by Do et al. . The CONTRAfold prediction program uses a trade-off parameter γ between sensitivity and PPV, and thus we report predictions for γ ranging from 2 to 20. When the target of one measure is fixed to the value obtained with "SimFold Turner99", the other is similar as well, showing that on this data set, CONTRAfold 1.1 gives similar average prediction accuracy as "SimFold Turner99". The remaining points of Figure 4 are described in the following section.
Improving RNA energy models
More importantly, RNA STRAND can facilitate approaches for improving the free energy models underlying energy-based RNA secondary structure prediction software [6,7]. In this context, it can be very useful to exploit training data consisting of RNA sequences with known secondary structures, and the size and variety of such data are key for obtaining good results.
Figure 4 shows the average sensitivity and PPV of various programs measured on the 2518 structures mentioned in the previous section, and trained on various training sets.
"CONTRAfold 1.1 151Rfam" was trained on a small set of 151 structures from various Rfam families , while "CONTRAfold 2.0 STRAND1.3" was trained on 3427 pre-processed structures (i.e., split and restricted) of average length 178 nucleotides from version 1.3 of the RNA STRAND database, as used by Andronescu et al. . The figure shows that using the much larger set in the latter case gives an improvement of roughly 7% in prediction accuracy.
To demonstrate even further the importance of using a large and comprehensive set of known RNA secondary structures for obtaining high-quality free energy parameters, we have used the current version of RNA STRAND v2.0 to obtain a new training set of 2246 structures of average length 246 nucleotides. Using the Maximum Likelihood parameter estimation method described by Andronescu et al. , which is similar to CONTRAfold , we have improved the average accuracy of prediction even further, as shown by the data point labelled "SimFold STRAND2.0" in Figure 4. This gives an improvement of 8% in average sensitivity and 10% in average PPV compared to the Turner99 parameters, when measured on our test set of 2518 structures. (Note that, since CONTRAfold and SimFold use different energy models and prediction algorithms, it is more appropriate to make comparisons between different versions of each, than it is to compare CONTRAfold versus SimFold).
These results provide clear evidence for the key role of large and carefully assembled sets of RNA secondary structures, such as provided by RNA STRAND, in the context of determining RNA free energy models. In the future, we are planning to use the RNA STRAND data to train free energy parameters for pseudoknotted structures. Existing energy models for RNA secondary structure prediction methods with pseudoknots are often ad-hoc [25,29], and we believe that by using data-driven methods in conjunction with the 1957 non-redundant RNA STRAND entries representing RNAs with pseudoknots, it will be possible to obtain improved energy models for pseudoknotted structure prediction.
Other uses of RNA STRAND
The numerous search criteria supported by the RNA STRAND web interface allow users to select and study molecules with specific structural features. For example, Tyagi and Mathews studied the computational prediction accuracy of helical coaxial stacking in multi-loops. RNA STRAND v2.0 conveniently allows the selection and download of 189 non-redundant entries with all-atom structures that have at least one multi-loop. Other examples include the use of naturally occuring pseudoknotted structures that can be used to evaluate computational methods to render a pseudoknotted RNA secondary structure pseudoknot free , or to evaluate RNA secondary structure visualisation tools .
In recent work on the role of RNA structure in splicing, Rogic et al. needed to identify thermodynamically stable stems that maximally shorten the distance between mRNA donor sites and branchpoint sequences. Since the optimal free energy of such stems is unknown, Rogic et al. wished to determine the most probable ranges of possible free energies for uninterrupted stems. By selecting all molecules on the RNA STRAND web site, they obtained distributions of estimated stem free energies (obtained with parameters by Xia et al. ), which were used to support a new model for the role of RNA secondary stucture in mRNA splicing.
In addition, RNA STRAND can facilitate the design of optical melting experiments , whose goal is to better understand the thermodynamics of RNA structure formation, and to improve RNA secondary structure prediction accuracy. When designing optical melting experiments, usually a set of known RNA secondary structures is first assembled to determine what types of structural motifs that have not been previously studied appear frequently in naturally occuring RNAs [34,35]. The RNA STRAND web interface, as well as the abundance of reliable RNA structures in the RNA STRAND database, can be very useful in this context. For example, a significant number of multi-loops (16% in all non-redundant RNA STRAND entries) have five or more branches, but, to the best of our knowledge, optical melting experiments only exist for multi-loops with up to four branches [36,37]. Moreover, 30% of the internal loops in all non-redundant RNA STRAND entries have seven or more unpaired bases, and 13% have an absolute asymmetry (i.e., absolute difference between the number of unpaired bases on each side) of at least three, while only limited optical melting experiments exist to cover these cases [38,39].
Conclusion
In this paper, we presented RNA STRAND, a new database for RNA secondary structure data that provides access to detailed information on known secondary structures as well as statistical analyses of structural aspects of various types of RNAs. We believe that such information will be useful in the context of understanding RNA structure and function; in particular, we expect it to further facilitate the development and evaluation of energy models for secondary structure prediction. Our database is flexible and extensible; it provides a convenient web interface to its major functions and supports searches according to many criteria, including properties of secondary structure elements. The database is publicly accessible and supports the submission of new RNA structures by the research community. We are committed to keeping RNA STRAND up-to-date with new structures that are added to the eight databases of provenance, and we invite submissions of all types of RNA secondary structures, which will help to further expand the database and increase its usefulness.
In the future, we intend to add RNA secondary structures obtained from the SHAPE technique [40,41], and also to provide further search options such as searches by specific structural motifs.
Availability and requirements
RNA STRAND is publicly available at The RNA Secondary Structure Analyser, as well as the database tables, are available upon request from the authors.
Authors' contributions
MA collected the data, implemented conversion and validation scripts, implemented the MySQL database and part of the PHP scripts, performed the statistical analyses and helped to draft the manuscript. VB implemented the vast majority of the PHP scripts and most of the RNA Secondary Structure Analyser. HHH and AC conceived the project, specified the design of the RNA Secondary Structure Analyser, supervised MA's and VB's work, and helped to write the manuscript. All authors checked the accuracy of the database and web interface, read and approved the final manuscript.
Acknowledgments
Acknowledgements
We thank Baharak Rastegari, Yinglei S. Zhao, Mohammad Safari and Jack Jia, who provided an efficient computer program for pseudoknotted structure parsing; Robin Gutell, Christian Zwieb, James Brown and Mathias Sprinzl for providing data and help with the CRW, SRPDB & tmRDB, RNase P DB and Sprinzl tRNA DB, respectively; Simon Moxon and Jennifer Daub for help using the Rfam database; Robert Giegerich and David Mathews for useful discussions; Alex Brown for help with the web interface; and Farheen Rawji for her work on an early version of the RNA Secondary Structure Analyser.
Funding for this work was provided by: Mathematics of Information Technology and Complex Systems Network of Centres of Excellence (to AC and HH); Natural Sciences and Engineering Research Council of Canada Discovery Grant Program (238788 to HH and 217192 to AC).
Contributor Information
Mirela Andronescu, Email: andrones@cs.ubc.ca.
Vera Bereg, Email: verab@cs.ubc.ca.
Holger H Hoos, Email: hoos@cs.ubc.ca.
Anne Condon, Email: condon@cs.ubc.ca.
References
Cannone J, Subramanian S, Schnare M, Collett J, D'Souza L, Du Y, Feng B, Lin N, Madabusi L, Müller K, Pande N, Shang Z, Yu N, Gutell R. The comparative RNA web (CRW) site: an online database of comparative sequence and structure information for ribosomal, intron, and other RNAs. BMC Bioinformatics. 2002;3:2. doi: 10.1186/1471-2105-3-2. [Correction: BMC Bioinformatics 3:15] [DOI] [PMC free article] [PubMed] [Google Scholar]
Sprinzl M, Vassilenko K. Compilation of tRNA sequences and sequences of tRNA genes. Nucleic Acids Res. 2005;33:D139–40. doi: 10.1093/nar/gki012. [DOI] [PMC free article] [PubMed] [Google Scholar]
Brown J. The Ribonuclease P Database. Nucleic Acids Res. 1999;27:314. doi: 10.1093/nar/27.1.314. [DOI] [PMC free article] [PubMed] [Google Scholar]
Andersen ES, Rosenblad MA, Larsen N, Westergaard JC, Burks J, Wower IK, Wower J, Gorodkin J, Samuelsson T, Zwieb C. The tmRDB and SRPDB resources. Nucleic Acids Res. 2006;34:D163–8. doi: 10.1093/nar/gkj142. [DOI] [PMC free article] [PubMed] [Google Scholar]
Griffiths-Jones S, Moxon S, Marshall M, Khanna A, Eddy S, Bateman A. Rfam: annotating non-coding RNAs in complete genomes. Nucleic Acids Res. 2005;33:D121–4. doi: 10.1093/nar/gki081. [DOI] [PMC free article] [PubMed] [Google Scholar]
Andronescu M, Condon A, Hoos HH, Mathews DH, Murphy KP. Efficient parameter estimation for RNA secondary structure prediction. Bioinformatics. 2007;23:i19–i28. doi: 10.1093/bioinformatics/btm223. [DOI] [PubMed] [Google Scholar]
Do CB, Woods DA, Batzoglou S. CONTRAfold: RNA secondary structure prediction without physics-based models. Bioinformatics. 2006;22:e90–e98. doi: 10.1093/bioinformatics/btl246. [DOI] [PubMed] [Google Scholar]
van Batenburg FH, Gultyaev AP, Pleij CW. PseudoBase: structural information on RNA pseudoknots. Nucleic Acids Res. 2001;29:194–195. doi: 10.1093/nar/29.1.194. [DOI] [PMC free article] [PubMed] [Google Scholar]
Gan HH, Fera D, Zorn J, Shiffeldrim N, Tang M, Laserson U, Kim N, Schlick T. RAG: RNA-As-Graphs database-concepts, analysis, and features. Bioinformatics. 2004;20:1285–1291. doi: 10.1093/bioinformatics/bth084. [DOI] [PubMed] [Google Scholar]
Westbrook J, Feng Z, Chen L, Yang H, Berman H. The Protein Data Bank and structural genomics. Nucleic Acids Res. 2003;31:489–491. doi: 10.1093/nar/gkg068. [DOI] [PMC free article] [PubMed] [Google Scholar]
Berman HM, Olson WK, Beveridge DL, Westbrook J, Gelbin A, Demeny T, Hsieh SH, Srinivasan AR, Schneider B. The nucleic acid database. A comprehensive relational database of three-dimensional structures of nucleic acids. Biophys J. 1992;63:751–759. doi: 10.1016/S0006-3495(92)81649-1. [DOI] [PMC free article] [PubMed] [Google Scholar]
Murthy VL, Rose GD. RNABase: an annotated database of RNA structures. Nucleic Acids Res. 2003;31:502–504. doi: 10.1093/nar/gkg012. [DOI] [PMC free article] [PubMed] [Google Scholar]
Tamura M, Hendrix DK, Klosterman PS, Schimmelman NR, Brenner SE, Holbrook SR. SCOR: Structural Classification of RNA, version 2.0. Nucleic Acids Res. 2004;32:D182–4. doi: 10.1093/nar/gkh080. [DOI] [PMC free article] [PubMed] [Google Scholar]
Nagaswamy U, Larios-Sanz M, Hury J, Collins S, Zhang Z, Zhao Q, Fox GE. NCIR: a database of non-canonical interactions in known RNA structures. Nucleic Acids Res. 2002;30:395–397. doi: 10.1093/nar/30.1.395. [DOI] [PMC free article] [PubMed] [Google Scholar]
Sarver M, Zirbel CL, Stombaugh J, Mokdad A, Leontis NB. FR3D: finding local and composite recurrent structural motifs in RNA 3D structures. J Math Biol. 2008;56:215–252. doi: 10.1007/s00285-007-0110-x. [DOI] [PMC free article] [PubMed] [Google Scholar]
Rocheleau L, Pelchat M. The Subviral RNA Database: a toolbox for viroids, the hepatitis delta virus and satellite RNAs research. BMC Microbiol. 2006;6:24. doi: 10.1186/1471-2180-6-24. [DOI] [PMC free article] [PubMed] [Google Scholar]
Yang H, Jossinet F, Leontis N, Chen L, Westbrook J, Berman H, Westhof E. Tools for the automatic identification and classification of RNA base pairs. Nucleic Acids Res. 2003;31:3450–3460. doi: 10.1093/nar/gkg529. [DOI] [PMC free article] [PubMed] [Google Scholar]
Apostolico A, Atallah MJ, Hambrusch SE. New clique and independent set algorithms for circle graphs. Discrete Applied Mathematics. 1996;32:1–24. [Google Scholar]
Smit S, Rother K, Heringa J, Knight R. From knotted to nested RNA structures: a variety of computational methods for pseudoknot removal. RNA. 2008;14:410–416. doi: 10.1261/rna.881308. [DOI] [PMC free article] [PubMed] [Google Scholar]
Steffen P, Voss B, Rehmsmeier M, Reeder J, Giegerich R. RNAshapes: an integrated RNA analysis package based on abstract shapes. Bioinformatics. 2006;22:500–503. doi: 10.1093/bioinformatics/btk010. [DOI] [PubMed] [Google Scholar]
Xia T, SantaLucia J, Burkard M, Kierzek R, Schroeder S, Jiao X, Cox C, Turner D. Thermodynamic parameters for an expanded nearest-neighbor model for formation of RNA duplexes with Watson-Crick base pairs. Biochemistry. 1998;37:14719–14735. doi: 10.1021/bi9809425. [DOI] [PubMed] [Google Scholar]
Mathews D, Sabina J, Zuker M, Turner D. Expanded sequence dependence of thermodynamic parameters improves prediction of RNA secondary structure. J Mol Biol. 1999;288:911–940. doi: 10.1006/jmbi.1999.2700. [DOI] [PubMed] [Google Scholar]
Jabbari H, Condon A, Pop A, Pop C, Zhao Y. HFold: RNA Pseudoknotted Secondary Structure Prediction Using Hierarchical Folding. Workshop on Algorithms in Bioinformatics. 2007. pp. 323–334.
Ren J, Rastegari B, Condon A, Hoos HH. HotKnots: heuristic prediction of RNA secondary structures including pseudoknots. RNA. 2005;11:1494–1504. doi: 10.1261/rna.7284905. [DOI] [PMC free article] [PubMed] [Google Scholar]
Rivas E, Eddy SR. A dynamic programming algorithm for RNA structure prediction including pseudoknots. J Mol Biol. 1999;285:2053–2068. doi: 10.1006/jmbi.1998.2436. [DOI] [PubMed] [Google Scholar]
Zuker M. Mfold web server for nucleic acid folding and hybridization prediction. Nucleic Acids Res. 2003;31:3406–3415. doi: 10.1093/nar/gkg595. [DOI] [PMC free article] [PubMed] [Google Scholar]
Andronescu M, Zhang ZC, Condon A. Secondary structure prediction of interacting RNA molecules. J Mol Biol. 2005;345:987–1001. doi: 10.1016/j.jmb.2004.10.082. [DOI] [PubMed] [Google Scholar]
Dirks R, Bois J, Schaeffer J, Winfree E, Pierce N. Thermodynamic analysis of interacting nucleic acid strands. SIAM Rev. 2007;49:65–88. [Google Scholar]
Dirks RM, Pierce NA. A partition function algorithm for nucleic acid secondary structure including pseudoknots. J Comput Chem. 2003;24:1664–1677. doi: 10.1002/jcc.10296. [DOI] [PubMed] [Google Scholar]
Andronescu M. Master's thesis. Dept. of Computer Science, University of British Columbia; 2003. Algorithms for predicting the Secondary Structure of pairs and combinatorial sets of nucleic acid strands. [Google Scholar]
Tyagi R, Mathews DH. Predicting helical coaxial stacking in RNA multibranch loops. RNA. 2007;13:939–951. doi: 10.1261/rna.305307. [DOI] [PMC free article] [PubMed] [Google Scholar]
Byun Y, Han K. PseudoViewer: web application and web service for visualizing RNA pseudoknots and secondary structures. Nucleic Acids Res. 2006;34:W416–422. doi: 10.1093/nar/gkl210. [DOI] [PMC free article] [PubMed] [Google Scholar]
Rogic S, Montpetit B, Hoos HH, Mackworth AK, Ouellette FB, Hieter P. Correlation between the secondary structure of pre-mRNA introns and the efficiency of splicing in Saccharomyces cerevisiae. BMC Genomics. 2008;9:355. doi: 10.1186/1471-2164-9-355. [DOI] [PMC free article] [PubMed] [Google Scholar]
Badhwar J, Karri S, Cass CK, Wunderlich EL, Znosko BM. Thermodynamic characterization of RNA duplexes containing naturally occurring 1 × 2 nucleotide internal loops. Biochemistry. 2007;46:14715–14724. doi: 10.1021/bi701024w. [DOI] [PubMed] [Google Scholar]
Davis AR, Znosko BM. Thermodynamic characterization of single mismatches found in naturally occurring RNA. Biochemistry. 2007;46:13425–13436. doi: 10.1021/bi701311c. [DOI] [PubMed] [Google Scholar]
Diamond J, Turner D, Mathews D. Thermodynamics of three-way multibranch loops in RNA. Biochemistry. 2001;40:6971–6981. doi: 10.1021/bi0029548. [DOI] [PubMed] [Google Scholar]
Mathews D, Turner D. Experimentally derived nearest-neighbor parameters for the stability of RNA three- and four-way multibranch loops. Biochemistry. 2002;41:869–880. doi: 10.1021/bi011441d. [DOI] [PubMed] [Google Scholar]
Peritz A, Kierzek R, Sugimoto N, Turner D. Thermodynamic study of internal loops in oligoribonucleotides: symmetric loops are more stable than asymmetric loops. Biochemistry. 1991;30:6428–6436. doi: 10.1021/bi00240a013. [DOI] [PubMed] [Google Scholar]
Chen G, Turner DH. Consecutive GA pairs stabilize medium-size RNA internal loops. Biochemistry. 2006;45:4025–4043. doi: 10.1021/bi052060t. [DOI] [PMC free article] [PubMed] [Google Scholar]
Merino EJ, Wilkinson KA, Coughlan JL, Weeks KM. RNA structure analysis at single nucleotide resolution by selective 2'-hydroxyl acylation and primer extension (SHAPE) J Am Chem Soc. 2005;127:4223–4231. doi: 10.1021/ja043822v. [DOI] [PubMed] [Google Scholar]
Wilkinson KA, Merino EJ, Weeks KM. Selective 2'-hydroxyl acylation analyzed by primer extension (SHAPE): quantitative RNA structure analysis at single nucleotide resolution. Nat Protoc. 2006;1:1610–1616. doi: 10.1038/nprot.2006.249. [DOI] [PubMed] [Google Scholar]
Leontis NB, Westhof E. Geometric nomenclature and classification of RNA base pairs. RNA. 2001;7:499–512. doi: 10.1017/s1355838201002515. [DOI] [PMC free article] [PubMed] [Google Scholar]
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823 | https://www.youtube.com/watch?v=p_vh_uiZ2QI | Calculating Moles in a Balanced Equation with the Mole Ratio
Mr. Causey
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Calculating Moles in a Balanced Equation with the Mole Ratio. Mr. Causey shows you step by step how to determine the number of moles produced or consumed in a chemical reaction. The key to these problems is the mole ratio.
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Mr. Causey shows you step by step how to determine the number of moles produced and consumed in a chemical reaction. You will write and balance a chemical equation and then determine the moles of the reactants and the products.
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Intro [Music] Howdy everyone. Mr. Kosy here with another problem of the week and let's get started. This week you're going to need your calculator. So make sure you get one out. I use the TI 84. And let's look at today's Mole Ratio problem. We have this chemical equation and you want to make sure it's balanced. And for the part one, we want to know how many moles of magnesium chloride would be used up if 8 moles of aluminum chloride were produced. Now, working with a problem like this, it's all about moles. So, all you need to do is work with the mole ratio. Now, if you're not sure what mole ratios are, go back to my channel. You can go to mrcazy.com or you can go to your uh chemcoach.com and get to my YouTube and at my YouTube find the mole ratios and Avagadro's number videos and watch them and learn about mole ratios. So let's go to the magic blackboard and let's write out our Equation equation. Now that's a chemical equation. It consists of chemical formulas. And there's our moles of magnesium chloride. And there's our moles of aluminum chloride. And we see that our mole ratio is uh 3:2 or 2:3 depending on what we need to cancel out. Now let's look at the equation. It says to start with 8 moles of aluminum chloride. So write that down and then multiply that by the mole ratio. And the mole ratio that we want to use is the 3:2 because we want aluminum chloride to cancel out. And that will leave us with magnesium chloride. Now let's go through and cancel out 2 and 8 cuz that'll give us four. That's just a little bit of reducing. And then 4 3 is 12 moles of magnesium chloride. And we've got it. 8 moles of aluminum chloride will come from 12 moles of magnesium chloride. Let's look at part two. How many moles of magnesium sulfide are produced in 1.4 moles of aluminum sulfide? Or if uh 1.4 4 moles of aluminum sulfate are consumed. Again, it's the mole ratio. And so, uh, making sure that everything's balanced, which we already have, let's go to the magic blackboard, get the mole ratio for this second part. And we have 1 3. And the mole ratio is going to be 1:3 or 3 1. Okay? And it depends on what we want to cancel out. Let's do the equation. The equation, we're going to have 1.4 moles of aluminum sulfide and we'll multiply it by 3 over 1 because we want aluminum sulfide to cancel out. Now, if I wanted magnesium sulfide to cancel, I would just flip it over and that equals 4.12 moles of magnesium sulfide. Not bad. The big key to this whole thing right here was using mole ratio. And uh that's it. All right. If you have any questions, be sure to send an email to mrcy@mrcausy.com. And remember to check out mrcasworld.com for PowerPoint videos and much much more. And of course you can uh subscribe to my YouTube channel. Don't forget to do that. Studies have shown that it increases your [Music] IQ. All righty, everyone. Happy islands. [Music] |
824 | https://www.flourishkh.com/downloads/DG4/DG4_PYSA_ans_2nd08.pdf | Discovering Geometry An Investigative Approach Practice Your Skills with Answers Teacher’s Materials Project Editor: Elizabeth DeCarli Project Administrator: Brady Golden Coordinating Writer: Jennifer North Morris Contributors: David Rasmussen, Ralph Bothe, Judy Hicks, Michael Serra Accuracy Checker: Dudley Brooks Production Editor: Holly Rudelitsch Copyeditor: Jill Pellarin Editorial Production Manager: Christine Osborne Production Supervisor: Ann Rothenbuhler Production Coordinator: Jennifer Young Text Designers: Jenny Somerville, Garry Harman Composition, Technical Art, Prepress: ICC Macmillan Inc.
Cover Designer: Jill Kongabel Printer: Data Reproductions Textbook Product Manager: James Ryan Executive Editor: Casey FitzSimons Publisher: Steven Rasmussen ©2008 by Kendall Hunt Publishing. All rights reserved.
Cover Photo Credits: Background image: Doug Wilson/Westlight/Corbis.
Construction site image: Sonda Dawes/The Image Works. All other images: Ken Karp Photography.
Limited Reproduction Permission The publisher grants the teacher whose school has adopted Discovering Geometry, and who has received Discovering Geometry: An Investigative Approach, Practice Your Skills with Answers as part of the Teaching Resources package for the book, the right to reproduce material for use in his or her own classroom. Unauthorized copying of Discovering Geometry: An Investigative Approach, Practice Your Skills with Answers constitutes copyright infringement and is a violation of federal law.
All registered trademarks and trademarks in this book are the property of their respective holders.
Kendall Hunt Publishing 4050 Westmark Drive PO Box 1840 Dubuque, IA 52004-1840 www.kendallhunt.com Printed in the United States of America 10 9 8 7 6 5 4 3 2 13 12 11 10 09 08 ISBN 978-1-55953-894-7 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi Chapter 1 Lesson 1.1: Building Blocks of Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Lesson 1.2: Poolroom Math . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Lesson 1.3: What’s a Widget? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Lesson 1.4: Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Lesson 1.5: Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Lesson 1.6: Special Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Lesson 1.7: Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Lesson 1.8: Space Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Lesson 1.9: A Picture Is Worth a Thousand Words . . . . . . . . . . . . . . . . . . . 9 Chapter 2 Lesson 2.1: Inductive Reasoning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Lesson 2.2: Finding the nth Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Lesson 2.3: Mathematical Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Lesson 2.4: Deductive Reasoning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Lesson 2.5: Angle Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Lesson 2.6: Special Angles on Parallel Lines . . . . . . . . . . . . . . . . . . . . . . . 15 Chapter 3 Lesson 3.1: Duplicating Segments and Angles . . . . . . . . . . . . . . . . . . . . 16 Lesson 3.2: Constructing Perpendicular Bisectors . . . . . . . . . . . . . . . . . . 17 Lesson 3.3: Constructing Perpendiculars to a Line . . . . . . . . . . . . . . . . . 18 Lesson 3.4: Constructing Angle Bisectors . . . . . . . . . . . . . . . . . . . . . . . . . 19 Lesson 3.5: Constructing Parallel Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Lesson 3.6: Construction Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Lesson 3.7: Constructing Points of Concurrency . . . . . . . . . . . . . . . . . . . 22 Lesson 3.8: The Centroid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Chapter 4 Lesson 4.1: Triangle Sum Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Lesson 4.2: Properties of Isosceles Triangles . . . . . . . . . . . . . . . . . . . . . . 25 Lesson 4.3: Triangle Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Lesson 4.4: Are There Congruence Shortcuts? . . . . . . . . . . . . . . . . . . . . . 27 Lesson 4.5: Are There Other Congruence Shortcuts? . . . . . . . . . . . . . . . 28 Lesson 4.6: Corresponding Parts of Congruent Triangles . . . . . . . . . . . 29 Lesson 4.7: Flowchart Thinking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Lesson 4.8: Proving Special Triangle Conjectures . . . . . . . . . . . . . . . . . . 31 iii Contents iv Chapter 5 Lesson 5.1: Polygon Sum Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Lesson 5.2: Exterior Angles of a Polygon . . . . . . . . . . . . . . . . . . . . . . . . . 33 Lesson 5.3: Kite and Trapezoid Properties . . . . . . . . . . . . . . . . . . . . . . . . 34 Lesson 5.4: Properties of Midsegments . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Lesson 5.5: Properties of Parallelograms . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Lesson 5.6: Properties of Special Parallelograms . . . . . . . . . . . . . . . . . . 37 Lesson 5.7: Proving Quadrilateral Properties . . . . . . . . . . . . . . . . . . . . . . 38 Chapter 6 Lesson 6.1: Tangent Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Lesson 6.2: Chord Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 Lesson 6.3: Arcs and Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Lesson 6.4: Proving Circle Conjectures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 Lesson 6.5: The Circumference/Diameter Ratio . . . . . . . . . . . . . . . . . . . . 43 Lesson 6.6: Around the World . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 Lesson 6.7: Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 Exploration: Intersecting Secants, Tangents, and Chords . . . . . . . . . . . 46 Chapter 7 Lesson 7.1: Transformations and Symmetry . . . . . . . . . . . . . . . . . . . . . . . 47 Lesson 7.2: Properties of Isometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 Lesson 7.3: Compositions of Transformations . . . . . . . . . . . . . . . . . . . . . 49 Lesson 7.4: Tessellations with Regular Polygons . . . . . . . . . . . . . . . . . . . 50 Lessons 7.5–7.8: Tessellations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Chapter 8 Lesson 8.1: Areas of Rectangles and Parallelograms . . . . . . . . . . . . . . . 52 Lesson 8.2: Areas of Triangles, Trapezoids, and Kites . . . . . . . . . . . . . . . 53 Lesson 8.3: Area Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Lesson 8.4: Areas of Regular Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Lesson 8.5: Areas of Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 Lesson 8.6: Any Way You Slice It . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Lesson 8.7: Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 Chapter 9 Lesson 9.1: The Theorem of Pythagoras . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Lesson 9.2: The Converse of the Pythagorean Theorem . . . . . . . . . . . . 60 Lesson 9.3: Two Special Right Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Lesson 9.4: Story Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 Lesson 9.5: Distance in Coordinate Geometry . . . . . . . . . . . . . . . . . . . . . 63 Lesson 9.6: Circles and the Pythagorean Theorem . . . . . . . . . . . . . . . . . 64 v Chapter 10 Lesson 10.1: The Geometry of Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Lesson 10.2: Volume of Prisms and Cylinders . . . . . . . . . . . . . . . . . . . . . 66 Lesson 10.3: Volume of Pyramids and Cones . . . . . . . . . . . . . . . . . . . . . . 67 Lesson 10.4: Volume Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 Lesson 10.5: Displacement and Density . . . . . . . . . . . . . . . . . . . . . . . . . . 69 Lesson 10.6: Volume of a Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Lesson 10.7: Surface Area of a Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 Chapter 11 Lesson 11.1: Similar Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Lesson 11.2: Similar Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Lesson 11.3: Indirect Measurement with Similar Triangles . . . . . . . . . . 74 Lesson 11.4: Corresponding Parts of Similar Triangles . . . . . . . . . . . . . 75 Lesson 11.5: Proportions with Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 Lesson 11.6: Proportions with Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 Lesson 11.7: Proportional Segments Between Parallel Lines . . . . . . . . 78 Chapter 12 Lesson 12.1: Trigonometric Ratios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 Lesson 12.2: Problem Solving with Right Triangles . . . . . . . . . . . . . . . . 80 Lesson 12.3: The Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 Lesson 12.4: The Law of Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 Lesson 12.5: Problem Solving with Trigonometry . . . . . . . . . . . . . . . . . . 83 Chapter 13 Lesson 13.1: The Premises of Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . 84 Lesson 13.2: Planning a Geometry Proof . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Lesson 13.3: Triangle Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 Lesson 13.4: Quadrilateral Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Lesson 13.5: Indirect Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 Lesson 13.6: Circle Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Lesson 13.7: Similarity Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 vi Introduction The author and editors of Discovering Geometry: An Investigative Approach are aware of the importance of students developing geometry skills along with acquiring concepts through investigation. The student book includes many skill-based exercises. These Practice Your Skills worksheets provide problems similar to the introductory exercises in each lesson of Discovering Geometry. Like those exercises, these worksheets allow students to practice and reinforce the important procedures and skills developed in the lessons. Some of these problems provide non contextual skills practice. Others give students an opportunity to apply geometry concepts in fairly simple, straightforward contexts. Some are more complex problems that are broken down into small steps.
You might assign the Practice Your Skills worksheet for every lesson, or only for those lessons your students find particularly difficult. Or, you may wish to assign the worksheets on an individual basis, only to those students who need extra help. One worksheet has been provided for nearly every lesson. There are no worksheets for Chapter 0, and the optional tessellation lessons have been combined into two worksheets. To save you the time and expense of copying pages, you can give students the inexpensive Practice Your Skills Student Workbook, which does not have answers. Though the copyright allows you to copy pages from Practice Your Skills with Answers for use with your students, the consumable Practice Your Skills Student Workbook should not be copied. Students, parents, and mentors can also download the student worksheets from flourishkh.com.
Lesson 1.1 • Building Blocks of Geometry Name Period Date For Exercises 1–7, complete each statement. PS 3 cm.
1. The midpoint of PQ is __.
2. NQ ________________.
3. Another name for NS is ________________.
4. S is the ________________ of SQ .
5. P is the midpoint of ________________.
6. NS ________________.
7. Another name for SN is ________________.
8. Name all pairs of congruent segments in KLMN. Use the congruence symbol to write your answer.
9. M(4, 8) is the midpoint of DE . D has coordinates (6, 1). Find the coordinates of E.
For Exercises 10 and 11, use a ruler to draw each figure. Label the figure and mark the congruent parts.
N L K M 8 cm 8 cm O P N S Q Discovering Geometry Practice Your Skills CHAPTER 1 1 10. AB and CD with M as the midpoint of both AB and CD . AB 6.4 cm and CD 4.0 cm. A, B, and C are not collinear.
11. AB and CD . C is the midpoint of AB , with AC 1.5 cm. D, not on AB , is the midpoint of AE , with AD 2BC.
12. Sketch six points A, B, C, D, E, and F, no three of which are collinear. Name the lines defined by these points. How many lines are there?
13. In the figure below, {B, C, H, E} is a set of four coplanar points. Name two other sets of four coplanar points. How many sets of four coplanar points are there?
Cube A B C G H E F D ©2008 Kendall Hunt Publishing Lesson 1.2 • Poolroom Math Name Period Date For Exercises 1–5, use the figure at right to complete each statement.
1. A is the ________________ of BAE.
2. AD is the ________________ of BAE.
3. AD is a ________________ of DAE.
4. If mBAC 42°, then mCAE ________________.
5. DAB ________________.
For Exercises 6–9, use your protractor to find the measure of each angle to the nearest degree.
6. mPRO 7. mORT 8. mO 9. mRTO For Exercises 10–12, use your protractor to draw and then label each angle with the given measure.
10. mMNO 15° 11. mRIG 90° 12. mz 160° For Exercises 13–15, find the measure of the angle formed by the hands at each time.
13. 3:00 14. 4:00 15. 3:30 For Exercises 16 and 17, mark each figure with all the given information.
16. mADB 90°, AD BD, DAB DBA 17. mRPQ 90°, QR TZ, RT QZ, Q T Q P Z T R A D C B 12 1 5 2 4 11 7 10 8 6 3 9 R P O T A B C D E A 2 CHAPTER 1 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 1.3 • What’s a Widget?
Name Period Date For Exercises 1–9, match each term with one of the items (a to i) below.
a.
b.
c.
d.
e.
f.
g.
h.
i.
1. _____ Vertical angles 2. _____ Obtuse angle 3. _____ Right angle 4. _____ Complementary angles 5. _____ Congruent angles 6. _____ Linear pair of angles 7. _____ Bisected angle 8. _____ Perpendicular lines 9. _____ Congruent segments 10. If mP 13°, mQ 77°, and Q and R are complementary, what can you conclude about P and R? Explain your reasoning.
For Exercises 11–13, sketch, label, and mark a figure showing each property.
11. 1 2, 2 3 12. PQ PR 13. BAC XAY, CX BC ?
Q P R ?
?
?
90° 2 1 ?
Discovering Geometry Practice Your Skills CHAPTER 1 3 ©2008 Kendall Hunt Publishing Lesson 1.4 • Polygons Name Period Date For Exercises 1–8, complete the table.
For Exercises 9 and 10, sketch and label each figure. Mark the congruences.
Polygon name Number of sides Number of diagonals 1. Triangle 2.
2 3.
5 4. Hexagon 5. Heptagon 6.
8 7.
35 8.
12 4 CHAPTER 1 Discovering Geometry Practice Your Skills 9. Concave pentagon PENTA, with external diagonal ET , and TA PE .
10. Equilateral quadrilateral QUAD, with Q U.
For Exercises 11–14, sketch and use hexagon ABCDEF.
11. Name the diagonals from A.
12. Name a pair of consecutive sides.
13. Name a pair of consecutive angles.
14. Name a pair of non-intersecting diagonals.
For Exercises 15–18, use the figures at right.
MNOPQ RSTUV 15. mN _____ 16. VR _____ 17. mP _____ 18. ON _____ 19. The perimeter of a regular pentagon is 31 cm. Find the length of each side.
16.1 8.4 7.2 S R U V T O N M P Q 82° 58° 61° ©2008 Kendall Hunt Publishing Lesson 1.5 • Triangles Name Period Date For Exercises 1–5, use the figure at right. Name a pair of 1. Parallel segments 2. Perpendicular segments 3. Congruent segments 4. Supplementary angles 5. Linear angles For Exercises 6 and 7, sketch, label, and mark each figure.
6. Isosceles obtuse triangle TRI with vertex angle T.
7. Scalene right triangle SCA with midpoints L, M, and N on SC , CA , and SA , respectively.
For Exercises 8 and 9, use your geometry tools to draw each figure.
A B F E D C G H I Discovering Geometry Practice Your Skills CHAPTER 1 5 9. Scalene right triangle RGH.
8. Acute isosceles triangle ACD with vertex angle A measuring 40°.
For Exercises 10–12, use the graph at right.
10. Locate F so that ABF is a right triangle.
11. Locate D so that ABD is an isosceles triangle.
12. Locate G so that ABG is scalene and not a right triangle.
y x B (8, 0) A (0, 0) C (8, 3) ©2008 Kendall Hunt Publishing Lesson 1.6 • Special Quadrilaterals Name Period Date For Exercises 1–6, sketch, label, and mark each figure.
6 CHAPTER 1 Discovering Geometry Practice Your Skills 1. Parallelogram PGRA 2. Square SQRE For Exercises 7–10, name each polygon in the figure. Assume that the grid is square.
7. Square 8. Parallelogram 9. Rhombus 10. Kite For Exercises 11–13, use the graph at right.
11. Locate D so that ABCD is a rectangle.
12. Locate E so that ABCE is a trapezoid.
13. Locate G so that points A, B, C, and G determine a parallelogram that is not a rectangle.
J G D F E A B C H I y x B (8, 0) A (0, 0) C (8, 3) 3. Rhombus RHOM with acute H.
4. Trapezoid TRAP with TR AP , RE PA , and P, E, and A collinear.
6. Rectangle RANG with perimeter 2a 4b 5. Kite KITE with EK KI and obtuse K.
©2008 Kendall Hunt Publishing Lesson 1.7 • Circles Name Period Date For Exercises 1–4, use the figure at right.
1. mQR _____ 2. mPR _____ 3. mPQR _____ 4. mQPR _____ 5. Sketch a circle with an inscribed pentagon.
6. Sketch a circle with a circumscribed quadrilateral.
P Q R O 48° Discovering Geometry Practice Your Skills CHAPTER 1 7 7. A circle with center (3, 2) goes through (2, 2). Give the coordinates of three other points on the circle.
8. Use a compass, protractor, and straightedge to draw circle O with diameter AB ; radius OC with OC AB ; OD , the angle bisector of AOC, with D on the circle; chords AC and BC ; and a tangent at D.
11. Use your compass to construct two circles with different radii that intersect in two points. Label the centers P and Q and the points of intersection A and B. Construct quadrilateral PAQB. What type of quadrilateral is it?
9. Use a compass to construct a circle. Label the center P. Sketch two parallel tangents.
Connect the points of tangency. What do you notice about the chord?
10. Use your compass and protractor to make an arc with measure 50°, an arc with measure 180°, and an arc with measure 290°. Label each arc with its measure.
y x (2, 2) (3, 2) ©2008 Kendall Hunt Publishing Lesson 1.8 • Space Geometry Name Period Date For Exercises 1 and 2, draw each figure.
1. A prism with a rectangular base.
2. A cylinder with base diameter greater than height.
For Exercises 3 and 4, sketch the three-dimensional figure formed by folding each net into a solid. Name the solid.
3.
4.
For Exercises 5 and 6, sketch the section formed when each solid is sliced by the plane as shown.
5.
6.
7. The prism below is built with 1-cm 8. Find the lengths of x and y.
cubes. How many cubes are completely hidden from sight, as seen from this angle? y x 2 3 4 4 8 CHAPTER 1 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 1.9 • A Picture Is Worth a Thousand Words Name Period Date Read and reread each problem carefully, determining what information you are given and what it is that you trying to find.
1. A pair of parallel interstate gas and power lines run 10 meters apart and are equally distant from relay station A. The power company needs to locate a gas-monitoring point on one of the lines exactly 12 meters from relay station A. Draw a diagram showing the locus of possible locations.
2. The six members of the Senica High School math club are having a group photo taken for the yearbook. The photographer has asked the club to submit the height of each member so that he can quickly arrange them in order. The math club sent him the following information. Anica is 4 inches taller than Bruce. Charles is the same height as Ellen but an inch taller than Anica. Fred is midway between Bruce and Dora. Dora is 2 inches taller than Anica. Help out the photographer by arranging the club members in order from tallest to shortest.
3. Create a Venn diagram showing the relationships among triangles, acute triangles, isosceles triangles, and scalene triangles.
4. Sketch a possible net for each solid.
a.
b.
c.
Discovering Geometry Practice Your Skills CHAPTER 1 9 ©2008 Kendall Hunt Publishing Lesson 2.1 • Inductive Reasoning Name Period Date For Exercises 1–7, use inductive reasoning to find the next two terms in each sequence.
1. 4, 8, 12, 16, _____, _____ 2. 400, 200, 100, 50, 25, _____, _____ 3. 1 8 , 2 7 , 1 2 , 4 5 , _____, _____ 4. 5, 3, 2, 1, 1, 0, _____, _____ 5. 360, 180, 120, 90, _____, _____ 6. 1, 3, 9, 27, 81, _____, _____ 7. 1, 5, 14, 30, 55, _____, _____ For Exercises 8–10, use inductive reasoning to draw the next two shapes in each picture pattern.
8.
9.
10.
For Exercises 11–13, use inductive reasoning to test each conjecture.
Decide if the conjecture seems true or false. If it seems false, give a counterexample.
11. The square of a number is larger than the number.
12. Every multiple of 11 is a “palindrome,” that is, a number that reads the same forward and backward.
13. The difference of two consecutive square numbers is an odd number.
y x (3, 1) y x (–1, 3) y x (–3, –1) 10 CHAPTER 2 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 2.2 • Finding the nth Term Name Period Date For Exercises 1–4, tell whether the rule is a linear function.
1.
2.
3.
4.
For Exercises 5 and 6, complete each table.
5.
6.
For Exercises 7–9, find the function rule for each sequence. Then find the 50th term in the sequence.
7.
8.
9.
10. Use the figures to complete the table.
11. Use the figures above to complete the table. Assume that the area of the first figure is 1 square unit.
n 1 2 3 4 5 . . .
n . . .
50 Area of figure 1 4 16 . . .
. . .
n 1 2 3 4 5 . . .
n . . .
50 Number of triangles 1 5 9 . . .
. . .
n 1 2 3 4 5 6 . . .
n . . .
50 h(n) 6.5 7 7.5 8 8.5 9 . . .
. . .
n 1 2 3 4 5 6 . . .
n . . .
50 g(n) 6 1 4 9 14 19 . . .
. . .
n 1 2 3 4 5 6 . . .
n . . .
50 f(n) 9 13 17 21 25 29 . . .
. . .
n 1 2 3 4 5 g(n) 8n 2 n 1 2 3 4 5 f(n) 7n 12 n 1 2 3 4 5 j(n) 3 2 1 1 2 0 1 2 n 1 2 3 4 5 h(n) 9 6 2 3 9 n 1 2 3 4 5 g(n) 14 11 8 5 2 n 1 2 3 4 5 f(n) 8 15 22 29 36 Discovering Geometry Practice Your Skills CHAPTER 2 11 ©2008 Kendall Hunt Publishing Lesson 2.3 • Mathematical Modeling Name Period Date 1. Draw the next figure in this pattern.
a. How many small squares will there be in the 10th figure?
b. How many in the 25th figure?
c. What is the general function rule for this pattern?
2. If you toss a coin, you will get a head or a tail. Copy and complete the geometric model to show all possible results of three consecutive tosses.
a. How many sequences of results are possible? b. How many sequences have exactly one tail? c. Assuming a head or a tail is equally likely, what is the probability of getting exactly one tail in three tosses?
3. If there are 12 people sitting at a round table, how many different pairs of people can have conversations during dinner, assuming they can all talk to each other? What geometric figure can you use to model this situation?
4. Tournament games and results are often displayed using a geometric model. Two examples are shown below. Sketch a geometric model for a tournament involving 5 teams and a tournament involving 6 teams.
Each team must have the same chance to win. Try to have as few games as possible in each tournament. Show the total number of games in each tournament. Name the teams a, b, c . . . and number the games 1, 2, 3 . . . .
a b 1 c d 2 3 4 teams, 3 games (single elimination) 3 teams, 3 games (round robin) a b c 1 3 2 H H H HHH T T HHT 12 CHAPTER 2 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 2 13 Lesson 2.4 • Deductive Reasoning Name Period Date 1. ABC is equilateral. Is ABD equilateral? Explain your answer.
What type of reasoning, inductive or deductive, do you use when solving this problem?
2. A and D are complementary. A and E are supplementary.
What can you conclude about D and E? Explain your answer.
What type of reasoning, inductive or deductive, do you use when solving this problem?
3. Which figures in the last group are whatnots? What type of reasoning, inductive or deductive, do you use when solving this problem?
4. Solve each equation for x. Give a reason for each step in the process.
What type of reasoning, inductive or deductive, do you use when solving these problems?
a. 4x 3(2 x) 8 2x b. 19 2( 5 3x 1) x 2 5. A sequence begins 4, 1, 6, 11 . . .
a. Give the next two terms in the sequence. What type of reasoning, inductive or deductive, do you use when solving this problem?
b. Find a rule that generates the sequence. Then give the 50th term in the sequence. What type of reasoning, inductive or deductive, do you use when solving this problem?
Which are whatnots?
d.
e.
f.
a.
b.
c.
Not whatnots Whatnots A C B D ©2008 Kendall Hunt Publishing 14 CHAPTER 2 Discovering Geometry Practice Your Skills Lesson 2.5 • Angle Relationships Name Period Date For Exercises 1–6, find each lettered angle measure without using a protractor.
1.
2.
3.
4.
5.
6.
For Exercises 7–10, tell whether each statement is always (A), sometimes (S), or never (N) true.
7. _____ The sum of the measures of two acute angles equals the measure of an obtuse angle.
8. _____ If XAY and PAQ are vertical angles, then either X, A, and P or X, A, and Q are collinear.
9. _____ If two angles form a linear pair, then they are complementary.
10. _____ If a statement is true, then its converse is true.
For Exercises 11–15, fill in each blank to make a true statement.
11. If one angle of a linear pair is obtuse, then the other is ____________.
12. If A B and the supplement of B has measure 22°, then mA ________________.
13. If P is a right angle and P and Q form a linear pair, then mQ is ________________.
14. If S and T are complementary and T and U are supplementary, then U is a(n) ________________ angle.
15. Switching the “if” and “then” parts of a statement changes the statement to its ________________.
b a c 110° 100° 25° e b a d c 70° e b a d c 138° 132° a b d c 40° 70° 38° 15° a a b c 112° ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 2 15 Lesson 2.6 • Special Angles on Parallel Lines Name Period Date For Exercises 1–3, use your conjectures to find each angle measure.
1.
2.
3.
For Exercises 4–6, use your conjectures to determine whether 1 2, and explain why. If not enough information is given, write “cannot be determined.” 4.
5.
6.
7. Find each angle measure.
8. Find x.
9. Find x and y.
5x 2 4y 2 182 4x x 3x 160° 44° 78° 64° f e d b c a 48° 48° 1 2 95° 25° 1 2 118° 62° 1 2 54° b a a b c d 65° a b c 54° ©2008 Kendall Hunt Publishing Lesson 3.1 • Duplicating Segments and Angles Name Period Date In Exercises 1–3, use the segments and angles below. Complete the constructions on a separate piece of paper.
1. Using only a compass and straightedge, duplicate each segment and angle. There is an arc in each angle to help you.
2. Construct a line segment with length 3PQ 2RS.
3. Duplicate the two angles so that the angles have the same vertex and share a common side, and the nonshared side of one angle falls inside the other angle. Then use a protractor to measure the three angles you created. Write an equation relating their measures.
4. Use a compass and straightedge to construct an isosceles triangle with two sides congruent to AB and base congruent to CD .
5. Repeat Exercise 4 with patty paper and a straightedge.
6. Construct an equilateral triangle with sides congruent to CD .
C D A B C D P Q R S B A 16 CHAPTER 3 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 3.2 • Constructing Perpendicular Bisectors Name Period Date For Exercises 1–6, construct the figures on a separate sheet of paper using only a compass and a straightedge.
1. Draw a segment and construct its perpendicular bisector.
2. Construct two congruent segments that are the perpendicular bisectors of each other. Form a quadrilateral by connecting the four endpoints.
What type of quadrilateral does this seem to be?
3. Duplicate AB . Then construct a segment with length 5 4 AB.
4. Draw a segment; label it CM . CM is a median of ABC. Construct ABC. Is ABC unique? If not, construct a different triangle, AB C, also having CM as a median.
5. Draw a segment; label it PQ . PQ is a midsegment of ABC. Construct ABC. Is ABC unique? If not, construct a different triangle, AB C, also having PQ as a midsegment.
6. Construct a right triangle. Label it ABC with right angle B. Construct median BD . Compare BD, AD, and CD.
7. Complete each statement as fully as possible.
a. L is equidistant from ________________.
b. M is equidistant from ________________.
c. N is equidistant from ________________.
d. O is equidistant from ________________.
D C B A E L M N O A B Discovering Geometry Practice Your Skills CHAPTER 3 17 ©2008 Kendall Hunt Publishing Lesson 3.3 • Constructing Perpendiculars to a Line Name Period Date For Exercises 1–5, decide whether each statement is true or false. If the statement is false, explain why or give a counterexample.
1. In a triangle, an altitude is shorter than either side from the same vertex.
2. In a triangle, an altitude is shorter than the median from the same vertex.
3. In a triangle, if a perpendicular bisector of a side and an altitude coincide, then the triangle is isosceles.
4. Exactly one altitude lies outside a triangle.
5. The intersection of the perpendicular bisectors of the sides lies inside the triangle.
For Exercises 6 and 7, use patty paper. Attach your patty paper to your worksheet.
6. Construct a right triangle. Construct the altitude from the right angle to the opposite side.
7. Mark two points, P and Q. Fold the paper to construct square PQRS.
Use your compass and straightedge and the definition of distance to complete Exercises 8 and 9 on a separate sheet of paper.
8. Construct a rectangle with sides equal in length to AB and CD .
9. Construct a large equilateral triangle. Let P be any point inside the triangle. Construct WX equal in length to the sum of the distances from P to each of the sides. Let Q be any other point inside the triangle. Construct YZ equal in length to the sum of the distances from Q to each side. Compare WX and YZ.
A B C D 18 CHAPTER 3 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 3 19 Lesson 3.4 • Constructing Angle Bisectors Name Period Date 1. Complete each statement as fully as possible.
a. M is equidistant from _______________ .
b. P is equidistant from __.
c. Q is equidistant from __ .
d. R is equidistant from ___ .
2. If the converse of the Angle Bisector Conjecture is true, what can you conclude about this figure?
3. If BE bisects ABD, find x and mABE.
4. Draw an obtuse angle. Use a compass and straightedge to construct the angle bisector. Draw another obtuse angle and fold to construct the angle bisector.
5. Draw a large triangle on patty paper. Fold to construct the three angle bisectors. What do you notice?
For Exercises 6 and 7, construct a figure with the given specifications using a straightedge and compass or patty paper. Use additional sheets of paper to show your work.
6. Using only your compass and straightedge, construct an isosceles right triangle.
7. Construct right triangle RGH with right angle R. Construct median RM , perpendicular MN from M to RG , and perpendicular MO from M to RH . Compare RN and GN, and compare RO and HO.
B A E D C 2x 10° 3x 20° A B P C M P R Q 1 4 5 3 2 ©2008 Kendall Hunt Publishing 20 CHAPTER 3 Discovering Geometry Practice Your Skills Lesson 3.5 • Constructing Parallel Lines Name Period Date For Exercises 1–6, construct a figure with the given specifications using a straightedge and compass or patty paper. Use additional sheets of paper to show your work.
1. Draw a line and a point not on the line. Use a compass and straightedge to construct a line through the given point parallel to the given line.
2. Repeat Exercise 1, but draw the line and point on patty paper and fold to construct the parallel line.
3. Use a compass and straightedge to construct a parallelogram.
4. Use patty paper and a straightedge to construct an isosceles trapezoid.
5. Construct a rhombus with sides equal in length to AB and having an angle congruent to P.
6. Construct trapezoid ZOID with ZO and ID as nonparallel sides and AB as the distance between the parallel sides.
Z O I D A B P A B ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 3 21 Lesson 3.6 • Construction Problems Name Period Date For Exercises 1–5, construct a figure with the given specifications using either a compass and straightedge or patty paper. Use additional sheets of paper to show your work.
1. Construct kite KITE using these parts.
2. Construct a rectangle with perimeter the length of this segment.
3. Construct a rectangle with this segment as its diagonal.
4. Draw obtuse OBT. Construct and label the three altitudes OU , BS , and TE .
5. Construct a triangle congruent to ABC. Describe your steps.
In Exercises 6–8, construct a triangle using the given parts. Then, if possible, construct a different (noncongruent) triangle using the same parts.
6.
7.
8.
S3 S2 S1 A B S1 A S1 S2 A B C I T I K I ©2008 Kendall Hunt Publishing Lesson 3.7 • Constructing Points of Concurrency Name Period Date For Exercises 1 and 2, make a sketch and explain how to find the answer.
1. A circular revolving sprinkler needs to be set up to water every part of a triangular garden. Where should the sprinkler be located so that it reaches all of the garden, but doesn’t spray farther than necessary?
2. You need to supply electric power to three transformers, one on each of three roads enclosing a large triangular tract of land. Each transformer should be the same distance from the power-generation plant and as close to the plant as possible. Where should you build the power plant, and where should you locate each transformer?
For Exercises 3–5, construct a figure with the given specifications using a compass and straightedge. Use additional sheets of paper to show your work.
3. Draw an obtuse triangle. Construct the inscribed and the circumscribed circles.
4. Construct an equilateral triangle. Construct the inscribed and the circumscribed circles. How does this construction differ from Exercise 3?
5. Construct two obtuse, two acute, and two right triangles. Locate the circumcenter of each triangle. Make a conjecture about the relationship between the location of the circumcenter and the measure of the angles.
22 CHAPTER 3 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 3.8 • The Centroid Name Period Date For Exercises 1–3, use additional sheets of paper to show your work.
1. Draw a large acute triangle. Construct the centroid.
2. Construct a regular hexagon and locate its center of gravity.
3. Use a ruler and compass to find the center of gravity of a sheet-metal triangle with sides measuring 6 cm, 8 cm, and 10 cm. How far is the center from each vertex, to the nearest tenth of a centimeter?
4. ABC has vertices A(9, 12), B(3, 2), and C(3, 2). Find the coordinates of the centroid.
5. PL 24, QC 10, and KC 7. Find PC, CL, QM, and CR.
6. Identify each statement as describing the incenter, circumcenter, orthocenter, or centroid.
a. __ The point equally distant from the three sides of a triangle.
b. __ The center of gravity of a thin metal triangle.
c. __ The point equidistant from the three vertices.
d. __ The intersection of the perpendicular bisectors of the sides of a triangle.
e. __ The intersection of the altitudes of a triangle.
f. __ The intersection of the angle bisectors of a triangle.
g. __ The intersection of the medians of a triangle.
P Q L R M C K C (3, 2) A (9, 12) y x B (3, 2) Discovering Geometry Practice Your Skills CHAPTER 3 23 ©2008 Kendall Hunt Publishing 24 CHAPTER 4 Discovering Geometry Practice Your Skills Lesson 4.1 • Triangle Sum Conjecture Name Period Date In Exercises 1–9, determine the angle measures.
1. p __, q _ 2. x , y 3. a , b 4. r , s , 5. x , y 6. y t 7. s 8. m 9. mP 10. Find the measure of QPT.
11. Find the sum of the measures of the marked angles.
12. Use the diagram to explain why 13. Use the diagram to explain why A and B are complementary.
mA mB mC mD.
A B E D C A C B P b c c a a m 35 s 76 y 30 4x 100 x 7x 85 x 31 y 100 s t r 79 50 23 a b 28 17 53 x y 82 98 q p 31 Q R S P T ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 4 25 Lesson 4.2 • Properties of Isosceles Triangles Name Period Date In Exercises 1–3, find the angle measures.
1. mT 2. mG 3. x In Exercises 4–6, find the measures.
4. mA , perimeter 5. The perimeter of LMO 6. The perimeter of QRS is of ABC is 536 m. LM , 344 cm. mQ , mM QR 7. a. Name the angle(s) congruent to DAB.
b. Name the angle(s) congruent to ADB.
c. What can you conclude about AD and BC ? Why?
8. x , y _ 9. PR QR and QS RS.
10. Use the diagram to explain If mRSQ 120°, what is why PQR is isosceles.
mQPR?
P Q R T S 70 55 P R S Q 4y 2x y 79 x A B C D R S Q y y 31 cm 68 M O L x 30 x 163 m 210 m 13 cm a 7 cm A a B C 39 102 x 110 N G A T R I 58 ©2008 Kendall Hunt Publishing 26 CHAPTER 4 Discovering Geometry Practice Your Skills Lesson 4.3 • Triangle Inequalities Name Period Date In Exercises 1 and 2, determine whether it is possible to draw a triangle with sides of the given measures. If it is possible, write yes. If it is not possible, write no and make a sketch demonstrating why it is not possible.
1. 16 cm, 30 cm, 45 cm 2. 9 km, 17 km, 28 km 3. If 17 and 36 are the lengths of two sides of a triangle, what is the range of possible values for the length of the third side?
In Exercises 4–6, arrange the unknown measures in order from greatest to least.
4.
5.
6.
7. x 8. x 9. What’s wrong with this picture?
10. Explain why PQS is isosceles.
In Exercises 11 and 12, use a compass and straightedge to construct a triangle with the given sides. If it is not possible, explain why not.
11.
12.
Q R R P Q P B C C A B A x 2x P R S Q 120 160 C A B x 158 142 66 x 28 40 71 a c d b 61 32 b c a a b c 20 18 13 ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 4 27 Lesson 4.4 • Are There Congruence Shortcuts?
Name Period Date In Exercises 1–3, name the conjecture that leads to each congruence.
1. PAT IMT 2. SID JAN 3. TS bisects MA , MT AT , and MST AST In Exercises 4–9, name a triangle congruent to the given triangle and state the congruence conjecture. If you cannot show any triangles to be congruent from the information given, write “cannot be determined” and redraw the triangles so that they are clearly not congruent.
4. M is the midpoint of AB 5. KITE is a kite with KI TI.
6. ABC _ and PQ .
APM KIE 7. MON _ 8. SQR 9. TOP In Exercises 10–12, use a compass and a straightedge or patty paper and a straightedge to construct a triangle with the given parts. Then, if possible, construct a different (noncongruent) triangle with the same parts. If it is not possible, explain why not.
10.
11.
12.
X Y X Z X B C C A B T S U T S U y x G D O T P 2 4 6 8 10 2 4 6 8 10 Q T U R S T N O M T I E K P M Q B A B A X Z C Y M T S A 6 6 8 8 9 9 J I N A D S M I A T P ©2008 Kendall Hunt Publishing 28 CHAPTER 4 Discovering Geometry Practice Your Skills Lesson 4.5 • Are There Other Congruence Shortcuts?
Name Period Date In Exercises 1–6, name a triangle congruent to the given triangle and state the congruence conjecture. If you cannot show any triangles to be congruent from the information given, write “cannot be determined” and explain why.
1. PIT _ 2. XVW 3. ECD 4. PS is the angle bisector 5. ACN _ 6. EFGH is a parallelogram.
of QPR.
GQ EQ.
PQS EQL 7. The perimeter of QRS is 350 cm.
8. The perimeter of TUV is 95 cm.
Is QRS MOL? Explain.
Is TUV WXV? Explain.
In Exercises 9 and 10, construct a triangle with the given parts. Then, if possible, construct a different (noncongruent) triangle with the same parts.
If it is not possible, explain why not.
9.
10.
C A A B P Q P Q x 40 x 25 2x 10 T U V X W L Q R S M O x 125 70 x 55 2x 15 E H G Q K L F P N A R C S P Q R D E B C A Z Y X V W P O T I ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 4 29 Lesson 4.6 • Corresponding Parts of Congruent Triangles Name Period Date 1. Give the shorthand name for each of the four triangle congruence conjectures.
In Exercises 2–5, use the figure at right to explain why each congruence is true. WXYZ is a parallelogram.
2. WXZ YZX 3. WZX YXZ 4. WZX YXZ 5. W Y For Exercises 6 and 7, mark the figures with the given information. To demonstrate whether the segments or the angles indicated are congruent, determine that two triangles are congruent. Then state which conjecture proves them congruent.
6. M is the midpoint of WX and 7. ABC is isosceles and CD is the bisector YZ . Is YW ZX ? Why?
of the vertex angle. Is AD BD ? Why?
In Exercises 8 and 9, use the figure at right to write a paragraph proof for each statement.
8. DE CF 9. EC FD 10. TRAP is an isosceles trapezoid with TP RA and PTR ART.
Write a paragraph proof explaining why TA RP .
T R A P D C B A E F C B A D Y W X Z M Z Y W X ©2008 Kendall Hunt Publishing 30 CHAPTER 4 Discovering Geometry Practice Your Skills Lesson 4.7 • Flowchart Thinking Name Period Date Complete the flowchart for each proof.
1. Given: PQ SR and PQ SR Show: SP QR Flowchart Proof 2. Given: Kite KITE with KE KI Show: KT bisects EKI and ETI Flowchart Proof 3. Given: ABCD is a parallelogram Show: A C Flowchart Proof ABCD is a parallelogram ____ AB CD __ Definition of _ Same segment ___ __ ___ ___ A D B C KET _ __ __ __ KITE is a kite __ Definition of bisect KE KI __ ETK ITK ___ __ T K I E PQS __ Given PQ SR QS _ SP QR __ ___ __ __ __ R S Q P ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 4 31 Lesson 4.8 • Proving Special Triangle Conjectures Name Period Date In Exercises 1–3, use the figure at right.
1. CD is a median, perimeter ABC 60, and AC 22. AD _ 2. CD is an angle bisector, and mA 54°. mACD 3. CD is an altitude, perimeter ABC 42, mACD 38°, and AD 8.
mB , CB _ 4. EQU is equilateral.
5. ANG is equiangular mE and perimeter ANG 51.
AN 6. ABC is equilateral, ACD is isosceles with base AC , perimeter ABC 66, and perimeter ACD 82.
Perimeter ABCD _ 7. Complete a flowchart proof for this conjecture: In an isosceles triangle, the altitude from the vertex angle is the median to the base.
Given: Isosceles ABC with AC BC and altitude CD Show: CD is a median Flowchart Proof 8. Write a flowchart proof for this conjecture: In an isosceles triangle, the median to the base is also the angle bisector of the vertex angle.
Given: Isosceles ABC with AC BC and median CD Show: CD bisects ACB A B C D • • • A __ ADC BDC ADC and BDC are right angles __ ____ Definition of altitude Given __ ___ CD is an altitude AC BC A B C D C D A B A B C D ©2008 Kendall Hunt Publishing 32 CHAPTER 5 Discovering Geometry Practice Your Skills Lesson 5.1 • Polygon Sum Conjecture Name Period Date In Exercises 1 and 2, find each lettered angle measure.
1. a , b _, c , 2. a , b _, c , d , e _ d , e , f _ 3. One exterior angle of a regular polygon measures 10°. What is the measure of each interior angle? How many sides does the polygon have?
4. The sum of the measures of the interior angles of a regular polygon is 2340°. How many sides does the polygon have?
5. ABCD is a square. ABE is an equilateral 6. ABCDE is a regular pentagon. ABFG triangle.
is a square.
x x 7. Use a protractor to draw pentagon ABCDE with mA 85°, mB 125°, mC 110°, and mD 70°. What is mE?
Measure it, and check your work by calculating.
D F C B A x E G x E A B D C 85 44 a c b d f e a e d c b 97 26 ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 5 33 Lesson 5.2 • Exterior Angles of a Polygon Name Period Date 1. How many sides does a regular polygon have if each exterior angle measures 30°?
2. How many sides does a polygon have if the sum of the measures of the interior angles is 3960°?
3. If the sum of the measures of the interior angles of a polygon equals the sum of the measures of its exterior angles, how many sides does it have?
4. If the sum of the measures of the interior angles of a polygon is twice the sum of its exterior angles, how many sides does it have?
In Exercises 5–7, find each lettered angle measure.
5. a _, b 6. a , b _ 7. a , b , c _ 8. Find each lettered angle measure.
9. Construct an equiangular quadrilateral that is not regular.
a d b c 150 95 a b c _ d 82 134 72 a b c 3x 2x x a b 116 82 a b ©2008 Kendall Hunt Publishing 34 CHAPTER 5 Discovering Geometry Practice Your Skills Lesson 5.3 • Kite and Trapezoid Properties Name Period Date In Exercises 1–4, find each lettered measure.
1. Perimeter 116. x 2. x _, y 3. x , y _ 4. x , y 5. Perimeter PQRS 220. PS _ 6. b 2a 1. a In Exercises 7 and 8, use the properties of kites and trapezoids to construct each figure. Use patty paper or a compass and a straightedge.
7. Construct an isosceles trapezoid given base AB , B, and distance between bases XY.
8. Construct kite ABCD with AB , BC , and BD .
9. Write a paragraph or flowchart proof of the Converse of the Isosceles Trapezoid Conjecture. Hint: Draw AE parallel to TP with E on TR .
Given: Trapezoid TRAP with T R Show: TP RA T P A R A B B C D B B X Y A B 34 M b L K N a S P T Q R 4x 1 2x 3 4 78 41 x y 137 22 x y x y 56 x 28 ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 5 35 Lesson 5.4 • Properties of Midsegments Name Period Date In Exercises 1–3, each figure shows a midsegment.
1. a , b _, 2. x , y , 3. x _, y , c z _ z 4. X, Y, and Z are midpoints. Perimeter PQR 132, RQ 55, and PZ 20.
Perimeter XYZ PQ _ ZX 5. MN is the midsegment. Find the 6. Explain how to find the width of the lake from coordinates of M and N. Find the A to B using a tape measure, but without slopes of AB and MN .
using a boat or getting your feet wet.
7. M, N, and O are midpoints. What type of quadrilateral is AMNO? How do you know? Give a flowchart proof showing that ONC MBN.
8. Give a paragraph or flowchart proof.
Given: PQR with PD DF FH HR and QE EG GI IR Show: HI FG DE PQ P D E F G H I R Q B M O N C A Lake A B x M N y C (20, 10) B (9, –6) A (4, 2) Z Y X P Q R x z y 29 11 41 13 x y z 16 21 14 b a c 37 54 ©2008 Kendall Hunt Publishing 36 CHAPTER 5 Discovering Geometry Practice Your Skills Lesson 5.5 • Properties of Parallelograms Name Period Date In Exercises 1–7, ABCD is a parallelogram.
1. Perimeter ABCD 2. AO 11, and BO 7.
3. Perimeter ABCD 46.
AC _, BD AB , BC _ 4. a , b , 5. Perimeter ABCD 119, and 6. a _, b , c BC 24. AB _ c 7. Perimeter ABCD 16x 12. AD 8. Ball B is struck at the same instant by two forces, F1 and F2. Show the resultant force on the ball.
9. Find each lettered angle measure.
10. Construct a parallelogram with diagonals AC and BD .
Is your parallelogram unique? If not, construct a different (noncongruent) parallelogram.
A C B D A D F h e d i k j b a g E f G B C 81 38 38 26 c g _ h i j _ k a b _ c d e _ f B F2 F1 A C B D 4x 3 63 A B C D a c b 26 27 68 A C B D b c a B A C D 19 62 110 A C B D 2x 7 x 9 A D B O C D B A C 26 cm 15 cm ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 5 37 Lesson 5.6 • Properties of Special Parallelograms Name Period Date 1. PQRS is a rectangle and 2. KLMN is a square and 3. ABCD is a rhombus, OS 16.
NM 8.
AD 11, and DO 6.
OQ mOKL _ OB mQRS mMOL _ BC PR Perimeter KLMN _ mAOD In Exercises 4–11, match each description with all the terms that fit it.
a. Trapezoid b. Isosceles triangle c. Parallelogram d. Rhombus e. Kite f. Rectangle g. Square h. All quadrilaterals 4. Diagonals bisect each other.
6. _ Diagonals are congruent.
8. Opposite sides are congruent.
10. Both diagonals bisect angles.
In Exercises 12 and 13, graph the points and determine whether ABCD is a trapezoid, parallelogram, rectangle, or none of these.
12. A(4, 1), B(0, 3), C(4, 0), D(1, 5) 13. A(0, 3), B(1, 2), C(3, 4), D(2, 1) 14. Construct rectangle ABCD with diagonal AC and CAB.
A CAB A C –5 5 –5 5 –5 5 –5 5 5. _ Diagonals are perpendicular.
7. Measures of interior angles sum to 360°.
9. Opposite angles are congruent.
11. _ Diagonals are perpendicular bisectors of each other.
C D O B A K L O N M P S Q O R ©2008 Kendall Hunt Publishing 38 CHAPTER 5 Discovering Geometry Practice Your Skills Lesson 5.7 • Proving Quadrilateral Properties Name Period Date Write or complete each flowchart proof.
1. Given: ABCD is a parallelogram and AP QC Show: AC and PQ bisect each other Flowchart Proof 2. Given: Dart ABCD with AB BC and CD AD Show: A C 3. Show that the diagonals of a rhombus divide the rhombus into four congruent triangles.
Given: Rhombus ABCD Show: ABO CBO CDO ADO 4. Given: Parallelogram ABCD, BY AC , DX AC Show: DX BY A D C B Y X A D B O C C B A D APR __ ___ __ Definition of bisect Given Given AIA Conjecture AR DC AB APR CQR AC and QP bisect each other __ C B P A D R Q ©2008 Kendall Hunt Publishing Lesson 6.1 • Tangent Properties Name Period Date 1. Rays r and s are tangents. w 2. AB is tangent to both circles and mAMC 295°. mBQX _ 3. PQ is tangent to two externally tangent noncongruent circles, M and N.
a. mNQP , mMPQ b. What kind of quadrilateral is MNQP? Explain your reasoning.
4. AT is tangent to circle P. Find the 5. PA , PB , PC , and PD are tangents.
equation of AT .
Explain why PA PD .
6. Circle A has diameter 16.4 cm. Circle B has diameter 6.7 cm.
a. If A and B are internally tangent, what is the distance between their centers?
b. If A and B are externally tangent, what is the distance between their centers?
7. Construct a circle, P. Pick a point, A, on the circle. Construct a tangent through A. Pick a point, T, on the tangent. Construct a second tangent to the circle through T.
A B C D P y x A (3, 9) T P M N P Q P A B Q X C M s r w 54 Discovering Geometry Practice Your Skills CHAPTER 6 39 ©2008 Kendall Hunt Publishing 40 CHAPTER 6 Discovering Geometry Practice Your Skills Lesson 6.2 • Chord Properties Name Period Date In Exercises 1–6, find each unknown or write “cannot be determined.” 1. a _, b , 2. w , v _ 3. z c 4. w _, x , 5. w , x _, 6. x , y y _ y 7. AB AC . AMON is a 8. What’s wrong with 9. Find the coordinates of ___.
this picture?
P and M.
Justify your answer.
10. mAB mABC mBAC _ mACB 2 6 6 N M A C B O y x P M B (5, –2) A (3, 6) x y 50 100 8 cm 66 w x y x w y 35 100 z v w 6 6 b c a 95 11. Trace part of a circle onto patty paper. Fold to find the center. Explain your method.
C O A B 49 107 ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 6 41 Lesson 6.3 • Arcs and Angles Name Period Date 1. mXM 80° 2. AB is a tangent.
mXNM x _ mXN y mMN _ z 3. a 4. a _ b b c _ c 5. AB and AC are tangents.
6. AD is a tangent. AC is a diameter.
x mA _ mAB mC mCB _ 7. mAD 8. p mD _ q mAB r _ mDAB s 9. Find the lettered angle and arc measures. AT and AZ are tangents.
a _ b c d _ e f g _ h j k _ m n 25 50 a g h j k b f n m c d e 1 2 T Z A 87 29 r q p s 80 70 A B D 54 B O A C D 40 A B C x 100 140 a b c 70 b c a 60 120 B A x z y 80 O N M X Y ©2008 Kendall Hunt Publishing 42 CHAPTER 6 Discovering Geometry Practice Your Skills Lesson 6.4 • Proving Circle Conjectures Name Period Date In Exercises 1–4, complete each proof with a paragraph or a flowchart.
1. Given: Circles O and P are externally tangent, with common tangents CD and AB Show: AB bisects CD at X 2. Given: Circle O with diameter AB and chord AD . OE AD .
Show: DE BE 3. Given: PQ and RS are tangent to both circles.
Show: PQ RS .
4. Prove the converse of the Chord Arcs Conjecture: If two arcs in a circle are congruent, then their chords are congruent. Hint: Draw radii.
Given: AB CD Show: AB CD O A B C D M X N P S R Q O A B E D A P D X C O B ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 6 43 Lesson 6.5 • The Circumference/Diameter Ratio Name Period Date In Exercises 1–4, leave your answers in terms of .
1. If r 10.5 cm, find C.
2. If C 25 cm, find r.
3. What is the circumference of a circle whose 4. What is the diameter of a circle whose radius is 30 cm?
circumference is 24 cm?
In Exercises 5–9, round your answer to the nearest 0.1 unit. Use the symbol to show that your answer is an approximation.
5. If d 9.6 cm, find C.
6. If C 132 cm, find d and r.
7. A dinner plate fits snugly in a square box with perimeter 48 inches.
What is the circumference of the plate?
8. Four saucers are part of the same set as the dinner plate in Exercise 7.
Each has a circumference of 15.7 inches. Will they fit, side by side, in the same square box? If so, how many inches will there be between the saucers for padding?
9. AT and AS are tangents. AT 12 cm.
10. How can you use a large carpenter’s What is the circumference of circle O?
square to find the circumference of a tree?
11. In order to increase the circumference of a circle from 16 cm to 20 cm, by how much must the diameter increase?
T A S O ©2008 Kendall Hunt Publishing 44 CHAPTER 6 Discovering Geometry Practice Your Skills Lesson 6.6 • Around the World Name Period Date 1. Alfonzo’s Pizzeria bakes olive pieces in the outer crust of its 20-inch (diameter) pizza. There is at least one olive piece per inch of crust.
How many olive pieces will you get in one slice of pizza? Assume the pizza is cut into eight slices.
2. To use the machine at right, you turn the crank, which turns the pulley wheel, which winds the rope and lifts the box. Through how many rotations must you turn the crank to lift the box 10 feet?
3. A satellite in geostationary orbit stays over the same spot on Earth.
The satellite completes one orbit in the same time that Earth rotates once about its axis (23.93 hours). If the satellite’s orbit has radius 4.23 107 m, calculate the satellite’s orbital speed (tangential velocity) in meters per second.
4. You want to decorate the side of a cylindrical can by coloring a rectangular piece of paper and wrapping it around the can. The paper is 19 cm by 29 cm. Find the two possible diameters of the can to the nearest 0.01 cm. Assume the paper fits exactly.
5. As you sit in your chair, you are whirling through space with Earth as it moves around the sun. If the average distance from Earth to the sun is 1.4957 1011 m and Earth completes one revolution every 364.25 days, what is your “sitting” speed in space relative to the sun?
Give your answer in km/h, rounded to the nearest 100 km/h.
Box 10 ft 7.5 in.
©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 6 45 Lesson 6.7 • Arc Length Name Period Date In Exercises 1–10, leave your answers in terms of .
1. Length of AB _ 2. The circumference is 24 3. The length of EF is 5.
and mCD 60°. Length Radius of CD 4. Length of XY _ 5. The radius is 20. Length 6. The circumference is 25.
of AB Length of AB 7. The diameter is 40. Length 8. The length of XY is 14.
9. Length of AB _ of AC Diameter 10. A circle has an arc with measure 80° and length 88. What is the diameter of the circle?
X Y 80 A C 50 A B T 110 36 B A A B 40 Y X 10 70 C D E F 30 120 6 B A ©2008 Kendall Hunt Publishing 46 CHAPTER 6 Discovering Geometry Practice Your Skills Exploration • Intersecting Secants, Tangents, and Chords Name Period Date 1. x _ 2. FC is tangent to circle A at point C.
mDC _, mED 3. ED and EC are tangents.
4. CE is a tangent, mBC 150° mDC , mDEC mBCE , mBAC 5. x , y , z 6. x , y , z 7. AB and AC are tangents.
8. AB is a tangent, mABC 75° x , y , z x , y A C y x B 85 97 z x y A B C A O 127 34 z x y 60 72 39 44 y z x 70 60 79 B A E C B D E C 246 A F D C E 140 35 x 44 86 ©2008 Kendall Hunt Publishing Lesson 7.1 • Transformations and Symmetry Name Period Date In Exercises 1–3, perform each transformation.
1. Reflect TRI across line .
2. Rotate PARL 270° clockwise 3. Translate PENTA by about Q.
the given vector.
4. ABCDE and its reflected image, A B C D E, are shown below.
Use construction tools to locate the line of reflection, . Explain your method.
In Exercises 5–8, identify the type(s) of symmetry in each figure.
5. Equilateral triangle 6. Rectangle 7. Isosceles triangle 8. Square In Exercises 9–12, draw each polygon and identify the type(s) of symmetry in each. Draw all lines of reflection and mark centers of rotation.
9. Rhombus 10. Parallelogram 11. Isosceles trapezoid 12. Square A E C D E B A D C B T A P E N Q P R L A T I R Discovering Geometry Practice Your Skills CHAPTER 7 47 ©2008 Kendall Hunt Publishing Lesson 7.2 • Properties of Isometries Name Period Date In Exercises 1–3, draw the image according to the rule and identify the type of transformation.
1. (x, y) →(x, y) 2. (x, y) →(x 4, y 6) 3. (x, y) →(4 x, y) In Exercises 4 and 5, the Harbour High Geometry Class is holding a Fence Race. Contestants must touch each fence at some point as they run from S to F. Use your geometry tools to draw the shortest possible race path.
4.
5.
In Exercises 6–8, complete the ordered pair rule that transforms each triangle to its image. Identify the transformation. Find all missing coordinates.
6. (x, y) →(, _) 7. (x, y) →(, ) 8. (x, y) →(_, ) y x T S R(3, 0) T(0, 7) S(3, 3) R y x R(4, 5) R P P(–7, –3) Q(–2, 1) Q(2, 1) y x C A(–5, –4) B(–5, 2) B A(8, 2) (5, 2) C S F Fence 2 Fence 1 S F Fence 2 –4 –2 4 6 y x –2 –4 6 4 2 –4 4 8 y x –8 –4 4 –2 x –4 –6 –6 –4 –2 2 6 4 2 4 6 y 48 CHAPTER 7 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 7.3 • Compositions of Transformations Name Period Date In Exercises 1–8, name the single transformation that can replace the composition of each set of multiple transformations.
1. Translation by 4, 1, followed by 2, 3, followed by 8, 7 2. Rotation 60° clockwise, followed by 80° counterclockwise, followed by 25° counterclockwise all about the same center of rotation 3. Reflection across vertical line m, followed by reflection across vertical line n, where n is 8 units to the right of m 4. Reflection across vertical line p, followed by reflection across horizontal line q 5. Reflection across vertical line n, followed by reflection across vertical line m, where n is 8 units to the right of m 6. Reflection across horizontal line q, followed by reflection across vertical line p 7. Translation by 6, 0, followed by reflection across the y-axis 8. Reflection across the y-axis, followed by translation by 6, 0 In Exercises 9–11, copy the figure onto your paper and use your geometry tools to perform the given transformation.
9. Locate P, the reflected image across OR , and P, the reflected image of Pacross OT . Find mROT and give a single transformation that maps P to P.
10. Locate P, the reflected image across k, and P, the reflected image of Pacross . Find the distance between and k and give a single transformation that maps P to P.
11. Draw five glide-reflected images of the triangle.
P k O P R T Discovering Geometry Practice Your Skills CHAPTER 7 49 ©2008 Kendall Hunt Publishing Lesson 7.4 • Tessellations with Regular Polygons Name Period Date 1. Find n.
2. Find n.
3. What is a regular tessellation? Sketch an example to illustrate your explanation.
4. What is a 1-uniform tiling? Sketch an example of a 1-uniform tiling that is not a regular tessellation.
5. Give the numerical name for the tessellation at right.
6. Use your geometry tools to draw the 4.82 tessellation.
Regular pentagon Regular n-gon Square Regular 10-gon Equilateral triangle Regular n-gon 50 CHAPTER 7 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lessons 7.5–7.8 • Tessellations Name Period Date 1. Trace the quadrilateral at right (or draw a similar one). Make the outline dark. Set another piece of paper on top of the quadrilateral and, by tracing, create a tessellation. (Hint: Trace vertices and use a straightedge to connect them.) 2. On dot paper, draw a small concave quadrilateral (vertices on dots).
Allow no more than three dots inside the figure. Tessellate the entire paper with your quadrilateral. Color and shade your tessellation.
3. In non-edge-to-edge tilings, the vertices of the polygons do not have to coincide, as in these wooden deck patterns. Use graph paper to create your own non-edge-to-edge tiling.
4. Use your geometry tools to draw a parallelogram. Draw squares on each side. Create a tessellation by duplicating your parallelogram and squares.
Discovering Geometry Practice Your Skills CHAPTER 7 51 ©2008 Kendall Hunt Publishing 52 CHAPTER 8 Discovering Geometry Practice Your Skills Lesson 8.1 • Areas of Rectangles and Parallelograms Name Period Date In Exercises 1–4, find the area of the shaded region.
1.
2.
3.
4.
5. Rectangle ABCD has area 2684 m2 and width 44 m. Find its length.
6. Draw a parallelogram with area 85 cm2 and an angle with measure 40°.
Is your parallelogram unique? If not, draw a different one.
7. Find the area of PQRS.
8. Find the area of ABCDEF.
9. Dana buys a piece of carpet that measures 20 square yards. Will she be able to completely cover a rectangular floor that measures 12 ft 6 in. by 16 ft 6 in.? Explain why or why not.
A(4, –5) B(10, –5) C(10, 2) (4, 2) F D(21, 7) (15, 7)E y x P(–4, –3) S(–4, 5) R(7, –1) Q(7, –9) y x 13 cm 9 cm 2 cm 2 cm 1.5 cm 8 cm 4 cm 4 cm 5 cm 17 cm 1 cm 4 cm 8 cm 4 cm 12 cm ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 8 53 Lesson 8.2 • Areas of Triangles, Trapezoids, and Kites Name Period Date In Exercises 1–4, solve for the unknown measures.
1. Area 64 ft2, h .
2. Area .
3. Area 126 in2 4. AB 6 cm, AC 8 cm, and BC 10 cm.
b .
Find AD.
5. Find the area of the shaded region.
6. TP is tangent to circles M and N. TP 16 cm. The radius of N is 7 cm and the radius of M is 4 cm. Find the area of NMPT.
7. Find the area of TRI.
8. ABCD is a parallelogram, ABDE is a kite, AD 18 cm, and BE 10 cm. Find the area of ABCDE.
A E B D C I (12, 12) (0, 6) R T (6, 0) y x T N M P 12 cm 20 cm 4 cm A C B D 9 in.
16 in.
b 2 cm 2 cm 8 cm 2 cm 8 ft h ©2008 Kendall Hunt Publishing Lesson 8.3 • Area Problems Name Period Date 1. A bundle of hardwood flooring contains 14 1 2 ft2 and costs $39.90.
a. How many square feet of flooring is needed to cover the kitchen and family room? Exclude the fireplace, hearth, and tiled area.
b. You should buy 5% extra flooring to account for waste.
How many bundles of flooring should you buy? What will be the cost? 2. Bert’s Bigtime Bakery has baked the world’s largest chocolate cake. It is a rectangular sheet cake that is 600 cm by 400 cm by 180 cm high. Bert wants to apply frosting to the four sides and the top. How many liters of frosting does he need if 1 liter of frosting covers about 1200 cm2?
3. For a World Peace Day celebration the students at Cabot Junior/Senior High School are making a 6 m-by-8 m flag. Each of the six grades will create a motif to honor the people of the six inhabited continents.
Sketch three possible ways to divide the flag: one into six congruent triangles; one into six triangles with equal area but none congruent; and one into six congruent trapezoids. Give measurements or markings on your sketches so each class knows it has equal area.
4. Kit and Kat are building a kite for the big kite festival. Kit has already cut his sticks for the diagonals. He wants to position P so that he will have maximum kite area. He asks Kat for advice. What should Kat tell him?
P 13 ft 3 ft 3 ft 4 ft 6 ft 18 ft 18 ft Kitchen Tile Family room Fireplace and hearth 4 ft 7 ft 14 ft 54 CHAPTER 8 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 8 55 Lesson 8.4 • Areas of Regular Polygons Name Period Date In Exercises 1–3, the polygons are regular.
1. s 12 cm 2. s 4.2 cm 3. a 6 cm a 14.5 cm A 197 cm2 A 130.8 cm2 A _ a p 4. In a regular n-gon, s 4.8 cm, a 7.4 cm, and A 177.6 cm2. Find n.
5. Draw a regular pentagon so that it has 6. Use a compass and straightedge to perimeter 20 cm. Use the Regular Polygon construct a regular octagon and Area Conjecture and a centimeter ruler to its apothem. Use a centimeter ruler to find its approximate area.
measure its side length and apothem, and use the Regular Polygon Area Conjecture to find its approximate area.
7. Find the area of the shaded region between the square and the regular octagon. s 5 cm. r 3 cm.
s r r a s a s 12-gon a s ©2008 Kendall Hunt Publishing Lesson 8.5 • Areas of Circles Name Period Date In Exercises 1–4, write your answers in terms of .
1. If r 9 cm, A _.
2. If d 6.4 cm, A .
3. If A 529 cm2, r .
4. If C 36 cm, A _.
In Exercises 5–8, round your answers to the nearest 0.01 unit.
5. If r 7.8 cm, A .
6. If A 136.46, C .
7. If d 3.12, A _.
8. If C 7.85, A .
For Exercises 9 and 10, refer to the figure of a circle inscribed in an equilateral triangle. Round your answers to the nearest 0.1 unit.
9. Find the area of the inscribed circle.
10. Find the area of the shaded region.
In Exercises 11 and 12, find the area of the shaded region. Write your answers in terms of .
11. ABCD is a square.
12. The three circles are tangent.
5 cm A C (0, 8) D (8, 0) B x y 14.0 cm a 4.04 cm a 56 CHAPTER 8 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 8 57 Lesson 8.6 • Any Way You Slice It Name Period Date In Exercises 1–6, find the area of the shaded region. Write your answers in terms of and rounded to the nearest 0.01 cm2.
1.
2.
3.
4.
5.
6.
7. Shaded area is 40 cm2.
8. Shaded area is 54 cm2.
9. Shaded area is 51 cm2.
Find r.
Find x.
The diameter of the larger circle is 20 cm. Find r.
r 12 cm x 144 r 45 12 cm 8 cm 3 cm 135 8 cm 2 cm 4 cm 120 4 cm 5 cm 30 ©2008 Kendall Hunt Publishing Lesson 8.7 • Surface Area Name Period Date In Exercises 1–8, find the surface area of each solid. All quadrilaterals are rectangles, and all measurements are in centimeters. Round your answers to the nearest 0.1 cm2.
1.
2.
3.
4.
5. Base is a regular hexagon.
6.
s 6, a 5.2, and l 9.
7. Both bases are squares.
8. A square hole in a round peg 9. Ilsa is building a museum display case. The sides and bottom will be plywood and the top will be glass. Plywood comes in 4 ft-by-8 ft sheets. How many sheets of plywood will she need to buy? Explain. Sketch a cutting pattern that will leave her with the largest single piece possible.
3 ft 2 ft 3 ft 2 ft 1 _ 2 1 ft 1 _ 2 13 14 4 12 6 8 a s l 10 4 5 3 4 3 3 13 13 8.5 12.4 6 5 12 13 7 6 2 58 CHAPTER 8 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 9 59 Lesson 9.1 • The Theorem of Pythagoras Name Period Date Give all answers rounded to the nearest 0.1 unit.
1. a 2. p _ 3. x 4. Area 39 in2 5. Find the area.
6. Find the coordinates of C h and the radius of circle A.
7. Find the area.
8. RS 3 cm. Find RV.
9. Base area 16 cm2 and slant height 10. Given PQR, with mP 90°, PQ 20 in., 3 cm. What’s wrong with this picture?
and PR 15 in., find the area of PQR, the length of the hypotenuse, and the altitude to the hypotenuse.
R S T U V 7.7 cm 13.4 cm 6.8 cm x y A (7, –1) B (11, –4) C 6 ft 7 ft 6 in.
h 26 ft 6 ft 24 ft x 14 cm p 8 cm 21 cm a 75 cm 72 cm ©2008 Kendall Hunt Publishing 60 CHAPTER 9 Discovering Geometry Practice Your Skills Lesson 9.2 • The Converse of the Pythagorean Theorem Name Period Date All measurements are in centimeters. Give answers rounded to the nearest 0.01 cm.
In Exercises 1–4, determine whether a triangle with the given side lengths is a right triangle.
1. 76, 120, 98 2. 221, 204, 85 3. 5.0, 1.4, 4.8 4. 80, 82, 18 5. Find the area of ABC.
6. What’s wrong with this picture?
7. Find x. Explain your method.
8. Find the area of ABCD.
In Exercises 9–11, determine whether ABCD is a rectangle and justify your answer. If not enough information is given, write “cannot be determined.” 9. AB 3, BC 4, and AC 6.
10. AB 3, BC 4, DA 4, and AC 5.
11. AB 3, BC 4, CD 3, DA 4, and AC BD.
A B D C A B C D 22 32 6 8 C A B D 7 25 45 24 x 5 6.5 6 95 C B 7 A ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 9 61 Lesson 9.3 • Two Special Right Triangles Name Period Date Give your answers in exact form unless otherwise indicated.
All measurements are in centimeters.
In Exercises 1–3, find the unknown lengths.
1. a __ 2. a _, b 3. a , b 4. Find the area of rectangle 5. Find the perimeter and 6. AC , AB _, ABCD.
area of KLMN.
and area ABC .
7. Find the area of an isosceles trapezoid if the bases have lengths 12 cm and 18 cm and the base angles have measure 60°.
In Exercises 8 and 9, find the coordinates of C.
8.
9.
10. Sketch and label a figure to demonstrate that 18 is equivalent to 32 .
x y (0, 12) (12, 0) 210 C x y (0, 1) (1, 0) 120 C A B C 60 45 30 K L M N 45 30 7 12 16 60 D C B A 6 a b 30 12 b a 60 3 a 14 ©2008 Kendall Hunt Publishing Lesson 9.4 • Story Problems Name Period Date 1. A 20 ft ladder reaches a window 18 ft high. How far is the foot of the ladder from the base of the building? How far must the foot of the ladder be moved to lower the top of the ladder by 2 ft?
2. Robin and Dovey have four pet pigeons that they train to race. They release the birds at Robin’s house and then drive to Dovey’s to collect them. To drive from Robin’s to Dovey’s, because of one-way streets, they go 3.1 km north, turn right and go 1.7 km east, turn left and go 2.3 km north, turn right and go 0.9 km east, turn left and go 1.2 km north, turn left and go 4.1 km west, and finally turn left and go 0.4 km south. How far do the pigeons have to fly to go directly from Robin’s house to Dovey’s house?
3. Hans needs to paint the 18 in.-wide trim around the roof eaves and gable ends of his house with 2 coats of paint. A quart can of paint covers 175 ft2 and costs $9.75. A gallon can of paint costs $27.95. How much paint should Hans buy? Explain.
4. What are the dimensions of the largest 30°-60°-90° triangle that will fit inside a 45°-45°-90° triangle with leg length 14 in.? Sketch your solution.
18 ft 28 ft 42 ft 18 in.
9.5 ft 62 CHAPTER 9 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 9 63 Lesson 9.5 • Distance in Coordinate Geometry Name Period Date In Exercises 1–3, find the distance between each pair of points.
1. (5, 5), (1, 3) 2. (11, 5), (5, 7) 3. (8, 2), (7, 6) In Exercises 4 and 5, use the distance formula and the slope of segments to identify the type of quadrilateral. Explain your reasoning.
4. A(2, 1), B(3, 2), C(8, 1), D(3, 4) 5. T(3, 3), U(4, 4), V(0, 6), W(5, 1) For Exercises 6 and 7, use ABC with coordinates A(4, 14), B(10, 6), and C(16, 14).
6. Determine whether ABC is scalene, isosceles, or equilateral. Find the perimeter of the triangle.
7. Find the midpoints M and N of AB and AC , respectively. Find the slopes and lengths of MN and BC . How do the slopes compare? How do the lengths compare?
8. Find the equation of the circle with center (1, 5) and radius 2.
9. Find the center and radius of the circle whose equation is x2 (y 2)2 25.
10. P is the center of the circle. What’s wrong with this picture?
P(10, 1) A(4, 6) C(16, –3) B(5, –5) y x ©2008 Kendall Hunt Publishing Lesson 9.6 • Circles and the Pythagorean Theorem Name Period Date In Exercises 1 and 2, find the area of the shaded region in each figure. All measurements are in centimeters. Write your answers in terms of and rounded to the nearest 0.1 cm2.
1. AO 5. AC 8.
2. Tangent PT , QM 12, mP 30° 3. AP 63 cm. Radius of circle O 37 cm.
How far is A from the circumference of the circle?
4. Two perpendicular chords with lengths 12.2 cm and 8.8 cm have a common endpoint. What is the area of the circle?
5. ABCD is inscribed in a circle. AC is a diameter. If AB 9.6 cm, BC 5.7 cm, and CD 3.1 cm, find AD.
6. Find ST.
7. The coordinate of point M is 2 3 , 1 2 .
Find the measure of AOM.
A O M (0, 1) (1, 0) y x P T S O R 6 3 30° P O A P T Q M S A C B O 64 CHAPTER 9 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 10 65 Lesson 10.1 • The Geometry of Solids Name Period Date For Exercises 1–14, refer to the figures below.
1. The cylinder is (oblique, right).
2. OP is ___ of the cylinder.
3. TR is __ of the cylinder.
4. Circles O and P are __ of the cylinder.
5. PQ is __ of the cylinder.
6. The cone is (oblique, right).
7. Name the base of the cone.
8. Name the vertex of the cone.
9. Name the altitude of the cone.
10. Name a radius of the cone.
11. Name the type of prism.
12. Name the bases of the prism.
13. Name all lateral edges of the prism.
14. Name an altitude of the prism.
In Exercises 15–17, tell whether each statement is true or false. If the statement is false, give a counterexample or explain why it is false.
15. The axis of a cylinder is perpendicular to the base.
16. A rectangular prism has four faces.
17. The bases of a trapezoidal prism are trapezoids.
For Exercises 18 and 19, draw and label each solid. Use dashed lines to show the hidden edges.
18. A right triangular prism with height 19. An oblique trapezoidal pyramid equal to the hypotenuse E A F G H I C D B J B C A O P Q R T ©2008 Kendall Hunt Publishing 66 CHAPTER 10 Discovering Geometry Practice Your Skills Lesson 10.2 • Volume of Prisms and Cylinders Name Period Date In Exercises 1–3, find the volume of each prism or cylinder.
All measurements are in centimeters. Round your answers to the nearest 0.01.
1. Right triangular prism 2. Right trapezoidal prism 3. Regular hexagonal prism In Exercises 4–6, use algebra to express the volume of each solid.
4. Right rectangular prism 5. Right cylinder; 6. Right rectangular prism base circumference p and half of a cylinder 7. You need to build a set of solid cement steps for the entrance to your new house. How many cubic feet of cement do you need?
8 in.
6 in.
3 ft y 2x 3x h 2x 3 4y x 4 10 8 6 3 5 6 10 14 6 ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 10 67 Lesson 10.3 • Volume of Pyramids and Cones Name Period Date In Exercises 1–3, find the volume of each solid. All measurements are in centimeters. Round your answers to two decimal places.
1. Rectangular pyramid; OP 6 2. Right hexagonal pyramid 3. Half of a right cone In Exercises 4–6, use algebra to express the volume of each solid.
4.
5.
6. The solid generated by spinning ABC about the axis In Exercises 7–9, find the volume of each figure and tell which volume is larger.
7.
A.
B.
8.
A.
B.
9.
A.
B.
9 x 3 x 2 5 3 2 3 5 6 12 4 8 B A 2y 3x C 2a b 25x 7x 30x 25 14 9 6 8 5 P O ©2008 Kendall Hunt Publishing Lesson 10.4 • Volume Problems Name Period Date 1. A cone has volume 320 cm3 and height 16 cm. Find the radius of the base. Round your answer to the nearest 0.1 cm.
2. How many cubic inches are there in one cubic foot? Use your answer to help you with Exercises 3 and 4.
3. Jerry is packing cylindrical cans with diameter 6 in. and height 10 in.
tightly into a box that measures 3 ft by 2 ft by 1 ft. All rows must contain the same number of cans. The cans can touch each other. He then fills all the empty space in the box with packing foam. How many cans can Jerry pack in one box? Find the volume of packing foam he uses. What percentage of the box’s volume is filled by the foam?
4. A king-size waterbed mattress measures 72 in. by 84 in. by 9 in. Water weighs 62.4 pounds per cubic foot. An empty mattress weighs 35 pounds. How much does a full mattress weigh?
5. Square pyramid ABCDE, shown at right, is cut out of a cube with base ABCD and shared edge DE . AB 2 cm. Find the volume and surface area of the pyramid.
6. In Dingwall the town engineers have contracted for a new water storage tank. The tank is cylindrical with a base 25 ft in diameter and a height of 30 ft. One cubic foot holds about 7.5 gallons of water. About how many gallons will the new storage tank hold?
7. The North County Sand and Gravel Company stockpiles sand to use on the icy roads in the northern rural counties of the state. Sand is brought in by tandem trailers that carry 12 m3 each. The engineers know that when the pile of sand, which is in the shape of a cone, is 17 m across and 9 m high they will have enough for a normal winter. How many truckloads are needed to build the pile?
A B C D E 68 CHAPTER 10 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 10 69 Lesson 10.5 • Displacement and Density Name Period Date 1. A stone is placed in a 5 cm-diameter graduated cylinder, causing the water level in the cylinder to rise 2.7 cm. What is the volume of the stone?
2. A 141 g steel marble is submerged in a rectangular prism with base 5 cm by 6 cm. The water rises 0.6 cm. What is the density of the steel?
3. A solid wood toy boat with a mass of 325 g raises the water level of a 50 cm-by-40 cm aquarium 0.3 cm. What is the density of the wood?
4. For Awards Night at Baddeck High School, the math club is designing small solid silver pyramids. The base of the pyramids will be a 2 in.-by-2 in. square. The pyramids should not weigh more than 2 1 2 pounds. One cubic foot of silver weighs 655 pounds. What is the maximum height of the pyramids?
5. While he hikes in the Gold Country of northern California, Sid dreams about the adventurers that walked the same trails years ago.
He suddenly kicks a small bright yellowish nugget. Could it be gold?
Sid quickly makes a balance scale using his walking stick and finds that the nugget has the same mass as the uneaten half of his 330 g nutrition bar. He then drops the stone into his water bottle, which has a 2.5 cm radius, and notes that the water level goes up 0.9 cm. Has Sid struck gold? Explain your reasoning. (Refer to the density chart in Lesson 10.5 in your book.) ©2008 Kendall Hunt Publishing Lesson 10.6 • Volume of a Sphere Name Period Date In Exercises 1–6, find the volume of each solid. All measurements are in centimeters. Write your answers in exact form and rounded to the nearest 0.1 cm3.
1.
2.
3.
4.
5.
6. Cylinder with hemisphere taken out of the top 7. A sphere has volume 221 5 6 cm3. What is its diameter?
8. The area of the base of a hemisphere is 225 in2. What is its volume?
9. Eight wooden spheres with radii 3 in. are packed snugly into a square box 12 in. on one side. The remaining space is filled with packing beads. What is the volume occupied by the packing beads? What percentage of the volume of the box is filled with beads?
10. The radius of Earth is about 6378 km, and the radius of Mercury is about 2440 km. About how many times greater is the volume of Earth than that of Mercury?
9 4 6 6 6 2 90° 6 3 6 70 CHAPTER 10 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills CHAPTER 10 71 Lesson 10.7 • Surface Area of a Sphere Name Period Date In Exercises 1–4, find the volume and total surface area of each solid.
All measurements are in centimeters. Round your answers to the nearest 0.1 cm.
1.
2.
3.
4.
5. If the surface area of a sphere is 48.3 cm2, find its diameter.
6. If the volume of a sphere is 635 cm3, find its surface area.
7. Lobster fishers in Maine often use spherical buoys to mark their lobster traps. Every year the buoys must be repainted. An average buoy has a 12 in. diameter, and an average fisher has about 500 buoys. A quart of marine paint covers 175 ft2. How many quarts of paint does an average fisher need each year?
3 3 3 8 5 7 4 7.2 ©2008 Kendall Hunt Publishing Lesson 11.1 • Similar Polygons Name Period Date All measurements are in centimeters.
1. HAPIE NWYRS 2. QUAD SIML AP _ SL EI MI _ SN mD YR _ mU mA In Exercises 3–6, decide whether or not the figures are similar. Explain why or why not.
3. ABCD and EFGH 4. ABC and ADE 5. JKON and JKLM 6. ABCD and AEFG 7. Draw the dilation of ABCD by a scale 8. Draw the dilation of DEF by a scale factor factor of 1 2 . What is the ratio of the of 2. What is the ratio of the area of the perimeter of the dilated quadrilateral dilated triangle to the area of the original to the perimeter of the original triangle?
quadrilateral?
x y 5 5 B(4, 6) C(3, 3) A(0, 2) D(4, 1) E(4, 1) D(1, 2) F(2, 1) x y 5 5 4 4 4 7 7 3 3 A E B D C G F J K O L M N 14 12 6 8 10 5 20 C E D B A 8 4 3 3 2 2 60° 60° 60° 120° 120° 120° 5 15 9 3 D E F G H A B C A U Q D 20 25 13 75° S I M L 8 85° 120° N W E 5 6 S R Y 21 18 24 I P 4 A H 72 CHAPTER 11 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 11.2 • Similar Triangles Name Period Date All measurements are in centimeters.
1. TAR MAC MC _ 2. XYZ QRS 3. ABC EDC 4. TRS TQP Q A TS _ QR CD QP _ QS AB For Exercises 5 and 6, refer to the figure at right.
5. Explain why CAT and DAG are similar.
6. CA _ In Exercises 7–9, identify similar triangles and explain why they are similar.
7.
8.
9.
M O N K L R Q T S P B A C E D 48 12 16 T A G D C 17 30 16 8 Q P S R T 6 B A E D C 201 _ 4 221 _ 2 9 20 12 28 Y X Z 8 Q S R 3 2 7 R T A M C Discovering Geometry Practice Your Skills CHAPTER 11 73 ©2008 Kendall Hunt Publishing Lesson 11.3 • Indirect Measurement with Similar Triangles Name Period Date 1. At a certain time of day, a 6 ft man casts a 4 ft shadow. At the same time of day, how tall is a tree that casts an 18 ft shadow?
2. Driving through the mountains, Dale has to go up and over a high mountain pass. The road has a constant incline for 7 3 4 miles to the top of the pass. Dale notices from a road sign that in the first mile he climbs 840 feet. How many feet does he climb in all?
3. Sunrise Road is 42 miles long between the edge of Moon Lake and Lake Road and 15 miles long between Lake Road and Sunset Road. Lake Road is 29 miles long. Find the length of Moon Lake.
4. Marta is standing 4 ft behind a fence 6 ft 6 in. tall. When she looks over the fence, she can just see the top edge of a building. She knows that the building is 32 ft 6 in. behind the fence. Her eyes are 5 ft from the ground. How tall is the building? Give your answer to the nearest half foot.
5. You need to add 5 supports under the ramp, in addition to the 3.6 m one, so that they are all equally spaced. How long should each support be? (One is drawn in for you.) 3.6 m 9.0 m Support Ramp 4 ft 32 ft 6 in.
6 ft 6 in.
5 ft Fence Building Moon Lake Sunset Road Sunrise Road Lake Road 74 CHAPTER 11 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 11.4 • Corresponding Parts of Similar Triangles Name Period Date All measurements are in centimeters.
1. ABC PRQ. M and N 2. The triangles are similar.
are midpoints. Find h and j.
Find the length of each side of the smaller triangle to the nearest 0.01.
3. ABC WXY 4. Find x and y.
WX AD DB _ YZ XZ 5. Find a, b, and c.
6. Find CB, CD, and AD.
7 C A B D 25 c b a 4 3 6 28 24 14 A B C D 16 12 W X Y Z 10 5 x y 8 10 12 8 3 M j h C A B N Q P R 1.2 1.5 3.2 2.4 Discovering Geometry Practice Your Skills CHAPTER 11 75 ©2008 Kendall Hunt Publishing Lesson 11.5 • Proportions with Area Name Period Date All measurements are in centimeters unless otherwise indicated.
1. ABC DEF. Area of ABC 15 cm2.
2.
4 9 .
Area of DEF _ a 3.
4.
_ 5. RECT ANGL 6. The ratio of the corresponding midsegments of two similar trapezoids is 4:5. What is the ratio of their areas?
7. The ratio of the areas of two similar pentagons is 4:9. What is the ratio of their corresponding sides?
8. If ABCDE FGHIJ, AC 6 cm, FH 10 cm, and area of ABCDE 320 cm2, then area of FGHIJ .
9. Stefan is helping his mother retile the kitchen floor. The tiles are 4-by-4-inch squares. The kitchen is square, and the area of the floor is 144 square feet. Assuming the tiles fit snugly (don’t worry about grout), how many tiles will be needed to cover the floor?
C E R T 10 A L G N 4 Area of RECT Area of ANGL 12 P O 2 3 5 U L E R G Q S A Area of circle P Area of circle O Area of square SQUA Area of square LRGE 6 cm P O a A D C F E B 5 cm 3 cm Area of circle O Area of circle P 76 CHAPTER 11 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 11.6 • Proportions with Volume Name Period Date All measurements are in centimeters unless otherwise indicated.
In Exercises 1 and 2, decide whether or not the two solids are similar.
1.
2.
3. The triangular prisms are similar and the ratio of a to b is 5 2 .
Volume of large prism 250 cm3 Volume of smaller prism _ 4. The right cylinders are similar and r 10 cm.
Volume of large cylinder 64 cm Volume of small cylinder 8 cm R _ 5. The corresponding heights of two similar cylinders is 2:5. What is the ratio of their volumes?
6. A rectangular prism aquarium holds 64 gallons of water. A similarly shaped aquarium holds 8 gallons of water. If a 1.5 ft2 cover fits on the smaller tank, what is the area of a cover that will fit on the larger tank?
R r a b 2 1.5 6 8 20 16 12 32 3 5 4 8 Discovering Geometry Practice Your Skills CHAPTER 11 77 ©2008 Kendall Hunt Publishing Lesson 11.7 • Proportional Segments Between Parallel Lines Name Period Date All measurements are in centimeters.
1. x 2. Is XY BC ?
3. Is XY MK ?
4. NE 5. PR _ 6. a PQ b _ RI 7. RS 8. x _ 9. p EB y _ q 2 3 4 4 q p 9 12 12 x y 15 R T S E B 10 15 30 5 A T M b a n 15 5 3 4.5 R P Q I T 3 5 10 N P M A 8 12 12.5 E M K O Y X 40 30 48 60 2 3 3 4.5 B A Y X C P T A x 8 6 9 78 CHAPTER 11 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 12.1 • Trigonometric Ratios Name Period Date In Exercises 1–4, give each answer as a fraction in terms of p, q, and r.
1. sin P 2. cos P _ 3. tan P 4. sin Q In Exercises 5–8, give each answer as a decimal accurate to the nearest 0.001.
5. sin T _ 6. cos T 7. tan T 8. sin R _ For Exercises 9–11, solve for x. Express each answer accurate to the nearest 0.01.
9. cos 64° 2 x 8 10. sin 24° 12 x .1 11. tan 51° 14 x .8 For Exercises 12–14, find the measure of each angle to the nearest degree.
12. sin A 0.9455 13. tan B 4 3 14. cos C 0.8660 For Exercises 15–17, write a trigonometric equation you can use to solve for the unknown value. Then find the value to the nearest 0.1.
15. w 16. x 17. y _ For Exercises 18–20, find the value of each unknown to the nearest degree.
18. a 19. t 20. z _ 12 cm 25 cm z 15 in.
11 in.
t 14 cm 26 cm a 73 cm 17° y 14 cm 28° x 28 cm 40° w 8 6 T R P r Q R p q Discovering Geometry Practice Your Skills CHAPTER 12 79 ©2008 Kendall Hunt Publishing Lesson 12.2 • Problem Solving with Right Triangles Name Period Date For Exercises 1–3, find the area of each figure to the nearest square unit.
1. Area 2. Area 3. Area _ For Exercises 4–9, find each unknown to the nearest tenth of a unit.
4. Area 88 cm2 5. y 6. a x _ 7. PS and PT are tangents.
8. Right cone 9. Right rectangular prism Diameter mABC _ In Exercises 10–12, give each answer to the nearest tenth of a unit.
10. A ladder 7 m long stands on level ground and makes a 73° angle with the ground as it rests against a wall. How far from the wall is the base of the ladder?
11. To see the top of a building 1000 feet away, you look up 24° from the horizontal. What is the height of the building?
12. A guy wire is anchored 12 feet from the base of a pole. The wire makes a 58° angle with the ground. How long is the wire?
24 in.
C A B 14 in.
10 in.
13 ft 5 ft 65° S 22 cm T P 16 cm x a 8 in.
8 in.
14 in.
14 in.
72° y 17 ft 28 ft 140° 13 in.
28° 28 ft 50° 2.0 cm 80 CHAPTER 12 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 12.3 • The Law of Sines Name Period Date Discovering Geometry Practice Your Skills CHAPTER 12 81 In Exercises 1–3, find the area of each figure to the nearest square unit.
1. Area 2. Area 3. Area _ In Exercises 4–6, find each length to the nearest centimeter. All lengths are in centimeters.
4. m 5. p 6. q _ In Exercises 7–9, find the measure of each angle to the nearest degree.
7. mB 8. mP 9. mK _ mC mQ mM _ 10. A large helium balloon is tethered to the ground by two taut lines. One line is 100 feet long and makes an 80° angle with the ground. The second line makes a 40° angle with the ground. How long is the second line, to the nearest foot? How far apart are the tethers?
K L M 105 106 78° P R Q 26 32 48° 29 16 A B C 81° 21 q 67° 76° 12 p 32° 22° 17 m 40° 51° 55° 14 ft 5 ft 8 ft 70° 8.7 m 15 cm 21 cm 17° ©2008 Kendall Hunt Publishing Lesson 12.4 • The Law of Cosines Name Period Date In Exercises 1–3, find each length to the nearest centimeter. All lengths are in centimeters.
1. t 2. b 3. w _ In Exercises 4–6, find each angle measure to the nearest degree.
4. mA 5. mA 6. mS _ mB mP mU _ mC mS mV _ 7. A circle with radius 12 in. has radii drawn to the endpoints of a 5 in. chord. What is the measure of the central angle?
8. A parallelogram has side lengths 22.5 cm and 47.8 cm. One angle measures 116°. What is the length of the shorter diagonal?
9. The diagonals of a parallelogram are 60 in. and 70 in. and intersect at an angle measuring 64°. Find the length of the shorter side of the parallelogram.
S V U 265 192 201 A S P 45 48 30 A C B 70 51 62 O M w W 87 114 12° b O B 152 118 25° Y P t U T 21 16 39° 82 CHAPTER 12 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 12.5 • Problem Solving with Trigonometry Name Period Date 1. While floating down a river with a 2.75 mi/h current, Alicia decides to swim directly toward the river bank. She can swim 0.75 mi/h in still water. What is the actual speed at which she moves toward the bank?
At what angle will she approach the bank, measured with respect to the bank?
2. Find the measure of each angle to the nearest hundredth of a degree.
3. Two fire watchtowers 8.4 km apart spot a fire at the same time. Tower 1 reports the fire at a 36° angle measure from its line of sight to Tower 2.
Tower 2 reports a 68° angle measure between the fire and Tower 1.
How far is the fire from each tower?
4. Two airplanes leave O’Hare Airport in Chicago at the same time. One plane flies 280 mi/h at bearing 55°. The other plane flies 350 mi/h at bearing 128°. How far apart are the two planes after 2 hours 15 minutes?
5. Carla needs to fence her triangular plot of land. The angle between the two shorter sides measures 83°. The shortest side is 122 ft and the longest is 215 ft. How much fencing does Carla need? What is the area of her plot of land?
x y B(4, 7) a b c A(15, 9) C(10, 0) Discovering Geometry Practice Your Skills CHAPTER 12 83 ©2008 Kendall Hunt Publishing Lesson 13.1 • The Premises of Geometry Name Period Date 1. Provide the missing property of equality or arithmetic as a reason for each step to solve the equation.
Solve for x: 5(x 4) 2x 17 Solution: 5(x 4) 2x 17 a. __ 5x 20 2x 17 b. __ 3x 20 17 c. __ 3x 37 d. __ x 3 3 7 e. __ In Exercises 2–4, identify each statement as true or false. If the statement is true, tell which definition, property, or postulate supports your answer. If the statement is false, give a counterexample.
2. If AM BM, then M is the midpoint of AB .
3. If P is on AB and D is not, then mAPD mBPD 180°.
4. If PQ ST and PQ KL , then ST KL.
5. Complete the flowchart proof.
Given: AB CD , AP CQ , PB QD Show: AB CD Flowchart Proof __ _ _ _ ABP CDQ CA Postulate Given AB CD AB CD __ _ _ _ __ _ AP CQ P B A Q D C 84 CHAPTER 13 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 13.2 • Planning a Geometry Proof Name Period Date For these exercises, you may use theorems added to your theorem list through the end of Lesson 13.2.
In Exercises 1–3, write a paragraph proof or a flowchart proof for each situation.
1. Given: AB CD , AP CQ Show: PAB QCD 2. Given: PQ ST , QPR STU Show: PR UT 3. Given: Noncongruent, nonparallel segments AB, BC, and AC Show: x y z 180° x A a b c C z y B R Q P U S T B P A C Q D Discovering Geometry Practice Your Skills CHAPTER 13 85 ©2008 Kendall Hunt Publishing Lesson 13.3 • Triangle Proofs Name Period Date Write a proof for each situation. You may use theorems added to your theorem list through the end of Lesson 13.3.
1. Given: XY ZY , XZ WY 2. Given: CD AC , BD AB , CD BD Show: WXY WZY Show: ABD ACD 3. Given: MN QM , NO QM , 4. Given: AB BC , ACB ECD, P is the midpoint of MO AB BD Show: QMN RON Show: BD CE C B E D A Q M P O R N D C A B W Y X Z M 86 CHAPTER 13 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 13.4 • Quadrilateral Proofs Name Period Date Discovering Geometry Practice Your Skills CHAPTER 13 87 In Exercises 1–6, write a proof of each conjecture on a separate piece of paper. You may use theorems added to your theorem list through the end of Lesson 13.4.
1. The diagonals of a parallelogram bisect each other. (Parallelogram Diagonals Theorem) 2. If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. (Converse of the Parallelogram Diagonals Theorem) 3. The diagonals of a rhombus bisect each other and are perpendicular.
(Rhombus Diagonals Theorem) 4. If the diagonals of a quadrilateral bisect each other and are perpendicular, then the quadrilateral is a rhombus. (Converse of the Rhombus Diagonals Theorem) 5. If the base angles on one base of a trapezoid are congruent, then the trapezoid is isosceles. (Converse of the Isosceles Trapezoid Theorem) 6. If the diagonals of a trapezoid are congruent, then the trapezoid is isosceles. (Converse of the Isosceles Trapezoid Diagonals Theorem) In Exercises 7–9, decide if the statement is true or false. If it is true, prove it. If it is false, give a counterexample.
7. A quadrilateral with one pair of parallel sides and one pair of congruent angles is a parallelogram.
8. A quadrilateral with one pair of congruent opposite sides and one pair of parallel sides is a parallelogram.
9. A quadrilateral with one pair of parallel sides and one pair of congruent opposite angles is a parallelogram.
©2008 Kendall Hunt Publishing Lesson 13.5 • Indirect Proof Name Period Date 1. Complete the indirect proof of the conjecture: In a triangle the side opposite the larger of two angles has a greater measure.
Given: ABC with mA mB Show: BC AC Proof: Assume ___ Case 1: BC AC If BC AC, then ABC is __ by __.
By ___, A B, which contradicts __.
So, BC AC.
Case 2: BC AC If BC AC, then it is possible to construct point D on CA such that CD CB , by the Segment Duplication Postulate. Construct DB , by the Line Postulate. DBC is ____. Complete the proof.
In Exercises 2–5, write an indirect proof of each conjecture.
2. Given: AD AB , DC BC Show: DAC BAC 3. If two sides of a triangle are not congruent, then the angles opposite them are not congruent.
4. If two lines are parallel and a third line in the same plane intersects one of them, then it also intersects the other.
C B D A D 4 3 2 1 B C A B C A 88 CHAPTER 13 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Lesson 13.6 • Circle Proofs Name Period Date Write a proof for each conjecture or situation. You may use theorems added to your theorem list through the end of Lesson 13.6.
1. If two chords in a circle are congruent, then their arcs are congruent.
2. Given: Regular pentagon ABCDE inscribed in circle O, with diagonals AC and AD Show: AC and AD trisect BAE 3. Given: Two circles externally tangent at R, common external tangent segment TS Show: TRS is a right angle 4. Given: Two circles internally tangent at T with chords TD and TB of the larger circle intersecting the smaller circle at C and A Show: AC BD A C D B T S R T E D C A B O Discovering Geometry Practice Your Skills CHAPTER 13 89 ©2008 Kendall Hunt Publishing Write a proof for each situation. You may use theorems added to your theorem list through the end of Lesson 13.7.
1. Given: ABC with A BCD Show: BC2 AB BD 2. The diagonals of a trapezoid divide each other into segments with lengths in the same ratio as the lengths of the bases.
3. In a right triangle the product of the lengths of the two legs equals the product of the lengths of the hypotenuse and the altitude to the hypotenuse.
4. If a quadrilateral has one pair of opposite right angles and one pair of opposite congruent sides, then the quadrilateral is a rectangle.
A C B D Lesson 13.7 • Similarity Proofs Name Period Date 90 CHAPTER 13 Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing 17.
LESSON 1.3 • What’s a Widget?
1. d 2. c 3. e 4. i 5. f 6. b 7. h 8. a 9. g 10. They have the same measure, 13°. Because mQ 77°, its complement has measure 13°.
So mR 13°, which is the same as mP.
11.
12.
13.
LESSON 1.4 • Polygons 9.
10.
U Q A D P E N T A Polygon Number of Number of name sides diagonals 1. Triangle 3 0 2. Quadrilateral 4 2 3. Pentagon 5 5 4. Hexagon 6 9 5. Heptagon 7 14 6. Octagon 8 20 7. Decagon 10 35 8. Dodecagon 12 54 A X B Y C R P Q 3 2 1 Q P Z T R LESSON 1.1 • Building Blocks of Geometry 1. S 2. 9 cm 3. SN 4. endpoint 5. NS 6. PQ 7. SP 8. KN KL , NM LM , NO LO 9. E(14, 15) 10.
11.
12. AB , AC , AD , AE , AF , BC , BD , BE , BF , CD , CE , CF , DE , DF , EF (15 lines) 13. Possible coplanar set: {C, D, H, G}; 12 different sets LESSON 1.2 • Poolroom Math 1. vertex 2. bisector 3. side 4. 126° 5. DAE 6. 133° 7. 47° 8. 63° 9. 70° 10.
11.
12.
13. 90° 14. 120° 15. 75° 16.
A D C B z 160° I R G 90° O M N 15° A D B F E C 1.5 cm 1.5 cm 3 cm 3 cm A B C E D 2.0 cm 3.2 cm A M B C D Discovering Geometry Practice Your Skills ANSWERS 91 A N S W E R S ©2008 Kendall Hunt Publishing For Exercises 6–10, 12, and 13, answers may vary.
Possible answers are shown.
6.
7. ACFD 8. EFHG 9. BFJD 10. BFHD 11. D(0, 3) 12. E(0, 5) 13. G(16, 3) LESSON 1.7 • Circles 1. 48° 2. 132° 3. 228° 4. 312° 5.
6.
7. (8, 2); (3, 7); (3, 3) 8.
9. The chord goes through the center, P. (It is a diameter.) 10.
50° 180° 290° P B A O C D R A N G a b a b b b 11. AC , AD , AE 12. Possible answer: AB and BC 13. Possible answer: A and B 14. Possible answer: AC and FD 15. 82° 16. 7.2 17. 61° 18. 16.1 19. 6.2 cm LESSON 1.5 • Triangles For Exercises 1–7, answers will vary. Possible answers are shown.
1. AB GH 2. EF BI 3. CG FH 4. DEG and GEF 5. DEG and GEF 6.
7.
8.
9.
For Exercises 10–12, answers may vary. Possible answers are shown.
10. F(8, 2) 11. D(4, 3) 12. G(10, 2) LESSON 1.6 • Special Quadrilaterals 1.
2.
3.
4.
5.
T K E I P E R T A M H R O S Q R E P A R G R H G C D A 40° L C M A N S I T R 92 ANSWERS Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing 4. Possible answers: a.
b.
c.
LESSON 2.1 • Inductive Reasoning 1. 20, 24 2. 12 1 2 , 6 1 4 3. 5 4 , 2 4. 1, 1 5. 72, 60 6. 243, 729 7. 91, 140 8.
9.
10.
11. False; 1 2 2 1 4 12. False; 11 10 110, 11 12 132 13. True LESSON 2.2 • Finding the nth Term 1. Linear 2. Linear 3. Not linear 4. Linear 5.
6.
7. f(n) 4n 5; f(50) 205 8. f(n) 5n 11; f(50) 239 9. f(n) 1 2 n 6; f(50) 31 n 1 2 3 4 5 g(n) 10 18 26 34 42 n 1 2 3 4 5 f(n) 5 2 9 16 23 y x (3, 1) y x (1, –3) 11. Kite LESSON 1.8 • Space Geometry 1.
2.
3. Rectangular prism 4. Pentagonal prism 5.
6.
7. 18 cubes 8. x 2, y 1 LESSON 1.9 • A Picture Is Worth a Thousand Words 1.
2. Dora, Ellen, Charles, Anica, Fred, Bruce 3.
Triangles Acute triangles Isosceles triangles Scalene triangles D 1 1 4 3 3 E C A F B 10 m 5 5 12 m Power Gas Possible locations A P B A Q Discovering Geometry Practice Your Skills ANSWERS 93 ©2008 Kendall Hunt Publishing 4. Answers will vary. Possible answers: LESSON 2.4 • Deductive Reasoning 1. No. Explanations will vary. Sample explanation: Because ABC is equilateral, AB BC. Because C lies between B and D, BD BC, so BD is not equal to AB. Thus ABD is not equilateral, by deductive reasoning.
2. Answers will vary. mE mD (mE mD 90°); deductive 3. a, e, f; inductive 4. Deductive a. 4x 3(2 x) 8 2x The original equation.
4x 6 3x 8 2x Distributive property.
x 6 8 2x Combining like terms.
3x 6 8 Addition property of equality.
3x 2 Subtraction property of equality.
x 2 3 Division property of equality.
a b c d e 1 2 3 4 5 6 f 6 teams, 6 games b a c 1 2 3 d e f 4 5 6 7 6 teams, 7 games a b d e c 2 3 4 5 1 6 10 8 9 7 5 teams, 10 games 10.
11.
LESSON 2.3 • Mathematical Modeling 1.
a. 240 b. 1350 c. f(n) 2n(n 2), or f(n) 2n2 4n 2.
a. 8 sequences b. 3 sequences have 1 tail.
c. 3 8 3. 66 different pairs. Use a dodecagon showing sides and diagonals.
H HHH HHT H T H T H T H H T T T H T HTH HTT THH THT TTH TTT n 1 2 3 4 5 . . .
n . . .
50 Area of figure 1 4 16 64 256 . . .
4n1 . . .
449 n 1 2 3 4 5 . . .
n . . .
50 Number of triangles 1 5 9 13 17 . . .
4n 3 . . . 197 94 ANSWERS Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing LESSON 3.1 • Duplicating Segments and Angles 1.
2. XY 3PQ 2RS 3. Possible answer: 4.
128° 35° 93° 5.
6.
LESSON 3.2 • Constructing Perpendicular Bisectors 1.
2. Square C D D B D C C D B X Y B A P Q R S b. 19 2( 5 3x 1) x 2 The original equation.
19 2(3x 1) 5(x 2) Multiplication property of equality.
19 6x 2 5x 10 Distributive property.
21 6x 5x 10 Combining like terms.
21 11x 10 Addition property of equality.
11 11x Subtraction property of equality.
1 x Division property of equality.
5. a. 16, 21; inductive b. f(n) 5n 9; 241; deductive LESSON 2.5 • Angle Relationships 1. a 68°, b 112°, c 68° 2. a 127° 3. a 35°, b 40°, c 35°, d 70° 4. a 90°, b 90°, c 42°, d 48°, e 132° 5. a 20°, b 70°, c 20°, d 70°, e 110° 6. a 70°, b 55°, c 25° 7. Sometimes 8. Always 9. Never 10. Sometimes 11. acute 12. 158° 13. 90° 14. obtuse 15. converse LESSON 2.6 • Special Angles on Parallel Lines 1. a 54°, b 54°, c 54° 2. a 115°, b 65°, c 115°, d 65° 3. a 72°, b 126° 4. 1 2 5. 1 2 6. cannot be determined 7. a 102°, b 78°, c 58°, d 122°, e 26°, f 58° 8. x 80° 9. x 20°, y 25° Discovering Geometry Practice Your Skills ANSWERS 95 ©2008 Kendall Hunt Publishing LESSON 3.3 • Constructing Perpendiculars to a Line 1. False. The altitude from A coincides with the side so it is not shorter.
2. False. In an isosceles triangle, an altitude and median coincide so they are of equal length.
3. True 4. False. In an acute triangle, all altitudes are inside. In a right triangle, one altitude is inside and two are sides. In an obtuse triangle, one altitude is inside and two are outside. There is no other possibility so exactly one altitude is never outside.
5. False. In an obtuse triangle, the intersection of the perpendicular bisectors is outside the triangle.
6.
7.
8.
9. WX YZ W Y X Z P Q S R P Q A B C 3. XY 5 4 AB 4. ABC is not unique.
5. ABC is not unique.
6. BD AD CD 7. a. A and B b. A, B, and C c. A and B and from C and D (but not from B and C) d. A and B and from D and E A C B D C P A Q A B B C C A A M B B X W Y A M B 96 ANSWERS Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing 5.
6. Possible answer: LESSON 3.6 • Construction Problems 1. Possible answer: 2. Possible answer: A U Q S Perimeter I T E K A Z D B I O O M R H LESSON 3.4 • Constructing Angle Bisectors 1. a. 1 and 2 b. 1, 2, and 3 c. 2, 3, and 4 d. 1 and 2 and from 3 and 4 2. AP is the bisector of CAB 3. x 20°, mABE 50° 4.
5. They are concurrent.
6.
7. RN GN and RO HO LESSON 3.5 • Constructing Parallel Lines 1.
2.
3.
4.
A R P T m P S R Q P P O H M G N R Discovering Geometry Practice Your Skills ANSWERS 97 ©2008 Kendall Hunt Publishing LESSON 3.7 • Constructing Points of Concurrency 1. Circumcenter 2. Locate the power-generation plant at the incenter.
Locate each transformer at the foot of the perpendi-cular from the incenter to each side.
3.
4. Possible answer: In the equilateral triangle, the centers of the inscribed and circumscribed circles are the same. In the obtuse triangle, one center is outside the triangle.
5. Possible answer: In an acute triangle, the circum-center is inside the triangle. In a right triangle, it is on the hypotenuse. In an obtuse triangle, the circumcenter is outside the triangle. (Constructions not shown.) 3. Possible answers: 4. Possible answer: 5. Possible answer: 6. Possible answer: 7. Possible answer: 8.
S2 S3 S1 90° B B B A B S1 S1 A A S1 S2 A S2 S1 A B C T E O B S U R E C T T R E C 98 ANSWERS Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing LESSON 4.2 • Properties of Isosceles Triangles 1. mT 64° 2. mG 45° 3. x 125° 4. mA 39°, perimeter of ABC 46 cm 5. LM 163 m, mM 50° 6. mQ 44°, QR 125 7. a. DAB ABD BDC BCD b. ADB CBD c. AD BC by the Converse of the AIA Conjecture.
8. x 21°, y 16° 9. mQPR 15° 10. mPRQ 55° by VA, which makes mP 55° by the Triangle Sum Conjecture. So, PQR is isosceles by the Converse of the Isosceles Triangle Conjecture.
LESSON 4.3 • Triangle Inequalities 1. Yes 2. No 3. 19 x 53 4. b a c 5. b c a 6. a c d b 7. x 76° 8. x 79° 9. The interior angle at A is 60°. The interior angle at B is 20°. But now the sum of the measures of the triangle is not 180°.
10. By the Exterior Angles Conjecture, 2x x mPQS. So, mPQS x. So, by the Converse of the Isosceles Triangle Conjecture, PQS is isosceles.
11. Not possible. AB BC AC 12.
LESSON 4.4 • Are There Congruence Shortcuts?
1. SAA or ASA 2. SSS 3. SSS 4. BQM (SAS) 5. TIE (SSS) Q P R 28 km 17 km 9 km LESSON 3.8 • The Centroid 1.
2.
3. CP 3.3 cm, CQ 5.7 cm, CR 4.8 cm 4. (3, 4) 5. PC 16, CL 8, QM 15, CR 14 6. a. Incenter b. Centroid c. Circumcenter d. Circumcenter e. Orthocenter f. Incenter g. Centroid LESSON 4.1 • Triangle Sum Conjecture 1. p 67°, q 15° 2. x 82°, y 81° 3. a 78°, b 29° 4. r 40°, s 40°, t 100° 5. x 31°, y 64° 6. y 145° 7. s 28° 8. m 72 1 2 ° 9. mP a 10. mQPT 135° 11. 720° 12. The sum of the measures of A and B is 90° because mC is 90° and all three angles must be 180°. So, A and B are complementary.
13. mBEA mCED because they are vertical angles. Because the measures of all three angles in each triangle add to 180°, if equal measures are subtracted from each, what remains will be equal.
Q P C R 10 cm 8 cm 6 cm G C Discovering Geometry Practice Your Skills ANSWERS 99 ©2008 Kendall Hunt Publishing 9. All triangles will be congruent by ASA. Possible triangle: 10. All triangles will be congruent by SAA. Possible procedure: Use A and C to construct B and then copy A and B at the ends of AB .
LESSON 4.6 • Corresponding Parts of Congruent Triangles 1. SSS, SAS, ASA, SAA 2. YZ WX , AIA Conjecture 3. WZ XY , AIA Conjecture 4. ASA 5. CPCTC 6. YWM ZXM by SAS. YW ZX by CPCTC.
7. ACD BCD by SAS. AD BD by CPCTC.
8. Possible answer: DE and CF are both the distance between DC and AB . Because the lines are parallel, the distances are equal. So, DE CF .
9. Possible answer: DE CF (see Exercise 8).
DEF CFE because both are right angles, EF FE because they are the same segment. So, DEF CFE by SAS. EC FD by CPCTC.
10. Possible answer: It is given that TP RA and PTR ART, and TR RT because they are the same segment. So PTR ART by SAS and TA RP by CPCTC.
LESSON 4.7 • Flowchart Thinking 1. (See flowchart proof at bottom of page 101.) 2. (See flowchart proof at bottom of page 101.) 3. (See flowchart proof at bottom of page 101.) LESSON 4.8 • Proving Special Triangle Conjectures 1. AD 8 2. mACD 36° 3. mB 52°, CB 13 4. mE 60° 5. AN 17 6. Perimeter ABCD 104 C B A Q P R 6. Cannot be determined, as shown by the figure.
7. TNO (SAS) 8. Cannot be determined, as shown by the figure.
9. DOG (SAS) 10. Only one triangle because of SSS.
11. Two possible triangles.
12. Only one triangle because of SAS.
LESSON 4.5 • Are There Other Congruence Shortcuts?
1. Cannot be determined 2. XZY (SAA) 3. ACB (ASA or SAA) 4. PRS (ASA) 5. NRA (SAA) 6. GQK (ASA or SAA) 7. Yes, QRS MOL by SSS.
8. No, corresponding sides TV and WV are not congruent.
Z X Y B C A B C A U T S Q T U R S A C Y X Z B 100 ANSWERS Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing 3. 170°; 36 sides 4. 15 sides 5. x 105° 6. x 18° 7. mE 150° LESSON 5.2 • Exterior Angles of a Polygon 1. 12 sides 2. 24 sides 3. 4 sides 4. 6 sides 5. a 64°, b 138 2 3 ° 6. a 102°, b 9° 7. a 156°, b 132°, c 108° 8. a 135°, b 40°, c 105°, d 135° 9.
B A D C E 150° 85° 125° 110° 70° 7. (See flowchart proof at bottom of page 102.) 8. Flowchart Proof LESSON 5.1 • Polygon Sum Conjecture 1. a 103°, b 103°, c 97°, d 83°, e 154° 2. a 92°, b 44°, c 51°, d 85°, e 44°, f 136° Same segment AC BC Given CD CD AD BD Definition of median CD is a median Given ADC BDC SSS Conjecture CD bisects ACB Definition of bisect ACD BCD CPCTC Discovering Geometry Practice Your Skills ANSWERS 101 1.
2.
3.
ABCD is a parallelogram Given ABD CDB AIA Conjecture ADB CBD AIA Conjecture AD CB Definition of parallelogram Definition of parallelogram AB CD BDA DBC ASA Conjecture BD DB Same segment A C CPCTC KE KI Given Same segment KT KT TE TI Definition of kite KITE is a kite Given KET KIT SSS Conjecture KT bisects EKI and ETI Definition of bisect EKT IKT CPCTC ETK ITK CPCTC PQS RSQ AIA Conjecture Same segment QS QS PQ SR Given PQS RSQ SAS Conjecture SP QR CPCTC PQ SR Given Lesson 4.7, Exercises 1, 2, 3 ©2008 Kendall Hunt Publishing 7. AMNO is a parallelogram. By the Triangle Mid-segment Conjecture, ON AM and MN AO .
Flowchart Proof 8. Paragraph proof: Looking at FGR, HI FG by the Triangle Midsegment Conjecture. Looking at PQR, FG PQ for the same reason. Because FG PQ , quadrilateral FGQP is a trapezoid and DE is the midsegment, so it is parallel to FG and PQ . Therefore, HI FG DE PQ .
SAS Conjecture ONC MBN CA Conjecture NMB A CA Conjecture CON A Both congruent to A CON NMB Definition of midpoint MB AB 1 _ 2 Definition of midpoint OC AC 1 _ 2 Midsegment Conjecture MN AC 1 _ 2 Midsegment Conjecture ON AB 1 _ 2 Both congruent to AC OC MN 1 _ 2 Both congruent to AB 1 _ 2 ON MB A M P N B LESSON 5.3 • Kite and Trapezoid Properties 1. x 30 2. x 124°, y 56° 3. x 64°, y 43° 4. x 12°, y 49° 5. PS 33 6. a 11 7.
8.
9. Possible answer: Paragraph proof: Draw AE PT with E on TR .
TEAP is a parallelogram. T AER because they are corresponding angles of parallel lines.
T R because it is given, so AER R, because both are congruent to T. Therefore, AER is isosceles by the Converse of the Isosceles Triangle Conjecture. TP EA because they are opposite sides of a parallelogram and AR EA because AER is isosceles. Therefore, TP RA because both are congruent to EA .
LESSON 5.4 • Properties of Midsegments 1. a 89°, b 54°, c 91° 2. x 21, y 7, z 32 3. x 17, y 11, z 6.5 4. Perimeter XYZ 66, PQ 37, ZX 27.5 5. M(12, 6), N(14.5, 2); slope AB 1.6, slope MN 1.6 6. Pick a point P from which A and B can be viewed over land. Measure AP and BP and find the midpoints M and N. AB 2MN.
A C D B D C B A 102 ANSWERS Discovering Geometry Practice Your Skills Given AC BC ADC and BDC are right angles Definition of altitude CD is an altitude Given CPCTC AD BD CD is a median Definition of median A B Converse of IT ADC BDC Both are right angles ADC BDC SAA Conjecture CD CD Same segment Lesson 4.8, Exercise 7 ©2008 Kendall Hunt Publishing 4. c, d, f, g 5. d, e, g 6. f, g 7. h 8. c, d, f, g 9. c, d, f, g 10. d, g 11. d, g 12. None 13. Parallelogram 14.
LESSON 5.7 • Proving Quadrilateral Properties 1. (See flowchart proof at bottom of page.) 2. Flowchart Proof 3. Flowchart Proof ABO CBO CDO ADO SSS Conjecture AO CO Diagonals of parallelogram bisect each other BO DO AB CB CD AD Definition of rhombus Diagonals of a parallelogram bisect each other Given AB CB AD CD Same segment DB DB Given CPCTC A C SSS Conjecture ABD CBD A B D C LESSON 5.5 • Properties of Parallelograms 1. Perimeter ABCD 82 cm 2. AC 22, BD 14 3. AB 16, BC 7 4. a 51°, b 48°, c 70° 5. AB 35.5 6. a 41°, b 86°, c 53° 7. AD 75 8.
9. a 38°, b 142°, c 142°, d 38°, e 142°, f 38°, g 52°, h 12°, i 61°, j 81°, k 61° 10. No LESSON 5.6 • Properties of Special Parallelograms 1. OQ 16, mQRS 90°, PR 32 2. mOKL 45°, mMOL 90°, perimeter KLMN 32 3. OB 6, BC 11, mAOD 90° A B D C A B D C F2 F1 F1 F2 Resultant vector Discovering Geometry Practice Your Skills ANSWERS 103 AP CQ AC and QP bisect each other DC AB APR CQR PAR QCR APR CQR ASA Conjecture AR CR CPCTC PR QR CPCTC Given Definition of bisect Given AIA Conjecture AIA Conjecture Lesson 5.7, Exercise 1 ©2008 Kendall Hunt Publishing 3. z 45° 4. w 100°, x 50°, y 110° 5. w 49°, x 122.5°, y 65.5° 6. x 16 cm, y cannot be determined 7. Kite. Possible explanation: OM ON because congruent chords AB and AC are the same distance from the center. AM AN because they are halves of congruent chords. So, AMON has two pairs of adjacent congruent sides and is a kite.
8. The perpendicular segment from the center of the circle bisects the chord, so the chord has length 12 units. But the diameter of the circle is 12 units, and the chord cannot be as long as the diameter because it doesn’t pass through the center of the circle.
9. P(0,1), M(4, 2) 10. mAB 49°, mABC 253°, mBAC 156°, mACB 311° 11. Possible answer: Fold and crease to match the endpoints of the arc. The crease is the perpendicular bisector of the chord connecting the endpoints. Fold and crease so that one endpoint falls on any other point on the arc. The crease is the perpendicular bisector of the chord between the two matching points. The center is the intersection of the two creases.
LESSON 6.3 • Arcs and Angles 1. mXNM 40°, mXN 180°, mMN 100° 2. x 120°, y 60°, z 120° 3. a 90°, b 55°, c 35° 4. a 50°, b 60°, c 70° 5. x 140° 6. mA 90°, mAB 72°, mC 36°, mCB 108° 7. mAD 140°, mD 30°, mAB 60°, mDAB 200° 8. p 128°, q 87°, r 58°, s 87° 9. a 50°, b 50°, c 80°, d 50°, e 130°, f 90°, g 50°, h 50°, j 90°, k 40°, m 80°, n 50° Center 4. Flowchart Proof LESSON 6.1 • Tangent Properties 1. w 126° 2. mBQX 65° 3. a. mNQP 90°, mMPQ 90° b. Trapezoid. Possible explanation: MP and NQ are both perpendicular to PQ , so they are parallel to each other. The distance from M to PQ is MP, and the distance from N to PQ is NQ. But the two circles are not congruent, so MP NQ. Therefore, MN is not a constant distance from PQ and they are not parallel.
Exactly one pair of sides is parallel, so MNQP is a trapezoid.
4. y 1 3 x 10 5. Possible answer: Tangent segments from a point to a circle are congruent. So, PA PB , PB PC , and PC PD . Therefore, PA PD .
6. a. 4.85 cm b. 11.55 cm 7.
LESSON 6.2 • Chord Properties 1. a 95°, b 85°, c 47.5° 2. v cannot be determined, w 90° A T P AXD CYB SAA Conjecture DX BY CPCTC DAX BCY AIA Conjecture AD CB Opposite sides of parallelogram AXD CYB Both are 90° AD BC Definition of parallelogram 104 ANSWERS Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing 4. Flowchart Proof Construct radii AO , OB , OC , and OD .
LESSON 6.5 • The Circumference/Diameter Ratio 1. C 21 cm 2. r 12.5 cm 3. C 60 cm 4. d 24 cm 5. C 30.2 cm 6. d 42.0 cm, r 21.0 cm 7. C 37.7 in.
8. Yes; about 2.0 in.
9. C 75.4 cm 10. Press the square against the tree as shown. Measure the tangent segment on the square. The tangent segment is the same length as the radius. Use C 2r to find the circumference.
11. 4 cm Tree 2 in.
5 in.
AOB COD SAS Conjecture AO CO Radii of same circle BO DO Radii of same circle AB CD CPCTC AOB COD Definition of arc measure Given AB CD LESSON 6.4 • Proving Circle Conjectures 1. Flowchart Proof 2. Angles are numbered for reference.
Paragraph Proof It is given that OE AD , so 2 1 by the CA Conjecture. Because OA and OD are radii, they are congruent, so AOD is isosceles. Therefore 4 1 by the IT Conjecture. Both 2 and 4 are congruent to 1, so 2 4. By the AIA Conjecture, 4 3, so 2 3. The measure of an arc equals the measure of its central angle, so because their central angles are congruent, DE BE .
3. Flowchart Proof PQ RS Transitivity PX RX Tangent Segments Conjecture XQ XS Tangent Segments Conjecture PX XQ RX XS Addition Property of Equality PX XQ PQ Segment addition RX XS RS Segment addition PQ RS Definition of congruent segments A D O 1 2 3 4 E B XC XD Transitivity XA XD Tangent Segments Conjecture XC XA Tangent Segments Conjecture AB bisects CD at X Definition of segment bisector Discovering Geometry Practice Your Skills ANSWERS 105 ©2008 Kendall Hunt Publishing 3.
4. Possible answers: The two points where the figure and the image intersect determine . Or connect any two corresponding points and construct the perpendicular bisector, which is .
5. 3-fold rotational symmetry, 3 lines of reflection 6. 2-fold rotational symmetry 7. 1 line of reflection 8. 1 line of reflection 9. 2-fold rotational symmetry, 2 lines of reflection 10. 2-fold rotational symmetry 11. 1 line of reflection 12. 4-fold rotational symmetry, 4 lines of reflection C C C E C D E B A D C B T P E N T A A P E N LESSON 6.6 • Around the World 1. At least 7 olive pieces 2. About 2.5 rotations 3.
3085 m/s (about 3 km/s or just under 2 mi/s) 4. 6.05 cm or 9.23 cm 5. Sitting speed 107,500 km/h LESSON 6.7 • Arc Length 1. 4 2. 4 3. 30 4. 35 9 5. 80 9 6. 6.25 or 25 4 7. 10 9 0 8. 31.5 9. 22 10. 396 EXPLORATION • Intersection Secants, Tangents, and Chords 1. x 21° 2. mDC 70°, mED 150° 3. mDC 114°, mDEC 66° 4. mBCE 75°, mBAC 210° 5. x 80°, y 110°, z 141° 6. x 34°, y 150°, z 122° 7. x 112°, y 68°, z 53° 8. x 28°, y 34.5° LESSON 7.1 • Transformations and Symmetry 1.
2.
P R L A L Q P R A T I R R T I (2 1.4957 1011103) (364.25 24) (2 4.23 107) (60 60 23.93) 106 ANSWERS Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing 6. (x, y) →(x 13, y 6); translation; B(8, 8), C(8, 4) 7. (x, y) →(x, y); reflection across the y-axis; P(7, 3), R(4, 5) 8. (x, y) →(y, x); reflection across the line y x; T(7, 0), R(0, 3) LESSON 7.3 • Compositions of Transformations 1. Translation by 2, 5 2. Rotation 45° counterclockwise 3. Translation by 16, 0 4. Rotation 180° about the intersection of the two lines 5. Translation by 16, 0 6. Rotation 180° about the intersection of the two lines 7. Reflection across the line x 3 8. Reflection across the line x 3 9. mROT 50°; rotation 100° clockwise about O 10. Translation 2 cm along the line perpendicular to k and in the direction from k to .
11.
1 cm P k P P O P R P P T LESSON 7.2 • Properties of Isometries 1. Rotation 2. Translation 3. Reflection 4.
5.
S S S F Fence 1 Fence 2 S F Fence S S x y (0, –2) (4, –2) (0, 5) (4, 5) (–2, –2) (6, –2) x y (2, 0) (–1, –2) (–5, 4) (–2, 6) (–2, 4) (2, –2) x y (2, 1) (5, 1) (2, 3) (–2, –1) (–2, –3) (–5, –1) Discovering Geometry Practice Your Skills ANSWERS 107 ©2008 Kendall Hunt Publishing 108 ANSWERS Discovering Geometry Practice Your Skills 4. Sample answer: LESSON 8.1 • Areas of Rectangles and Parallelograms 1. 112 cm2 2. 7.5 cm2 3. 110 cm2 4. 81 cm2 5. 61 m 6. No. Possible answer: 7. 88 units2 8. 72 units2 9. No. Carpet area is 20 yd2 180 ft2. Room area is (21.5 ft)(16.5 ft) 206.25 ft2. Dana will be 26 1 4 ft2 short.
LESSON 8.2 • Areas of Triangles, Trapezoids, and Kites 1. 16 ft 2. 20 cm2 3. b 12 in.
4. AD 4.8 cm 5. 40 cm2 6. 88 cm2 7. 54 units2 8. 135 cm2 LESSON 8.3 • Area Problems 1. a. 549.5 ft2 b. 40 bundles; $1596.00 2. 500 L 3. Possible answer: 5 3 3 3 2 2 2 2 3 3 2 5 8 2 2 2 2.5 cm 40 34 cm 5 cm 40 17 cm LESSON 7.4 • Tessellations with Regular Polygons 1. n 15 2. n 20 3. Possible answer: A regular tessellation is a tessellation in which the tiles are congruent regular polygons whose edges exactly match.
4. Possible answer: A 1-uniform tiling is a tessellation in which all vertices are identical.
5. 3.42.63.6.3.6 6.
LESSONS 7.5–7.8 • Tessellations 1.
2. Sample answer: 3. Sample answer: ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills ANSWERS 109 6. 10 cm2 31.42 cm2 7. r 10 cm 8. x 135° 9. r 7 cm LESSON 8.7 • Surface Area 1. 136 cm2 2. 240 cm2 3. 558.1 cm2 4. 796.4 cm2 5. 255.6 cm2 6. 356 cm2 7. 468 cm2 8. 1055.6 cm2 9. 1 sheet: front rectangle: 3 1 1 2 4 1 2 ; back rectangle: 3 2 1 2 7 1 2 ; bottom rectangle: 3 2 6; side trapezoids: 22 8; total 26 ft2.
Area of 1 sheet 4 8 32 ft2. Possible pattern: LESSON 9.1 • The Theorem of Pythagoras 1. a 21 cm 2. p 23.9 cm 3. x 8 ft 4. h 14.3 in.
5. Area 19.0 ft2 6. C(11, 1); r 5 7. Area 49.7 cm2 8. RV 15.4 cm 9. If the base area is 16 cm2, then the radius is 4 cm.
The radius is a leg of the right triangle; the slant height is the hypotenuse. The leg cannot be longer than the hypotenuse.
10. Area 150 in2; hypotenuse QR 25 in.; altitude to the hypotenuse 12 in.
LESSON 9.2 • The Converse of the Pythagorean Theorem 1. No 2. Yes 3. Yes 4. Yes 5. Area 21.22 cm2 6. The top triangle is equilateral, so half its side length is 2.5. A triangle with sides 2.5, 6, and 6.5 is a right triangle because 2.52 62 6.52. So, the angle marked 95° should be 90°.
7. x 44.45 cm. By the Converse of the Pythagorean Theorem, ADC is a right triangle, and ADC is a right angle. ADC and BDC are supplementary, so BDC is also a right triangle. Use the Pythagorean Theorem to find x.
2 ft 2 ft 2 ft 3 ft 3 ft 3 ft 1 ft 1 _ 2 1 ft 1 _ 2 2 ft 1 _ 2 2 ft 1 _ 2 1 ft 1 _ 2 2 ft 1 _ 2 Back Bottom Left over Front Side Side 2 1 2 1 1 2 2 4. It is too late to change the area. The length of the diagonals determines the area.
LESSON 8.4 • Areas of Regular Polygons 1. A 696 cm2 2. a 7.8 cm 3. p 43.6 cm 4. n 10 5. s 4 cm, a 2.8 cm, A 28 cm2 6. Possible answer (s will vary): s 3.1 cm, a 3.7 cm, A 45.9 cm2 7. Approximately 31.5 cm2: area of square 36; area of square within angle 3 8 36 13.5; area of octagon 120; area of octagon within angle 3 8 120 45; shaded area 45 13.5 31.5 cm2 LESSON 8.5 • Areas of Circles 1. 81 cm2 2. 10.24 cm2 3. 23 cm 4. 324 cm2 5. 191.13 cm2 6. 41.41 cm 7. 7.65 cm2 8. 4.90 cm2 9. 51.3 cm2 10. 33.5 or 33.6 cm2 11. (64 128) square units 12. 25 cm2 LESSON 8.6 • Any Way You Slice It 1. 2 1 5 2 cm2 6.54 cm2 2. 32 3 cm2 33.51 cm2 3. 12 cm2 37.70 cm2 4. (16 32) cm2 18.27 cm2 5. 13.5 cm2 42.41 cm2 ©2008 Kendall Hunt Publishing 110 ANSWERS Discovering Geometry Practice Your Skills 2. About 6.4 km 3. 149.5 linear feet of trim must be painted, or 224.3 feet2. Two coats means 448.6 ft2 of coverage.
Just over 2 1 2 quarts of paint is needed. If Hans buys 3 quarts, he would have almost 1 2 quart left. It is slightly cheaper to buy 1 gallon and have about 1 1 2 quarts left. The choice is one of money versus conserving. Students may notice that the eaves extend beyond the exterior walls of the house and adjust their answer accordingly.
4. 14 in., 14 3 in. 8.08 in., 28 3 in. 16.17 in.
LESSON 9.5 • Distance in Coordinate Geometry 1. 10 units 2. 20 units 3. 17 units 4. ABCD is a rhombus: All sides 34 , slope AB 3 5 , slope BC 3 5 , so B is not a right angle, and ABCD is not a square.
5. TUVW is an isosceles trapezoid: TU and VW have slope 1, so they are parallel. UV and TW have length 20 and are not parallel (slope UV 1 2 , slope TW 2).
6. Isosceles; perimeter 32 units 7. M(7, 10); N(10, 14); slope MN 4 3 ; slope BC 4 3 ; MN 5; BC 10; the slopes are equal; MN 1 2 BC.
8. (x 1)2 (y 5)2 4 9. Center (0, 2), r 5 10. The distances from the center to the three points on the circle are not all the same: AP 61 , BP 61 , CP 52 14 30 0.4 km 4.1 km 1.2 km 0.9 km 2.3 km 1.7 km 3.1 km 8. 129.6 cm2 9. No. Because AB2 BC2 AC2, B of ABC is not a right angle.
10. Cannot be determined. The length of CD is unknown. One possible quadrilateral is shown.
11. Yes. Using SSS, ABC BAD CDA DCB. That means that the four angles of the quadrilateral are all congruent by CPCTC. Because the four angles must sum to 360° and they are all congruent, they must be right angles. So, ABCD is a rectangle.
LESSON 9.3 • Two Special Right Triangles 1. a 14 2 cm 2. a 12 cm, b 24 cm 3. a 12 cm, b 6 3 cm 4. 64 3 cm2 5. Perimeter 32 6 2 6 3 cm; area 60 18 3 cm2 6. AC 30 2 cm; AB 30 30 3 cm; area 450 450 3 cm2 7. 45 3 cm2 8. C 1 2 , 2 3 9. C(6 3 , 6) 10. Possible answer: LESSON 9.4 • Story Problems 1. The foot is about 8.7 ft away from the base of the building. To lower it by 2 ft, move the foot an additional 3.3 ft away from the base of the building.
3 3 2 2 18 2 2 A B C D 6 6 8 22 32 10 4.8 ©2008 Kendall Hunt Publishing Discovering Geometry Practice Your Skills ANSWERS 111 LESSON 10.3 • Volume of Pyramids and Cones 1. 80 cm3 2. 209.14 cm3 3. 615.75 cm3 4. V 840x3 5. V 8 3 a2b 6. V 4xy2 7. A: 128 cubic units, B: 144 cubic units.
B is larger.
8. A: 5 cubic units, B: 5 cubic units.
They have equal volumes.
9. A: 9x cubic units, B: 27x cubic units. B is larger.
LESSON 10.4 • Volume Problems 1. 4.4 cm 2. 1728 in3 3. 24 cans; 3582 in3 2.07 ft3; 34.6% 4. 2000.6 lb (about 1 ton) 5. Note that AE AB and EC BC . V 8 3 cm3; SA (8 4 2 ) cm2 13.7 cm2 6. About 110,447 gallons 7. 57 truckloads LESSON 10.5 • Displacement and Density All answers are approximate.
1. 53.0 cm3 2. 7.83 g/cm3 3. 0.54 g/cm3 4. 4.94 in.
5. No, it’s not gold (or at least not pure gold). The mass of the nugget is 165 g, and the volume is 17.67 cm3, so the density is 9.34 g/cm3. Pure gold has density 19.3 g/cm3.
LESSON 10.6 • Volume of a Sphere 1. 288 cm3, or about 904.8 cm3 2. 18 cm3, or about 56.5 cm3 3. 72 cm3, or about 226.2 cm3 4. 2 3 8 cm3, or about 29.3 cm3 5. 432 cm3, or about 1357.2 cm3 6. 30 3 4 cm3, or about 318.3 cm3 7. 11 cm 8. 2250 in3 7068.6 in3 9. 823.2 in3; 47.6% 10. 17.86 LESSON 10.7 • Surface Area of a Sphere 1. V 1563.5 cm3; S 651.4 cm2 2. V 184.3 cm3; S 163.4 cm2 LESSON 9.6 • Circles and the Pythagorean Theorem 1. (25 24) cm2, or about 54.5 cm2 2. (72 3 24) cm2, or about 49.3 cm2 3. ( 5338 37) cm 36.1 cm 4. Area 56.57 cm 177.7 cm2 5. AD 115.04 cm 10.7 cm 6. ST 9 3 15.6 7. 150° LESSON 10.1 • The Geometry of Solids 1. oblique 2. the axis 3. the altitude 4. bases 5. a radius 6. right 7. Circle C 8. A 9. AC or AC 10. BC or BC 11. Right pentagonal prism 12. ABCDE and FGHIJ 13. AF , BG , CH , DI , EJ 14. Any of AF , BG , CH , DI , EJ or their lengths 15. False. The axis is not perpendicular to the base in an oblique cylinder.
16. False. A rectangular prism has six faces. Four are called lateral faces and two are called bases.
17. True 18.
19.
LESSON 10.2 • Volume of Prisms and Cylinders 1. 232.16 cm3 2. 144 cm3 3. 415.69 cm3 4. V 4xy(2x 3), or 8x2y 12xy 5. V 1 4 p2h 6. V 6 1 2 x2y 7. 6 ft3 ©2008 Kendall Hunt Publishing 6. CA 64 cm 7. ABC EDC. Possible explanation: A E and B D by AIA, so by the AA Similarity Conjecture, the triangles are similar.
8. PQR STR. Possible explanation: P S and Q T because each pair is inscribed in the same arc, so by the AA Similarity Conjecture, the triangles are similar.
9. MLK NOK. Possible explanation: MLK NOK by CA and K K because they are the same angle, so by the AA Similarity Conjecture, the two triangles are similar.
LESSON 11.3 • Indirect Measurement with Similar Triangles 1. 27 ft 2. 6510 ft 3. 110.2 mi 4. About 18.5 ft 5. 0.6 m, 1.2 m, 1.8 m, 2.4 m, and 3.0 m LESSON 11.4 • Corresponding Parts of Similar Triangles 1. h 0.9 cm; j 4.0 cm 2. 3.75 cm, 4.50 cm, 5.60 cm 3. WX 13 5 7 13.7 cm; AD 21 cm; DB 12 cm; YZ 8 cm; XZ 6 6 7 6.9 cm 4. x 5 1 0 3 3.85 cm; y 8 1 0 3 6.15 cm 5. a 8 cm; b 3.2 cm; c 2.8 cm 6. CB 24 cm; CD 5.25 cm; AD 8.75 cm LESSON 11.5 • Proportions with Area 1. 5.4 cm2 2. 4 cm 3. 2 9 5 4. 3 1 6 5. 2 4 5 6. 16:25 7. 2:3 8. 888 8 9 cm2 9. 1296 tiles LESSON 11.6 • Proportions with Volume 1. Yes 2. No 3. 16 cm3 4. 20 cm 5. 8:125 6. 6 ft2 LESSON 11.7 • Proportional Segments Between Parallel Lines 1. x 12 cm 2. Yes 3. No 4. NE 31.25 cm 5. PR 6 cm; PQ 4 cm; RI 12 cm 6. a 9 cm; b 18 cm 3. V 890.1 cm3; S 486.9 cm2 4. V 34.1 cm3; S 61.1 cm2 5. About 3.9 cm 6. About 357.3 cm2 7. 9 quarts LESSON 11.1 • Similar Polygons 1. AP 8 cm; EI 7 cm; SN 15 cm; YR 12 cm 2. SL 5.2 cm; MI 10 cm; mD 120°; mU 85°; mA 80° 3. Yes. All corresponding angles are congruent. Both figures are parallelograms, so opposite sides within each parallelogram are equal. The corresponding sides are proportional 1 5 5 9 3 .
4. Yes. Corresponding angles are congruent by the CA Conjecture. Corresponding sides are proportional 2 4 = 3 6 = 4 8 .
5. No. 1 6 8 2 8 2 .
6. Yes. All angles are right angles, so corresponding angles are congruent. The corresponding side lengths have the ratio 4 7 , so corresponding side lengths are proportional.
7. 1 2 8. 4 to 1 LESSON 11.2 • Similar Triangles 1. MC 10.5 cm 2. Q X; QR 4.8 cm; QS 11.2 cm 3. A E; CD 13.5 cm; AB 10 cm 4. TS 15 cm; QP 51 cm 5. AA Similarity Conjecture x y 5 5 D(2, 4) E(8, 2) F(4, 2) x y 4 4 C(1.5, 1.5) D(2, 0.5) A(0, 1) B(2, 3) 112 ANSWERS Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing 7. About 24° 8. About 43.0 cm 9. About 34.7 in.
LESSON 12.5 • Problem Solving with Trigonometry 1. About 2.85 mi/h; about 15° 2. mA 50.64°, mB 59.70°, mC 69.66° 3. About 8.0 km from Tower 1, 5.1 km from Tower 2 4. About 853 miles 5. About 530 ft of fencing; about 11,656 ft2 LESSON 13.1 • The Premises of Geometry 1. a. Given b. Distributive property c. Subtraction property d. Addition property e. Division property 2. False 3. False 4. True; transitive property of congruence and definition of congruence 5.
LESSON 13.2 • Planning a Geometry Proof Proofs may vary.
1. Flowchart Proof ABP CDQ CA Postulate PAB QCD Third Angle Theorem APQ CQD CA Postulate Given AB CD Given AP CQ ABP CDQ CA Postulate APB CQD CA Postulate ABP CDQ ASA Postulate CPCTC AB CD AB CD Given AP CQ Given PB QD Given B P D A B A M 7. RS 22.5 cm, EB 20 cm 8. x 20 cm; y 7.2 cm 9. p 1 3 6 5.3 cm; q 8 3 2.6 cm LESSON 12.1 • Trigonometric Ratios 1. sin P p r 2. cos P q r 3. tan P p q 4. sin Q q r 5. sin T 0.800 6. cos T 0.600 7. tan T 1.333 8. sin R 0.600 9. x 12.27 10. x 29.75 11. x 18.28 12. mA 71° 13. mB 53° 14. mC 30° 15. sin 40° 2 w 8 ; w 18.0 cm 16. sin 28° 1 x 4 ; x 7.4 cm 17. cos 17° 7 y 3 ; y 76.3 cm 18. a 28° 19. t 47° 20. z 76° LESSON 12.2 • Problem Solving with Right Triangles 1. Area 2 cm2 2. Area 325 ft2 3. Area 109 in2 4. x 54.0° 5. y 31.3° 6. a 7.6 in.
7. Diameter 20.5 cm 8. 45.2° 9. 28.3° 10. About 2.0 m 11. About 445.2 ft 12. About 22.6 ft LESSON 12.3 • The Law of Sines 1. Area 46 cm2 2. Area 24 m2 3. Area 45 ft2 4. m 14 cm 5. p 17 cm 6. q 13 cm 7. mB 66°, mC 33° 8. mP 37°, mQ 95° 9. mK 81°, mM 21° 10. Second line: about 153 ft, between tethers: about 135 ft LESSON 12.4 • The Law of Cosines 1. t 13 cm 2. b 67 cm 3. w 34 cm 4. mA 76°, mB 45°, mC 59° 5. mA 77°, mP 66°, mS 37° 6. mS 46°, mU 85°, mV 49° Discovering Geometry Practice Your Skills ANSWERS 113 ©2008 Kendall Hunt Publishing 2. Proof: Statement Reason 1. CD BD 1. Given 2. BD AB 2. Given 3. CD AC 3. Given 4. AD is bisector 4. Converse of Angle of CAB Bisector Theorem 5. CAD BAD 5. Definition of angle bisector 6. ACD is a right 6. Definition of angle perpendicular 7. ABD is a right 7. Definition of angle perpendicular 8. ACD ABD 8. Right Angles Are Congruent Theorem 9. ABD ACD 9. SAA Theorem 3. Flowchart Proof 4. Proof: Statement Reason 1. AB BC 1. Given 2. ABC is isosceles 2. Definition of isosceles triangle 3. A ACB 3. IT Theorem 4. ACB DCE 4. Given NMO NOP IT Theorem QMN RON Supplements of Congruent Angles Theorem MNO is isosceles Definition of isosceles triangle Linear Pair Postulate QMN and NMO are supplementary Linear Pair Postulate RON and NOP are supplementary Transitivity MN NO Given MN QM Given NO QM 2. Flowchart Proof 3. Flowchart Proof LESSON 13.3 • Triangle Proofs Proofs may vary.
1. Flowchart Proof WXY WZY SAS Theorem XYM ZYM Definition of angle bisector XZY is isosceles Definition of isosceles triangle YM is angle bisector of XYZ Isosceles Triangle Vertex Angle Theorem Reflexive property WY WY Given XZ WY Given XY ZY YM is the altitude from vertex Y Definition of altitude and vertex angle a b c 180° Triangle Sum Theorem x b c 180° Substitution x y c 180° Substitution x y z 180° Substitution a x VA Theorem b y VA Theorem c z VA Theorem ABC Given PQR TSU AIA Theorem Converse of AEA Theorem PR UT QRP TUS Third Angle Theorem Given PQ ST QPR STU Given 114 ANSWERS Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing 3. AM CM 3. Given 4. AM CM 4. Definition of congruence 5. DMA BMC 5. VA Theorem 6. AMD CMB 6. SAS Postulate 7. DAC BCA 7. CPCTC 8. AD BC 8. Converse of AIA Theorem 9. DMC BMA 9. VA Theorem 10. DMC BMA 10. SAS Postulate 11. CDB ABD 11. CPCTC 12. DC AB 12. Converse of AIA Theorem 13. ABCD is a 13. Definition of parallelogram parallelogram 3. Given: ABCD is a rhombus Show: AC and BD bisect each other at M and AC BD Flowchart Proof DAM BAM Rhombus Angles Theorem ADM ABM SAS Postulate AMD AMB CPCTC Definition of perpendicular AC BD AMD and AMB are supplementary Linear Pair Postulate Definition of rhombus AD AB ABCD is a rhombus Given Definition of rhombus ABCD is a parallelogram AC and BD bisect each other Parallelogram Diagonals Theorem Reflexive property AM AM Congruent and Supplementary Theorem AMB is a right angle A D C B M 5. A DCE 5. Transitivity 6. AB CE 6. Converse of CA Postulate 7. ABD CED 7. CA Postulate 8. AB BD 8. Given 9. ABD is a right 9. Definition of angle perpendicular 10. CED is a right 10. Definition of right angle angle, transitivity 11. BD CE 11. Definition of perpendicular LESSON 13.4 • Quadrilateral Proofs Proofs may vary.
1. Given: ABCD is a parallelogram Show: AC and BD bisect each other at M Flowchart Proof 2. Given: DM BM, AM CM Show: ABCD is a parallelogram Proof: Statement Reason 1. DM BM 1. Given 2. DM BM 2. Definition of congruence A D C B M BDC DBA AIA Theorem CAB ACD AIA Theorem ABM CDM ASA Postulate CPCTC DM BM CPCTC AM CM AC and BD bisect each other at M Definition of bisect, definition of congruence Opposite Sides Theorem CD AB Definition of parallelogram AB CD ABCD is a parallelogram Given A D C B M Discovering Geometry Practice Your Skills ANSWERS 115 ©2008 Kendall Hunt Publishing 6. Given: ABCD is a trapezoid with AB CD and AC BD Show: ABCD is isosceles Proof: Statement Reason 1. ABCD is a trapezoid 1. Given with AB CD 2. Construct BE AC 2. Parallel Postulate 3. DC and BE intersect 3. Line Intersection at F Postulate 4. ABFC is a 4. Definition of parallelogram parallelogram 5. AC BF 5. Opposite Sides Congruent Theorem 6. AC BD 6. Given 7. BF BD 7. Transitivity 8. DFB is isosceles 8. Definition of isosceles triangle 9. DFB FDB 9. IT Theorem 10. CAB DFB 10. Opposite Angles Theorem 11. FDB DBA 11. AIA Theorem 12. CAB DBA 12. Transitivity 13. AB AB 13. Reflexive property 14. ACB BDA 14. SAS Postulate 15. AD BC 15. CPCTC 16. ABCD is isosceles 16. Definition of isosceles trapezoid D A B C E F 4. Given: AC and BD bisect each other at M and AC BD Show: ABCD is a rhombus Flowchart Proof (See flowchart at bottom of page.) 5. Given: ABCD is a trapezoid with AB CD and A B Show: ABCD is isosceles Proof: Statement Reason 1. ABCD is a trapezoid 1. Given with AB CD 2. Construct CE AD 2. Parallel Postulate 3. AECD is a 3. Definition of parallelogram parallelogram 4. AD CE 4. Opposite Sides Congruent Theorem 5. A BEC 5. CA Postulate 6. A B 6. Given 7. BEC B 7. Transitivity 8. ECB is isosceles 8. Converse of IT Theorem 9. EC CB 9. Definition of isosceles triangle 10. AD CB 10. Transitivity 11. ABCD is isosceles 11. Definition of isosceles trapezoid D A B E C A D C B M 116 ANSWERS Discovering Geometry Practice Your Skills ADM ABM SAS Postulate Reflexive property AM AM CPCTC AD AB DMA and BMA are right angles Definition of perpendicular Given AC BD Converse of the Parallelogram Diagonals Theorem ABCD is a parallelogram Definition of rhombus ABCD is a rhombus AC and BD bisect each other at M Given All 4 sides are congruent Transitivity Definition of bisect, definition of congruence DM BM Opposite Sides Theorem AB DC DMA BMA Right Angles Congruent Theorem Opposite Sides Theorem AD BC Lesson 13.4, Exercise 4 ©2008 Kendall Hunt Publishing Therefore the assumption, BC AC, is false, so BC AC.
2. Paragraph Proof: Assume DAC BAC It is given that AD AB . By the reflexive property AC AC . So by SAS, ADC ABC. Then DC BC by CPCTC. But this contradicts the given that DC BC . So DAC BAC.
3. Given: ABC with AB BC Show: C A Paragraph Proof: Assume C A If C A, then by the Converse of the IT Theorem, ABC is isosceles and AB BC . But this contradicts the given that AB BC . Therefore, C A.
4. Given: Coplanar lines k, , and m, k , and m intersecting k Show: m intersects Paragraph Proof: Assume m does not intersect If m does not intersect , then by the definition of parallel, m . But because k , by the Parallel Transitivity Theorem, k m. This contradicts the given that m intersects k. Therefore, m intersects .
LESSON 13.6 • Circle Proofs 1. Given: Circle O with AB CD Show: AB CD Flowchart Proof Definition of congruence, definition of arc measure, transitivity AB CD OAB ODC SSS Postulate AOB DOC CPCTC AB CD Given OA OD Definition of circle, definition of radii Construct OA, OB, OC, OD Line Postulate OB OC Definition of circle, definition of radii B C D O A k m B C A 7. False 8. False 9. True Given: ABCD with AB CD and A C Show: ABCD is a parallelogram Flowchart Proof LESSON 13.5 • Indirect Proof Proofs may vary.
1. Assume BC AC Case 1: If BC AC, then ABC is isosceles, by the definition of isosceles. By the IT Theorem, A B, which contradicts the given that mA mB. So, BC AC.
Case 2: DBC is isosceles.
By the Exterior Angle Theorem, m1 m2 m4, so m1 m4.
By the Angle Sum Postulate, m2 m3 mABC, so m3 mABC. But DBC is isosceles, so m4 m3 by the IT Theorem.
So, by transitivity, m1 m4 m3 mABC, or m1 mABC, which contradicts the given that mA mB. So, BC AC.
B C D 4 3 2 1 A A C Given D B Supplements of Congruent Angles Theorem ABCD is a parallelogram Converse of Opposite Angles Theorem AB CD Given A and D are supplementary Interior Supplements Theorem C and B are supplementary Interior Supplements Theorem A D C B Discovering Geometry Practice Your Skills ANSWERS 117 ©2008 Kendall Hunt Publishing LESSON 13.7 • Similarity Proofs 1. Flowchart Proof 2. Given: Trapezoid ABCD with AB CD , and AC and BD intersecting at E Show: D BE E C AE E D AB C Flowchart Proof CDE ABE AA Similarity Postulate BDC DBA AIA Theorem DCA CAB AIA Theorem Definition of similarity DE BE CE AE DC AB AB CD Given D A B C E A BCD Given B B Reflexive property ABC CBD AA Similarity Postulate BC2 ABBD Multiplication property Definition of similar triangle AB BC BC BD 2. Paragraph Proof: Chords BC , CD , and DE are congruent because the pentagon is regular. By the proof in Exercise 1, the arcs BC , CD , and DE are congruent and therefore have the same measure.
mEAD 1 2 mDE by the Inscribed Angles Inter-cepting Arcs Theorem. Similarly, mDAC 1 2 mDC and mBAC 1 2 mBC . By transitivity and algebra, the three angles have the same measure. So, by the definition of trisect, the diagonals trisect BAE.
3. Paragraph Proof: Construct the common internal tangent RU (Line Postulate, definition of tangent).
Label the intersection of the tangent and TS as U.
TU RU SU by the Tangent Segments Theorem. TUR is isosceles by definition because TU RU . So, by the IT Theorem, T TRU.
Call this angle measure x. SUR is isosceles because RU SU , and by the IT Theorem, S URS.
Call this angle measure y. The angle measures of TRS are then x, y, and (x y). By the Triangle Sum Theorem, x y (x y) 180°. By algebra (combining like terms and dividing by 2), x y 90°. But mTRS x y, so by transitivity and the definition of right angle, TRS is a right angle.
4. Paragraph Proof: Construct tangent TP (Line Postulate, definition of tangent). PTD and TAC both have the same intercepted arc, TC . Similarly, PTD and TBD have the same intercepted arc, TD . So, by transitivity, the Inscribed Angles Inter-cepting Arcs Theorem, and algebra, TAC and TBD are congruent. Therefore, by the Converse of the CA Postulate, AC BD .
T P C A B D S U R T 118 ANSWERS Discovering Geometry Practice Your Skills ©2008 Kendall Hunt Publishing Proof: Statement Reason 1. Construct DB 1. Line Postulate 2. A and C are 2. Given right angles 3. A C 3. Right Angles Are Congruent Theorem 4. AB DC 4. Given 5. DB DB 5. Reflexive property 6. DBA BDC 6. HL Congruence Theorem 7. DBA BDC 7. CPCTC 8. mDBA mBDC 8. Definition of congruence 9. mADB mDBA 9. Triangle Sum mA 180° Theorem 10. mA 90° 10. Definition of right angle 11. mADB mDBA 11. Subtraction property 90° 12. mADB mBDC 12. Substitution 90° 13. mADB mBDC 13. Angle Addition mADC Postulate 14. mADC 90° 14. Transitivity 15. mC 90° 15. Definition of right angle 16. mA mABC 16. Quadrilateral Sum mC mADC Theorem 360° 17. mABC 90° 17. Substitution property and subtraction property 18. A ABC C 18. Definition of ADC congruence 19. ABCD is a rectangle 19. Four Congruent Angles Rectangle Theorem Discovering Geometry Practice Your Skills ANSWERS 119 3. Given: ABC with ACB right, CD AB Show: AC BC AB CD Flowchart Proof 4. Given: ABCD with right angles A and C, AB DC Show: ABCD is a rectangle B C D A ADC ACB Right Angles Are Congruent Theorem A A Reflexive property ADC is right Definition of perpendicular ACB is right Given CD AB Given ACB ADC AA Similarity Postulate Definition of similarity AC CD AB BC ACBC ABCD Multiplication property A D C B ©2008 Kendall Hunt Publishing |
825 | https://wiki.ncscpartners.org/index.php/Core_Content_Connectors_by_Common_Core_State_Standards:_Mathematics_Number_and_Quantity | Core Content Connectors by Common Core State Standards: Mathematics Number and Quantity - NCSC Wiki
Core Content Connectors by Common Core State Standards: Mathematics Number and Quantity
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Contents
1Mathematics High School—Number and Quantity Overview
1.1The Real Number System
1.2Quantities
1.3The Complex Number System
1.4Vector and Matrix Quantities
Mathematics High School—Number and Quantity Overview
The Real Number System
Extend the properties of exponents to rational exponents
Use properties of rational and irrational numbers.
Quantities
Reason quantitatively and use units to solve problems.
The Complex Number System
Perform arithmetic operations with complex numbers
Represent complex numbers and their operations on the complex plane
Use complex numbers in polynomial identities and equations.
Vector and Matrix Quantities
Represent and model with vector quantities.
Perform operations on vectors.
Perform operations on matrices and use matrices in applications.
The Real Number SystemHSN-RN
Extend the properties of exponents to rational exponents.
1. Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 5(⅓) to be the cube root of 5 because we want (5(⅓))³ = 5(⅓)³ to hold, so (5(⅓))³ must equal 5.
CCCs linked to HSN-RN.A.1 None
2. Rewrite expressions involving radicals and rational exponents using the properties of exponents.
CCCs linked to HSN-RN.A.2 HS.NO.1a1 Simplify expressions that include exponents.
HS.NO.1a2 Explain the influence of an exponent on the location of a decimal point in a given number.
HS.NO1a3 Convert a number expressed in scientific notation.
H.NO.2c2 Rewrite expressions that include rational exponents.
Use properties of rational and irrational numbers.
3. Explain why the sum or product of two rational numbers is rational; that the sum of a rational number and an irrational number is irrational; and that the product of a nonzero rational number and an irrational number is irrational.
CCCs linked to HSN-RN.B.3 H.NO.2b1 Explain the pattern for the sum or product for combinations of rational and irrational numbers.
QuantitiesHSN-Q
Reason quantitatively and use units to solve problems.
1. Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.
CCCs linked to HSN-Q.A.1 H.ME.1a1 Determine the necessary unit(s) to use to solve real-world problems.
H.ME.1a2 Solve real-world problems involving units of measurement
2. Define appropriate quantities for the purpose of descriptive modeling.
CCCs linked to HSN-Q.A.2 None
3. Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.
CCCs linked to HSN-Q.A.3 None
The Complex Number SystemHSN-CN
Perform arithmetic operations with complex numbers.
1. Know there is a complex number i such that i ² = –1, and every complex number has the form a+ bi with a and b real.
CCCs linked to HSN-CN.A.1 None
2. Use the relation i ² = –1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers.
CCCs linked to HSN-CN.A.2 None
3. Find the conjugate of a complex number; use conjugates to find moduli and quotients of complex numbers.
CCCs linked to HSN-CN.A.3 None
Represent complex numbers and their operations on the complex plane.
4. Represent complex numbers on the complex plane in rectangular and polar form (including real and imaginary numbers), and explain why the rectangular and polar forms of a given complex number represent the same number.
CCCs linked to HSN-CN.B.4 None
5. Represent addition, subtraction, multiplication, and conjugation of complex numbers geometrically on the complex plane; use properties of this representation for computation. For example,(–1 + √3 i)³ = 8 because(–1 + √3 i) has modulus 2 and argument 120°.
CCCs linked to HSN-CN.B.5 None
6. Calculate the distance between numbers in the complex plane as the modulus of the difference, and the midpoint of a segment as the average of the numbers at its endpoints.
CCCs linked to HSN-CN.B.6 None
Use complex numbers in polynomial identities and equations.
7. Solve quadratic equations with real coefficients that have complex solutions.
CCCs linked to HSN-CN.C.7 None
8. Extend polynomial identities to the complex numbers. For example, rewrite x ² + 4 as(x+ 2 i)(x– 2 i).
CCCs linked to HSN-CN.C.8 None
9. Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials.
CCCs linked to HSN-CN.C.9 None
Vector and Matrix QuantitiesHSN-VM
Represent and model with vector quantities.
1. Recognize vector quantities as having both magnitude and direction. Represent vector quantities by directed line segments, and use appropriate symbols for vectors and their magnitudes (e.g., v, |v|, \v\, v).
CCCs linked to HSN-VM.A.1 None
2. Find the components of a vector by subtracting the coordinates of an initial point from the coordinates of a terminal point.
CCCs linked to HSN-VM.A.2 None
3. Solve problems involving velocity and other quantities that can be represented by vectors.
CCCs linked to HSN-VM.A.3 None
Perform operations on vectors.
4. Add and subtract vectors.
a. Add vectors end-to-end, component-wise, and by the parallelogram rule. Understand that the magnitude of a sum of two vectors is typically not the sum of the magnitudes.
b. Given two vectors in magnitude and direction form, determine the magnitude and direction of their sum.
c. Understand vector subtraction v – w as v + (–w), where –w is the additive inverse of w, with the same magnitude as w and pointing in the opposite direction. Represent vector subtraction graphically by connecting the tips in the appropriate order, and perform vector subtraction component-wise.
CCCs linked to HSN-VM.B.4 None
5. Multiply a vector by a scalar.
a. Represent scalar multiplication graphically by scaling vectors and possibly reversing their direction; perform scalar multiplication component-wise, e.g., as c(v x, v y) = (cv x, cv y).
b. Compute the magnitude of a scalar multiple cv using \cv\ = |c|v. Compute the direction of cv knowing that when |c|v ≠ 0, the direction of cv is either along v (for c> 0) or against v (for c< 0).
CCCs linked to HSN-VM.B.5 None
Perform operations on matrices and use matrices in applications.
6. Use matrices to represent and manipulate data, e.g., to represent payoffs or incidence relationships in a network.
CCCs linked to HSN-VM.C.6 None
7. Multiply matrices by scalars to produce new matrices, e.g., as when all of the payoffs in a game are doubled.
CCCs linked to HSN-VM.C.7 None
8. Add, subtract, and multiply matrices of appropriate dimensions.
CCCs linked to HSN-VM.C.8 None
9. Understand that, unlike multiplication of numbers, matrix multiplication for square matrices is not a commutative operation, but still satisfies the associative and distributive properties.
CCCs linked to HSN-VM.C.9 None
10. Understand that the zero and identity matrices play a role in matrix addition and multiplication similar to the role of 0 and 1 in the real numbers. The determinant of a square matrix is nonzero if and only if the matrix has a multiplicative inverse.
CCCs linked to HSN-VM.C.10 None
11. Multiply a vector (regarded as a matrix with one column) by a matrix of suitable dimensions to produce another vector. Work with matrices as transformations of vectors.
CCCs linked to HSN-VM.C.11 None
12. Work with 2 × 2 matrices as transformations of the plane, and interpret the absolute value of the determinant in terms of area
CCCs linked to HSN-VM.C.12 None
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826 | https://obgyn.onlinelibrary.wiley.com/doi/full/10.1002/uog.28011 | OP17.02: Quality of life after uterine artery embolisation versus hysterectomy for symptomatic adenomyosis: one‐year outcome from the QUESTA study - Trommelen - 2024 - Ultrasound in Obstetrics & Gynecology - Wiley Online Library
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Ultrasound in Obstetrics & Gynecology
Ultrasound in Obstetrics & Gynecology
Volume 64, Issue S1 p. 104
Abstracts
Free Access
OP17.02: Quality of life after uterine artery embolisation versus hysterectomy for symptomatic adenomyosis: one-year outcome from the QUESTA study
L. Trommelen,
L. Trommelen
Gynecology, Amsterdam University Medical Center (UMC), VUmc Campus, Amsterdam, Netherlands
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J.F. Huirne,
J.F. Huirne
Obstetrics and Gynecology, Amsterdam UMC, AMC Campus, Amsterdam, Netherlands
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P. Lohle,
P. Lohle
Radiology, Elisabeth-TweeSteden Ziekenhuis, Tilburg, Netherlands
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W.K. Hehenkamp,
W.K. Hehenkamp
Obstetrics and Gynecology, Amsterdam UMC, AMC Campus, Amsterdam, Netherlands
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L. Trommelen,
L. Trommelen
Gynecology, Amsterdam University Medical Center (UMC), VUmc Campus, Amsterdam, Netherlands
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J.F. Huirne,
J.F. Huirne
Obstetrics and Gynecology, Amsterdam UMC, AMC Campus, Amsterdam, Netherlands
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P. Lohle,
P. Lohle
Radiology, Elisabeth-TweeSteden Ziekenhuis, Tilburg, Netherlands
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W.K. Hehenkamp,
W.K. Hehenkamp
Obstetrics and Gynecology, Amsterdam UMC, AMC Campus, Amsterdam, Netherlands
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First published: 09 September 2024
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Objectives
To demonstrate non-inferiority of uterine artery embolisation (UAE) on health-related quality of life (HRQOL) in comparison to hysterectomy in the treatment of symptomatic adenomyosis, one year after treatment.
Methods
Twelve Dutch hospitals included patients with symptomatic adenomyosis eligible for hysterectomy. Patients were offered UAE as an alternative treatment option. Primary endpoint was the change in HRQOL between UAE and hysterectomy, measured with standardised WHO-QOL-Bref (physical, psychological, social and environmental domain) and SF-12 (mental and physical) questionnaires. UAE is considered non-inferior to hysterectomy when HRQOL does not differ more than 5 points. Secondary endpoints were a subsequent hysterectomy after UAE and changes in WHO-QOL100 facet Pain and Discomfort and facet Sexual Activity. Linear mixed models were used for the within and between group analyses, adjusted for relevant covariates.
Results
Of 101 patients, 51 underwent a hysterectomy and 50 a UAE. Both hysterectomy and UAE had a significant positive effect on all HRQOL scores, one year after intervention (all p < .05). There was no significant difference between the effect of UAE versus hysterectomy on HRQOL (all p > .05). Mean HRQOL difference of UAE versus hysterectomy at 1 year did not exceed 5 points. Both hysterectomy and UAE had a significant positive effect on facet Pain and Discomfort and facet Sexual Activity (both p < .001). However, hysterectomy had a greater effect on Pain and Discomfort than UAE (p = .002). Six patients (12%) underwent a subsequent hysterectomy in the first year after receiving UAE.
Conclusions
UAE for symptomatic adenomyosis is considered non-inferior to hysterectomy one year after treatment in terms of health-related quality of life and could therefore be an effective and less-invasive alternative in the treatment for symptomatic adenomyosis. Nevertheless, when patients seek for a definitive solution hysterectomy remains the treatment of choice.
Volume 64, Issue S1
Supplement: Abstracts of the 34th World Congress on Ultrasound in Obstetrics and Gynecology, 15–18 September 2024, Budapest, Hungary
September 2024
Page 104
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Quality of life 1 year after uterine artery embolization vs hysterectomy for symptomatic adenomyosis (QUESTA study)
Lisa M. Trommelen,Annika Semmler,Annefleur M. de Bruijn,Marissa Harmsen,Marieke Smink,Petra F. Janssen,Ilse van Rooij,Jeroen van Bavel,Peggy Geomini,Jacques W. M. Maas,Celine M. Radder,Paul van Kesteren,Janet Kwee,Erica Bakkum,Marleen de Lange,Robert A. de Leeuw,Freek Groenman,Velja Mijatovic,Anne Timmermans,Rutger Lely,Armand Lamers,Douwe Vos,Gretel van Hoecke,Otto Elgersma,Huib A. A. M. van Vliet,Lonneke S. F. Yo,Andries R. H. Twijnstra,Frank W. Jansen,Catharina S. P. van Rijswijk,Han Kruimer,Carroll M. E. S. N. Tseng,Sjors Coppus,Mark Arntz,Aloys F. J. Wust,Joost G. A. M. Blomjous,Laurens van Boven,Alexander Venmans,Jos W. R. Twisk,Judith A. F. Huirne,Paul N. M. Lohle,Wouter J. K. Hehenkamp, Acta Obstetricia et Gynecologica Scandinavica
Long‐term durability of uterine artery embolisation for treatment of symptomatic adenomyosis
Jolande Ma,Bevan Brown,Eisen Liang, Australian and New Zealand Journal of Obstetrics and Gynaecology
Pelvic organ prolapse after hysterectomy: A 10‐year national follow‐up study
Tea Kuittinen,Sari Tulokas,Päivi Rahkola-Soisalo,Tea Brummer,Jyrki Jalkanen,Eija Tomas,Juha Mäkinen,Jari Sjöberg,Päivi Härkki,Maarit Mentula, Acta Obstetricia et Gynecologica Scandinavica
Uterine artery embolization for symptomatic uterine fibroids
Maja Pakiz,Igor But, International Journal of Gynecology & Obstetrics
Identifying the indications for laparoscopically assisted vaginal hysterectomy: a prospective, randomised comparison with abdominal hysterectomy in patients with symptomatic uterine fibroids
Maria Maddalena Ferrari,Nicola Berlanda,Raffaella Mezzopane,Guglielmo Ragusa,Michela Cavallo,Giorgio Pardi, BJOG: An International Journal of Obstetrics & Gynaecology
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827 | https://www.youtube.com/watch?v=1VooGDT2xmQ | End Behavior of a Polynomial Function from a Graph: Even or Odd Degree / Sign of Coefficient
Mathispower4u
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This video explains how to determine the end behavior of a polynomial function based upon a graph to determine if the degree is odd or even as well as the sign of the leading coefficient.
Transcript:
0 seconds Determine if the degree of each polynomial function is odd or even based upon the graph. By determining the end behavior 6 seconds of a polynomial function, we can determine whether the degree is even or odd, as well as whether A, the leading 12 seconds coefficient is positive or negative. Referring to our notes, if the polynomial function 18 seconds has an even degree in the leading coefficient A is positive, that for both the left and right-hand behavior, 25 seconds F of X approaches positive infinity. And if the degree is even, the leading coefficient is negative, then for both the left and right-hand behavior, F of X approaches negative 38 seconds infinity. If the degree is odd and the leading coefficient is positive, then for the left-hand behavior, as X approaches negative infinity, F of X approaches negative infinity. 50 seconds And for the right-hand behavior, as X approaches positive infinity, F of X approaches 57 seconds positive infinity. If the degree is odd and the leading coefficient is negative, then for the left-hand behavior, as X approaches negative infinity, F of X approaches positive infinity. 1 minute, 10 seconds And for the right-hand behavior, as X approaches infinity, F of X approaches negative infinity. 1 minute, 17 seconds Going back to our problem, for the first polynomial function, notice for both the left and right-hand behavior, 1 minute, 24 seconds F of X increases without bound. And F of X approaches positive infinity. 1 minute, 31 seconds In this case, we know the polynomial has an even degree, and the leading coefficient A is positive. 1 minute, 37 seconds I know the question doesn't ask about the sign of A, but I'm going to go ahead and include that information. And now let's jump down to this last graph. Notice for both the left-hand right-hand behavior, in this case, F of 1 minute, 51 seconds X decreases without bound and approaches negative infinity. This indicates the 1 minute, 57 seconds polynomial has an even degree, but now A is negative or less than zero. And now it's going through 2 minutes, 3 seconds the graph of this polynomial here, for the left-hand behavior, as X approaches and negative infinity, F of X decreases without bound and approaches negative infinity. And for the right-hand 2 minutes, 16 seconds behavior, as X approaches, positive infinity, F of X increases without bound. 2 minutes, 22 seconds And F of X approaches positive infinity. Notice this in behavior indicates the polynomial has an odd degree, and the leading coefficient A is positive. And finally for the 2 minutes, 35 seconds last graph, for the left-hand behavior, as X approaches negative infinity, F of X approaches positive infinity. And for the right-hand behavior, as X approaches 2 minutes, 48 seconds positive infinity, F of X approaches negative infinity. This indicates the polynomial has an odd degree, and the leading coefficient A is negative. 2 minutes, 59 seconds I hope you found this helpful. |
828 | https://bestpractice.bmj.com/topics/en-us/36 | Skip to main content
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Central hypothyroidism
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Last reviewed:
5 Aug 2025
Last updated:
17 Jan 2023
Summary
Symptoms of central hypothyroidism are similar to those of primary hypothyroidism (including fatigability, cold intolerance, weight gain, and depression), and may or may not be accompanied by symptoms of hypopituitarism, such as those of hypogonadism (i.e., decreased libido, infertility) and secondary adrenal insufficiency (i.e., weakness, nausea, vomiting).
Signs of central hypothyroidism on physical examination would also be similar to those of primary hypothyroidism and may include dry skin, hair loss, bradycardia, and delayed deep tendon reflexes.
Specific signs suggestive of a sellar or parasellar mass include papilledema and bitemporal hemianopia.
Diagnostic evaluation of central hypothyroidism includes serum thyroid-stimulating hormone (TSH) and free thyroxine (T4) concentrations. In central hypothyroidism, free T4 is low and TSH may be low, normal, or minimally elevated. Magnetic resonance imaging may reveal a sellar or parasellar mass or infiltrative disorder.
Treatment of central hypothyroidism is by thyroid hormone replacement (levothyroxine).
Complications of treatment may include thyrotoxicosis and osteoporosis.
Definition
Hypothyroidism is a condition resulting from the deficiency of thyroid hormones, which leads to a generalized slowing of metabolic processes. Central hypothyroidism arises from an anatomic or functional disorder of the pituitary gland and/or the hypothalamus. This decreases thyroid-stimulating hormone secretion, leading in turn to decreased thyroid hormone synthesis and release.Beck-Peccoz P, Rodari G, Giavoli C, et al. Central hypothyroidism: a neglected thyroid disorder. Nat Rev Endocrinol. 2017 Oct;13(10):588-98.
Central hypothyroidism may be due to congenital, neoplastic, inflammatory, infiltrative, traumatic, or iatrogenic etiologies.
History and exam
Key diagnostic factors
weakness
fatigue
cold intolerance
Full details
Other diagnostic factors
decreased memory
constipation
muscle cramps
weight gain
depression
dry, coarse skin
oligomenorrhea/amenorrhea
bradycardia
reduced body and scalp hair
delayed relaxation of deep tendon reflexes
hearing impairment
impassive facial expression
diabetes insipidus (DI)
headache
diplopia
decreased peripheral vision
atrophic breasts
galactorrhea
moon facies
buffalo hump
Full details
Risk factors
presence of pituitary mass lesions
multiple endocrine neoplasia (MEN) type I
head and neck irradiation
traumatic brain injury (TBI)
age between 5-15 years and 45-60 years (craniopharyngiomas)
age 20-50 years (prolactinomas)
age 40-80 years (nonfunctioning pituitary adenomas)
sarcoidosis
histiocytosis
hemochromatosis
Sheehan syndrome (postpartum pituitary necrosis)
lymphocytic hypophysitis
family history of central hypothyroidism
anticonvulsant drugs
Full details
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Diagnostic tests
1st tests to order
serum free thyroxine (FT4)
serum thyroid-stimulating hormone (TSH)
Full details
Tests to consider
pituitary MRI
head CT
serum prolactin (PRL)
fasting morning serum cortisol
serum testosterone
serum gonadotropins
genetic analyses
Full details
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Treatment algorithm
ONGOING
all patients
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Contributors
Authors
Angela M. Leung, MD, MSc
Associate Professor of Medicine
UCLA David Geffen School of Medicine
VA Greater Los Angeles Healthcare System
Los Angeles
CA
Disclosures
AML serves on the board of directors of the American Thyroid Association (ATA); speaks for China Merck; consulted for Vertice Pharma; and received article process charges paid on behalf of IBSA Institut Biochimique for an invited review article.
Acknowledgements
Dr Angela M. Leung would like to gratefully acknowledge Dr Jacqueline Gilbert and Dr Rasa Zarnegar, previous contributors to this topic.
Disclosures
JG and RZ declare that they have no competing interests.
Peer reviewers
James Lee, MD
Assistant Professor
Department of Endocrine Surgery
Columbia University
New York
NY
Disclosures
JL declares that he has no competing interests.
James V. Hennessey, MD
Director of Clinical Endocrinology
Beth Israel Deaconess Medical Center
Boston
MA
Disclosures
JVH declares that he has no competing interests.
Anthony Weetman, MD, DSc, FRCP, FMedSci
Sir Arthur Hall Professor of Medicine/Pro Vice Chancellor
University of Sheffield
Sheffield
UK
Disclosures
AW declares that he has no competing interests.
Peer reviewer acknowledgements
BMJ Best Practice topics are updated on a rolling basis in line with developments in evidence and guidance. The peer reviewers listed here have reviewed the content at least once during the history of the topic.
Disclosures
Peer reviewer affiliations and disclosures pertain to the time of the review.
References
Our in-house evidence and editorial teams collaborate with international expert contributors and peer reviewers to ensure that we provide access to the most clinically relevant information possible.
Key articles
Persani L, Brabant G, Dattani M, et al. 2018 European Thyroid Association (ETA) guidelines on the diagnosis and management of central hypothyroidism. Eur Thyroid J. 2018 Oct;7(5):225-37.
Full text
Abstract
Garber JR, Cobin RH, Gharib H, et al. Clinical practice guidelines for hypothyroidism in adults: cosponsored by the American Association of Clinical Endocrinologists and the American Thyroid Association. Endocr Pract. 2012 Nov-Dec;18(6):988-1028.
Abstract
Jonklaas J, Bianco AC, Bauer AJ, et al. Guidelines for the treatment of hypothyroidism: prepared by the American Thyroid Association Task Force on Thyroid Hormone Replacement. Thyroid. 2014 Dec;24(12):1670-751.
Full text
Abstract
Reference articles
A full list of sources referenced in this topic is available to users with access to all of BMJ Best Practice.
Differentials
Primary hypothyroidism
Nonthyroidal illness
Iodine deficiency
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#### Guidelines
Hypothalamic-pituitary and growth disorders in survivors of childhood cancer
Guidelines on the diagnosis and management of central hypothyroidism
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#### Patient information
Underactive thyroid
Underactive thyroid: questions to ask your doctor
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829 | https://online.stat.psu.edu/stat462/node/185/ | 10.10 - Other Regression Pitfalls | STAT 462
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10.10 - Other Regression Pitfalls
Overfitting
When building a regression model, we don't want to include unimportant or irrelevant predictors whose presence can overcomplicate the model and increase our uncertainty about the magnitudes of the effects for the important predictors (particularly if some of those predictors are highly collinear). Such "overfitting" can occur the more complicated a model becomes and the more predictor variables, transformations, and interactions are added to a model. It is always prudent to apply a sanity check to any model being used to make decisions. Models should always make sense, preferably grounded in some kind of background theory or sensible expectation about the types of associations allowed between variables. Predictions from the model should also be reaosnable (over-complicated models can give quirky results that may not reflect reality).
Excluding Important Predictor Variables
However, there is potentially greater risk from excluding important predictors than from including unimportant ones. The linear association between two variables ignoring other relevant variables can differ both in magnitude and direction from the association that controls for other relevant variables. Whereas the potential cost of including unimportant predictors might be increased difficulty with interpretation and reduced prediction accuracy, the potential cost of excluding important predictors can be a completely meaningless model containing misleading associations. Results can vary considerably depending on whether such predictors are (inappropriately) excluded or (appropriately) included. These predictors are sometimes called confounding or lurking variables, and their absence from a model can lead to incorrect decisions and poor decision-making.
Extrapolation
"Extrapolation" beyond the "scope of the model" occurs when one uses an estimated regression equation to estimate a mean µ Y or to predict a new response y new for x values not in the range of the sample data used to determine the estimated regression equation. In general, it is dangerous to extrapolate beyond the scope of model. The following example illustrates why this is not a good thing to do.
Researchers measured the number of colonies of grown bacteria for various concentrations of urine (ml/plate). The scope of the model — that is, the range of the x values — was 0 to 5.80 ml/plate. The researchers obtained the following estimated regression equation:
Using the estimated regression equation, the researchers predicted the number of colonies at 11.60 ml/plate to be 16.0667 + 1.61576(11.60) or 34.8 colonies. But when the researchers conducted the experiment at 11.60 ml/plate, they observed that the number of colonies decreased dramatically to about 15.1 ml/plate:
The moral of the story is that the trend in the data as summarized by the estimated regression equation does not necessarily hold outside the scope of the model.
Missing Data
Real-world datasets frequently contain missing values, so that we do not know the values of particular variables for some of the sample observations. For example, such values may be missing because they were impossible to obtain during data collection. Dealing with missing data is a challenging task. Missing data has the potential to adversely affect a regression analysis by reducing the total usable sample size. The best solution to this problem is to try extremely hard to avoid having missing data in the first place. When there are missing values that are impossible or too costly to avoid, one approach is to replace the missing values with plausible estimates, known as imputation. Another (easier) approach is to consider only models that contain predictors with no (or few) missing values. This may be unsatisfactory, however, because even a predictor variable with a large number of missing values can contain useful information.
Power and Sample Size
In small datasets, a lack of observations can lead to poorly estimated models with large standard errors. Such models are said to lack statistical power because there is insufficient data to be able to detect significant associations between the response and predictors. So, how much data do we need to conduct a successful regression analysis? A common rule of thumb is that 10 data observations per predictor variable is a pragmatic lower bound for sample size. However, it is not so much the number of data observations that determines whether a regression model is going to be useful, but rather whether the resulting model satisfies the LINE conditions. In some circumstances, a model applied to fewer than 10 data observations per predictor variable might be perfectly fine (if, say, the model fits the data really well and the LINE conditions seem fine), while in other circumstances a model applied to a few hundred data points per predictor variable might be pretty poor (if, say, the model fits the data badly and one or more conditions are seriously violated). For another example, in general we’d need more data to model interaction compared to a similar model without the interaction. However, it is difficult to say exactly how much data would be needed. It is possible that we could adequately model interaction with a relatively small number of observations if the interaction effect was pronounced and there was little statistical error. Conversely, in datasets with only weak interaction effects and relatively large statistical error, it might take a much larger number of observations to have a satisfactory model. In practice, we have methods for assessing the LINE conditions, so it is possible to consider whether an interaction model approximately satisfies the assumptions on a case-by-case basis. In conclusion, there is not really a good standard for determining sample size given the number of predictors, since the only truthful answer is, “It depends.” In many cases, it soon becomes pretty clear when working on a particular dataset if we are trying to fit a model with too many predictor terms for the number of sample observations (results can start to get a little odd and standard errors greatly increase). From a different perspective, if we are designing a study and need to know how much data to collect, then we need to get into sample size and power calculations, which rapidly become quite complex. Some statistical software packages will do sample size and power calculations, and there is even some software specifically designed to do just that. When designing a large, expensive study, it is recommended that such software be used or to get advice from a statistician with sample size expertise.
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Lesson 1: Statistical Inference Foundations
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Lesson 6: MLR Assumptions, Estimation & Prediction
Lesson 7: Transformations & Interactions
Lesson 8: Categorical Predictors
Lesson 9: Influential Points
Lesson 10: Regression Pitfalls
10.1 - Nonconstant Variance and Weighted Least Squares
10.2 - Autocorrelation and Time Series Methods
10.3 - Regression with Autoregressive Errors
10.4 - Multicollinearity
10.5 - Uncorrelated Predictors
10.6 - Highly Correlated Predictors
10.7 - Detecting Multicollinearity Using Variance Inflation Factors
10.8 - Reducing Data-based Multicollinearity
10.9 - Reducing Structural Multicollinearity
10.10 - Other Regression Pitfalls
Lesson 11: Model Building
Lesson 12: Logistic, Poisson & Nonlinear Regression
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How can we prove that the square root of 3 is irrational?
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David Joyce
Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979) · Upvoted by
Justin Rising
, PhD in statistics · Author has 9.9K answers and 67.6M answer views
·Updated 9y
Originally Answered: How can we prove that the squareroot of 3 is irrational? ·
There are many ways to show irrationality. Here's one that works to show that the square root of any integer that is not a perfect square is irrational. In other words, if √n n is not an integer, then it's irrational.
Proof: First note that the prime factorization of a positive integer m m is a product of powers of prime numbers, m=p e 1 1⋯p e k k,m=p 1 e 1⋯p k e k, then m 2=p 2 e 1 1⋯p 2 e k k.m 2=p 1 2 e 1⋯p k 2 e k. That means that the sum of the powers in the prime factorization is even.
Suppose that the positive rational number m n m n is in lowest terms, that is to say, m m and n n
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There are many ways to show irrationality. Here's one that works to show that the square root of any integer that is not a perfect square is irrational. In other words, if √n n is not an integer, then it's irrational.
Proof: First note that the prime factorization of a positive integer m m is a product of powers of prime numbers, m=p e 1 1⋯p e k k,m=p 1 e 1⋯p k e k, then m 2=p 2 e 1 1⋯p 2 e k k.m 2=p 1 2 e 1⋯p k 2 e k. That means that the sum of the powers in the prime factorization is even.
Suppose that the positive rational number m n m n is in lowest terms, that is to say, m m and n n are relatively prime. Then m 2 n 2 m 2 n 2 is also in lowest terms and the sum of the powers of the prime factorizations of both the numerator and denominator are even.
That implies that if (m/n)2(m/n)2 is an integer, then n=1 n=1 and the integer is the perfect square m 2.m 2. Q.E.D.
Thus, √2,√3,√5,√6,√7,√8,√10 2,3,5,6,7,8,10, etc. are all irrational.
More generally, if the k t h k t h root of an integer n n is not an integer, then it's irrational.
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·9mo
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Lakshya Jain
Loves Math · Author has 342 answers and 2.3M answer views
·10y
Originally Answered: How can we prove that the squareroot of 3 is irrational? ·
First notice that 3 is a prime number. Thus if a 2 a 2 is divible by 3, a a must also be divisible by 3.
Proof by contradiction:
Let's assume that √3 3 is rational. i.e
√3=m/n 3=m/n, where m and n have no common factors.
Multiplying both side by n n and squaring we get.
3 n 2=m 2 3 n 2=m 2
Since lhs is divisible by 3, rhs should also be divisible by 3. Thus, we can write m m as 3 p 3 p.
3 n 2=(3 p)2 3 n 2=(3 p)2
=>3 n 2=9 p 2 3 n 2=9 p 2
=>n 2=3 p 2 n 2=3 p 2
Since rhs is divisible by 3, lhs should also be divisible by 3. Thus n n is divisible by 3.
Both n,m n,m are divisible by 3, contradicting our original assumption that m and n are co-prime.
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First notice that 3 is a prime number. Thus if a 2 a 2 is divible by 3, a a must also be divisible by 3.
Proof by contradiction:
Let's assume that √3 3 is rational. i.e
√3=m/n 3=m/n, where m and n have no common factors.
Multiplying both side by n n and squaring we get.
3 n 2=m 2 3 n 2=m 2
Since lhs is divisible by 3, rhs should also be divisible by 3. Thus, we can write m m as 3 p 3 p.
3 n 2=(3 p)2 3 n 2=(3 p)2
=>3 n 2=9 p 2 3 n 2=9 p 2
=>n 2=3 p 2 n 2=3 p 2
Since rhs is divisible by 3, lhs should also be divisible by 3. Thus n n is divisible by 3.
Both n,m n,m are divisible by 3, contradicting our original assumption that m and n are co-prime.
Thus, √3 3 is irrational. Hence Proved.
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Sridhar Ramesh
Mathematician/Logician/All-Around Great Guy · Upvoted by
Justin Rising
, PhD in statistics and
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 954 answers and 6.6M answer views
·10y
Originally Answered: How can we prove that the squareroot of 3 is irrational? ·
David Joyce's answer is a very good one. However, here's an even more general observation which you may find worthwhile even outside the context of this particular problem:
Observation: Suppose P(x)=c k x k+c k−1 x k−1+...+c 0 x 0 P(x)=c k x k+c k−1 x k−1+...+c 0 x 0 is a polynomial with integer coefficients, and P(n/d)=0 P(n/d)=0 for some lowest-terms rational n/d n/d. Then the polynomial's leading coefficient c k c k must be divisible by the root's (lowest-terms) denominator d d. (In other words, the root is some integer over c k c k)
Proof: Consider 0=P(n/d)d k 0=P(n/d)d k=c k n k+c k−1 n k−1 d 1+...+c 0 n 0 d k=c k n k+c k−1 n k−1 d 1+...+c 0 n 0 d k. Shuffling ter
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David Joyce's answer is a very good one. However, here's an even more general observation which you may find worthwhile even outside the context of this particular problem:
Observation: Suppose P(x)=c k x k+c k−1 x k−1+...+c 0 x 0 P(x)=c k x k+c k−1 x k−1+...+c 0 x 0 is a polynomial with integer coefficients, and P(n/d)=0 P(n/d)=0 for some lowest-terms rational n/d n/d. Then the polynomial's leading coefficient c k c k must be divisible by the root's (lowest-terms) denominator d d. (In other words, the root is some integer over c k c k)
Proof: Consider 0=P(n/d)d k 0=P(n/d)d k=c k n k+c k−1 n k−1 d 1+...+c 0 n 0 d k=c k n k+c k−1 n k−1 d 1+...+c 0 n 0 d k. Shuffling terms around, we find that c k n k c k n k is divisible by d d. And as n n and d d were, by stipulation, relatively prime, it follows that c k c k itself must be divisible by d d.
Applying this to the particular polynomial P(x)=x 2−3 P(x)=x 2−3, we see that any rational root of this polynomial must be an integer; thus, as it has no integer roots, we conclude it has no rational roots either.
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Dean Rubine
Former Faculty at Carnegie Mellon School Of Computer Science (1991–1994) · Author has 10.5K answers and 23M answer views
·9y
Originally Answered: How do you prove that the square root of 3 is an irrational number? ·
Let's assume √3 3 is rational. This means there exist whole numbers p p and q q with no common factors such that
p q=√3 p q=3
p 2=3 q 2 p 2=3 q 2
We know p 2 p 2 is multiple of 3. What does that tell us about p p?
A whole number n n has a remainder of 0, 1 or 2 when divided by 3, so either n=3 k,n=3 k+1 n=3 k,n=3 k+1 or n=3 k−1 n=3 k−1 for some integer k k (I changed 2 to -1 to make the math easier -- you could work it with 2).
So the possibilities for n 2 n 2 are (3 k)2=9 k 2(3 k)2=9 k 2, (3 k+1)2=9 k 2+6 k+1(3 k+1)2=9 k 2+6 k+1 or (3 k−1)2=9 k 2−6 k+1(3 k−1)2=9 k 2−6 k+1. Only the first one is a multiple of three, the others giving remainder 1 when divided by 3. So we've proved a
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Let's assume √3 3 is rational. This means there exist whole numbers p p and q q with no common factors such that
p q=√3 p q=3
p 2=3 q 2 p 2=3 q 2
We know p 2 p 2 is multiple of 3. What does that tell us about p p?
A whole number n n has a remainder of 0, 1 or 2 when divided by 3, so either n=3 k,n=3 k+1 n=3 k,n=3 k+1 or n=3 k−1 n=3 k−1 for some integer k k (I changed 2 to -1 to make the math easier -- you could work it with 2).
So the possibilities for n 2 n 2 are (3 k)2=9 k 2(3 k)2=9 k 2, (3 k+1)2=9 k 2+6 k+1(3 k+1)2=9 k 2+6 k+1 or (3 k−1)2=9 k 2−6 k+1(3 k−1)2=9 k 2−6 k+1. Only the first one is a multiple of three, the others giving remainder 1 when divided by 3. So we've proved a lemma:
Lemma: If n 2 n 2 is a multiple of 3, n n must be a multiple of 3.
From our lemma, we can write p=3 k p=3 k for some whole number k k.
3 q 2=p 2=(3 k)2=9 k 2 3 q 2=p 2=(3 k)2=9 k 2
q 2=3 k 2 q 2=3 k 2
Applying our lemma again, we find q q is also a multiple of 3. Thus our claim that p p and q q had no common factors is false. We've found a contradiction, so our original assumption of rationality is false. Therefore:
√3 3 is irrational.
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To prove that √3 3 is irrational, we can use a proof by contradiction. Here’s how the argument goes:
Assume the opposite: Suppose √3 3 is rational. This means that it can be expressed as a fraction of two integers a a and b b (where a a and b b have no common factors other than 1 and b≠0 b≠0):
√3=a b 3=a b
2. Square both sides: Squaring both sides of the equation gives:
3=a 2 b 2 3=a 2 b 2
Rearranging this leads to:
a 2=3 b 2 a 2=3 b 2
3. Analyze the implications: From the equation a 2=3 b 2 a 2=3 b 2, we can see that a 2 a 2 is a multiple of 3. This implies that a a itself must also be a multiple of 3
Continue Reading
To prove that √3 3 is irrational, we can use a proof by contradiction. Here’s how the argument goes:
Assume the opposite: Suppose √3 3 is rational. This means that it can be expressed as a fraction of two integers a a and b b (where a a and b b have no common factors other than 1 and b≠0 b≠0):
√3=a b 3=a b
2. Square both sides: Squaring both sides of the equation gives:
3=a 2 b 2 3=a 2 b 2
Rearranging this leads to:
a 2=3 b 2 a 2=3 b 2
3. Analyze the implications: From the equation a 2=3 b 2 a 2=3 b 2, we can see that a 2 a 2 is a multiple of 3. This implies that a a itself must also be a multiple of 3 (since the square of a non-multiple of 3 cannot be a multiple of 3). Therefore, we can express a a as:
a=3 k a=3 k
for some integer k k.
4. Substitute back: Substituting a=3 k a=3 k back into the equation a 2=3 b 2 a 2=3 b 2 gives:
(3 k)2=3 b 2(3 k)2=3 b 2
Simplifying this results in:
9 k 2=3 b 2⇒3 k 2=b 2 9 k 2=3 b 2⇒3 k 2=b 2
5. Conclusion about b b: From 3 k 2=b 2 3 k 2=b 2, we see that b 2 b 2 is also a multiple of 3, which implies that b b must also be a multiple of 3.
6. Contradiction: At this point, we have shown that both a a and b b are multiples of 3, which contradicts our original assumption that a a and b b have no common factors other than 1 (since they both have at least 3 as a common factor).
Since our assumption that √3 3 is rational leads to a contradiction, we conclude that √3 3 must be irrational.
Upvote ·
Ege Erdil
Very interested in mathematics · Upvoted by
Horst H. von Brand
, PhD Computer Science & Mathematics, Louisiana State University (1987) · Author has 273 answers and 535.5K answer views
·9y
Originally Answered: How do you prove that the square root of 3 is an irrational number? ·
Generally speaking, if q∈Z q∈Z is not a perfect nth power, id est if there are no integers a,n a,n such that q=a n q=a n, then the number q 1/n q 1/n is irrational. I appeal to a powerful criterion about the rational roots of a polynomial for the proof:
Theorem. (Rational root test / Z Z is integrally closed) Let P(x)=∑i≤n a i x i P(x)=∑i≤n a i x i be a nonconstant polynomial with unit leading coefficient and integer coefficients. Then, any q∈Q q∈Q such that P(q)=0 P(q)=0 is an integer and divides a 0 a 0.
Proof. The proof is very direct. Assume that q=u/v q=u/v is in its primitive form, id est gcd(u gcd(u
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Generally speaking, if q∈Z q∈Z is not a perfect nth power, id est if there are no integers a,n a,n such that q=a n q=a n, then the number q 1/n q 1/n is irrational. I appeal to a powerful criterion about the rational roots of a polynomial for the proof:
Theorem. (Rational root test / Z Z is integrally closed) Let P(x)=∑i≤n a i x i P(x)=∑i≤n a i x i be a nonconstant polynomial with unit leading coefficient and integer coefficients. Then, any q∈Q q∈Q such that P(q)=0 P(q)=0 is an integer and divides a 0 a 0.
Proof. The proof is very direct. Assume that q=u/v q=u/v is in its primitive form, id est gcd(u,v)=1 gcd(u,v)=1. Then, by direct substitution into the polynomial, we have
∑i≤n a i(u v)i=0∑i≤n a i(u v)i=0
or, multiplying through by v n v n,
u n+a n−1 u n−1 v+…+a 0 v n=0 u n+a n−1 u n−1 v+…+a 0 v n=0
Every term in the equality except u n u n is divisible by v v, therefore so is u n u n. However, this contradicts the coprimality of u u and v v unless v=1 v=1, as by unique factorization they must share at least one prime factor otherwise. Therefore, q=u q=u is an integer. It is obvious that it must divide a 0 a 0.
QED.
Now, consider the polynomial equation x n−q=0 x n−q=0. By the above theorem, if any rational roots exist at all, they must be integers. However, as we assumed q q is not a perfect nth power, we know that no such integers exist. Therefore, this polynomial cannot have rational roots. As q 1/n q 1/n is a root, it cannot be rational.
Now, simply note that 3 3 is not the square of any integer, and we are done.
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Ryan Gossiaux
Studied Mathematics&Computer Science at University of Oxford (Graduated 2015) · Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) and
Justin Rising
, PhD in statistics · Author has 143 answers and 2M answer views
·10y
Originally Answered: How can we prove that the squareroot of 3 is irrational? ·
I'm fond of this one-liner: x 2−3 x 2−3 is irreducible over Q Q by Eisenstein's criterion, so in particular it has no root in Q Q.
I think it's a worse proof than David Joyce's, which makes it a lot clearer why this is true. But it's hard to resist the temptation of cute one-liners.
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Benjamin Canals
PhD in theoretical physics, senior scientist, perpetual student · Author has 257 answers and 412.1K answer views
·10y
Originally Answered: How can we prove that the squareroot of 3 is irrational? ·
I take n a positive integer. If √n n is an integer, it is rational. If not, I assume √n n belongs to Q Q. I build the set P = {p∈N p∈N such that p√n∈N p n∈N}. Then P is not empty given my assumption. This set is a subset of N N, so it has a smallest element p 0 p 0.
Let’s consider the new element p 1=p 0√n−p 0∗E(√n)p 1=p 0 n−p 0∗E(n) with E(√n)E(n) the larger integer smaller than √n n. Then, p 1√n=p 0 n−p 0 E(√n)√n p 1 n=p 0 n−p 0 E(n)n is also an integer.
Therefore, p 1 p 1 belongs to P. Because p 1<p 0 p 1<p 0, we reach a contradiction by definition o
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I take n a positive integer. If √n n is an integer, it is rational. If not, I assume √n n belongs to Q Q. I build the set P = {p∈N p∈N such that p√n∈N p n∈N}. Then P is not empty given my assumption. This set is a subset of N N, so it has a smallest element p 0 p 0.
Let’s consider the new element p 1=p 0√n−p 0∗E(√n)p 1=p 0 n−p 0∗E(n) with E(√n)E(n) the larger integer smaller than √n n. Then, p 1√n=p 0 n−p 0 E(√n)√n p 1 n=p 0 n−p 0 E(n)n is also an integer.
Therefore, p 1 p 1 belongs to P. Because p 1<p 0 p 1<p 0, we reach a contradiction by definition of p 0 p 0.
So my first assumption is proven false, and as soon as √n n is not an integer, it is irrational. In particular, it works for n=3.
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Ravi Sharma
Former Group A Officer From Indian Railways (1973–2009) · Author has 14.6K answers and 3.6M answer views
·May 2
Originally Answered: How can you prove that √3 is an irrational number? ·
WHAT IS AN IRRATIONAL NUMBER?
A NUMBER WHICH CAN NOT BE PUT IN THE FORM OF P/Q IS AN IRRATIONAL NUMBER. P & Q HAVE TO BE COPRIMES.
SUPPOSE THAT ✓3 IS A RATIONAL NUMBER. HENCE IT CAN BE PUT IN THE FORM OF P/Q, WHERE P AND Q ARE COPRIMES.
LET P/Q= ✓3
OR P²/Q²= 3, P²= 3Q²
P² IS MULTIPLE OF 3, HENCE P IS ALSO MULTIPLE OF 3
LET P= 3M
(3M)²= 3Q², Q²= 3M²
Q² IS MULTIPLE OF 3, HENCE Q IS ALSO MULTIPLE OF 3.
P IS MULTIPLE OF 3, Q IS ALSO MULTIPLE OF 3.
P AND Q ARE NOT COPRIMES.
✓3 CAN NOT BE PUT IN THE FORM OF P/Q, WHERE P AND Q ARE COPRIMES.
OUR SUPPOSITION THAT ✓ 3 IS RATIONAL NUMBER IS WRONG.
✓3 IS AN IRRATIONAL
Continue Reading
WHAT IS AN IRRATIONAL NUMBER?
A NUMBER WHICH CAN NOT BE PUT IN THE FORM OF P/Q IS AN IRRATIONAL NUMBER. P & Q HAVE TO BE COPRIMES.
SUPPOSE THAT ✓3 IS A RATIONAL NUMBER. HENCE IT CAN BE PUT IN THE FORM OF P/Q, WHERE P AND Q ARE COPRIMES.
LET P/Q= ✓3
OR P²/Q²= 3, P²= 3Q²
P² IS MULTIPLE OF 3, HENCE P IS ALSO MULTIPLE OF 3
LET P= 3M
(3M)²= 3Q², Q²= 3M²
Q² IS MULTIPLE OF 3, HENCE Q IS ALSO MULTIPLE OF 3.
P IS MULTIPLE OF 3, Q IS ALSO MULTIPLE OF 3.
P AND Q ARE NOT COPRIMES.
✓3 CAN NOT BE PUT IN THE FORM OF P/Q, WHERE P AND Q ARE COPRIMES.
OUR SUPPOSITION THAT ✓ 3 IS RATIONAL NUMBER IS WRONG.
✓3 IS AN IRRATIONAL NUMBER
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Here’s what nobody tells you about owning a dog (and why I send all my fellow pet owners to Lemonade for insurance):
The most expensive part usually isn’t the food, the toys, or the training. It’s the moment something unexpected happens at 2am — and you're staring down a $5,000 vet bill that you didn’t plan for.
That moment happened to my friend when his pup started having seizures. The diagnostics alone were $2,400. Add meds, consults, and overnight monitoring… it spiraled fast.
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Edward James
M.S. in Mathematics (college major), Portland State University (Graduated 2020) · Upvoted by
Joe Schlessinger
, MA Mathematics, University of California, Berkeley · Author has 189 answers and 480.6K answer views
·7y
Related
Is there any simple way to prove root 2 is irrational?
I wouldn’t use the word simple, because it might not be simple for someone with zero experience in proof writing, but there is a very common proof. It goes like this:
Theorem: √2 2 is irrational.
Proof:
Assume, by way of contradiction, that √2 2 is rational.
Then there exist integers p p and q q such that √2=p q 2=p q, where p p and q q do not share any factors. We know that these integers exist because any rational can be represented by a ratio of integers that are coprime (do not share factors).
Squaring both sides gives us 2=p 2 q 2 2=p 2 q 2 then multiplying by q 2 q 2 gives us 2 q 2=p 2 2 q 2=p 2.
That
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I wouldn’t use the word simple, because it might not be simple for someone with zero experience in proof writing, but there is a very common proof. It goes like this:
Theorem: √2 2 is irrational.
Proof:
Assume, by way of contradiction, that √2 2 is rational.
Then there exist integers p p and q q such that √2=p q 2=p q, where p p and q q do not share any factors. We know that these integers exist because any rational can be represented by a ratio of integers that are coprime (do not share factors).
Squaring both sides gives us 2=p 2 q 2 2=p 2 q 2 then multiplying by q 2 q 2 gives us 2 q 2=p 2 2 q 2=p 2.
That last equation shows us that p 2 p 2 is even and that also makes p p an even number. So, there exist an integer k k such that p=2 k p=2 k. If we substitute that back into the last equation, we get 2 q 2=4 k 2 2 q 2=4 k 2 and then divide by 2 2 and you have q 2=2 k 2 q 2=2 k 2.
We have now shown that q 2 q 2 is even and that means q q is even. This means that both p p and q q can be divided by 2 2, but this contradicts the fact that they don’t share a factor.
This means that √2 2 can not be represented by the ratio of integers. Therefore, our assumption that it is rational is false.
QED
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Michel Paul
an altitude to a unit hypotenuse divides it into sin^2+cos^2 · Author has 839 answers and 921.5K answer views
·3y
Originally Answered: How do you prove that √3 is an irrational number? ·
Suppose that √3 3 is rational, and see what happens.
Suppose that √3=p q 3=p q, where p p and q q are relatively prime integers.
It follows that 3 q 2=p 2 3 q 2=p 2, so p 2 p 2 is divisible by 3 3.
This means that p p is also divisible by 3 3, as the prime factorization of a squared integer must contain two of each prime factor.
Therefore, p 2=(3 r)2 p 2=(3 r)2 for some integer r=1 3 p r=1 3 p.
This means that 3 q 2=9 r 2 3 q 2=9 r 2, or q 2=3 r 2 q 2=3 r 2, and that means q 2 q 2 is also divisible by 3 3.
This leads to an infinite regress and contradicts our initial supposition that p p and q q share no common factors.
Note that the same reasonin
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Suppose that √3 3 is rational, and see what happens.
Suppose that √3=p q 3=p q, where p p and q q are relatively prime integers.
It follows that 3 q 2=p 2 3 q 2=p 2, so p 2 p 2 is divisible by 3 3.
This means that p p is also divisible by 3 3, as the prime factorization of a squared integer must contain two of each prime factor.
Therefore, p 2=(3 r)2 p 2=(3 r)2 for some integer r=1 3 p r=1 3 p.
This means that 3 q 2=9 r 2 3 q 2=9 r 2, or q 2=3 r 2 q 2=3 r 2, and that means q 2 q 2 is also divisible by 3 3.
This leads to an infinite regress and contradicts our initial supposition that p p and q q share no common factors.
Note that the same reasoning applies for √any prime number any prime number.
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9 2
Trần Quốc Anh
Works at Diễn Đàn MathScope
·9y
Originally Answered: What is the proof that the square root of 3 is irrational? ·
Suppose that √3=p q 3=p q, where p,q p,q are integers, then we have 3 q 2=p 2 3 q 2=p 2.
This is wrong because the left side has odd number of prime factors, while the right hand has an even number of prime factors in its unique factorization into primes.
This proof can be generalized to replace 3 with any non-square number.
Let n n be an integer. We have the unique factorization of n=p a 1 1 p a 2 2...p a m m n=p 1 a 1 p 2 a 2...p m a m, where p i p i's are different primes.
Then n 2=p 2 a 1 1 p 2 a 2 2...p 2 a m m n 2=p 1 2 a 1 p 2 2 a 2...p m 2 a m
So n 2 n 2 has 2(a 1+a 2+...+a m)2(a 1+a 2+...+a m), which is an even number, prime factors.
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9 3
9 5
Bharath Keshav
Talent Acquisition @ Directi · Author has 97 answers and 566K answer views
·9y
Related
How can we prove that the square root of 2 is irrational?
The easiest way is to try and see if √2 can be represented in a/b form where a and b are whole numbers and b not equal to zero. Here in case of √2 it cannot be represented in that form. Below is the detailed explanation
Let's suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero.
We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplifya/b furthe
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The easiest way is to try and see if √2 can be represented in a/b form where a and b are whole numbers and b not equal to zero. Here in case of √2 it cannot be represented in that form. Below is the detailed explanation
Let's suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero.
We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplifya/b further.
From the equality √2 = a/b it follows that 2 =a2/b2, or a2 = 2 · b2. So the square of a is an even number since it is two times something.
From this we know that a itself is also an even number. Why? Because it can't be odd; ifa itself was odd, then a · a would be odd too. Odd number times odd number is always odd. Check it if you don't believe me!
Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols,a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.
If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:
2=(2k)2/b22=4k2/b22b2=4k2b2=2k2
This means that b2 is even, from which follows again that b itself is even. And that is a contradiction!!!
WHY is that a contradiction? Because we started the whole process assuming that a/bwas simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.
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831 | https://physics.howard.edu/sites/physics.coas.howard.edu/files/2019-07/12-resonance.pdf | General Physics Lab Handbook by D.D.Venable, A.P.Batra, T.H¨ ubsch, D.Walton & M.Kamal Resonance of Air Columns 1. Theory Imagine two waves, say simple harmonic motion (SHM, i.e., sine/cosine) waves of identical wavelength and amplitude traveling in opposite directions with equal speeds.
The net displacement of the medium at any point and at any time is determined by applying the superposition principle which states that the net displacement is given by the algebraic sum of the two individual displacements. The resulting wave pattern will then have points, separated by one-half wavelength, where the displacement is always zero. These points are called nodes.
Midway between this nodes, the particles of the medium located at the antinodes, vibrate with maximum displacement.
N A N N A N A N N A N A N A N Fig. 1: The three lowest frequency for transverse vibrations of a string clamped at the ends.
We can visualize transverse standing waves on a string, of length L, fixed at both ends. These waves can be established by plucking the string at some point and are caused by continual reflection of the traveling waves at the boundaries, in this case the two fixed ends. The boundary conditions demand that at each end there must be a node. We can therefore fit an integral number of half-wavelengths into the length L of the string as shown in Fig. 1.
Even though the wave shape is not moving, we can associate a velocity with the standing wave which is the same as that of the traveling wave in the same medium. It can be deduced from Fig. 1, that the wavelengths of the traveling waves that combine to give the standing waves are given by L = n(λn) 2 so that λn = 2L n , where n is an integer.
The corresponding frequencies are therefore given by fn = v λn = nv 2L = nf0 with f0 = v 2L, where v is the speed of transverse traveling waves on the string.
The frequencies are known as resonance frequencies of this system. The lowest fre-quency of the system, having the longest wavelength, is called the fundamental frequency and the mode of vibration as the fundamental mode or the first harmonic. The modes of Resonance of Air Columns: page 1 General Physics Lab Handbook by D.D.Venable, A.P.Batra, T.H¨ ubsch, D.Walton & M.Kamal vibration with progressively higher frequencies, are called second harmonic (n = 2), third harmonic (n = 3), etc.
Longitudinal standing waves of air columns can also be set up in closed (open at one end, closed at the other) and in open (open at both ends) organ pipes. A longitudinal wave, such as a sound wave, is a wave of density variation. The fixed or closed and of the pipe cannot have any longitudinal motion and is therefore a node of the density wave.
The free or open end of the pipe is a position of maximum longitudinal displacement and is therefore an antinode of the density wave.
N A N A N A N A N A N A Fig. 2: The longitudinal vibrations of air in a pipe, closed at the left and open at the right.
In a closed pipe of length L the boundary conditions require that the closed end be a node and the open end an antinode. We can therefore fit an odd multiple of quarter-wavelengths into the length L of the pipe as shown in Fig. 2.
The transverse wave pattern for the standing wave for the standing waves are displayed in this figure. The wavelengths of the traveling waves that combine to give the standing waves are given by L = nλ 4 , i.e., λ = 4L n where n = 1, 3, 5, ... The corresponding resonance frequencies are therefore given by fn = v λn = nv 4L = nf0 with f0 = v 4L, where v is the speed of longitudinal traveling waves(sound) in air. It is to be noted that the closed organ pipe will support only odd harmonics. The frequency of the first harmonic or the fundamental frequency of equals ( v 4L).
For an organ pipe in which both ends of the pipe are open the situation is the same as for a string with its two ends fixed except that both the ends are now antinodes for any standing harmonic wave. The wave patterns therefore will yield the resonance frequencies as fn = n v 2L = nf0 with f0 = v 2L. It is seen that the open organ pipe will support all harmonics.
For a given frequency, the first resonance corresponds to the length of the closed-end air column equal to λ 4 . The second resonance corresponds to 3λ 4 . The difference between these positions would correspond to half a wave length. The sound velocity is determined from v = fλ where f is the tuning fork frequency in Hertz. At room temperature T (in degrees of Celsius), the theoretical value of sound velocity (in m/s) is v = 331+0.61T.
Resonance of Air Columns: page 2 General Physics Lab Handbook by D.D.Venable, A.P.Batra, T.H¨ ubsch, D.Walton & M.Kamal First Resonance Second Resonance λ 4 3 λ 4 Water Water Fig. 3: The longitudinal vibrations of an air column, in a pipe with water at the bottom.
2. Experiment Object: To determine the speed of sound in air using resonance of a close-end air column.
Apparatus: Resonance tube apparatus, tuning fork, rubber mallet and a Macintosh Com-puter.
Procedure: 1. Strike a tuning fork of frequency 512Hz with a rubber mallet and hold it at about an inch above the open end of the resonance tube with its prongs horizontal. Adjust the water level starting from its highest level. Gradually increase the length of the air column by lowering the can to find the first position of resonance, where the sound coming out of the air column is loudest. You may have to strike the fork several times and move the water column up and down to precisely locate the resonance position.
3. Continue this procedure to find second (and if possible, the third) position of reso-nance. Record these lengths as l1 and l2.
4. Repeat the experiment with a tuning fork of different frequency.
Calculations: Using the computer spreadsheet perform the following calculation.
1. Calculate the speed of sound from the formula v = λf = 2(l2 −l1)f, where f is the frequency of the tuning fork.
2. If you get only the first resonance, and not the second resonance then, to calculate the speed of the sound, use v = 4fl1.
3. Compare the calculated speed of sound with the theoretical value from the formula v = 331 + 0.61T, where T is the room temperature in Celsius degrees.
Questions: 1. Sketch the air column vibrations for third resonance. What is the difference between the positions of the third and the first resonance in terms of wave length?
2. Should the velocity of sound in air depend upon the frequency of the tuning fork?
Resonance of Air Columns: page 3 General Physics Lab Handbook by D.D.Venable, A.P.Batra, T.H¨ ubsch, D.Walton & M.Kamal 3. Are there “anti-resonances” where the sound coming out of the air column reaches minimum? What is the length of air columns for these “anti-resonances”?
4. Why does the resonance position correspond to the “loudest sound”?
5. Do you see any advantage of v = f(l2 −l1) over v = 4fl1?
Resonance of Air Columns: page 4 |
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This article is part of the Research TopicReview of Hyperbaric Therapy & Hyperbaric Oxygen Therapy in the Treatment of Neurological Disorders According to Dose of Pressure and HyperoxiaView all 14 articles
Hyperbaric oxygen therapy for thalamic pain syndrome: case report
John Benjamin SladeNathan KwanPeter LennoxRussell Gray
David Grant USAF Medical Center, Travis Air Force Base, Fairfield, CA, United States
Thalamic pain syndrome is a distressing type of central post-stroke pain (CPSP) that occurs in up to 10% of cases following a cerebrovascular accident, typically with a delayed onset of signs and symptoms, and is often chronic or even life-long. Thalamic pain syndrome, as is the case for other CPSPs, is difficult to treat, and the response is typically moderate at best. Central pain also occurs after vascular insults in parts of the CNS other than the thalamus. Only a few patients present with the classic “Dejerine and Roussy syndrome,” so the term CPSP is preferred for describing neuropathic pain after stroke. There are no pathognomonic features of this syndrome. The thalamus probably has a substantial role in some patients with central pain, either as a pain generator or by abnormal processing of ascending input. Long-term post-stroke pain disorders can reduce the quality of life, affect mood, sleep, and social functioning, and can lead to suicide. Hemi-body pain is common in patients with thalamic lesions. Hyperbaric oxygen has known physiologic and pharmacologic effects with documented benefits in brain-related hemorrhages, acute and chronic stroke, traumatic brain injury, mild cognitive impairment, neurodegenerative diseases, and neuroprotection, but has never been reported as a treatment for thalamic pain syndrome. A 55-year-old man with a history of migraines suffered a right thalamic lacunar infarction following a brain angiogram to investigate a suspected AVM found on prior imaging that resulted in immediate left-sided weakness and numbness, evolving to severe chronic pain and subsequent stiffness. Diagnosed with thalamic pain syndrome, multiple pharmacologic therapies provided only partial relief for a year after the stroke. The patient’s symptoms resolved and quality of life markedly improved with hyperbaric oxygen therapy, as assessed by multiple validated questionnaires, thus it may be a treatment option for thalamic pain syndrome.
Introduction
This is the first known case report of the use of hyperbaric oxygen (HBO2) for the treatment of thalamic pain syndrome, also known as Dejerine–Roussy syndrome, a type of central post-stroke pain (CPSP), also known as post-stroke thalamic pain (PS-TP) (1).
Thalamic pain syndrome is a debilitating disease process that has limited treatment options.
A pure sensory stroke is a well-defined clinical entity with predominant hemisensory symptoms without other major neurological signs (2). The prognosis is typically poor. The character and severity of the pain will be persistent and often unchanging.
The prevalence of thalamic pain syndrome following a stroke is up to 10% of cases, and the onset of symptoms is often delayed, with the patient not experiencing significant pain until months or years after the stroke.
There are multiple reports of treatments for thalamic pain syndrome, including tricyclic antidepressants, anticonvulsants and opioid analgesics, intravenous morphine, lidocaine, and propofol (3), naloxone infusions (4), and Cilostazol (5). Proposed non-pharmacological treatments include neurostimulation (motor cortex, deep brain, and transcranial magnetic stimulation) for treatment-resistant cases of CPSP (3, 5), mirror therapy (6), ipsilateral stellate ganglion block (7), botulinum neurotoxin A injection (8), precentral gyrus stimulation, caloric vestibular stimulation, transcranial direct current stimulation, and bee venom acupuncture point injection (9).
HBO2 has been reported to be an effective treatment for multiple brain-related diagnoses, including TBI, acute stroke, neurodegenerative disease, and cognitive impairment. Proposed mechanisms of action include amelioration of oxidative stress, activating endogenous antioxidant activity, modulation of neuroinflammation, inhibiting apoptosis, stimulating pathways of neuroprotection, and modulating cerebral blood flow and brain metabolism (10).
In a randomized, prospective trial, the authors conclude that HBO2 during the degenerative (or acute post-stroke) stage could increase post-injury damage but that elevated oxygen during the regenerative stage would supply the energy needed for the innate brain repair processes. The study convincingly demonstrates that HBO2 can induce significant neurological improvement and that neuroplasticity can be operatively activated in chronic late-stage stroke patients long after the acute brain insult (11).
Case description
This is the case of a 55-year-old man with thalamic pain syndrome. His past medical history includes asthma, CAD, diabetes type 2, hyperlipidemia, hypertension, and migraines. His past surgeries include gastric bypass, lumbar, and cervical fusions. A twin brother died in 2016 with end-stage renal disease and multiple strokes.
An MRI performed to evaluate migraines was significant for a focal prominence of the anterior portion of the inferior sagittal sinus, thought to represent a venous varix or focal tortuosity; it did not show arteriovenous shunting. The patient was referred to interventional radiology for further evaluation, with a cerebral angiogram conducted on 30 December 2021. The patient suffered an acute post-procedural thromboembolic right thalamic lacunar infarction, resulting in left-sided weakness and numbness including grip and foot drop. The patient was hospitalized and treated appropriately for an acute thromboembolic stroke with IV tenecteplase (TNK). Echo was normal without PFO, LDL 73, and A1c 5.8. The patient was placed on aspirin 325. A brain CT on 3 January 22 showed a 1 cm right thalamus lesion consistent with an acute infarct. Subsequent MRIs showed multifocal posterior circulation strokes, including the right MCA territory, notably the right thalamus (Figure 1).
Figure 1
Figure 1. MRI January 2022 showing right thalamic lacunar infarction.
The patient’s grip weakness and foot drop resolved completely in the subacute post-stroke stage, but numbness persisted. The patient developed severe left-sided burning hyperalgesia with an aftersensation described as a painful left hemi-body stiffness, which worsened 9 months post-cerebral vascular accident (post-CVA). In the ensuing year after the CVA, the patient experienced chronic, recurrent, frequent symptoms of intense acute left anterior chest burning pain that variably included the left face, followed predictably by a left hemi-body tightness/stiffness sensation that initially occurred 8–10 times/month, lasting an average of 2–3 days, rated 10/10 on the pain scale. Pain was paroxysmal burning or tingling, sometimes from the left ear distally.
The patient was evaluated in the ED over 10 times during that year for acute episodes and diagnosed with thalamic pain syndrome. Acute coronary syndrome was considered and ruled out.
Medical therapy included an anticonvulsant (gabapentin 300–600 gm t.i.d.), that variably provided partial relief approximately 1 h after administration, and a serotonin–noradrenaline reuptake inhibitor (SNRI, duloxetine 30–60 mg daily). The patient was initially treated with muscle relaxants, Methocarbamol, then baclofen (dose), and later cyclobenzaprine 10 mg t.i.d. PRN, all of which were subsequently discontinued due to marginal reported benefit. Pain medications included Norco 2/325 q6h PRN, Ibuprofen 600 mg q.i.d., PRN, and Tylenol. The patient reported symptomatic improvement with two short courses of Valium given during emergency room visits but was never prescribed for maintenance therapy.
The patient was managed by neurology as an inpatient, but not evaluated by outpatient neurology until 1 year post-CVA (Tables 1, 2).
Table 1
Table 1. Outcome measurements and assessments.
Table 2
Table 2. Timeline with relevant data from the episode of care.
Diagnostic assessment
Criteria for the diagnosis of CPSP include pain in an area of the body corresponding to the CNS lesion, a history suggestive of stroke and pain onset at or after stroke onset, confirmation of the CNS lesion by imaging, and other causes of pain excluded or considered highly unlikely (3).
Diagnosis of PS-TP based on contralateral burning pain and spasticity after an imaging-demonstrated right thalamic ischemic stroke lesion that occurred at the time of a cerebral angiogram performed for a migraine workup, with associated showering of emboli.
One year post-CVA, the patient’s symptoms remained unabated, with partial, temporary relief from gabapentin 300–600 mg TID, duloxetine 60 mg daily, a muscle relaxant (baclofen, then Flexeril 10 mg TID PRN), and Tylenol. At the initiation of HBO2 treatments, the patient described at least daily pain episodes lasting approximately 1 h that began with a burning sensation in the left anterior chest, progressing to cramp-like pain involving his entire left side. The pain was consistently triggered by airplane flights, with no other obvious aggravating or relieving factors. It variably kept the patient awake at night. HBO2 treatments were administered at 2 atmospheres absolute (ATA) for 90 min each, the protocol routinely used at our multiplace hyperbaric chamber facility for non-emergent diagnoses and used by our group in prior research protocols for the treatment of traumatic brain injury. The initial plan was for 40 treatments based on the report in post-stroke patients to be effective for inducing late neuroplasticity (11) and that showed improvements in NIHSS, activities of daily living (ADL), and quality of life (QoL). However, due to family circumstances and personal obligations, the patient had to take three breaks in the therapy (41, 90, and 28 days), resulting in a treatment course of 100 HBO2 sessions over 11+ months, extended primarily due to ongoing progressive improvement.
The patient continued gabapentin 300–600 mg TID throughout the course of HBO2, required less Tylenol (initially multiple times daily to several times/week), replaced baclofen with Flexeril, and eventually stopped both.
After the initial 10 HBO2 treatments, the patient noted significantly reduced pain during the acute episodes, rated at 4–6/10 compared to his baseline of 10/10. The patient also noted the reduced duration of the acute episodes, lasting hours versus days, and decreased intensity of the left arm and leg stiffness that followed each pain episode. The frequency decreased to only two episodes in the first 3 weeks of HBO2 therapy. After 33 treatments, there was a 41-day gap in therapy due to a planned trip. On resuming HBO2, there was a slight loss of prior improvements, and at treatment #34, the patient had one of his worst pain flare episodes with severe left hemi-body pain and usual post-pain stiffness. The patient continued taking gabapentin 300 mg TID (earlier was 600 mg TID), rare use of Tylenol (earlier was 325 mg TID), duloxetine 30–60 mg/day, and Flexeril.
After 37 treatments, the patient was no longer having acute exacerbations and was only experiencing baseline hemi-body stiffness with no pain. The patient continued with gabapentin and duloxetine with the rare use of Tylenol.
After 56 treatments, there was a 90-day break in HBO2 therapy for another trip. On return, there was again a slight loss of prior improvements, and the decision was made to continue with a total of 80 treatments, at which time the patient had planned a trip lasting 28 days. Notably, the patient consistently had minor flares in symptoms for 2–3 days after airline flights. After the 28-day break, the patient felt that he was continuing to improve, and was given a final of 20 HBO2 sessions, for a total of 100 sessions over a period of 11 months. At the end of the 100 treatments, the patient was completely asymptomatic. He continued with gabapentin at 300 mg TID and duloxetine at 30 mg/day.
Considering that the patient was seen multiple times by various providers during the 11 months between the CVA and HBO2 and had stable, severe symptoms (no significant or sustained improvement), he was encouraged to maintain a stable medication regimen to best assess his response to HBO2 therapy.
Discussion
Thalamic pain was first described by Dejerine and Roussy as, “… severe, persistent, paroxysmal, often intolerable, pains on the hemiplegic side, not yielding to any analgesic treatment” (12). Thalamic pain is characterized by constant or intermittent pain in the hemi-body, contralateral to the thalamic lesion, and is a severe and treatment-resistant type of CPSP that occurs after a cerebrovascular lesion results in thalamic stroke.
Treatment remains difficult, and the prognosis is typically poor. Possible pathogenesis includes central imbalance, central disinhibition, central sensitization, other thalamic adaptive changes, and local inflammatory responses (1).
PS-TP may develop immediately after a stroke, as in this case report. Up to 40% occur within a month of the stroke, over 40% have symptoms between 1 and 12 months, and sometimes develop 1 to 6 years post-stroke (2, 13, 14).
Right-sided stroke lesions were more commonly associated with the development of PS-TP than left-sided lesions (15).
In a series of patients with thalamic infarcts, only lesions in the ventral posterior part of the thalamus caused CPSP (16).
CPSP can develop after both hemorrhagic and ischemic lesions of the CNS. A lesion in the CNS results in anatomical, neurochemical, excitotoxic, and inflammatory changes, all of which might trigger an increase in neuronal excitability (17) that might lead to chronic pain.
Treatments for thalamic pain syndrome
Treatment of CPSP is usually a combination of several drugs. Commonly recommended first and second lines of pharmacological therapies include traditional medications, e.g., tricyclic antidepressants, SSRIs, SNRIs, anticonvulsants, and opioid analgesics. Anticonvulsants have several mechanisms of action, including reducing neuronal hyperexcitability (3).
Non-pharmacological interventions, such as transcranial magnetic or direct current brain stimulations, vestibular caloric stimulation, epidural motor cortex stimulation (MCS), and deep brain stimulation (DBS), were effective in some case studies and can be recommended in the management of therapy-resistant PS-TP (1). Intracranial neurostimulation for pain relief is most frequently delivered by stimulating the motor cortex, the sensory thalamus, or the periaqueductal and periventricular gray matter. The stimulation of these sites through MCS and DBS has proven effective for treating a number of neuropathic and nociceptive pain states that are not responsive or amenable to other therapies or types of neurostimulation (18).
Recent data suggest that the therapy may have value for treating post-stroke pain, central pain syndromes, and peripheral deafferentation pain (19, 20). An “insertion effect” has been reported, in which the placement of DBS leads provides pain relief even when the leads are not activated (21).
Cilostazol, a selective and potent inhibitor of phosphodiesterase (PDE) 3A, likely increases the level of cAMP and has been reported to be effective (5).
Mexiletine, an orally active antiarrhythmic agent, produced improvement in eight of nine thalamic pain syndrome patients over 4 weeks (22).
In a case report, multidrug-resistant thalamic pain was successfully alleviated by modulating the contralateral thalamic pathway with ipsilateral stellate ganglion block (7).
A patient who developed CPSP 5 years after a right lenticular-capsular thalamic stroke was treated with 2 weeks of mirror therapy with a reduction in pain that was maintained at the 1-year follow-up (6).
Two cases of thalamic syndrome were treated with weekly, increasing doses of naloxone infusions successfully relieved pain for up to 6 months (4).
There is a growing body of evidence showing benefits with HBO2 in post-stroke patients, including post-stroke memory impairment (23), cognitive impairment (24), traumatic brain injury (25, 26), and chronic neuropathic pain (27).
A prospective, randomized, controlled trial of 74 patients, all of whom suffered a stroke 6–36 months prior to treatment with HBO2. The treatment protocol was 40 sessions (5 days/week) at 2 ATA. The authors concluded that HBO2 can lead to significant neurological improvements in stroke patients, even in chronic late stages. Improvements were measured by neurologic evaluation with NIHSS, brain functional imaging (SPECT), and QoL evaluations. The improvements support the idea that neuroplasticity can be activated months to years after stroke with proper stimulation, such as HBO2. This and other studies reveal that many aspects of the brain remain plastic even in adulthood (11). HBO2 can initiate cellular and vascular repair mechanisms and improve cerebral vascular flow. At the cellular level, HBO2 can improve neuronal and glial cell mitochondrial function, improve blood–brain barrier and inflammatory reactions, reduce apoptosis, alleviate oxidative stress, increase levels of neurotrophins, and nitric oxide, and upregulate axon guidance agents. HBO2 may promote the neurogenesis of endogenous neural stem cells (28).
In a review of four studies, the authors concluded that HBO2 is effective for ischemic strokes (26).
HBO2 can exert neuroprotective effects through multiple pathways in acute ischemic stroke (AIS), with consequent salvage of neurological function.
The three phases of potential HBO2 benefit in ischemic stroke are pretreatment, early (limited to 12 h from the time of onset of AIS), and recovery from neurological damage in the chronic phase (29).
In a large retrospective 2020 study including chronic stage post-stroke patients at least 3 months post-injury (median 1.5 ± 3.3-year post-stroke), HBO2 was found to induce significant improvements in all cognitive function domains. In 50 patients (30.86%), the stroke was subcortical. There were no significant differences in HBO2 effects on subcortical strokes compared to cortical strokes (30).
In a clinical study, six patients 1–2 years post-stroke with a stable baseline were treated with HBO2 at 2 ATA X for 60 min for a total of 40 treatments over a 12-week period. The authors reported improvements in cognition, gait velocity, upper extremity mobility, sleep, and overall recovery that were maintained for up to 3 months (31).
A review of HBO2 in acute stroke discussed physiological effects that include ameliorating AIS-induced brain tissue hypoxia, stabilizing the blood–brain barrier, decreasing intracranial pressure, and relieving cerebral edema (32).
HBO2 alleviates neuroinflammation through effects that include reduced inflammatory enzymes and inhibition of neutrophil infiltration and matrix metalloproteinases (MMP-9).
HBO2 can inhibit apoptosis and have neuroprotective effects. Mechanisms include increased expression of neurotrophic and nerve growth factors, mobilization and migration of pluripotent mesenchymal stem cells, proliferation of astrocytes, inhibition of the secretion of microglial cells, and improving outcomes for neurological and motor functions.
The timing of HBO2 in AIS may be an important factor in efficacy. Studies suggest that the ideally, HBO2 should be initiated within the first 3-5 hours post ischemic stroke (33, 34) and has been shown to be beneficial in the chronic phase (11).
HBO2 was shown in a retrospective study of 162 patients to induce significant improvements in all cognitive domains, even in the post-stroke late chronic stage. There were no significant differences in the HBO2 effect on cortical strokes compared to subcortical strokes. Participants were treated at 2 ATA for 90 min (with 5-min air breaks every 20 min) for 40–60 sessions. The median time from stroke to HBO2 was 1.5 ± 3.3 years. Neuroplasticity is induced by two main physiological effects: increasing tissue oxygenation and repeated oxygen level fluctuations, which increase HIF-1a. This, in turn, triggers regenerative processes in the metabolically injured brain areas (30).
HBO2 can improve cognitive functions in patients with mild cognitive impairment. Elevated humanin levels (mitochondrion-derived neuroprotective peptide) were reported in vascular dementia patients treated with HBO2. Maintenance of HBO2 will likely be needed for progressive neurodegenerative diseases (24).
Mechanisms of the neuroprotective effects of HBO2 in stroke patients include stabilizing the blood–brain barrier, improving metabolism, reducing brain edema, alleviating post-stroke neuroinflammation, inhibiting post-stroke apoptosis, improving anoxic area microcirculation, and suppressing ischemia–reperfusion injury (35).
The fact that the patient consistently experienced exacerbations of symptoms during and for 1–2 days after each airline flight strongly suggests that the neurological injury is sensitive to hypoxia, hypobaric pressure, or both. This same observation has been made in other clinical scenarios and was the basis for an experiment on the effect of commercial airline cabin pressure/hypoxic exposures incurred on evacuation flights of injured U.S. military members from Iraq. That experiment in acute brain-injured animals demonstrated amplification of the acute brain injury by hypobaric exposure to 8,000 feet altitude despite the correction of hypoxia (36). The patient’s sensitivity to hypobaric exposure at cabin altitude suggests that the opposite, exposure to increased pressure achieved with HBO2 therapy, is consistent with the observed clinical improvements.
Study limitations
Baseline assessments were retrospective, based on the patient and his spouse’s recollection of symptoms during the year prior to initiation of HBO2 therapy.
It is unknown whether the patient would have responded differently without the interruptions in the HBO2 therapy.
A definite limitation in the case report is the short (1 month) follow-up. Fortuitously, the symptomatic responses noted during the multiple patient-imposed breaks in therapy, might offer some insight, and imply that the patient could require periodic HBO2 to maintain benefit, as has been suggested in the HBO2 treatment of other brain pathologies.
Questionnaires were completed for only 80 out of the 100 treatments. However, the patient was entirely asymptomatic at the end of HBO2 therapy and had no adverse effects or complications from HBO2 treatments.
Study strengths
Multiple other pharmacologic and non-pharmacologic therapies have reported benefits in thalamic pain syndrome, but none have been attempted in this patient. There were no significant changes in the pharmacologic treatments during the study except for a reduction in pain medications. This allowed for clinical improvements to be attributed to the HBO2 intervention.
Reports of HBO2 mechanisms of action and efficacy in stroke and other brain pathologies support the clinical benefit realized in this case report and are consistent with a regenerative improvement in the chronic stages (1-year post-CVA in this case). HBO2 therapy is a potential treatment option for thalamic pain syndrome. The specific mechanisms of HBO2 therapy need to be further studied and explored.
Patient perspective
I am glad that my wife suggested trying the HBO2 therapy. My symptoms started at the time of the cerebral angiogram (December 2021) but became most intense approximately 9 months later (September 2022), with only partial relief from medications. With each pain episode, I worried whether it would get worse. The year of HBO2 really relaxed my body, and it is the best I have felt since September 2022. The symptoms improved with each new course of HBO2. Overall, the entire year of HBO2 treatments was the best I have felt since onset; the episodes were less painful, shorter, and did not keep me from doing what I wanted to do.
Data availability statement
The original contributions presented in the study are included in the article/supplementary material, further inquiries can be directed to the corresponding author.
Ethics statement
Written informed consent was obtained from the individual(s) for the publication of any potentially identifiable images or data included in this article.
Author contributions
JS: Writing – original draft, Writing – review & editing. NK: Writing – original draft. PL: Writing – original draft, Writing – review & editing. RG: Writing – review & editing.
Funding
The author(s) declare that no financial support was received for the research, authorship, and/or publication of this article.
Conflict of interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Publisher’s note
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.
References
Ri, S. The management of poststroke thalamic pain: update in clinical practice. Diagnostics. (2022) 12:1439. doi: 10.3390/diagnostics12061439
PubMed Abstract | Crossref Full Text | Google Scholar
Kim, JS. Pure sensory stroke: clinical-radiological correlates of 21 cases. Stroke. (1992) 23:983–7. doi: 10.1161/01.STR.23.7.983
PubMed Abstract | Crossref Full Text | Google Scholar
Klit, H, Finnerup, NB, and Jensen, TS. Central post-stroke pain: clinical characteristics, pathophysiology, and management. Lancet Neurol. (2009) 8:857–68. doi: 10.1016/S1474-4422(09)70176-0
PubMed Abstract | Crossref Full Text | Google Scholar
Ray, DAA, and Tai, YMA. Increasing doses of naloxone hydrochloride by infusion to treat pain due to the thalamic syndrome. Br Med J. (1988) 296:969–70. doi: 10.1136/bmj.296.6627.969-a
PubMed Abstract | Crossref Full Text | Google Scholar
Haruta-Tsukamoto, A, Funahashi, H, Miyahara, Y, Matsuo, T, Toshikazu Nishimorimi, T, and Ishida, Y. Alleviatiion of thalamic pain by cilostazol administration: a case report. Clin Case Reports. (2018) 6:380–4. doi: 10.1002/ccr3.1363
PubMed Abstract | Crossref Full Text | Google Scholar
Corbetta, D, Sarasso, E, Agosta, F, Filippi, M, and Gatti, R. Mirror therapy for an adult with central post-stroke pain: a case report. Arch Physiother. (2018) 8:4. doi: 10.1186/s40945-018-0047-y
PubMed Abstract | Crossref Full Text | Google Scholar
Liao, C, Yang, M, Liu, P, Zhong, W, and Zhang, W. Thalamic pain alleviated by stellate ganglion block: a case report. Medicine. (2017) 96:e6058. doi: 10.1097/MD.0000000000006058
PubMed Abstract | Crossref Full Text | Google Scholar
Wissel, J, and Ri, S. Assessment, goal setting, and botulinum neurotoxin a therapy in the management of post-stroke spastic movement disorder: updated perspectives on best practice. Expert Rev Neurother. (2022) 22:27–42. doi: 10.1080/14737175.2021.2021072
PubMed Abstract | Crossref Full Text | Google Scholar
Xu, X-M, Luo, H, Rong, B, Zheng, X-M, Wang, F, Zhang, S-J, et al. Nonpharmacological therapies for central poststroke pain. Medicine. (2020) 99:e22611. doi: 10.1097/MD.0000000000022611
Crossref Full Text | Google Scholar
Ding, Z, Tong, WC, Lu, X, and Peng, H. Hyperbaric oxygen therapy in acute ischemic stroke: a review. Interv Neurol. (2013) 2:201–11. doi: 10.1159/000362677
PubMed Abstract | Crossref Full Text | Google Scholar
Efrati, S, Fishlev, G, Bechor, Y, Volkov, O, Bergan, J, and Kliakhandler, K. Hyperbaric oxygen induces late neuroplasticity in post-stroke patients – randomized, prospective trial. PLoS One. (2013) 8:e53716. doi: 10.1371/journal.pone.0053716
Crossref Full Text | Google Scholar
Dejerine, J, and Roussy, G. Dejerine-Roussy syndrome. Rev Neurol. (1906) 14:521–32.
Google Scholar
Liampas, A, Velidakis, N, Georgiou, T, Vadaloluca, A, Varrassi, G, Hadjigeorgios, GM, et al. Prevalence and management challenges in central post-stroke neuropathic pain: a systematic review and Meta-anaylsis. Adv Ther. (2020) 37:3278–91. doi: 10.1007/s12325-020-01388-w
Crossref Full Text | Google Scholar
Bowsher, D. Central post-stroke (‘thalamic syndrome’) and other central pains. Am J Hosp Palliat Med. (1999) 16:593–7. doi: 10.1177/104990919901600408
PubMed Abstract | Crossref Full Text | Google Scholar
Nasreddine, ZS, and Saver, JL. Pain after thalamic stroke: right diencephalic predominance and clinical features in 180 patients. Neurology. (1998) 51:927–8. doi: 10.1212/WNL.51.3.927-b
PubMed Abstract | Crossref Full Text | Google Scholar
Bogousslavsky, J, Regli, F, and Uske, A. Thalamic infarcts: clinical syndromes, etiology, and prognosis. Neurology. (1988) 38:837–48. doi: 10.1212/WNL.38.6.837
Crossref Full Text | Google Scholar
Yezierski, RP. Spinal cord injury: a model of central neuropathic pain. Neurosignals. (2005) 14:182–93. doi: 10.1159/000087657
Crossref Full Text | Google Scholar
Nandi, D, Smith, H, Owen, S, Joint, C, Stein, J, and Aziz, T. Peri-ventricular grey stimulation versus motor cortex stimulation for post stroke neuropathic pain. J Clin Neurosci. (2002) 9:557–61. doi: 10.1054/jocn.2001.1042
PubMed Abstract | Crossref Full Text | Google Scholar
Sokal, P, Harat, M, Zielinski, P, Furtak, J, Paczkowski, D, and Rusinek, M. Motor cortex stimulation in patients with chronic pain. Adv Clin Exp Med. (2015) 24:289–96. doi: 10.17219/acem/40452
Crossref Full Text | Google Scholar
Bittar, RG, Kay-Purkayastha, I, Owen, SL, Bear, RE, Green, A, Wang, S, et al. Deep brain stimulation for pain relief: a meta-analysis. J Clin Neurosci. (2005) 12:515–9. doi: 10.1016/j.jocn.2004.10.005
PubMed Abstract | Crossref Full Text | Google Scholar
Levy, R, Deer, TR, and Henderson, J. Intracranial neurostimulation for pain control: a review. Pain Physician. (2010) 13:157–65. ISSN 1533-3159. doi: 10.36076/ppj.2010/13/157
PubMed Abstract | Crossref Full Text | Google Scholar
Awerbuch, GI, and Sandyk, R. Mexiletine for thalamic pain syndrome. Int J Neurosci. (1990) 55:129–33. doi: 10.3109/00207459008985960
PubMed Abstract | Crossref Full Text | Google Scholar
Boussi-Gross, R, Golan, H, Volkov, O, Bechor, Y, Hoofien, D, Beeri, MS, et al. Improvement of memory impairments in poststroke patients by hyperbaric oxygen therapy. Neuropsychology. (2015) 29:610–21. doi: 10.1037/neu0000149
PubMed Abstract | Crossref Full Text | Google Scholar
Gottfried, I, Schottlender, N, and Ashery, U. Hyperbaric oxygen treatment – from mechanisms to cognitive improvement. Biomol Ther. (2021) 11:1520. doi: 10.3390/biom11101520
PubMed Abstract | Crossref Full Text | Google Scholar
Chen, Y, Wang, L, You, W, Huang, F, Jiang, Y, Sun, L, et al. Hyperbaric oxygen therapy promotes consciousness, cognitive function, and prognosis recovery in patients following traumatic brain injury through various pathways. Front Neurol. 13:929386. doi: 10.3389/fneur.2022.929386
Crossref Full Text | Google Scholar
Iqbal, J, Naeem, A, Jahangir, K, Ali, Y, Mashkoor, Y, Ashraf, A, et al. Hyperbaric oxygen and outcomes following the brain injury: a systematic review. J Neurol Rees Rev Rep. (2023) 5:1–11. doi: 10.47363/JNRRR/2023(5)178
Crossref Full Text | Google Scholar
Schiavo, S, Debacker, J, Djaiani, C, Bhatia, A, Englesakis, M, and Katznelson, R. Mechanistic rationale and clinical efficacy of hyperbaric oxygen therapy in chronic neuropathic pain: an evidence-based narrative review. Pain Res Manag. (2021) 2021:8817504. doi: 10.1155/2021/8817504
PubMed Abstract | Crossref Full Text | Google Scholar
Yang, YJ, Wang, XL, Yu, XH, Wang, X, Xie, M, and Liu, CT. Hyperbaric oxygen induces endogenous neural stem cells to proliferate and differentiate in hypoxic-ischemic brain damage in neonatal rats. Undersea Hyperb Med. (2008) 35:113–29.
PubMed Abstract | Google Scholar
Yan, Y, Zhang, Z, An, X, Fan, W, Liang, J, Luo, B, et al. The application and perspective of hyperbaric oxygen therapy in acute ischemic stroke: from the bench to a starter? Front Neurol. 13:928802. doi: 10.3389/fneur.2022.928802
Crossref Full Text | Google Scholar
Hadanny, A, Rittblat, M, Bitterman, M, May-Raz, I, Suzin, G, Boussi-Gross, R, et al. Hyperbaric oxygen therapy improves neurocognitive functions of post-stroke patients – a retrospective analysis. Restor Neurol Neurosci. (2020) 38:93–107. doi: 10.3233/RNN-190959
PubMed Abstract | Crossref Full Text | Google Scholar
Rosario, ER, Kaplan, SE, Khonsari, S, Vazquez, G, Solanki, N, Lane, M, et al. The effect of hyperbaric oxygen therapy on functional impairments caused by ischemic stroke. Neurol Res Int. (2018) 2018:3172679. doi: 10.1155/2018/3172679
PubMed Abstract | Crossref Full Text | Google Scholar
Veltkamp, R, Siebing, DA, Sun, L, Heiland, S, Bieber, K, Marti, HH, et al. Hyperbaric oxygen reduces blood-brain barrier damage and edema after transient focal cerebral ischemia. Stroke. (2005) 36:1679–83. doi: 10.1161/01.STR.0000173408.94728.79
PubMed Abstract | Crossref Full Text | Google Scholar
McCormick, JG, Houle, TT, Saltzman, HA, Whaley, RC, and Roy, RC. Treatment of acute stroke with hyperbaric oxygen: time window for efficacy. Undersea Hyperb Med. (2011) 38:321–34.
PubMed Abstract | Google Scholar
Sanchez, EC. Mechanisms of action of hyperbaric oxygenation in stroke: a review. Crit Care Nurs Q. (2013) 36:290–8. doi: 10.1097/CNQ.0b013e318294e9e3
Crossref Full Text | Google Scholar
Zhai, W, Sun, L, Yu, Z, and Chen, G. Hyperbaric oxygen therapy in experimental and clinical stroke. Med Gas Res. (2016) 6:111–8. doi: 10.4103/2045-9912.184721
Crossref Full Text | Google Scholar
Skovira, JW, Kabadi, SV, Wu, J, Zhao, Z, DuBose, J, Rosenthal, R, et al. Simulated aeromedical evacuation exacerbates experimental brain injury. J Neurotrauma. (2016) 33:1292–302. doi: 10.1089/neu.2015.4189
PubMed Abstract | Crossref Full Text | Google Scholar
Keywords: thalamic pain syndrome, hyperbaric oxygen, case report, central post-stroke pain, post-stroke thalamic pain
Citation: Slade JB, Kwan N, Lennox P and Gray R (2024) Hyperbaric oxygen therapy for thalamic pain syndrome: case report. Front. Neurol. 15:1364716. doi: 10.3389/fneur.2024.1364716
Received: 02 January 2024; Accepted: 21 February 2024;
Published: 13 March 2024.
Edited by:
Paul Gregory Harch, Louisiana State University, United States
Reviewed by:
Jiying Zhou, First Affiliated Hospital of Chongqing Medical University, China
Judy Wilson, University of Texas at Arlington, United States
Copyright © 2024 Slade, Kwan, Lennox and Gray. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
Correspondence: John Benjamin Slade, jslade1515@aol.com
Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher. |
833 | https://erikdemaine.org/papers/SlidingCoins_G4G5/paper.pdf | Sliding-Coin Puzzles Erik D. Demaine Martin L. Demaine∗ In what ways can an arrangement of coins be reconfigured by a sequence of moves where each move slides one coin and places it next to at least two other coins? Martin Gardner publicized this family of sliding-coin puzzles (among others) in 1966. Recently, a general form of such puzzles was solved both mathematically and algorithmically. We describe the known results on this problem, and show several examples in honor of Martin Gardner for the 5th Gathering for Gardner.
1 Puzzles Sliding-coin puzzles ask you to re-arrange a collection of coins from one configuration to another using the fewest possible moves. Coins are identical in size, but may be distinguished by labels; in some of our examples, the coins are labeled with the letters G, A, R, D, N, E, R.
For our purposes, a move involves sliding any coin to a new position that touches at least two other coins, without disturbing any other coins during the motion.
The rest of this section presents several sliding-coin puzzles.
1.1 Triangular Lattice We start with some basic puzzles that are on the triangular lattice in the sense that the center of every coin is at a vertex of the planar lattice of equilateral triangles. The restriction that a move must bring a coin to a new position that touches at least two other coins forces a puzzle to stay on the triangular lattice if it is originally on it.
Triangular Lattice Puzzles Classic Puzzles Puzzle 1: Rhombus to a circle (3 moves) Puzzle 2: Turn the pyramid upside-down (3 moves) ∗The Demaines are a father-and-son team. Erik D. Demaine, the father, is an assistant professor of computer science at MIT who likes anything algorithmic or geometric. Martin L. Demaine, the son, is a researcher in computer science at MIT who likes anything mathematical and artistic.
1 New Puzzles Puzzle 3: Pyramid to a line (7 moves). Source unknown.
A G R D E N G A R E D N Puzzle 4: Spread out the GARDEN (9 moves).
1.2 Square Lattice Next we give a few puzzles on the square lattice. Here the centers of the coins are at vertices of the planar lattice of squares, and we make the additional constraint that every move brings a coin to such a position. The restriction that a move must bring a coin to a new position that touches at least two other coins does not force the puzzle to stay on the square lattice, but this additional constraint does.
Square Lattice Puzzles Penta Puzzles Puzzle 5: Flip the L (4 moves) Puzzle 6: Rotate the L by 90◦ (8 moves) G N D A R G N D R A Puzzle 7: Fix the spelling of GRAND (8 moves) Hard Puzzles A G D E R R N D G A R N E R Puzzle 8: Flip the diagonal (hard) Puzzle 9: Flip the roman numeral V (very hard) 2 History Sliding-coin puzzles have long been popular. For example, the classic Puzzle 1 is described in Martin Gardner’s Scientific American article on “Penny Puzzles” , in Winning Ways , in Tokyo Puzzles , in Moscow Puzzles , and in The Penguin Book of Curious and Interesting 2 Puzzles . Langman shows all 24 ways to solve this puzzle in three moves. Puzzle 2 is another classic [2, 7, 8, 12]. Other puzzles are presented by Dudeney , Fujimura , and Brooke .
The historical puzzles described so far are all on the triangular lattice. Puzzles on the square lattice appear less often in the literature but have significantly more structure and can be more difficult. The only published example we are aware of is given by Langman , which is also described by Brooke , Bolt , and Wells ; see Puzzle 10. However, the second of these puzzles does not remain on the square lattice; it only starts on the square lattice, and the only restriction on moves is that the new position of a coin is adjacent to at least two other coins.
Puzzle 10: Re-arrange the H into the O in four moves while staying on the square lattice (and always moving adjacent to two other coins), and return to the H in six moves using both the triangular and square lattices.
3 Mathematics A paper by Helena Verrill and the present authors solves a large portion of the general sliding-coin puzzle-solving problem: given two configuration of coins, is it possible to re-arrange the first configuration into the second via a sequence of moves? One catch is that, for the results to apply, a move must be redefined to allow a coin to be picked up and placed instead of just slid on the table. another catch is that the solution does not say anything about the minimum number of moves required to solve a specific puzzle, though it does provide a polynomial upper bound on the number of moves required. (From this information we can also determine which puzzle requires roughly the most moves, among all puzzles.) Despite these catches, the results often apply directly to sliding-coin puzzles and tell us whether a given puzzle is solvable, and if so, how to solve it. The ability to tell whether a puzzle is solvable is ideal for puzzle design.
A surprising aspect of this work is that there is an efficient algorithm to solve most sliding-coin puzzles, which runs fast even for very large puzzles. In contrast, most other games and puzzles, when scaled up sufficiently large, are computationally intractable.
3.1 Triangular Lattice It turns out to be fairly easy to characterize which triangular-lattice puzzles are solvable.
Part of what makes this characterization easy is that most puzzles are solvable. Consider a puzzle with an initial configuration that differs from the goal configuration. There are a few basic restrictions for this puzzle to be solvable: 3 1. There must be at least one valid move from the initial configuration.
2. The number of coins must be the same for the initial and goal configurations.
3. At least one of the following four conditions must hold: (a) The final configuration contains a triangle of three mutually touching coins.
(b) The final configuration contains four connected coins.
(c) The final configuration contains three connected coins and two different touching coins (as in Puzzle 4).
(d) The puzzle is solvable by a single move.
4. If the coins are labeled and there are only three coins (a rather boring situation), then the goal configuration must follow the same 3-coloring of the triangular lattice.
After some thought, you will probably see why each of these conditions must hold for a puzzle to be solvable. What is more surprising, but beyond the scope of this article, is that these conditions are enough to guarantee that the puzzle is solvable. Interested readers are referred to for the proof.
3.2 Square Lattice Solvable square-lattice puzzles are trickier to characterize. Much more stringent conditions must hold.
For example, it is impossible for a configuration of coins to get outside its enclosing box. This property is quite different from the triangular lattice, where coins can travel arbitrary distances.
A notion that turns out to be particularly important with square lattices is the span of a configuration. Figure 1 shows an example. Suppose we had a bag full of extra coins, and we could place them onto the lattice at any empty position adjacent to at least two other coins.
If we repeat this process for as long as possible, we obtain the span of the configuration.
Figure 1: In general, the span consists of one or more rectangles separated by distance at least two empty spaces.
A key property of span is that it can never get larger by a sequence of moves. The span effectively represents all the possibly reachable positions in a configuration. So if you are to move the coins from one configuration to another, the span of the first configuration better contain the span of the second configuration.
This condition is not quite enough, though. In fact, the exact conditions are not known for when a square-lattice puzzle can be solved. However, most solvable configurations have some extra coins whose removal would not change the span of the configuration. The main 4 result of says that if a configuration has at least two extra coins, then it can reach any other configuration with the same or smaller span.
This result is complicated, leading to some puzzles with intricate solutions, as in Puzzles 8 and 9, for example. To work your way up to these difficult puzzles, we have provided some “warm up” puzzles that involve 5 coins, in the spirit of the “penta” theme of the Fifth Gathering for Gardner.
References Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy. A solitaire-like puzzle and some coin-sliding problems. In Winning Ways, volume 2, pages 755–756. Academic Press, London, 1982.
Brian Bolt. Invert the triangle. In The Amazing Mathematical Amusement Arcade, amuse-ment 53, page 30. Cambridge University Press, Cambridge, 1984.
Brian Bolt. A two touching transformation. In Mathematical Cavalcade, puzzle 20, page 10.
Cambridge University Press, Cambridge, 1991.
Maxey Brooke. Fun for the Money. Charles Scriber’s Sons, New York, 1963. Reprinted as Coin Games and Puzzles by Dover Publications, 1973.
Erik D. Demaine, Martin L. Demaine, and Helena Verrill. Coin-moving puzzles. In R. J.
Nowakowski, editor, More Games of No Chance, pages 405–431. Cambridge University Press, 2002. Collection of papers from the MSRI Combinatorial Game Theory Research Workshop, Berkeley, California, July 24–28, 2000.
Henry Ernest Dudeney. “The four pennies” and “The six pennies”. In 536 Puzzles & Curious Problems, problems 382–383, page 138. Charles Scribner’s Sons, New York, 1967.
Kobon Fujimura. “Coin pyramids,” “Four pennies,” “Six pennies,” and “Five coins”. In The Tokyo Puzzles, puzzles 23 and 25–27, pages 29–33. Charles Scribner’s Sons, New York, 1978.
Martin Gardner. Penny puzzles. In Mathematical Carnival, chapter 2, pages 12–26. Alfred A. Knopf, New York, 1975. Appeared in Scientific American, 214(2):112–118, February 1966, with solutions in 214(3):116–117, March 1966.
Boris A. Kordemsky. A ring of disks. In The Moscow Puzzles, problem 117, page 47. Charles Scribner’s Sons, New York, 1972.
Harry Langman. Curiosa 261: A disc puzzle. Scripta Mathematica, 17(1–2):144, March–June 1951.
Harry Langman. Curiosa 342: Easy but not obvious. Scripta Mathematica, 19(4):242, Decem-ber 1953.
David Wells. “Six pennies,” “OH-HO,” and “Inverted triangle”. In The Penguin Book of Curious and Interesting Puzzles, puzzle 305, 375, and 376, pages 101–102 and 125. Penguin Books, 1992.
5 A Solutions Solutions to Puzzles 1, 2, 3, 5, 6, and 7 were found by an exhaustive breadth-first search, and as a result we are sure that the solutions use the fewest possible moves. For Puzzle 4, it is conceivable that using more than two “rows” leads to a shorter solution; the solution below is the best among all two-row solutions. Puzzles 8 and 9 are left as challenges to the reader; we do not know the minimum number of moves required to solve them. The second half of Puzzle 10 was solved by hand, but the number of moves is minimum: the maximum overlap between the H and the O is four coins, three moves are necessary to enter the triangular lattice and return to the square lattice, and every move starting with Move 3 puts a coin in its final position.
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Solution to Puzzle 1: Three moves.
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Solution to Puzzle 2: Three moves.
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Solution to Puzzle 3: Seven moves.
E N D R A G R A G E N D A G R E N D A G R E N D G E R A N D E R A G N D R G A E N D G A E R N D G A R E N D E R G A N D 6.
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Solution to Puzzle 4: Nine moves.
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Solution to Puzzle 5: Four moves.
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Solution to Puzzle 6: Eight moves.
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G N D A R G N D A R G N D R A G N D R A G N A D R G N R A D G N D A R G N D A R G N D A R Solution to Puzzle 7: Eight moves.
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Solution to Puzzle 10: Four moves there and six moves back.
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ESPAÑOL AL DÍA
Uso de los pronombres «lo(s)», «la(s)», «le(s)». Leísmo, laísmo, loísmo
Para usar adecuadamente los pronombres átonos de 3.ª persona lo(s), la(s), le(s) según la norma culta del español general, debe tenerse en cuenta, en primer lugar, la función sintáctica que desempeña el pronombre y, en segundo lugar, el género y el número gramatical de la palabra a la que se refiere. En el siguiente cuadro se muestra la distribución de formas y funciones de estos pronombres:
Formas de los pronombres personales átonos de 3.ª persona
| | | | |
--- --- |
| | | SINGULAR | PLURAL |
| compl. directo | masculino | lo (tambiénle, cuando el referente es de persona) | los (tambiénles, cuando el referente es de persona) |
| femenino | la | las |
| compl. directo o atributo | neutro | lo
| compl. indirecto | | le (o seante otro pron. átono) | les (o seante otro pron. átono) |
| forma reflexiva o recíproca | | se | |
A continuación se expone de forma sucinta la norma que rige el empleo de estos pronombres:
• Cuando el pronombre desempeña la función de complemento directo, corresponde usar las formas lo, los para el masculino (singular y plural, respectivamente) y la, las para el femenino (singular y plural, respectivamente): :
¿Has visto a Juan? Sí, lo vi ayer.
¿Has visto a Juan y a los niños? Sí, los he visto en el parque.
Compré la medicina y se la di sin que nadie me viera.
¿Has recogido a las niñas? Sí, las recogí antes de ir al taller.
[Se considera válido, debido a su gran extensión entre hablantes cultos de ciertas zonas del ámbito hispanohablante, el uso de le(s) en función de complemento directo cuando el referente es de persona masculina: ¿Has visto a Jorge? Sí, le vi ayer en el parque; A tus primos les conozco desde hace años].
• Cuando el pronombre desempeña la función de complemento indirecto, corresponde usar las formas le,les(singular y plural, respectivamente), con independencia del género de la palabra a la que se refiera el pronombre:
Lepedí disculpas a mi madre.
Ledije a su hermana que viniera.
Lesdi un regalo a los niños.
A pesar de la aparente simplicidad del sistema, existen casos excepcionales o aparentemente excepcionales dentro de la norma, así como una enorme variedad en cuanto a los usos efectivos en las distintas zonas hispanohablantes. Si se desea información pormenorizada, pueden consultarse los artículos leísmo, laísmo y loísmo del Diccionario panhispánico de dudas, así como las entradas dedicadas a verbos que plantean problemas a los hablantes en cuanto a la selección de los pronombres átonos de tercera persona (avisar, ayudar, curar, disparar, escribir, llamar, molestar, obedecer, pegar, saludar, etc.)
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835 | https://www.quora.com/What-is-the-variable-acceleration-in-motion | What is the variable acceleration in motion? - Quora
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Physics
Define Motion
Variable Displacement
Basic Kinematics
Acceleration (physics)
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Describing Motion
Motion (physics)
Kinematics and Dynamics
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What is the variable acceleration in motion?
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Rohan Naik
Ph.D. in Quantum Mechanics&Physics, University of Mumbai (Graduated 2016) · Author has 505 answers and 1.6M answer views
·6y
The SUVAT equations from equations of motion can only be used when an object is moving with constant acceleration. When the acceleration varies, this is when we must use calculus.
The diagram above shows the relationship between acceleration a a a, velocity v v v and displacement x x x.
Velocity is the rate of change of displacement, therefore to obtain velocity from displacement you differentiate.
To find displacement from velocity, you integrate.
Acceleration is the rate of change of velocity. To obtain acceleration from velocity you differentiate. To obtain acceleration from displacement you differentia
Continue Reading
The SUVAT equations from equations of motion can only be used when an object is moving with constant acceleration. When the acceleration varies, this is when we must use calculus.
The diagram above shows the relationship between acceleration a a a, velocity v v v and displacement x x x.
Velocity is the rate of change of displacement, therefore to obtain velocity from displacement you differentiate.
To find displacement from velocity, you integrate.
Acceleration is the rate of change of velocity. To obtain acceleration from velocity you differentiate. To obtain acceleration from displacement you differentiate twice.
To get the displacement from acceleration you integrate twice, whereas to get the velocity from acceleration you integrate only once.
For more Information, you can also watch the below video.
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Johnny M
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·Updated Sep 9
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Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t.
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This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars.
I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend.
Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take.
But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious.
They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage.
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If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying.
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What is accelerated motion?
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John Yelton
D.Phil in Physics, University of Oxford (Graduated 1981) · Author has 3.3K answers and 1.3M answer views
·6y
Variable acceleration is when the acceleration is not constant.
Asking what is “the variable acceleration in motion” makes no sense. What motion?
Upvote ·
Oknotok Ko
Lives in Kolkata, West Bengal, India · Upvoted by
Florens de Wit
, M.S. Physics, Eindhoven University of Technology (1999)
·Updated 8y
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If an object is moving with constant speed, then what is the acceleration?
First of all what is acceleration? — Acceleration is the rate of change of velocity per unit time.
Now, we know that velocity is a vector quantity i.e we need both magnitude and direction to define it and so is acceleration. We can now say that in acceleration, the direction of motion is very important or to define acceleration we need both the magnitude of motion (here it is the speed) and it's direction.
Now coming to your question. There are two possible answers to this question.
Case 1 — When the motion a body is one dimensional i.e it is moving in a straight line. Let's assume that speed of
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First of all what is acceleration? — Acceleration is the rate of change of velocity per unit time.
Now, we know that velocity is a vector quantity i.e we need both magnitude and direction to define it and so is acceleration. We can now say that in acceleration, the direction of motion is very important or to define acceleration we need both the magnitude of motion (here it is the speed) and it's direction.
Now coming to your question. There are two possible answers to this question.
Case 1 — When the motion a body is one dimensional i.e it is moving in a straight line. Let's assume that speed of this body is constant and is X. Now as the speed is constant and there will be no change in direction (moving in a straight line) there will be no change in its velocity at any given point of time or change in velocity will be zero. Now change in velocity divided by the time period is acceleration and the change is zero and hence acceleration also is zero.
Case 2 — When the body is not moving in a straight line or is constantly changing its direction. In this case the speed still is constant but the direction is constantly changing, for example in case of a circular motion the direction of motion constantly changes (the direction of a moving body in a circular path at anypoint is the direction of the tangent of the circle at that point). Hence the velocity changes as the body moves along the circular or curved path.
Now, as I earlier mentioned that to define acceleration we need to know it's magnitude and it's direction, and acceleration is the rate of change of velocity per unit time — the speed remaining constant the direction constantly changing, the velocity changes at different time intervals or rate of change of velocity is ≠ 0 unlike in case 1. Since there is change in velocity (speed constant but direction is changing) there is some acceleration.
To sum everything up — in case of linear motion i.e motion in a straight line when the speed is constant and direction is not changing, velocity will be zero and hence acceleration would be zero. In case of a non-linear motion i.e motion not in a straight line, or in a circle or any path where the speed is constant but direction is constantly changing, velocity will not be zero and hence the acceleration would not be equal to zero. Either it would be greater or less than zero.
If you understood, then please hit the upvote button!
~Cheers
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M.A. Raheem
Updated 4y
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Are constant acceleration and uniform acceleration the same?
The confusion is because most of the text book says something like this, “ the equation of motions are derived for constant or uniform acceleration”.
The below figure should help you out, (although I have drawn it by hand, you can even see the shadow of my phone :-)).
Well, the “acceleration is constant” means, along the time it is not varying. As shown by the horizontal line, in the above image.
Acceleration is uniform implies either uniformly increasing or uniformly decreasing.
If you check the values, in the above image. The constant acceleration is the second table. In the second table the vel
Continue Reading
The confusion is because most of the text book says something like this, “ the equation of motions are derived for constant or uniform acceleration”.
The below figure should help you out, (although I have drawn it by hand, you can even see the shadow of my phone :-)).
Well, the “acceleration is constant” means, along the time it is not varying. As shown by the horizontal line, in the above image.
Acceleration is uniform implies either uniformly increasing or uniformly decreasing.
If you check the values, in the above image. The constant acceleration is the second table. In the second table the velocity value is increasing uniformaly i.e., for every 1 second it is increasing by 2 units. However, the acceleration value is remaining same.
As we can see in the Table 1, acceleration values are increasing by 1 unit per second, so the acceleration is increasing uniformly. However the velocity increment is non-uniform. In the Ist second the velocity increment is 2.5 m/s ( 2.5 -0). In the 2nd second the velocity increment is 3.5 m/s (6 - 2.5).
hope the answer is satisfactory.
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What acceleration is in the direction of motion?
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What is an example of motion with constant acceleration and variable speed?
What feature of motion is described by acceleration?
In which motion is acceleration not constant?
Timothy Dutton
Call Agent · Author has 271 answers and 995.5K answer views
·7y
Related
Is it possible that a body has acceleration in the opposite direction of its motion?
Yes - If you imagine a vehicle travelling along the road in the direction shown
Now lets apply the brakes.
We can represent the deceleration in one of two ways.
A negative acceleration in the direction of travel.
But with mathematics we can transform the equation by reversing the direction through 180 degrees, and in order to preserve the equation at the same time reversing the sign.
So as you can see, when the car is slowing down it is effectively accelerating in the opposite direction.
Another way is picture the bouncing ball below.
Gravity is always pulling the ball towards the earth. So when it’s
Continue Reading
Yes - If you imagine a vehicle travelling along the road in the direction shown
Now lets apply the brakes.
We can represent the deceleration in one of two ways.
A negative acceleration in the direction of travel.
But with mathematics we can transform the equation by reversing the direction through 180 degrees, and in order to preserve the equation at the same time reversing the sign.
So as you can see, when the car is slowing down it is effectively accelerating in the opposite direction.
Another way is picture the bouncing ball below.
Gravity is always pulling the ball towards the earth. So when it’s rising it’s accelerating in the opposite direction to it’s motion, meaning it slows down until it gets to the top of it’s bounce, where for an instant it has zero velocity before beginning to fall towards the ground.
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E. Dean Fowlkes
Private pilot with an instrument rating. · Author has 2.3K answers and 990.4K answer views
·4y
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Why is an object has a 0 acceleration but still in motion?
Because velocity and acceleration are two different things. Velocity is the speed of an object in a specific direction. Acceleration is a change in velocity over time. It can be a change in speed or direction or both.
For example, the speed of a car might be 61 mph. The velocity of a car might be 61 mph going in a specific straight line. The acceleration of a car might be 0 to 61 mph in a straight line in 5 seconds flat. Without that second time component (the “per hour” part being the first time component) you do not have acceleration.
So, to sum it up:
distance = position change
speed = distance
Continue Reading
Because velocity and acceleration are two different things. Velocity is the speed of an object in a specific direction. Acceleration is a change in velocity over time. It can be a change in speed or direction or both.
For example, the speed of a car might be 61 mph. The velocity of a car might be 61 mph going in a specific straight line. The acceleration of a car might be 0 to 61 mph in a straight line in 5 seconds flat. Without that second time component (the “per hour” part being the first time component) you do not have acceleration.
So, to sum it up:
distance = position change
speed = distance per time
velocity = direction and distance per time
acceleration = direction and distance per time per time
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Prem Kumar
Computer Engineer and Amateur Physicst · Author has 129 answers and 599.4K answer views
·Updated 7y
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Can there be acceleration if a body is moving with constant speed?
Yes a body can definitely have acceleration if it is moving with constant speed.
As speed is a scalar quantity, direction of movement doesn’t affect it. But acceleration is a vector quantity as it is the rate of change of velocity, which is itself a vector quantity.
So, take uniform circular motion as example.
It’s speed is constant but it has varying velocity as it changes it’s direction, so there is always a acceleration acting towards centre which is provided by centripetal force.
Continue Reading
Yes a body can definitely have acceleration if it is moving with constant speed.
As speed is a scalar quantity, direction of movement doesn’t affect it. But acceleration is a vector quantity as it is the rate of change of velocity, which is itself a vector quantity.
So, take uniform circular motion as example.
It’s speed is constant but it has varying velocity as it changes it’s direction, so there is always a acceleration acting towards centre which is provided by centripetal force.
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Manas Pratim Biswas
High School in Physics&Math Bio Chem, Springdale High School, Kalyani (Graduated 2019) · Author has 314 answers and 898K answer views
·6y
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In a uniform circular motion, acceleration is uniform or variable?
In CIRCULAR MOTION there are two types of accelaration...
Tangential acceleration,which is responsible for the change in magnitude given by a=dv/dt.
Centripetal acceleration,which is responsible for the change in direction given by a=v^2/r.
Now in UNIFORM CIRCULAR MOTION(UCM), the magnitude of velocity is constant but the direction varies, which means the angular velocity or speed is constant,hence the tangential acceleration is zero.
But the centripetal acceleration is always present which is responsible for the circular path as it is always directed towards the centre.
Since both these
Continue Reading
In CIRCULAR MOTION there are two types of accelaration...
Tangential acceleration,which is responsible for the change in magnitude given by a=dv/dt.
Centripetal acceleration,which is responsible for the change in direction given by a=v^2/r.
Now in UNIFORM CIRCULAR MOTION(UCM), the magnitude of velocity is constant but the direction varies, which means the angular velocity or speed is constant,hence the tangential acceleration is zero.
But the centripetal acceleration is always present which is responsible for the circular path as it is always directed towards the centre.
Since both these accelerations are perpendicular to each other,the net acceleration is given by
Net accln = sq.rt [ (tang. accln)^2+(centripetal accln)^2 ]
UCM depends only upon the centripetal component which in turn depends upon velocity and radius which are constant,so centripetal accln is constant and hence acceleration in UCM is constant.
If you want a more deeper understanding, continue reading.
Fun time, let us do a thought experiment that Sir Isaac Newton did in 1687,
This is a 'thought experiment', in which a cannon on top of a tall mountain is able to fire a cannonball horizontally at any chosen muzzle speed. The effects of air friction on the cannonball are ignored (or perhaps the mountain is high enough that the cannon is above the Earth's atmosphere).
(Newton’s Cannonball, an illustration of how objects can "fall" in a curve)
If the cannon fires its ball with a low initial speed, the trajectory of the ball curves downward and hits the ground (A). As the firing speed is increased, the cannonball hits the ground farther (B) away from the cannon, because while the ball is still falling towards the ground, the ground is increasingly curving away from it (see first point, above). All these motions are actually "orbits" in a technical sense – they are describing a portion of an elliptical path around the center of gravity – but the orbits are interrupted by striking the Earth.
If the cannonball is fired with sufficient speed, the ground curves away from the ball and It is now in what could an orbit. For any specific combination of height above the center of gravity an mass of the planet, there is one specific firing speed (unaffected by the mass of the ball, which is assumed to be very small relative to the Earth's mass) that produces a circular orbit, as shown in (C).
As the firing speed is increased beyond this, non-interrupted elliptic orbits are produced; one is shown in (D). If the initial firing is above the surface of the Earth as shown, there will also be non-interrupted elliptical orbits at slower firing speed; these will come closest to the Earth at the point half an orbit beyond, and directly opposite the firing point, below the circular orbit.
At a specific horizontal firing speed called escape velocity, dependent on the mass of the planet, an open orbit (E) is achieved that has a parabolic path. At even greater speeds the object will follow a range of hyperbolic trajectory. In a practical sense, both of these trajectory types mean the object is "breaking free" of the planet's gravity, and "going off into space" never to return.
I hope this helps.
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Adnan Adip
Studied at Pabna Zilla School · Author has 68 answers and 151.3K answer views
·3y
Related
What is the relationship between motion and acceleration?
In terms of Calculus acceleration is the derivative of motion. When an object moves with uniform velocity it has no acceleration. More clearly we can say the rate of change of velocity with time is acceleration. Since velocity has both magnitude and direction the change of velocity can occur in two ways. If you tied a stone with a string and rotated it over your head with uniform speed then the rotating stone would change its direction of motion continuously. That is change of velocity will occur and acceleration will take place. If your motion is linear then there is no scope of changing the
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In terms of Calculus acceleration is the derivative of motion. When an object moves with uniform velocity it has no acceleration. More clearly we can say the rate of change of velocity with time is acceleration. Since velocity has both magnitude and direction the change of velocity can occur in two ways. If you tied a stone with a string and rotated it over your head with uniform speed then the rotating stone would change its direction of motion continuously. That is change of velocity will occur and acceleration will take place. If your motion is linear then there is no scope of changing the direction. Then acceleration may occur only for the change of magnitude of the velocities to stop when the magnitude of the velocity increases then we say acceleration is taking place along the direction of the velocity. If the velocity decreases then we can say negative acceleration or acceleration is taking place.
The equation connecting velocity and acceleration is usually considered these ones.
Here u=initial velocity
V=final velocity
a=acceleration
t=time
In terms of Calculus.
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Manny
Student · Author has 545 answers and 591.6K answer views
·4y
Related
What feature of motion is described by acceleration?
The feature of motion that is described by acceleration is basically increasing or decreasing velocity of an object in motion which can be increasing or decreasing constantly or variably over the period of time and it also tells us that if the object is changing its direction of velocity or not.
Its a derivative of velocity with respect to time.
Like this :
a = dv/dt
Where dv is changing velocity and dt is changing time.
Velocity being a vector quantity (depending on direction) makes Acceleration a vector quantity too , so its varies with direction.
I hope my answer helps, if it does then please don
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The feature of motion that is described by acceleration is basically increasing or decreasing velocity of an object in motion which can be increasing or decreasing constantly or variably over the period of time and it also tells us that if the object is changing its direction of velocity or not.
Its a derivative of velocity with respect to time.
Like this :
a = dv/dt
Where dv is changing velocity and dt is changing time.
Velocity being a vector quantity (depending on direction) makes Acceleration a vector quantity too , so its varies with direction.
I hope my answer helps, if it does then please don’t forget to visit my space on Quora: ★ QUOSMOS ★
:)
Image credits: Khan Academy
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Alon Goldring
PhD in Nuclear Physics, Hebrew University of Jerusalem (Graduated 1991) · Author has 1.7K answers and 1.2M answer views
·1y
Related
What is the formula for calculating average speed and distance for an object with variable acceleration?
average speed is always (distance travelled) /time
Distance travelled in the most general case is the integral of the velocity whi...
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Mohd Ariz
Studied Human Biology&Physics
·4y
Related
What types of change in motion causes acceleration?
Acceleration can be caused by - :
Speed of body changes( increases or decreases) , but direction of motion doesn't change.i.e body moves in straight line.
Speed of body remains constant (fixed) but direction of motion of body changes.i.e its velocity changes.
Ex- Body moves with constant(fixed) speed on curved path.
If both 1 and 2 occurs.i.e speed as well as direction of motion of body changes.
Note - : Direction of motion simply means in which direction object is moving.
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Acceleration can be caused by - :
Speed of body changes( increases or decreases) , but direction of motion doesn't change.i.e body moves in straight line.
Speed of body remains constant (fixed) but direction of motion of body changes.i.e its velocity changes.
Ex- Body moves with constant(fixed) speed on curved path.
If both 1 and 2 occurs.i.e speed as well as direction of motion of body changes.
Note - : Direction of motion simply means in which direction object is moving.
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Peter Boetzkes
Ph.D. in Physics, University of Alberta (Graduated 1973) · Author has 4.8K answers and 2.5M answer views
·3y
Related
Does variable acceleration mean tangential acceleration is acting on the body?
It is possible, but not necessary. I take tangential acceleration to be acceleration tangent to the curve along which motion proceeds, and therefore acceleration in the direction of that motion, and results in change of speed. Thus constant-speed movement around a circle involves no tangent acceleration. Similarly, speed change along a straight line is tangential acceleration along a circle of inf
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It is possible, but not necessary. I take tangential acceleration to be acceleration tangent to the curve along which motion proceeds, and therefore acceleration in the direction of that motion, and results in change of speed. Thus constant-speed movement around a circle involves no tangent acceleration. Similarly, speed change along a straight line is tangential acceleration along a circle of infinite radius.
Any acceleration is a vector which has components (possible zero) in the direction of motion, and in the plane normal to the direction of motion. Therefore an acceleration vector that is constraine...
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In how many ways can 2 girls and 4 boys sit at a circular table so that no two girls sit together?
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Nick Shales
MPhys in Theoretical Physics, The University of York (Graduated 2004) · Author has 419 answers and 1.5M answer views
·Updated 7y
Seat the boys first in 4!/4=3!4!/4=3! ways (4!4! for linear seating arrangements then divide by 4 4 since there are 4 4 equivalence classes for circular permutations) place the 2 girls in the 4 gaps by making ordered choices of 2 of the 4 gaps in P(4,2)=4!/(4−2)!=4!/2!P(4,2)=4!/(4−2)!=4!/2! ways.
Arrangements=(4−1)!P(4,2)=3!4!2!=3⋅4!=72 Arrangements=(4−1)!P(4,2)=3!4!2!=3⋅4!=72
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Seat the boys first in 4!/4=3!4!/4=3! ways (4!4! for linear seating arrangements then divide by 4 4 since there are 4 4 equivalence classes for circular permutations) place the 2 girls in the 4 gaps by making ordered choices of 2 of the 4 gaps in P(4,2)=4!/(4−2)!=4!/2!P(4,2)=4!/(4−2)!=4!/2! ways.
Arrangements=(4−1)!P(4,2)=3!4!2!=3⋅4!=72 Arrangements=(4−1)!P(4,2)=3!4!2!=3⋅4!=72
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More answers below
In how many ways can 5 boys and 5 girls sit around a table, if no two girls were to sit side-by-side?
In how many ways are six boys and five girls seated in a circle that no two girls are sitting together?
In how many ways may 5 boys and 5 girls be arranged around a table so that no two girls sit together?
How can 6 boys and 4 girls be seated in a round table so that no two girls be seated together?
How many ways can 5 boys and 5 girls be seated in a circular table if; d) 3 particular boys must not sit together; e) all girls must sit together?
Doug Dillon
Ph.D. Mathematics · Author has 12.4K answers and 11.4M answer views
·8y
Place one girl, say Donna. That sort of fixes the table from rotation. Assuming that there are 6 spots at which to sit, the second girl may be seated in one of 3 spots. The 4 boys then can be seated in 4! ways giving us a total of 72 ways. Note that in most “sitting around table questions” we need to divide by a factorial to account for rotations but Donna dealt with that.
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·
Sep 6
Place the 4 boys around the circle first, then insert the 2 girls into the spaces between boys so that no two girls are adjacent.
Step-by-step:
Number of distinct circular arrangements of 4 boys = (4 − 1)! = 3! = 6.
Between these 4 boys there are 4 spaces (one between each adjacent pair). Choose 2 distinct spaces for the girls: C(4,2) = 6.
Arrange the 2 girls in the chosen spaces in 2! = 2 ways.
Multiply: 6 (boys) × 6 (choice of spaces) × 2 (order of girls) = 72.
Answer: 72.
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Michiel van der Blonk
3.5 million points on Khan Academy · Author has 1.5K answers and 3.7M answer views
·8y
Originally Answered: In how many ways 3 girls and 3 boys seated in a round table such that no 2 girls and 2 boys sitted together? ·
That depends on your definition of ‘how many’, ‘seated’, ‘boy’, ‘girl’ and ‘together’.
Combinatorics and because of that probability is a weird branch of mathematics. Everything depends on the meaning of regular English words, and whether you know something before or not, whereas in most of math these things are quite clear.
If we call boys B and girls G, then we can draw exactly 1 circle
B-G-B-G-B-G -> and circle to the beginning
That’s the only situation where this is the case.
Now if we call the boys and girls by ‘name’, e.g. B1, B2, B3, G1, G2, G3 then suddenly we have way more combinations:
B1
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That depends on your definition of ‘how many’, ‘seated’, ‘boy’, ‘girl’ and ‘together’.
Combinatorics and because of that probability is a weird branch of mathematics. Everything depends on the meaning of regular English words, and whether you know something before or not, whereas in most of math these things are quite clear.
If we call boys B and girls G, then we can draw exactly 1 circle
B-G-B-G-B-G -> and circle to the beginning
That’s the only situation where this is the case.
Now if we call the boys and girls by ‘name’, e.g. B1, B2, B3, G1, G2, G3 then suddenly we have way more combinations:
B1 - G1 - B2 - G2 - B3 - G3
B2 - G2 - B1 - G1 - B3 - G3
etc.
We also have to take into account if mirrors and rotations count as new or the same situation. Both will normally lead to different outcomes. Although in this case mirrors lead to already counted situations, because of the circular shape.
If this were a math question on an exam (which it probably is) I would point out all these differences, and then assume the boys and girls are named, otherwise the outcome “exactly 1” is just too easy.
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More answers below
In how many ways can four boys and 4 girls be seated at a round table if each girl is to be sit between two boys?
In how many ways can 8 boys and 4 girls sit on the round table if all girls sit together?
In how many ways can 7 boys sit alternatively with 7 girls around a round table so that there are no two boys or girls sitting next to each other?
In how many ways 3 girls and 3 boys seated in a round table such that no 2 girls and 2 boys sitted together?
In how many ways can 8 girls and 5 boys sit at a round table so that no two boys sit next to each other?
John Smith
Author has 84 answers and 52.3K answer views
·6y
Originally Answered: In how many different ways can 4 boys be seated among five girls around a circular table such that no two boys sit together? ·
First make your 5 girls sit down around. You will
have 5!/5 =4!=24 possibilities. Then take your four boys and a dummy (to fill the missing boy)
and make them sit around in between the girls.
you will have also 4!=24 possibilities. Now, the girls are sitting either in even numbered places or odd numbered. Thus 2 more. Multiplying
all numbers together you will get 24x24x2=1152 possibilities.
if all the girls are the same (as it is often the case) you will have only 48 possibilities.
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Doug Dillon
Ph.D. Mathematics · Author has 12.4K answers and 11.4M answer views
·7y
Originally Answered: In how many ways 3 girls and 3 boys seated in a round table such that no 2 girls and 2 boys sitted together? ·
Place one of the individuals, say a girl, in a seat at 12 o’clock. Then there are girls at 4 and 8 o’clock. There are boys at 2, 6 and 10 o’clock. The second and third girls have 2 and 1 choice for their seats. The boys can be seated in 6 ways. 2×6=12.2×6=12.
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John Zidianakis
BSc in Mathematics, University of Crete (Graduated 1995)
·3y
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In how many ways can 5 boys be seated among 8 girls around a circular desk, such that no two boys will sit next to each other?
Step 1: Put the 8 girls in a circle. This can be done in (8–1)! = 7! different ways
Step 2: Since 2 boys cannot sit next to each other that means that each boy is surrounded by 2 girls. Thus, 2 girls that sit next to each other form a ‘slot’ / seat where one boy (at most) can be placed/seated. Since we have 8 girls in a circle we have 8 such seats (8th seat formed by G1 and G8). In the second step, we select which of those seats we are going to use to place a boy. Since we have 8 seats and we need to select 5 (order doesn’t matter) this can be done in (8C5) different ways i.e. 8!/(5!3!)
Step
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Step 1: Put the 8 girls in a circle. This can be done in (8–1)! = 7! different ways
Step 2: Since 2 boys cannot sit next to each other that means that each boy is surrounded by 2 girls. Thus, 2 girls that sit next to each other form a ‘slot’ / seat where one boy (at most) can be placed/seated. Since we have 8 girls in a circle we have 8 such seats (8th seat formed by G1 and G8). In the second step, we select which of those seats we are going to use to place a boy. Since we have 8 seats and we need to select 5 (order doesn’t matter) this can be done in (8C5) different ways i.e. 8!/(5!3!)
Step 3: We put the 5 boys in the 5 seats (here order does matter) in 5! different ways
So the total combinations S1xS2xS3 = 7![8!/(5!3!)]5! = 7!8!/3! =33868800
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Anthony Madden
Writer for Betterbuck
·Updated Aug 15
What are the weirdest mistakes people make on the internet right now?
Here are a couple of the worst mistakes I’ve seen people make:
Not using an ad blocker
If you aren’t using an ad blocker yet, you definitely should be.
A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load.
No more YouTube ads, no more banner ads, no more pop-up ads, etc.
Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options.
Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your pr
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Here are a couple of the worst mistakes I’ve seen people make:
Not using an ad blocker
If you aren’t using an ad blocker yet, you definitely should be.
A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load.
No more YouTube ads, no more banner ads, no more pop-up ads, etc.
Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options.
Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your privacy, which is nice.
More often than not, it saves even more than 50% on load times - here’s a test I ran:
Using an ad blocker saved a whopping 6.5+ seconds of load time.
Here’s a link to Total Adblock, if you’re interested.
Not getting paid for your screentime
Apps like Freecash will pay you to test new games on your phone.
Some testers get paid as much as £270/game. Here are a few examples right now (from Freecash's website):
You don't need any kind of prior experience or degree or anything: all you need is a smartphone (Android or IOS).
If you're scrolling on your phone anyway, why not get paid for it?
I've used Freecash in the past - it’s solid. (They also gave me a £3 bonus instantly when I installed my first game, which was cool).
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Kavita Chawdhary
I know about Permutations and Combinations. · Author has 170 answers and 1.3M answer views
·8y
Related
In how many ways can 7 boys sit alternatively with 7 girls around a round table so that there are no two boys or girls sitting next to each other?
If we fix the position of one girl at any point (black dot) on the round table, then the other 6 girls (blue dots) can be arranged in 6! ways.
Once the girls sit down, there are 7! ways to arrange the boys (7 orange dots), giving 6!7!6!7! possible seating arrangements in which boys sit alternatively with the girls.
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If we fix the position of one girl at any point (black dot) on the round table, then the other 6 girls (blue dots) can be arranged in 6! ways.
Once the girls sit down, there are 7! ways to arrange the boys (7 orange dots), giving 6!7!6!7! possible seating arrangements in which boys sit alternatively with the girls.
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William Vaughn
Former Research Associate at University of Rochester (1983–2017) · Author has 426 answers and 308.4K answer views
·5y
Related
In how many ways 10 boys and 5 girls can sit around a circular table so that no two girls sit together?
First let's do the “distinguishable" case. It is more straightforward.
The 10 10 boys are seated around table with a single chair between each one, i.e. there are 10 10 empty chairs. There are 9!9! ways of doing this seating since we want to avoid permutations which are related by a simple rotation of the boys. Now count the number of ways we can select 5 5 empty chairs in which to seat the 5 5 girls (eventually removing the other 5 5 empty chairs.) This count is 10!5!5!10!5!5!. Since the girls are distinguishable, there are 5!5! ways of placing them among the 5 5 selected positions. The final count is \frac{9\frac{9
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First let's do the “distinguishable" case. It is more straightforward.
The 10 10 boys are seated around table with a single chair between each one, i.e. there are 10 10 empty chairs. There are 9!9! ways of doing this seating since we want to avoid permutations which are related by a simple rotation of the boys. Now count the number of ways we can select 5 5 empty chairs in which to seat the 5 5 girls (eventually removing the other 5 5 empty chairs.) This count is 10!5!5!10!5!5!. Since the girls are distinguishable, there are 5!5! ways of placing them among the 5 5 selected positions. The final count is 9!10!5!9!10!5!=10,973,471,200=10,973,471,200. In contrast, there around 87 87 billion circular permutations without restrictions.
Now for the “indistinguishable” case. There are far fewer permutations here, but they are trickier to count. It turns out that it's easier “fix" the girls and treat them as 5 5 boundaries for 5 5 bins in which to place the 10 10 boys. We need a minimum of 1 1 boy in each bin which implies a maximum of 6 6 boys in a bin. So we need to enumerate all the partitions of 10 10 into 5 5 parts with a maximum part of 6 6 and a minimum part of 1 1. And we then need to count the number of ways to permute any given partition, but fixing the largest part to eliminate the rotational equivalences. Here is the list I've come up with followed by the number of possible permutations:
6+1+1+1+1 6+1+1+1+1 4!4!4!4!=1=1
5+2+1+1+1 5+2+1+1+1 4!3!1!4!3!1!=4=4
4+3+1+1+1 4+3+1+1+1 4!3!1!4!3!1!=4=4
4+2+2+1+1 4+2+2+1+1 4!2!2!4!2!2!=6=6
3+3+2+1+1 3+3+2+1+1 4!2!1!1!4!2!1!1!=12=12
3+2+2+2+1 3+2+2+2+1 4!3!1!4!3!1!=4=4
2+2+2+2+2 2+2+2+2+2 4!4!4!4!=1=1
And the final total is 32 32.
EDIT: I said this was tricky. The actual answer is 26 26. I over counted the permutations for the 3+3+2+1+1 3+3+2+1+1 case. Here I needed to “fix" the 2 2 and not the 3 3. Then the number of permutations is 4!2!2!4!2!2!=6=6 not 12 12. So the total is 26 26. I've written a Mathematcia program which has verified this. In fact, it pointed to my error. The point is that one must “fix" a non-duplicated number when counting permutations. This is possible in all the above cases except the last which obviously has only one permutation.
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Edwin Asmus
Author has 462 answers and 412.4K answer views
·3y
Originally Answered: In how many ways 3 girls and 3 boys seated in a round table such that no 2 girls and 2 boys sitted together? ·
Divide them in two groups: boy-girl-boy and girl boy-girl. When you make a circle with the 2 groups you’ll end up in an arrangement where no girl sits next to a girl, and no boy is next to another boy.
Now between them, the boys and girls can change position in 3!=6 ways each.
Altogether: 66=36 ways.
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Marco Domizio
Addetto Alla Clientela at Banca Intesa Sanpaolo (1996–present)
·7y
Originally Answered: In how many ways can 5 boys and 5 girls take their seats in a round table so that no 2 girls will sit side by side? ·
Only 1 (m-f-m-f-……) if any girl and any boy counts only as a girl or as boy. Instead if you consider them as different boys and girls (i.e. Peter, Jack, John… and Jane, Susan, Beth), then the possible comvinations are 5x5x4x4x3x3x2x2x1x1=14,400 different combinations (any of 5 boys followe by any of 5 girls gives 25 different combination, followed by any of the remaining 4 boys, followed by any of the remaining 4 girls, and so on).
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Edwin Asmus
Author has 462 answers and 412.4K answer views
·Updated 4y
Related
How many ways can 5 boys and 5 girls be seated at a round table if 3 particular girls must not sit together?
Number the seats 1–10. Seat the 3 particular girls on seats 1–3 and see how many ways there are. Subtract that from the total of seating 10 persons.
As the table is circular it doesn’t change a thing if the girls would occupy any other seats, only the order of the persons counts.
Total #ways all can be seated is 10!/10=362.880
You have to divide by 10 as for each order of persons, there are 10 ways (ten times all persons can be moved one place, but the order is still the same).
Ways the 3 girls can sit together on is 3!=6
Ways the others can wrap around the girls=7!
Now you don’t divide by 7, beca
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Number the seats 1–10. Seat the 3 particular girls on seats 1–3 and see how many ways there are. Subtract that from the total of seating 10 persons.
As the table is circular it doesn’t change a thing if the girls would occupy any other seats, only the order of the persons counts.
Total #ways all can be seated is 10!/10=362.880
You have to divide by 10 as for each order of persons, there are 10 ways (ten times all persons can be moved one place, but the order is still the same).
Ways the 3 girls can sit together on is 3!=6
Ways the others can wrap around the girls=7!
Now you don’t divide by 7, because the 3 girls form 2 beginning/end points, like a row has a 2 end points.
Ways altogether with the 3 girls together is 67!=30.240
Total #ways - #ways altogether with the 3 girls together = 362.880 - 30.240 = 332.640 ways.
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In how many ways 6 boys and 6 girls sit around a table, if no two boys should sit together?
Assuming the fact that the table is circular if we arrange the boys first we notice that there are 6 boys sitting and there are 6 gaps between the boys. Now the boys can sit in a circular table in 5! = 120 ways. (apply circular permutation concept)
Now we start placing the girls one by one in the gaps between the 2 boys and the 6 girls can be placed in the 6 gaps in 6! ways. This way it is ensured that no 2 boys sit side by side.
Thus the total number of ways for this kind of sitting arrangement is 6!5! = 86400 ways.
Note that if the table had a different shape-for example rectangular then there
Continue Reading
Assuming the fact that the table is circular if we arrange the boys first we notice that there are 6 boys sitting and there are 6 gaps between the boys. Now the boys can sit in a circular table in 5! = 120 ways. (apply circular permutation concept)
Now we start placing the girls one by one in the gaps between the 2 boys and the 6 girls can be placed in the 6 gaps in 6! ways. This way it is ensured that no 2 boys sit side by side.
Thus the total number of ways for this kind of sitting arrangement is 6!5! = 86400 ways.
Note that if the table had a different shape-for example rectangular then there has to be a different approach as you cannot apply circular permutation blindly there.
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837 | https://www.cadence.com/en_US/home/explore/transient-thermal-analysis.html | What Is Transient Thermal Analysis? | Transient Thermal Analysis Overview | Cadence
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Transient Thermal Analysis
Glossary
Transient Thermal Analysis
What Is Transient Thermal Analysis?
Transient thermal analysis is the evaluation of how a system responds to fixed and varying boundary conditions over time. For fixed boundary conditions, the time to reach a steady state temperature can be evaluated, as well as how long operating conditions can be sustained before reaching a threshold temperature. For time-varying boundary conditions, transient analysis can show you the resulting thermal response of the system.
Frequently, transient analysis is used to evaluate transitions in operating conditions before the system returns to steady state. In such cases, it is typical to find the steady-state solution to one set of operating conditions and then perform a transient analysis using a different set of operating conditions. Transient analysis is also useful for evaluating control algorithms, using inputs such as time and temperature to control outputs such as power and fan speed.
Why Is Transient Thermal Analysis Important?
Transient thermal analysis is essential to electronics simulation. Whether you are testing control algorithms or cooling failures, the most critical processes are transient. Transient thermal analysis is how we learn about a system’s time-dependent response to varying conditions.
1. Transient Controls
Figure 1. Analyze control systems with transient analysis
Transient analysis is critical when designing and validating the control system. An engineer needs to know that the control system can keep temperatures below a certain threshold without too much of a performance impact. Standard control algorithms may throttle power to components and fans in order to manage heat dissipation within the system. If the device is being throttled too frequently, then they may need to improve the thermal management of the device as a whole. Control algorithms can be based on a temperature response, built in as a function of time during a known process, or some combination of the two. To accurately analyze controls, it’s important to have full control over the time grid to fully capture the heat transfer process.
2. Cooling Failure
Figure 2. A blade server with three fans that examines the system temperature response to a single fan failure
Another common reason for employing transient thermal analysis is to evaluate cooling failure. Cooling failure occurs when a fan or liquid cooling system fails. Thermal simulation allows an engineer to understand how the system copes with the failure. This may be relevant for power throttling or controls designed to prevent component overheat and eventually full system failure. The above example is a blade server with three fans and examines the system temperature response to a single fan failure. The resulting temperature response can be used to inform control algorithms designed to handle such a situation, or the controls can be directly implemented into the simulation. Failure simulations are typically initialized with a steady state solution with full cooling capacity. The relevant fan(s) can then be turned off before initializing a transient simulation to determine the system response.
3. Transient Processes
A final example of transient thermal analysis involves the analysis of known transient processes. Many electronic products operate with frequently known, power-intensive processes. A device might include several components following different power profiles over time. The temperature response to these processes may be relevant to failure, durability, and performance analysis. Another situation in which this sort of analysis is relevant is when a system is subject to changing ambient conditions. Devices with outdoor applications may be subject to known fluctuations in temperature and solar radiation that are critical to reliability evaluation.
Figure 3. The component operating power increases over time to a steady state value
The above example relates to a startup sequence in which the component operating power increases over time to a steady state value. Optimizing this process may increase power performance that contributes to overall system performance. For devices with a frequent use-case (e.g., a camera), analyzing the system behavior during the process can be more insightful than any steady state analysis.
Transient Modeling with Cadence
When solving a transient model in Cadence Celsius EC Solver, there are a few important points to consider. The first is related to the time step. It is better (and often quicker) to model many small time steps than a few large time steps for a given simulation length. Smaller time steps result in more accurate transient results, and each time step will converge more quickly than a longer time step due to the small change between each step.
Transient Thermal Analysis and Thermal Simulation
Users have come to expect more powerful and complex, yet smaller and more compact devices. Engineers must create innovative solutions to manage rising thermal demands now more than ever. Transient analysis is so important because it enables engineers to push the boundaries of thermal management.
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838 | https://en.wikipedia.org/wiki/Pingelap | Pingelap - Wikipedia
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1 History
2 Climate
3 Total color blindness
4 References
5 External links
Pingelap
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Coordinates: 6°13′5″N 160°42′10″E / 6.21806°N 160.70278°E / 6.21806; 160.70278
From Wikipedia, the free encyclopedia
Atoll in Pohnpei, Micronesia
| Pingelap Pingelap Atoll |
| Atoll |
| A satellite image of the Pingelap atoll at low-tide. The pale strip in the center is the airstrip. |
| Pingelap Location of Pingelap in the Pacific Pingelap Pingelap (Pacific Ocean) Show map of Federated States of Micronesia Show map of Pacific Ocean Show all |
| Coordinates: 6°13′5″N 160°42′10″E / 6.21806°N 160.70278°E / 6.21806; 160.70278 |
| Country | Micronesia |
| State | Pohnpei |
| Area |
| •Land | 1.8 km 2 (0.69 sq mi) |
| Dimensions |
| •Width | 4 km (2.5 mi) |
| Population |
| •Total | 250 |
Pingelap is an atoll in the Pacific Ocean, part of Pohnpei State of the Federated States of Micronesia, consisting of three islands: Pingelap Island, Sukoru and Daekae, linked by a reef system and surrounding a central lagoon, although only Pingelap Island is inhabited. The entire system has a land area of 1.8 km² (455 acres) at high-tide, and is less than 2.5 miles (4.0 km) at its widest point. The atoll has its own language, Pingelapese, spoken by most of the atoll's 250 residents.
History
[edit]
A map of the Pohnpei region. Pingelap is at the eastern edge of the map.
The first European observer of the islands was Captain Thomas Musgrave in the ship Sugar Cane. Captain MacAskill in Lady Barlow revisited them in 1809. Errors in measurement of their location resulted in the islands being separately named on charts in the 19th century as the Musgrave Islands and the MacAskill islands, within the Caroline archipelago.
Historically, the Ouwa or the Paramount Chief or King of Pikelap which is a hereditary title that granted supreme rule of the land, ruled the island of Pikelap which is now known as Pingelap. Japan seized the atoll in October 1914, following the start of World War I. The hereditary system remained in place during Japanese rule, although the title was renamed "Island Magistrate".
Japan used the southern part of Pingelap Island during hostilities in the Pacific Ocean theater of World War II for a supply base. Allied Forces later attacked it. The presence of foreign troops on the island led to the introduction of a number of infectious diseases, including gonorrhoea, tuberculosis and dysentery, which reduced the population from its pre-war level of around 1000 to 800, and decreased the fertility rate significantly.
The arrival of the U.S. Navy in 1945 resulted in the setting up of a democratically elected system alongside the traditional system, which gradually weakened in power. Universal primary education was provided for Pingelapese children and a limited health care scheme was set up to eradicate the diseases introduced during the war.
During the 1960s, the Peace Corps and U.S. Air Force settled on the main island. The U.S. Air Force constructed a missile watching station in the northeast of the island and a pier, with work beginning in 1978 on an airstrip, jutting into the lagoon, on the main island. The runway was finished in 1982, and currently Caroline Islands Air makes two or three flights daily to and from the atoll.
Climate
[edit]
Pingelap enjoys a tropical climate, with even, warm temperatures throughout the year.
Precipitation is generally plentiful, with heavy year-round rainfall.
| Pingelap, Pohnpei, Federated States of Micronesia elevation 8 feet (2.4 m) |
| Climate chart (explanation) |
| J F M A M J J A S O N D 13 89 74 10 90 75 13 90 74 12 90 74 13 90 75 13 90 74 13 90 74 13 90 74 14 90 74 12 90 74 13 90 74 13 89 75 █ Average max. and min. temperatures in °F █ Precipitation totals in inches Source: 1981-2010 provisional normals [dead link] |
| | show Metric conversion | | J | F | M | A | M | J | J | A | S | O | N | D | | 323 32 23 | 265 32 24 | 318 32 24 | 314 32 23 | 336 32 24 | 329 32 24 | 333 32 23 | 332 32 23 | 362 32 23 | 315 32 24 | 332 32 23 | 328 32 24 | | █ Average max. and min. temperatures in °C | | █ Precipitation totals in mm | |
Total color blindness
[edit]
A significant proportion of the population has complete achromatopsia due to total absence of working cones in their retinas, leaving them with only rods, a recessivegenetic disorder that causes total color blindness in sufferers. This condition is known on the island as maskun, meaning literally "not see" in Pingelapese.
Complete achromatopsia is normally a very rare condition, and its prevalence on the island has been traced back to a population bottleneck in 1775 after a catastrophic typhoon swept through the island, leaving only about 20 survivors. One of these, Doahkaesa Mwanenihsed (the ruler at that time), is now believed to have been a carrier for the underlying genetic condition, but the achromatopsia disorder did not appear until the fourth generation after the typhoon, by which time 2.7% of the Pingelapese were affected. Since achromatopsia is an autosomal recessive disorder, inbreeding between the descendants of Doahkaesa Mwanenised would result in an increased recessive allele frequency. By generation six, the incidence rose to approximately 4.9%, due to the founder effect and inbreeding, with all achromats on the island nowadays tracing their ancestry to Doahkaesa Mwanenihsed.
Today the atoll is still of particular interest to geneticists; due to the small gene pool and rapid population growth, the disorder is now prevalent in almost 10% of the population, with a further 30% being unaffected carriers. (By comparison, in the United States, only 1 in 33,000, or 0.003%, are affected). Leading neurologist Oliver Sacks's 1997 book The Island of the Colorblind references the island.
It is reported that one Pingelapese island sea-fisherman with this condition has difficulty seeing in bright sunlight, but at night can see in much fainter light than people with normal vision can; he uses this ability in a boat at night waving a large burning torch about to attract or confuse flying fish, which he then catches; the flying fish act as if the torch is the moon. When Sanne De Wilde was photographing the island, she said that red was the most common color the islanders claimed to "see". Despite green being one of the colors they are least able to recognize, many described it as their favorite color, which De Wilde attributed to their love of the jungle vegetation.
References
[edit]
Wikimedia Commons has media related to Pingelap.
^ Jump up to: abcdDamas, David (1994). Bountiful Island: A Study of Land Tenure on a Micronesian Atoll. Wilfrid Laurier University Press. ISBN0-88920-239-7.
^Damas, David (1985). "Pingelap Politics and American-Micronesian Relations". Ethnology. 24 (1): 43–55. doi:10.2307/3773489. JSTOR3773489.
^Findlay (1851), Vol. 2, p. 1076.
^Brigham (1900), Vol. 1, issue 2, p. 131.
^"Micronesian Diary: Pingelap, Phonpei". intangible.org. Retrieved 2007-06-13.
^Morton, N.E.; Hussels, I.E.; Lew, R.; Little, G.F. (1972). "Pingelap and Mokil Atolls: historical genetics". American Journal of Human Genetics. 24 (3): 277–289. PMC1762283. PMID4537352.
^ Jump up to: abHussels, I.E.; Mortons, N.E. (1972). "Pingelap and Mokil Atolls: achromatopsia". American Journal of Human Genetics. 24 (3): 304–309. PMC1762260. PMID4555088.
^Cabe, Paul R. (2004). "Inbreeding and Assortive Mating". Encyclopedia of genetics. Vol.2ed.
^"The Achromatopsia Group". Retrieved 2007-06-13.
^Sacks, Oliver (1997). The Island of the Colour-blind. Picador. ISBN0-330-35887-1.
^BBCTV2 program "Countdown to Life, the Extraordinary Making of You", part 3, 28 September 2015
^Alberge, Dalya (12 July 2017). "'Island of the Colorblind': Where the jungle is pink, the sea is gray and the light is far too bright". CNN. Retrieved 2020-04-13.
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Brigham, William Tufts (1900) An Index to the Islands of the Pacific Ocean: A Handbook to the Chart on the Walls of the Bernice Pauahi Bishop Museum of Polynesian Ethnology and Natural History. (Bishop Museum Press)
Findlay, A.G. (1851; reprinted 2013) A Directory for the Navigation of the Pacific Ocean, with Descriptions of Its Coasts, Islands, Etc.: From the Strait of Magalhaens to the Arctic Sea, and Those of Asia and Australia. (Cambridge University). ISBN9781108059732
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839 | https://www.totalmateria.com/en-us/articles/engineering-stress-strain-curve/ | chevron_left Articles
Engineering Stress-strain Curve: Part One
Abstract
When subjected to tensile forces, the engineering stress-strain curve is a critical representation of a material's mechanical properties. By plotting stress against strain, this curve helps in understanding key attributes such as material strength, elasticity, and ductility. The curve is divided into distinct regions, with the initial elastic zone where stress and strain are proportional, followed by plastic deformation where permanent changes occur. Critical parameters like yield strength, tensile strength, and elongation are extracted from this curve, providing valuable insights for material selection and engineering design. Factors such as material composition, heat treatment, and testing conditions all influence the shape and characteristics of the curve. This tool is essential in evaluating material behavior for applications requiring specific mechanical performance.
Introduction
The engineering stress-strain curve is a vital aspect of materials testing, providing crucial data for material selection and design. This curve is constructed by applying a uniaxial tensile force to a specimen while simultaneously measuring its elongation. The resulting curve is essential in determining several key properties of materials, including strength and ductility.
Figure 1. The engineering stress-strain curve
It is obtained by dividing the load by the original area of the cross-section of the specimen.
The strain used for the engineering stress-strain curve is the average linear strain, which is obtained by dividing the elongation of the gage length of the specimen, d, by its original length.
Stress and Strain Definitions
In the context of the engineering stress-strain curve, stress is defined as the applied load divided by the original cross-sectional area of the specimen. Strain is the elongation of the specimen divided by its original length. These two parameters allow the construction of the stress-strain curve, which can be used to assess the mechanical performance of a material.
Factors Affecting the Stress-Strain Curve
The shape and magnitude of the stress-strain curve depend on several factors:
Composition: Different alloying elements can significantly affect a material's mechanical properties.
Heat Treatment: Materials that have undergone different heat treatments will show distinct stress-strain behavior.
Prior History of Deformation: A material's previous processing or deformation history can influence its response to stress.
Testing Conditions: Variables such as strain rate and temperature can alter the curve's shape.
The Elastic Region and Plastic Deformation
In the elastic region of the curve, stress is proportional to strain, and the material will return to its original shape when the load is removed. Once the material exceeds its yield strength, it undergoes plastic deformation, which is permanent. As the load continues to increase, the material begins to strain-harden, meaning that more force is required for further deformation. However, eventually, the specimen begins to neck, which causes a localized reduction in cross-sectional area, and the stress begins to decrease until fracture occurs.
Tensile Strength
Tensile strength (or ultimate tensile strength) refers to the maximum load a material can withstand before failure.
Although this value is often quoted in engineering applications, it is not always the most relevant measure of material strength, particularly for ductile materials. For design purposes, yield strength is typically a more reliable parameter, as it reflects the material's ability to perform under more realistic loading conditions.
Yielding and Related Concepts
There are several ways to define the point at which a material begins to yield or undergo plastic deformation:
True Elastic Limit: The lowest stress at which permanent deformation occurs, measured at microscopic levels.
Proportional Limit: The highest stress at which stress is proportional to strain.
Elastic Limit: The maximum stress a material can handle without permanent strain when the load is removed.
Offset Yield Strength: The stress required to produce a specified amount of plastic deformation, commonly defined as the stress where the curve deviates from the elastic region by a set offset strain.
A good way of looking at offset yield strength is that after a specimen has been loaded to its 0.2 percent offset yield strength and then unloaded it will be 0.2 percent longer than before the test. The offset yield strength is often referred to in Great Britain as the proof stress, where offset values are either 0.1 or 0.5 percent. The yield strength obtained by an offset method is commonly used for design and specification purposes because it avoids the practical difficulties of measuring the elastic limit or proportional limit.
Some materials have essentially no linear portion to their stress-strain curve, for example, soft copper or gray cast iron. For these materials the offset method cannot be used and the usual practice is to define the yield strength as the stress to produce some total strain, for example, e = 0.005.
Conclusion
The engineering stress-strain curve is a fundamental tool in material testing, providing valuable information on material behavior under stress. It helps engineers determine the appropriate material for various applications, ensuring performance and safety. Understanding the different parameters of the curve, including tensile strength, yield strength, and ductility, is essential for making informed decisions in material selection and design.
September, 2004
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Total Materia Horizon includes a unique collection of stress-strain curves of metallic and nonmetallic materials. Both true and engineering stress curves are given, for various strain rates, heat treatments and working temperatures where applicable.
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References
S.Kumar BaJena: Testing of Materials, (MM 15 025) B.Tech, 6th Semester, The tension test, Veer Surendra SAI University of Technol Burla-768018; Accessed MAY 2020;
C. Moosbrugger: Representation of Stress-Strain Behavior, Atlas of Stress-strain Curves, ASM International, 2002, ISBN 087170739X and 9780871707390.
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840 | https://www.webqc.org/molecular-weight-of-agno3.html | AgNO3 (Silver nitrate) molar mass
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Molar Mass, Molecular Weight and Elemental Composition Calculator
Enter a chemical formula to calculate its molar mass and elemental composition:
Molar mass of AgNO 3 (Silver nitrate) is 169.8731 g/mol
Discover more
Ag Silver
Silver
Ag
Silver nitrate
Convert between AgNO 3 weight and moles
| Compound | Moles | Weight, g |
---
| AgNO 3 | | | | |
Elemental composition of AgNO 3
| Element | Symbol | Atomic weight | Atoms | Mass percent |
--- ---
| Silver | Ag | 107.8682 | 1 | 63.4993 |
| Nitrogen | N | 14.0067 | 1 | 8.2454 |
| Oxygen | O | 15.9994 | 3 | 28.2553 |
| Computing molar mass step by step |
| First, compute the number of each atom in AgNO 3: Ag: 1, N: 1, O: 3 Then, lookup atomic weights for each element in periodic table: Ag: 107.8682, N: 14.0067, O: 15.9994 Now, compute the sum of products of number of atoms to the atomic weight: Molar mass (AgNO 3) = ∑ Count i Weight i = Count(Ag) Weight(Ag) + Count(N) Weight(N) + Count(O) Weight(O) = 1 107.8682 + 1 14.0067 + 3 15.9994 = 169.8731 g/mol |
| Mass Percent Composition | Atomic Percent Composition |
--- |
| Ag Silver (63.50%) N Nitrogen (8.25%) O Oxygen (28.26%) | Ag Silver (20.00%) N Nitrogen (20.00%) O Oxygen (60.00%) |
| Mass Percent Composition |
| Ag Silver (63.50%) N Nitrogen (8.25%) O Oxygen (28.26%) |
| Atomic Percent Composition |
| Ag Silver (20.00%) N Nitrogen (20.00%) O Oxygen (60.00%) |
| Chemical structure |
| |
| Lewis structure |
| 3D molecular structure |
| Appearance |
| Silver nitrate appears as a white crystalline solid. |
Sample reactions for AgNO 3
| Equation | Reaction type |
--- |
| Cu + AgNO 3 = Cu(NO 3)2 + Ag | single replacement |
| AgNO 3 + NaCl = AgCl + NaNO 3 | double replacement |
| AgNO 3 + MgCl 2 = AgCl + Mg(NO 3)2 | double replacement |
| AgNO 3 + CaCl 2 = AgCl + Ca(NO 3)2 | double replacement |
| Zn + AgNO 3 = Zn(NO 3)2 + Ag | single replacement |
Related compounds
| Formula | Compound name |
--- |
| AgNO 2 | Silver nitrite |
| Ag 2 N 2 O 2 | Silver hyponitrite |
| Related |
| Oxidation state calculator |
| Compound properties |
Computing molar mass (molar weight)
To calculate molar mass of a chemical compound enter its formula and click 'Compute'. In chemical formula you may use:
Any chemical element. Capitalize the first letter in chemical symbol and use lower case for the remaining letters: Ca, Fe, Mg, Mn, S, O, H, C, N, Na, K, Cl, Al.
Functional groups: D, T, Ph, Me, Et, Bu, AcAc, For, Tos, Bz, TMS, tBu, Bzl, Bn, Dmg
parenthesis () or brackets [].
Common compound names.
Examples of molar mass computations: NaCl, Ca(OH)2, K4[Fe(CN)6], CuSO45H2O, nitric acid, potassium permanganate, ethanol, fructose, caffeine, water.
Molar mass calculator also displays common compound name, Hill formula, elemental composition, mass percent composition, atomic percent compositions and allows to convert from weight to number of moles and vice versa.
Computing molecular weight (molecular mass)
To calculate molecular weight of a chemical compound enter it's formula, specify its isotope mass number after each element in square brackets.
Examples of molecular weight computations: CO2, SO2.
Definitions
Molecular mass (molecular weight) is the mass of one molecule of a substance and is expressed in the unified atomic mass units (u). (1 u is equal to 1/12 the mass of one atom of carbon-12)
Molar mass (molar weight) is the mass of one mole of a substance and is expressed in g/mol.
Mole is a standard scientific unit for measuring large quantities of very small entities such as atoms and molecules. One mole contains exactly 6.022 ×10 23 particles (Avogadro's number)
Steps to calculate molar mass
Identify the compound: write down the chemical formula of the compound. For example, water is H 2 O, meaning it contains two hydrogen atoms and one oxygen atom.
Find atomic masses: look up the atomic masses of each element present in the compound. The atomic mass is usually found on the periodic table and is given in atomic mass units (amu).
Calculate molar mass of each element: multiply the atomic mass of each element by the number of atoms of that element in the compound.
Add them together: add the results from step 3 to get the total molar mass of the compound.
Example: calculating molar mass
Let's calculate the molar mass of carbon dioxide (CO 2):
Carbon (C) has an atomic mass of about 12.01 amu.
Oxygen (O) has an atomic mass of about 16.00 amu.
CO 2 has one carbon atom and two oxygen atoms.
The molar mass of carbon dioxide is 12.01 + (2 × 16.00) = 44.01 g/mol.
Lesson on computing molar mass
Practice what you learned:
Practice calculating molar mass
Weights of atoms and isotopes are from NIST article.
Related: Molecular weights of amino acids
molecular weights calculated today
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841 | https://math.stackexchange.com/questions/1441383/why-does-a-ac-ac | boolean algebra - Why does (A'+ AC) = (A'+C)? - Mathematics Stack Exchange
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Why does (A'+ AC) = (A'+C)?
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Why does (A'+ AC) = (A'+C)?
I can understand this via a truth table, but I cannot see why this works in boolean algebra. Conceptually I understand that A doesn't matter, but I can not seem to prove this. Can someone lend some advice?
boolean-algebra
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edited Sep 18, 2015 at 18:59
AdamAdam
asked Sep 18, 2015 at 18:54
AdamAdam
103 7 7 bronze badges
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Try this using Truth table.user249332 –user249332 2015-09-18 18:57:07 +00:00 Commented Sep 18, 2015 at 18:57
I understand the truth table. Would you like me to show you the truth table I have?Adam –Adam 2015-09-18 18:58:26 +00:00 Commented Sep 18, 2015 at 18:58
I am trying to see how to prove this, using the distributive law doesn't work for me.Adam –Adam 2015-09-18 19:00:47 +00:00 Commented Sep 18, 2015 at 19:00
See this answer, I think, this should satisfy you.user249332 –user249332 2015-09-18 19:06:54 +00:00 Commented Sep 18, 2015 at 19:06
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A′+A C=A′(1+C)+A C=A′+(A′+A)C=A′+C A′+A C=A′(1+C)+A C=A′+(A′+A)C=A′+C
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answered Sep 18, 2015 at 19:05
user249332 user249332
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How are we able to get A'(1+C)?Adam –Adam 2015-09-18 19:07:51 +00:00 Commented Sep 18, 2015 at 19:07
2 1+C=1 1+C=1, for whatever C C be (0 0 or 1 1). So, A′=A′×1=A′×(1+C)A′=A′×1=A′×(1+C). Now, did you got?user249332 –user249332 2015-09-18 19:10:12 +00:00 Commented Sep 18, 2015 at 19:10
That makes since, but how does A =(A'+A)?Adam –Adam 2015-09-18 19:14:50 +00:00 Commented Sep 18, 2015 at 19:14
I never told A=(A′+A)A=(A′+A), it is A′(1+C)+A C=A′+A′C+A C=A′+(A+A′)C A′(1+C)+A C=A′+A′C+A C=A′+(A+A′)C. Now, A+A′=1 A+A′=1. So, your result.user249332 –user249332 2015-09-18 19:18:06 +00:00 Commented Sep 18, 2015 at 19:18
I understand now, thank you very much for your help/Adam –Adam 2015-09-18 19:19:38 +00:00 Commented Sep 18, 2015 at 19:19
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In general, in boolean algebra we also have X+Y Z=(X+Y)(X+Z)X+Y Z=(X+Y)(X+Z).
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answered Sep 18, 2015 at 21:46
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842 | https://openstax.org/books/precalculus-2e/pages/chapter-1 | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Precalculus 2e
Chapter 1
Precalculus 2eChapter 1
Search for key terms or text.
Try It
1.1 Functions and Function Notation
1.
ⓐ yes
ⓑ yes. (Note: If two players had been tied for, say, 4th place, then the name would not have been a function of rank.)
2.
3.
4.
5.
6.
7.
8.
or
9.
ⓐ yes, because each bank account has a single balance at any given time
ⓑ no, because several bank account numbers may have the same balance
ⓒ no, because the same output may correspond to more than one input.
10.
ⓐ Yes, letter grade is a function of percent grade;
ⓑ No, it is not one-to-one. There are 100 different percent numbers we could get but only about five possible letter grades, so there cannot be only one percent number that corresponds to each letter grade.
11.
yes
12.
No, because it does not pass the horizontal line test.
1.2 Domain and Range
1.
2.
3.
4.
5.
ⓐ values that are less than or equal to –2, or values that are greater than or equal to –1 and less than 3;
ⓑ ;
ⓒ
6.
domain =[1950,2002] range = [47,000,000,89,000,000]
7.
domain: range:
8.
1.3 Rates of Change and Behavior of Graphs
1.
per year.
2.
3.
4.
The local maximum appears to occur at and the local minimum occurs at The function is increasing on and decreasing on
1.4 Composition of Functions
1.
No, the functions are not the same.
2.
A gravitational force is still a force, so makes sense as the acceleration of a planet at a distance r from the Sun (due to gravity), but does not make sense.
3.
and
4.
5.
ⓐ 8
ⓑ 20
6.
7.
Possible answer:
1.5 Transformation of Functions
1.
2.
The graphs of and are shown below. The transformation is a horizontal shift. The function is shifted to the left by 2 units.
3.
4.
5.
6.
ⓐ
| | | | | |
--- ---
| | -2 | 0 | 2 | 4 |
| | | | | |
2. ⓑ
| | | | | |
--- ---
| | -2 | 0 | 2 | 4 |
| | 15 | 10 | 5 | unknown |
7.
Notice: looks the same as .
8.
9.
| | 2 | 4 | 6 | 8 |
| | 9 | 12 | 15 | 0 |
10.
11.
so using the square root function we get
1.6 Absolute Value Functions
1.
2.
using the variable for passing,
3.
4.
or
5.
so the graph intersects the vertical axis at when and so the graph intersects the horizontal axis at and
6.
7.
or in interval notation, this would be
1.7 Inverse Functions
1.
2.
Yes
3.
Yes
4.
The domain of function is and the range of function is
5.
In 60 minutes, 50 miles are traveled.
To travel 60 miles, it will take 70 minutes.
6.
a. 3; b. 5.6
7.
8.
9.
1.1 Section Exercises
1.
A relation is a set of ordered pairs. A function is a special kind of relation in which no two ordered pairs have the same first coordinate.
3.
When a vertical line intersects the graph of a relation more than once, that indicates that for that input there is more than one output. At any particular input value, there can be only one output if the relation is to be a function.
5.
When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to only one input.
7.
function
9.
function
11.
function
13.
function
15.
function
17.
function
19.
function
21.
function
23.
function
25.
not a function
27.
29.
31.
33.
35.
ⓐ
ⓑ
37.
ⓐ
ⓑ or
39.
ⓐ
ⓑ
ⓒ
41.
not a function
43.
function
45.
function
47.
function
49.
function
51.
function
53.
ⓐ
ⓑ or
55.
not a function so it is also not a one-to-one function
57.
one-to-one function
59.
function, but not one-to-one
61.
function
63.
function
65.
not a function
67.
69.
71.
73.
75.
20
77.
79.
81.
83.
85.
87.
89.
ⓐ
ⓑ The number of cubic yards of dirt required for a garden of 100 square feet is 1.
91.
ⓐ The height of a rocket above ground after 1 second is 200 ft.
ⓑ the height of a rocket above ground after 2 seconds is 350 ft.
1.2 Section Exercises
1.
The domain of a function depends upon what values of the independent variable make the function undefined or imaginary.
3.
There is no restriction on for because you can take the cube root of any real number. So the domain is all real numbers, When dealing with the set of real numbers, you cannot take the square root of negative numbers. So -values are restricted for to nonnegative numbers and the domain is
5.
Graph each formula of the piecewise function over its corresponding domain. Use the same scale for the -axis and -axis for each graph. Indicate inclusive endpoints with a solid circle and exclusive endpoints with an open circle. Use an arrow to indicate or Combine the graphs to find the graph of the piecewise function.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
domain: range
29.
domain: range:
31.
domain: range:
33.
domain: range:
35.
domain: range:
37.
domain: range:
39.
domain:
41.
domain:
43.
domain:
45.
domain:
47.
49.
51.
53.
domain:
55.
window: range:
window: range:
57.
59.
Many answers. One function is
1.3 Section Exercises
1.
Yes, the average rate of change of all linear functions is constant.
3.
The absolute maximum and minimum relate to the entire graph, whereas the local extrema relate only to a specific region around an open interval.
5.
7.
3
9.
11.
13.
15.
17.
19.
increasing on decreasing on
21.
increasing on decreasing on
23.
local maximum: local minimum:
25.
absolute maximum at approximately absolute minimum at approximately
27.
a. –3000; b. –1250
29.
-4
31.
27
33.
–0.167
35.
Local minimum at decreasing on increasing on
37.
Local minimum at decreasing on increasing on
39.
Local maximum at local minima at and decreasing on increasing on
41.
A
43.
45.
2.7 gallons per minute
47.
approximately –0.6 milligrams per day
1.4 Section Exercises
1.
Find the numbers that make the function in the denominator equal to zero, and check for any other domain restrictions on and such as an even-indexed root or zeros in the denominator.
3.
Yes. Sample answer: Let Then and So
5.
domain:
domain:
domain:
domain:
7.
domain:
domain:
domain:
domain:
9.
domain:
domain:
domain:
domain:
11.
ⓐ 3
ⓑ
ⓒ
ⓓ
ⓔ
13.
15.
17.
19.
21.
ⓐ Text
ⓑ
23.
ⓐ
ⓑ c.
25.
27.
sample:
29.
sample:
31.
sample:
33.
sample:
35.
sample:
37.
sample:
39.
sample:
41.
sample:
43.
2
45.
5
47.
4
49.
0
51.
2
53.
1
55.
4
57.
4
59.
9
61.
4
63.
2
65.
3
67.
11
69.
0
71.
7
73.
75.
77.
79.
81.
2
83.
85.
False
87.
;
89.
91.
c
93.
and square inches
95.
square units
97.
ⓐ
ⓑ 3.38 hours
1.5 Section Exercises
1.
A horizontal shift results when a constant is added to or subtracted from the input. A vertical shifts results when a constant is added to or subtracted from the output.
3.
A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output.
5.
For a function substitute for in Simplify. If the resulting function is the same as the original function, then the function is even. If the resulting function is the opposite of the original function, then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even.
7.
9.
11.
The graph of is a horizontal shift to the left 43 units of the graph of
13.
The graph of is a horizontal shift to the right 4 units of the graph of
15.
The graph of is a vertical shift up 8 units of the graph of
17.
The graph of is a vertical shift down 7 units of the graph of
19.
The graph of is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of
21.
decreasing on and increasing on
23.
decreasing on
25.
27.
29.
31.
33.
35.
37.
39.
41.
43.
45.
47.
even
49.
odd
51.
even
53.
The graph of is a vertical reflection (across the -axis) of the graph of
55.
The graph of is a vertical stretch by a factor of 4 of the graph of
57.
The graph of is a horizontal compression by a factor of of the graph of
59.
The graph of is a horizontal stretch by a factor of 3 of the graph of
61.
The graph of is a horizontal reflection across the -axis and a vertical stretch by a factor of 3 of the graph of
63.
65.
67.
69.
The graph of the function is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units.
71.
The graph of is stretched vertically by a factor of 2, shifted horizontally 4 units to the right, reflected across the horizontal axis, and then shifted vertically 3 units up.
73.
The graph of the function is compressed vertically by a factor of
75.
The graph of the function is stretched horizontally by a factor of 3 and then shifted vertically downward by 3 units.
77.
The graph of is shifted right 4 units and then reflected across the vertical line
79.
81.
1.6 Section Exercises
1.
Isolate the absolute value term so that the equation is of the form Form one equation by setting the expression inside the absolute value symbol, equal to the expression on the other side of the equation, Form a second equation by setting equal to the opposite of the expression on the other side of the equation, Solve each equation for the variable.
3.
The graph of the absolute value function does not cross the -axis, so the graph is either completely above or completely below the -axis.
5.
First determine the boundary points by finding the solution(s) of the equation. Use the boundary points to form possible solution intervals. Choose a test value in each interval to determine which values satisfy the inequality.
7.
9.
11.
13.
15.
17.
19.
21.
No solution
23.
25.
27.
no -intercepts
29.
31.
33.
35.
37.
39.
41.
43.
45.
47.
49.
51.
53.
range:
55.
intercepts:
57.
59.
There is no solution for that will keep the function from having a -intercept. The absolute value function always crosses the -intercept when
61.
63.
1.7 Section Exercises
1.
Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that -values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no -values repeat and the function is one-to-one.
3.
Yes. For example, is its own inverse.
5.
Given a function solve for in terms of Interchange the and Solve the new equation for The expression for is the inverse,
7.
9.
11.
13.
domain of
15.
domain of
16.
ⓐ and
ⓑ This tells us that and are inverse functions
17.
19.
one-to-one
21.
one-to-one
23.
not one-to-one
25.
27.
29.
31.
33.
35.
37.
39.
41.
| | | | | | |
--- --- --- |
| | 1 | 4 | 7 | 12 | 16 |
| | 3 | 6 | 9 | 13 | 14 |
43.
45.
Given the Fahrenheit temperature, this formula allows you to calculate the Celsius temperature.
47.
The time for the car to travel 180 miles is 3.6 hours.
Review Exercises
1.
function
3.
not a function
5.
7.
one-to-one
9.
function
11.
function
13.
15.
17.
or
19.
21.
23.
25.
27.
increasing decreasing
29.
increasing constant
31.
local minimum local maximum
33.
Absolute Maximum: 10
35.
37.
39.
41.
43.
sample:
45.
47.
49.
51.
53.
55.
57.
even
59.
odd
61.
even
63.
65.
67.
69.
71.
73.
77.
The function is one-to-one.
78.
The function is not one-to-one.
79.
Practice Test
1.
The relation is a function.
3.
5.
The graph is a parabola and the graph fails the horizontal line test.
7.
9.
11.
13.
15.
17.
19.
and
21.
23.
25.
27.
29.
31.
33.
yes
35.
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843 | https://blog.csdn.net/mark_to_win/article/details/100580272 | 什么是导数和切线？以及他们的关系？_切线的斜率:设p0(x0,y0)是曲线y=f(x)上一点,则曲线y=f(x)在点p0(x0,y0)处-CSDN博客
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什么是导数和切线？以及他们的关系？
函数导数与几何意义
最新推荐文章于 2025-09-04 10:20:14 发布
原创 于 2019-09-06 15:33:30 发布·5.8k 阅读
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#什么是导数和切线#人工智能#马克java社区#马克-to-win
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本文探讨了函数在某一点的导数定义及其几何意义，即函数曲线在该点处切线的斜率。介绍了导数的概念，以及如何通过极限求得导数，强调了导数在数学分析中的核心作用。
（引自高等数学）设函数y=f（x）在点x0的某个邻域内有定义，当自变量x在x0处有增量Δx，相应地函数取得增量Δy=f（x0+Δx）-f（x0）；如果Δy与Δx之比当Δx→0时极限存在，则称函数y=f（x）在点x0处可导，并称这个极限为函数y=f（x）在点x0处的导数。马克-to-win @ 马克 java 社区： 所以说：函数y=f（x）在x0点的导数f'（x0）的几何意义：表示函数曲线在点P0（x0,f（x0））处的切线的斜率（导数的几何意义是该函数曲线在这一点上的切线斜率）。
直线斜率通常用直线与(横)坐标轴夹角的正切表示或两点的纵坐标之差与横坐标之差的比来表示，tan45=1， tan135=-1
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切线 和 倒数_ 导数 法求 切线
weixin_34997006的博客
02-05 1530
根据 导数 的几何意义，切 点 处 导数 等于 切线 斜率，这也作为我们求解函数 切线 方程的基础.求 切线，我们需要两个要素：①切 点 坐标，②切 点 斜率.1、已经切 点：直接通过求切 点 导数 来求 切线 斜率，然后用 点 斜式方程求解 切线 方程.【例1 - 1】( 2 0 19全国1卷理数13题 ) 曲线 在 点 处 的 切线 方程为_____.【分析】已知切 点，可以求切 点 导数 来得到 切线 斜率，再代入 点 斜式方程得解.【解析】求导 ,所以 ,代入 点...
函数的梯度方向 和 切线 方向_ 导数、方向 导数 与梯度
weixin_39801356的博客
01-17 6461
导数，方向 导数，切线、梯度是从高中就开始接触的概念，然而对这几个概念的认识不清，困惑了我很长时间，下面我将以图文并茂的形式，对这几个概念做详细的解释。1， 导数 定义：设 函数y = f ( x ) 在 点 x 0 的某个邻域内有定义，当自变量x在x 0 处 有增量Δx，( x 0+Δx ) 也在该邻域内时，相应地函数取得增量Δy = f ( x 0+Δx ) - f ( x 0 )；如果当Δx→0 时， Δy与Δx之比极限存在，则称函数y = f ( x ) 在 点 x 0 处...
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漫步微积分三——如何计算 切线 的 斜率 抛物线 _切线 斜率 公式
9-27
P = ( x 0,y 0 ) P = ( x 0,y 0 ) 是抛物线y = x2y = x2上的任意一个定 点,如图1所示。作为基本思想的第一个图例,给定抛物线上 一点 PP,计算 切线 的 斜率。首先,我们选择 曲线 上的一个临近 点 Q = ( x1,y1 ) Q = ( x1,y1 )。接下来,我们画出由这两 点 确定的割线PQPQ,割线的 斜率 明显是 : msec = slopeofPQ = y1−y 0 x1−x 0 ( 1 ) ( 1 ) ms...
函数的单调性与极值 点
9-26
如果函数y = f ( x ) y = f ( x ) 在[a,b][a,b]上单调增加 ( 单调减少 ),那么它的图形是一条沿xx轴正在上升 ( 下降 ) 的 曲线.可知,这时 曲线 上各 点 的 切线 斜率 是非负的 ( 非正的 ),即y′= f′( x )≥0 ( y′= f ( x )≤0 ) y′= f′( x )≥0 ( y′= f ( x )≤0 ),反之亦然. 可见,函数的单调性与其 导数 的符号有密切的 关系. 由此...
微积分——导数 和 切线 问题
ComputerInBook的专栏
01-10 4274
微积分 导数 和 切线 问题
数学概念： 导数 和 切线 方程
读万卷书，行万里路。
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导数 是微积分的重要基础概念。当函数y = f（x）的自变量x在 一点 x 0 上产生一个增量Δx, y 方向上的增量Δy与Δx的比值在Δx趋于 0 时的极限如果存在。那么这个极限就是x 0 处 的 导数。即为f‘（x 0）。 下面是不才今天早上做的一个简单的函数求导问题。错误之 处 请指出。 涉及到复合函数求导问题。我们可以拆分。比如 对于x^2+1 求开方。 f[g ( x )]中,设 g ( x ) = u,则f[g ( x )]=...
导数、微分、偏 导数、全微分、方向 导数、梯度的定义与 关系 方向 _导数 和...
9-27
通过函数的极限定义出 导数 ( 以一元函数为例 ) 函数f ( x ) 在 点 x 0 可微的充分必要条件是函数f ( x ) 在 点 x 0 处 可导 扩展到多元函数时,衍生出偏 导数 导数 定义 : 设 函数y = f ( x ) y = f ( x ) y = f ( x ) 在 点 x 0 x_ 0 x 0 的某个领域内有定义,如果ΔyΔx\frac{Δy}{Δx}ΔxΔy在当ΔxΔxΔx ->0 时极限存在,则称函数y = f ( x )...
牛顿迭代法求近似根与二分法求根_迭代法求近似根 和 二分法求近似根哪个...
9-23
设 r是f ( x ) = 0 的根,选取x 0 作为r初始近似值,过 点 ( x 0,f ( x 0 ) ) 做 曲线 y = f ( x ) 的 切线 L,L的方程为y = f ( x 0 )+f’( x 0 ) ( x - x 0 ),求出L与x轴交 点 的横坐标 x1 = x 0 - f ( x 0 )/f’( x 0 ),称x1为r的一次近似值。过 点 ( x1,f ( x1 ) ) 做 曲线 y = f ( x ) 的 切线,并求该 切线 与x轴交 点 的横坐标 x2 ...
利用 导数 求 曲线 的 切线 和 公 切线.pdf
11-01
"利用 导数 求 曲线 的 切线 和 公 切线" 本文主要讲述了利用 导数 求 曲线 的 切线 和 公 切线 的方法 和 应用。首先，我们介绍了 曲线 的 切线 方程的求法，包括利用 导数 求 切线 方程 和 公 切线 方程的方法。然后，我们讨论了 切线 的条数问题，...
2 0 届高考数学一轮复习讲义 ( 提高版 ) 专题2.12 导数 的 切线 方程（解析版）.docx
09-08
这些知识 点 是高中数学中关于 导数 和 切线 方程的核心内容，对理解 和 解决相关问题至关重要。掌握这些概念 和 方法，能有效提升解决实际问题的能力，尤其是在高考复习阶段，对提高数学成绩具有明显作用。
牛顿迭代公式_nr迭代
9-26
设 r是f ( x ) = 0 的根,选取x 0 作为r初始近似值,过 点 ( x 0,f ( x 0 ) ) 做 曲线 y = f ( x ) 的 切线 L,L的方程为y = f ( x 0 )+f'( x 0 ) ( x - x 0 ),求出L与x轴交 点 的横坐标 x1 = x 0 - f ( x 0 )/f'( x 0 ),称x1为r的一次近似值。 过 点 ( x1,f ( x1 ) ) 做 曲线 y = f ( x ) 的 切线,并求该 切线 与x轴交 点 的横坐标 x2 ...
切线 斜率 tan ( θ ) 就是 导数 y′ = f′( x ) = dy/dx
9-28
✅ 1. 几何定义 : 切线 角与 斜率 的 关系 我们考虑一条光滑 曲线 y = f ( x ) y = f ( x ) y = f ( x ),在某 点 P = ( x 0,f ( x 0 ) ) P = ( x_ 0, f ( x_ 0 ) ) P = ( x 0,f ( x 0) ) 处 有一条 切线。 定义 : 切线 角θ\thetaθ是指这条 切线 与x轴之间的夹角 ( 逆时针为正方向 )。
导数 和 梯度的区别 ? 最新发布
mayaohao的博客
09-04 825
维度 导数 ( Derivative ) 梯度 ( Gradient ) 适用函数单变量函数：_( y = f ( x )_)（仅 1 个自变量x）多变量函数：_( z = f ( x_1, x_2, ..., x_n )_)（_( n≥2_) 个自变量）数学本质标量 ( Scalar )：表示函数在某 点 的 “瞬时变化率”向量 ( Vec to r )：表示函数在某 点 “变化最快的方向 + 变化率大小”几何意义函数 曲线 在该 点 的 切线 斜率 函数曲面（或超曲面）在该 点 的 “最陡上升方向” 的向量符号表示。
切线、斜率、梯度 和 导数 以及其 关系
qq_39697468的博客
03-20 1782
导数：描述单变量函数在某 一点 的变化率，是函数在该 点 的 切线 斜率。梯度：描述多变量函数在某 一点 的变化率，是一个向量，其方向是函数增加最快的方向，大小表示变化率。关系：单变量函数的 导数 是梯度的特例，多变量函数的梯度是偏 导数 的集合。应用：在优化问题中，通过求导（单变量）或梯度（多变量）并令其为零，可以找到函数的极值 点。
【考研数学一·高数 ( 6 )】多元函数微分学_y = f ( x ) f ( x,y ) 多元微分...
9-16
设 函数f ( x,y ) f ( x,y ) f ( x,y ) 在区域DDD上有定义,P 0 ( x 0,y 0 )∈DP_ 0 ( x_ 0,y_ 0 )\in DP 0( x 0,y 0)∈D或为DDD边界上的 一点.如果对于任意给定的ϵ>0\epsilon>0 ϵ>0,总存在δ>0\delta>0 δ>0,当 点 P ( x,y )∈DP ( x,y )\in DP ( x,y )∈D,且满足 0<∣PP 0∣= ( x−x 0 ) 2+( y−y 0...
...渐进线、弧微分与曲率_渐近线的 斜率 是 曲线 导函数吗 ?
9-15
设 函数f ( x ) 在区间 ( a,b ) 内具有连续 导数,则 曲线 y = f ( x ) 在 点 ( x,y ) 处 的 切线 斜率 为f′( x ),因此 曲线 在 点 ( x,y ) 处 的 切线 方程为y−f ( x ) = f′( x ) ( x−x 0),即y = f′( x ) ( x−x 0)+f ( x 0)。 设 曲线 y = f ( x ) 上 一点 ( x 0,y 0),如果 曲线 在 点 ( x 0,y 0) 处 的 切线 平行于x轴,...
函数的梯度方向 和 切线 方向_偏 导数、方向 导数 与梯度
weixin_40007515的博客
12-23 1507
1. 多个变量的函数两个变量的函数 是一个定义域为 的子集，值域为 的子集的函数。值域为 的两个变量的函数 的图 ( graph ) 是所有 点 的集合，使得 多变量函数的定义与两个变量的函数类似。2. 偏 导数 函数相对于它的某个自变量的偏 导数，是假定其他自变量不变时，对该自变量的 导数。对于两个变量的函数 ，它的偏 导数 和 定义为： 和 其他表示方法有： 单变量函数中，函数的图是一个...
偏 导数 的几何意义
piglite的专栏
06-03 1万+
表示固定面上 一点 的 切线 斜率。 偏 导数 f’x ( x 0,y 0 ) 表示固定面上 一点 对 x 轴的 切线 斜率；偏 导数 f’y ( x 0,y 0 ) 表示固定面上 一点 对 y 轴的 切线 斜率。 高阶偏 导数：如果二元函数 z = f ( x,y ) 的偏 导数 f’x ( x,y ) 与 f’y ( x,y ) 仍然可导，那么这两个偏导函数的偏 导数 称为 z = f ( x,y ) 的二阶偏 导数。二元函数的二阶偏 导数 有四个：f"xx，f"xy，f"yx，f"yy。 注意： f"xy与f"yx的区别在于：前者是先对 x 求偏导，然后将所得的偏导函数再对 y 求偏导；
导数 切线 的 斜率
u014241071的博客
09-15 3083
导数 = 切线 的 斜率
曲线 切线 的定义 和 导数（极限）
weixin_30642305的博客
05-13 1699
那么一般的 曲线 的 切线 该怎么定义呢？且看下文！ _( P ( x{0},y_{0})_) _和_( Q ( x{0} + \Delta x,y_{0} + \Delta y )_) 分别是上图 _曲线 上不同的两 点（这意味着_(\Delta x \neq _0_)），Q可以选在P的右边也可以选在左边（这意味着_(\text{Δx}_) 可正可负），称通过PQ的直线为该 _曲线 的一条割线。在_(\text{Δx}_) 不断逼近于...
导数＝（切线）斜率？
暴雪123的博客
03-14 2014
应用 : 数学→瞬时速度 物理→加速度 和 力的方向
梯度下降算法_从中学数学到AI算法 0 1：切线、导数、偏 导数、梯度、梯度下降算法...
weixin_39889337的博客
11-18 414
内容导读：切线、导数、偏 导数、梯度、梯度下降算法，从中学、大学数学到 人工智能，这些概念是一脉相承的。本文将这些知识进行大串联。如果你是个中学生，读完本篇文章，你将会了解到，中学里学习的数学将来会在 人工智能 的哪些方面应用。如果你正在从事 人工智能 的学习研究，本文将有助于你系统地重温 人工智能 算法及其预备知识的前世今生。( 1 ) 中学数学里的相切、切线、导数 等概念。曲线 上某 点 A的 切线 是这样定义的：B是 曲线 上邻...
导数 和 梯度，切线 和 法向量 热门推荐
chivalry
04-10 4万+
转自：http ://www.cnblogs.com/jerrylead/archive/2 0 11/0 3/0 9/197828 0.html 作者讲得很清楚 记得在高中做数学题时，经常要求 曲线 的 切线。见到形如之类的函数，不管三七二十一直接求导得到，这就是 切线 的 斜率，然后就得到了 处 的 切线。 上大学又学习了曲面 切线 和 法向量的求法，求偏导是法向量，然后套公式求出 切线。 一个
Anaconda Prompt的用法
mark_to_win的专栏
09-08 2万+
Win dows/开始菜单,打开Anaconda Prompt，这个窗口 和 cmd窗口一样的，用命令“conda list”查看已安装的包，从这些库中我们可以发现NumPy,Matplotlib,Pandas，说明已经安装成功了! 下一步可以测试随anaconda一起安装的python：（还是在上一个窗口） 更多请见：https ://blog.csdn.net/qq_4365 0 923/ar...
什么叫训练模型？
mark_to_win的专栏
09-04 1万+
马克 - to - win @ 马克 java 社区：根据上面的计算过程可知，下图的神经网络模型果然比较符合现实情况。带游泳池的首先大概率属于高档房，其次价格也比较高。不带游泳池的 属于低档房，而且价格较低。 为什么同样的模型，判断出的结果不一样呢？ 1）w3比较大，为 0.8,w4比较小，为 0.0 0 1.意思就是贵的 设 施对于判断是否为高档很重要。马克 - to - win @ 马克 java 社区：对于低档几乎没用。所以找...
导数 与微分有什么本质区别？
03-10
好的，我现在需要帮用户解析 导数 和 微分的本质差异。首先，我得回忆一下自己学过的微积分知识，但可能不够准确，得查资料确认。导数 和 微分经常被一起提到，但它们的区别可能让初学者混淆。用户的问题是要了解它们的本质差异，所以得从数学定义、几何意义 和 应用层面来比较。 先从定义入手。导数 是函数在某 点 的变化率，也就是极限，导数 定义为极限值，而微分则是函数改变量的线性部分，记作dy = f'( x ) dx。这里可能需要确认 导数 是数值，微分是函数或线性映射。需要查资料核实这些定义，确保准确。 然后是几何解释。导数 是 切线 的 斜率，而微分是 切线 方程的纵坐标改变量。这 点 可能需要用图形说明，但文本回答里要描述清楚。用户可能需要直观的理解，所以几何意义的对比很重要。 接下来是数学属性。导数 关注的是 点 上的局部性质，而微分是函数的线性逼近。微分有线性主部的特性，而 导数 则是微分的系数。这部分需要明确两者的不同数学 处 理方式。 应用场景方面，导数 用于变化率、极值问题，微分用于近似计算 和 积分。需要举例说明，比如物理中的速度 和 微分在误差估计中的应用。可能需要搜索具体例子来支持这 一点。 还要注意微分的形式，dy = f'( x ) Δx，这里Δx自变量的微分dx等于Δx，所以常写成dy = f'( x ) dx。这部分要解释清楚微分与 导数 的 关系，即微分是 导数 乘以自变量的微分。 需要检查是否有混淆 点，比如微分有时指整个微分运算，或者微分形式。要明确微分是一个线性映射，而 导数 是这个映射的系数。 最后整理结构，分定义、几何意义、数学属性、应用四个部分，每部分简明对比。确保不出现第一人称，用客观描述。然后生成相关问题，可能需要关于微分应用、几何意义对比、微分在积分中的作用等。 现在开始搜索相关资料，确认各个 点 的正确性。比如，导数 的定义是极限，微分是线性近似，微分dy = f'( x ) dx的 关系。检查是否有权威来源支持这些说法。可能参考数学教科书或可靠的教育网站。 整合信息时，确保每个区别 点 清晰，引用来源但不集中引用。按照用户的要求，分 点 说明，换行分隔。避免步骤词汇，用方法或方面来组织。比如： 1. 定义差异 2. 几何意义对比 3. 数学属性不同 4. 应用场景区别 最后生成相关问题，比如 导数 和 微分在积分中的不同角色，微分在近似计算的具体例子，或者微分几何中的推广形式。确保问题基于回答内容，符合用户可能深入的方向。导数 和 微分的本质差异可通过以下四个维度解析： 定义核心差异 导数 是函数在某 点 处 的瞬时变化率，定义为极限 $f'( x_ 0 ) =\lim_{\Delta x _to_ 0}\frac{\Delta y}{\Delta x}$，本质是数值计算结果。微分是函数改变量的线性近似，表达式为 $dy = f'( x_ 0 ) dx$，包含自变量增量 $dx$ 和 对应线性映射 关系。 几何解释对比 导数 值表示 曲线 在该 点 切线 的 斜率，具有方向属性。微分对应 切线 纵坐标的改变量，当 $\Delta x$ 趋近 0 时，微分 $dy$ 与函数实际改变量 $\Delta y$ 的误差为高阶无穷小（$\Delta y = dy + o (\Delta x )$）。 数学属性区别 导数 作为算子具有局部性特征，仅反映单 点 特性。微分作为线性映射存在于某邻域内，满足可加性：$d ( u+v ) = du+dv$ 和 齐次性：$d ( cu ) = cdu$。微分形式保持 $dy = f'( x ) dx$ 的完整结构，而 导数 仅提取比例系数。 应用场景分化 导数 在运动学中表征瞬时速度，在优化问题中定位极值 点。微分用于工程计算的线性近似，如 $\sqrt{4.0 2} \approx 2 + \frac{1}{4}( 0.0 2 )$，以及微分方程建模。在多元函数中，导数 发展为梯度向量，微分则扩展为全微分形式。 python # _导数_ 与微分计算示例 import sympy as sp x _=_ sp.symbols _(_'x'_)_ f _=_ x2 df_dx _=_ sp.diff _(_ f, x _)_ # _导数_ 计算 dx _=_ _0_._0_ 1 dy _=_ df_dx.subs _(_ x, 2 _)_dx # 微分计算当x _=_ 2,dx _=_ _0_.时 print _(_ f"_导数_ _:_ {df_dx}, 微分值 _:_ {dy.evalf _(_ _)_}"_)_
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java中JDBC当中请给出一个DataSource的HelloWorld例子
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什么叫训练模型？
未尼从不加班:想问一下，train.py test.py一个是训练一个是测试，那么最终得出的map是训练结果还是测试结果哩？
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下载APP 程序员都在用的中文IT技术交流社区 公众号 专业的中文 IT 技术社区，与千万技术人共成长 视频号 关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部 |
844 | https://www.ms.uky.edu/~droyster/courses/summer16/ch07.pdf | 253 C H A P T E R 7 Systems of First Order Linear Equations 7.1 1. Introduce the variables x1 = u and x2 = u ′. It follows that x ′ 1 = x2 and x ′ 2 = u ′′ = −2u −0.5 u ′.
In terms of the new variables, we obtain the system of two first order ODEs x ′ 1 = x2 x ′ 2 = −2x1 −0.5 x2 .
3. First divide both sides of the equation by t2, and write u ′′ = −1 t u ′ −(1 −1 4t2 )u .
Set x1 = u and x2 = u ′. It follows that x ′ 1 = x2 and x ′ 2 = u ′′ = −1 t u ′ −(1 −1 4t2 )u .
We obtain the system of equations x ′ 1 = x2 x ′ 2 = −(1 −1 4t2 )x1 −1 t x2 .
254 Chapter 7. Systems of First Order Linear Equations 5. Let x1 = u and x2 = u ′; then u ′′ = x′ 2 . In terms of the new variables, we have x′ 2 + 0.25 x2 + 4 x1 = 2 cos 3t with the initial conditions x1(0) = 1 and x2(0) = −2 . The equivalent first order system is x′ 1 = x2 x′ 2 = −4 x1 −0.25 x2 + 2 cos 3t with the above initial conditions.
7.(a) Solving the first equation for x2 , we have x2 = x ′ 1 + 2x1 . Substitution into the second equation results in (x ′ 1 + 2x1)′ = x1 −2(x ′ 1 + 2x1). That is, x ′′ 1 + 4 x ′ 1 + 3 x1 = 0 . The resulting equation is a second order differential equation with con-stant coefficients. The general solution is x1(t) = c1e−t + c2e−3t. With x2 given in terms of x1 , it follows that x2(t) = c1e−t −c2e−3t.
(b) Imposing the specified initial conditions, we obtain c1 + c2 = 2, c1 −c2 = 3, with solution c1 = 5/2 and c2 = −1/2 . Hence x1(t) = 5 2e−t −1 2e−3t and x2(t) = 5 2e−t + 1 2e−3t .
(c) 10.(a) Solving the first equation for x2 , we obtain x2 = (x1 −x ′ 1)/2 . Substitution into the second equation results in (x1 −x ′ 1)′/2 = 3 x1 −2(x1 −x ′ 1). Rearranging the terms, the single differential equation for x1 is x ′′ 1 + 3 x ′ 1 + 2 x1 = 0.
(b) The general solution is x1(t) = c1e−t + c2e−2t. With x2 given in terms of x1 , it follows that x2(t) = c1e−t + 3c2e−2t /2. Invoking the specified initial conditions, c1 = −7 and c2 = 6 . Hence x1(t) = −7e−t + 6 e−2t and x2(t) = −7e−t + 9 e−2t .
7.1 255 (c) 11.(a) Solving the first equation for x2 , we have x2 = x ′ 1/2 . Substitution into the second equation results in x ′′ 1 /2 = −2 x1. The resulting equation is x ′′ 1 + 4 x1 = 0 .
(b) The general solution is x1(t) = c1 cos 2t + c2 sin 2t. With x2 given in terms of x1 , it follows that x2(t) = −c1 sin 2t + c2 cos 2t. Imposing the specified initial conditions, we obtain c1 = 3 and c2 = 4 . Hence x1(t) = 3 cos 2t + 4 sin 2t and x2(t) = −3 sin 2t + 4 cos 2t .
(c) 13. Solving the first equation for V , we obtain V = L · I ′. Substitution into the second equation results in L · I ′′ = −I C −L RC I ′ .
Rearranging the terms, the single differential equation for I is LRC · I ′′ + L · I ′ + R · I = 0 .
15. Let x = c1x1(t) + c2x2(t) and y = c1y1(t) + c2y2(t). Then x′ = c1x′ 1(t) + c2x′ 2(t) y′ = c1y′ 1(t) + c2y′ 2(t).
256 Chapter 7. Systems of First Order Linear Equations Since x1(t), y1(t) and x2(t), y2(t) are solutions for the original system, x′ = c1(p11x1(t) + p12y1(t)) + c2(p11x2(t) + p12y2(t)) y′ = c1(p21x1(t) + p22y1(t)) + c2(p21x2(t) + p22y2(t)).
Rearranging terms gives x′ = p11(c1x1(t) + c2x2(t)) + p12(c1y1(t) + c2y2(t)) y′ = p21(c1x1(t) + c2x2(t)) + p22(c1y1(t) + c2y2(t)), and so x and y solve the original system.
16. Based on the hypothesis, x ′ 1(t) = p11(t)x1(t) + p12(t)y1(t) + g1(t) x ′ 2(t) = p11(t)x2(t) + p12(t)y2(t) + g1(t) .
Subtracting the two equations, x ′ 1(t) −x ′ 2(t) = p11(t) [x ′ 1(t) −x ′ 2(t)] + p12(t) [y ′ 1(t) −y ′ 2(t)] .
Similarly, y ′ 1(t) −y ′ 2(t) = p21(t) [x ′ 1(t) −x ′ 2(t)] + p22(t) [y ′ 1(t) −y ′ 2(t)] .
Hence the difference of the two solutions satisfies the homogeneous ODE.
17. For rectilinear motion in one dimension, Newton’s second law can be stated as X F = m x ′′.
The resisting force exerted by a linear spring is given by Fs = k δ , in which δ is the displacement of the end of a spring from its equilibrium configuration. Hence, with 0 < x1 < x2 , the first two springs are in tension, and the last spring is in compression. The sum of the spring forces on m1 is F 1 s = −k1x1 −k2(x2 −x1) .
The total force on m1 is X F 1 = −k1x1 + k2(x2 −x1) + F1(t) .
Similarly, the total force on m2 is X F 2 = −k2(x2 −x1) −k3x2 + F2(t) .
18. One of the ways to transform the system is to assign the variables y1 = x1, y2 = x2, y3 = x′ 1, y4 = x ′ 2.
Before proceeding, note that x ′′ 1 = 1 m1 [−(k1 + k2)x1 + k2x2 + F1(t)] x ′′ 2 = 1 m2 [k2x1 −(k2 + k3)x2 + F2(t)] .
7.1 257 Differentiating the new variables, we obtain the system of four first order equations y ′ 1 = y3 y ′ 2 = y4 y ′ 3 = 1 m1 (−(k1 + k2)y1 + k2y2 + F1(t)) y ′ 4 = 1 m2 (k2y1 −(k2 + k3)y2 + F2(t)) .
19.(a) Taking a clockwise loop around each of the paths, it is easy to see that voltage drops are given by V1 −V2 = 0 , and V2 −V3 = 0 .
(b) Consider the right node. The current in is given by I1 + I2 . The current leaving the node is −I3 . Hence the current passing through the node is (I1 + I2) −(−I3).
Based on Kirchhoff’s first law, I1 + I2 + I3 = 0 .
(c) In the capacitor, C V ′ 1 = I1.
In the resistor, V2 = R I2 .
In the inductor, L I ′ 3 = V3 .
(d) Based on part (a), V3 = V2 = V1. Based on part (b), C V ′ 1 + 1 RV2 + I3 = 0 .
It follows that C V ′ 1 = −1 RV1 −I3 and L I ′ 3 = V1.
21. Let I1, I2, I3,and I4 be the current through the resistors, inductor, and capac-itor, respectively. Assign V1, V2, V3,and V4 as the respective voltage drops. Based on Kirchhoff’s second law, the net voltage drops, around each loop, satisfy V1 + V3 + V4 = 0, V1 + V3 + V2 = 0 and V4 −V2 = 0 .
Applying Kirchhoff’s first law to the upper-right node, I3 −(I2 + I4) = 0 .
Likewise, in the remaining nodes, I1 −I3 = 0 and I2 + I4 −I1 = 0 .
That is, V4 −V2 = 0, V1 + V3 + V4 = 0 and I2 + I4 −I3 = 0 .
Using the current-voltage relations, V1 = R1I1, V2 = R2I2, L I ′ 3 = V3, C V ′ 4 = I4.
258 Chapter 7. Systems of First Order Linear Equations Combining these equations, R1I3 + L I ′ 3 + V4 = 0 and C V ′ 4 = I3 −V4 R2 .
Now set I3 = I and V4 = V , to obtain the system of equations L I ′ = −R1I −V and C V ′ = I −V R2 .
23.(a) Let Q1(t) and Q2(t) be the amount of salt in the respective tanks at time t . Note that the volume of each tank remains constant. Based on conservation of mass, the rate of increase of salt, in any given tank, is given by rate of increase = rate in −rate out.
The rate of salt flowing into Tank 1 is rin = q1 oz gal 3 gal min + Q2 100 oz gal 1 gal min = 3 q1 + Q2 100 oz min .
The rate at which salt flows out of Tank 1 is rout = Q1 60 oz gal 4 gal min = Q1 15 oz min .
Hence dQ1 dt = 3 q1 + Q2 100 −Q1 15 .
Similarly, for Tank 2, dQ2 dt = q2 + Q1 30 −3Q2 100 .
The process is modeled by the system of equations Q ′ 1 = −Q1 15 + Q2 100 + 3 q1 Q ′ 2 = Q1 30 −3Q2 100 + q2 .
The initial conditions are Q1(0) = Q0 1 and Q2(0) = Q0 2 .
7.1 259 (b) The equilibrium values are obtained by solving the system −Q1 15 + Q2 100 + 3 q1 = 0 Q1 30 −3Q2 100 + q2 = 0 .
Its solution leads to QE 1 = 54 q1 + 6 q2 and QE 2 = 60 q1 + 40 q2 .
(c) The question refers to a possible solution of the system 54 q1 + 6 q2 = 60 60 q1 + 40 q2 = 50 .
It is possible to formally solve the system of equations, but the unique solution gives q1 = 7 6 oz gal and q2 = −1 2 oz gal, which is not physically possible.
(d) We can write q2 = −9 q1 + QE 1 6 q2 = −3 2 q1 + QE 2 40 , which are the equations of two lines in the q1-q2-plane: The intercepts of the first line are QE 1 /54 and QE 1 /6 . The intercepts of the second line are QE 2 /60 and QE 2 /40 . Therefore the system will have a unique solution, in the first quadrant, as long as QE 1 /54 ≤QE 2 /60 or QE 2 /40 ≤QE 1 /6 . That is, 10 9 ≤QE 2 QE 1 ≤20 3 .
260 Chapter 7. Systems of First Order Linear Equations 7.2 2.(a) A −2 B = 1 + i −2i −1 + 2i −6 3 + 2i −4 2 −i + 4i = 1 −i −7 + 2i −1 + 2i 2 + 3i .
(b) 3 A + B = 3 + 3i + i −3 + 6i + 3 9 + 6i + 2 6 −3i −2i = 3 + 4i 6i 11 + 6i 6 −5i .
(c) AB = (1 + i)i + 2(−1 + 2i) 3(1 + i) + (−1 + 2i)(−2i) (3 + 2i)i + 2(2 −i) 3(3 + 2i) + (2 −i)(−2i) = −3 + 5i 7 + 5i 2 + i 7 + 2i .
(d) BA = (1 + i)i + 3(3 + 2i) (−1 + 2i)i + 3(2 −i) 2(1 + i) + (−2i)(3 + 2i) 2(−1 + 2i) + (−2i)(2 −i) = 8 + 7i 4 −4i 6 −4i −4 .
3.(c,d) AT + BT = −2 1 2 1 0 −1 2 −3 1 + 1 3 −2 2 −1 1 3 −1 0 = −1 4 0 3 −1 0 5 −4 1 = (A + B)T .
4.(b) A = 3 + 2i 1 −i 2 + i −2 −3i .
(c) By definition, A∗= AT = (A)T = 3 + 2i 2 + i 1 −i −2 −3i .
5.
2(A + B) = 2 5 3 −2 0 2 5 2 2 3 = 10 6 −4 0 4 10 4 4 6 .
7.
Let A= (aij) and B= (bij) .
The given operations in (a)-(d) are performed elementwise. That is, (a) aij + bij = bij + aij .
7.2 261 (b) aij + (bij + cij) = (aij + bij) + cij .
(c) α(aij + bij) = α aij + α bij .
(d) (α + β) aij = α aij + β aij .
In the following, let A= (aij) , B= (bij) and C= (cij) .
(e) Calculating the generic element, (BC)ij = n X k=1 bik ckj .
Therefore [A(BC)]ij = n X r=1 air( n X k=1 brk ckj) = n X r=1 n X k=1 air brk ckj = n X k=1 ( n X r=1 air brk) ckj.
The inner summation is recognized as n X r=1 air brk = (AB)ik , which is the ik-th element of the matrix AB. Thus [A(BC)]ij = [(AB)C]ij.
(f) Likewise, [A(B + C)]ij = n X k=1 aik( bkj + ckj) = n X k=1 aik bkj + n X k=1 aik ckj = (AB)ij + (AC)ij .
8.(a) xT y= 2(−1 + i) + 2(3i) + (1 −i)(3 −i) = 4i .
(b) yT y= (−1 + i)2 + 22 + (3 −i)2 = 12 −8i .
(c) (x, y) = 2(−1 −i) + 2(3i) + (1 −i)(3 + i) = 2 + 2i .
(d) (y, y) = (−1 + i)(−1 −i) + 22 + (3 −i)(3 + i) = 16 .
9. Indeed, 5 + 3i = xT y = n X j=1 xj yj = yT x , and 3 −5i = (x, y) = n X j=1 xj yj = n X j=1 yj xj = n X j=1 yj xj = (y, x).
262 Chapter 7. Systems of First Order Linear Equations 11. First augment the given matrix by the identity matrix: [A | I ] = 3 −1 1 0 6 2 0 1 .
Divide the first row by 3 , to obtain 1 −1/3 1/3 0 6 2 0 1 .
Adding −6 times the first row to the second row results in 1 −1/3 1/3 0 0 4 −2 1 .
Divide the second row by 4 , to obtain 1 −1/3 1/3 0 0 1 −1/2 1/4 .
Finally, adding 1/3 times the second row to the first row results in 1 0 1/6 1/12 0 1 −1/2 1/4 .
Hence 3 −1 6 2 −1 = 1 12 2 1 −6 3 .
13. The augmented matrix is 1 1 −1 1 0 0 2 −1 1 0 1 0 1 1 2 0 0 1 .
Combining the elements of the first row with the elements of the second and third rows results in 1 1 −1 1 0 0 0 −3 3 −2 1 0 0 0 3 −1 0 1 .
Divide the elements of the second row by −3 , and the elements of the third row by 3 . Now subtracting the new second row from the first row yields 1 0 0 1/3 1/3 0 0 1 −1 2/3 −1/3 0 0 0 1 −1/3 0 1/3 .
Finally, combine the third row with the second row to obtain 1 0 0 1/3 1/3 0 0 1 0 1/3 −1/3 1/3 0 0 1 −1/3 0 1/3 .
7.2 263 Hence 1 1 −1 2 −1 1 1 1 2 −1 = 1 3 1 1 0 1 −1 1 −1 0 1 .
15. Elementary row operations yield 2 1 0 1 0 0 0 2 1 0 1 0 0 0 2 0 0 1 → 1 1/2 0 1/2 0 0 0 1 1/2 0 1/2 0 0 0 1 0 0 1/2 → 1 0 −1/4 1/2 −1/4 0 0 1 0 0 1/2 −1/4 0 0 1 0 0 1/2 → 1 0 −1/4 1/2 −1/4 0 0 1 0 0 1/2 −1/4 0 0 1 0 0 1/2 .
Finally, combining the first and third rows results in 1 0 0 1/2 −1/4 1/8 0 1 0 0 1/2 −1/4 0 0 1 0 0 1/2 , so A−1 = 1/2 −1/4 1/8 0 1/2 −1/4 0 0 1/2 .
16. Elementary row operations yield 1 −1 −1 1 0 0 2 1 0 0 1 0 3 −2 1 0 0 1 → 1 −1 −1 1 0 0 0 3 2 −2 1 0 0 1 4 −3 0 1 → 1 0 −1/3 1/3 1/3 0 0 1 2/3 −2/3 1/3 0 0 0 10/3 −7/3 −1/3 1 → 1 0 0 1/10 3/10 1/10 0 1 0 −1/5 2/5 −1/5 0 0 10/3 −7/3 −1/3 1 .
Finally, normalizing the last row results in 1 0 0 1/10 3/10 1/10 0 1 0 −1/5 2/5 −1/5 0 0 1 −7/10 −1/10 3/10 , so A−1 = 1/10 3/10 1/10 −1/5 2/5 −1/5 −7/10 −1/10 3/10 .
17. Elementary row operations on the augmented matrix yield the row-reduced form of the augmented matrix 1 0 −1/7 0 1/7 2/7 0 1 3/7 0 4/7 1/7 0 0 0 1 −2 −1 .
The left submatrix cannot be converted to the identity matrix. Hence the given matrix is singular.
264 Chapter 7. Systems of First Order Linear Equations 18. Elementary row operations on the augmented matrix yield 1 0 0 −1 1 0 0 0 0 −1 1 0 0 1 0 0 −1 0 1 0 0 0 1 0 0 1 −1 1 0 0 0 1 → 1 0 0 −1 1 0 0 0 0 −1 1 0 0 1 0 0 0 0 1 −1 1 0 1 0 0 1 −1 1 0 0 0 1 → 1 0 0 −1 1 0 0 0 0 1 −1 0 0 −1 0 0 0 0 1 −1 1 0 1 0 0 0 0 1 0 1 0 1 → 1 0 0 0 1 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 1 1 1 1 0 0 0 1 0 1 0 1 , so A−1 = 1 1 0 1 1 0 1 1 1 1 1 1 0 1 0 1 .
19. Elementary row operations on the augmented matrix yield 1 −1 2 0 1 0 0 0 −1 2 −4 2 0 1 0 0 1 0 1 3 0 0 1 0 −2 2 0 −1 0 0 0 1 → 1 −1 2 0 1 0 0 0 0 1 −2 2 1 1 0 0 0 1 −1 3 −1 0 1 0 0 0 4 −1 2 0 0 1 → 1 0 0 2 2 1 0 0 0 1 −2 2 1 1 0 0 0 0 1 1 −2 −1 1 0 0 0 4 −1 2 0 0 1 → 1 0 0 2 2 1 0 0 0 1 0 4 −3 −1 2 0 0 0 1 1 −2 −1 1 0 0 0 0 −5 10 4 −4 1 .
Normalizing the last row and combining it with the others results in 1 0 0 2 2 1 0 0 0 1 0 4 −3 −1 2 0 0 0 1 1 −2 −1 1 0 0 0 0 1 −2 −4/5 4/5 −1/5 → 1 0 0 0 6 13/5 −8/5 2/5 0 1 0 0 5 11/5 −6/5 4/5 0 0 1 0 0 −1/5 1/5 1/5 0 0 0 1 −2 −4/5 4/5 −1/5 , so A−1 = 6 13/5 −8/5 2/5 5 11/5 −6/5 4/5 0 −1/5 1/5 1/5 −2 −4/5 4/5 −1/5 .
20. Suppose that there exist matrices B and C, such that AB = I and CA = I .
Then CAB = IB = B, also, CAB = CI = C. This shows that B = C.
23. First note that x ′ = 1 0 et + 2 1 1 (et + t et) = 3et + 2t et 2et + 2t et .
7.3 265 We also have 2 −1 3 −2 x = 2 −1 3 −2 1 0 et + 2 −1 3 −2 2 2 (t et) = 2 3 et + 2 2 (t et) = 2et + 2t et 3et + 2t et .
It follows that 2 −1 3 −2 x + 1 −1 et = 3et + 2t et 2et + 2t et .
24. It is easy to see that x ′ = −6 8 4 e−t + 0 4 −4 e2t = −6e−t 8e−t + 4e2t 4e−t −4e−2t .
On the other hand, 1 1 1 2 1 −1 0 −1 1 x = 1 1 1 2 1 −1 0 −1 1 6 −8 −4 e−t + 1 1 1 2 1 −1 0 −1 1 0 2 −2 e2t = −6 8 4 e−t + 0 4 −4 e2t .
26. Differentiation, elementwise, results in Ψ ′ = et −2e−2t 3e3t −4et 2e−2t 6e3t −et 2e−2t 3e3t .
On the other hand, 1 −1 4 3 2 −1 2 1 −1 Ψ = 1 −1 4 3 2 −1 2 1 −1 et e−2t e3t −4et −e−2t 2e3t −et −e−2t e3t = et −2e−2t 3e3t −4et 2e−2t 6e3t −et 2e−2t 3e3t .
7.3 4. The augmented matrix is 1 2 −1 | 0 2 1 1 | 0 1 −1 2 | 0 .
266 Chapter 7. Systems of First Order Linear Equations Adding −2 times the first row to the second row and subtracting the first row from the third row results in 1 2 −1 | 0 0 −3 3 | 0 0 −3 3 | 0 .
Adding the negative of the second row to the third row results in 1 2 −1 | 0 0 −3 3 | 0 0 0 0 | 0 .
We evidently end up with an equivalent system of equations x1 + 2x2 −x3 = 0 −x2 + x3 = 0 .
Since there is no unique solution, let x3 = α , where α is arbitrary. It follows that x2 = α , and x1 = −α . Hence all solutions have the form x = α −1 1 1 .
5. The augmented matrix is 1 0 −1 | 0 3 1 1 | 0 −1 1 2 | 0 .
Adding −3 times the first row to the second row and adding the first row to the last row yields 1 0 −1 | 0 0 1 3 | 0 0 1 1 | 0 .
Now add the negative of the second row to the third row to obtain 1 0 −1 | 0 0 1 3 | 0 0 0 −2 | 0 .
We end up with an equivalent linear system x1 −x3 = 0 x2 + 3 x3 = 0 x3 = 0 .
Hence the unique solution of the given system of equations is x1 = x2 = x3 = 0 .
6. The augmented matrix is 1 2 −1 | −2 −2 −4 2 | 4 2 4 −2 | −4 .
7.3 267 Adding 2 times the first row to the second row and subtracting 2 times the first row from the third row results in 1 2 −1 | −2 0 0 0 | 0 0 0 0 | 0 .
We evidently end up with an equivalent system of equations x1 + 2x2 −x3 = −2.
Since there is no unique solution, let x2 = α , and x3 = β, where α, β are arbitrary.
It follows that x1 = −2 −2α + β . Hence all solutions have the form x = −2 −2α + β α β .
8. Write the given vectors as columns of the matrix X = 2 0 −1 1 1 2 0 0 0 .
It is evident that det(X) = 0. Hence the vectors are linearly dependent. In order to find a linear relationship between them, write c1x(1) + c2x(2) + c3x(3) = 0 . The latter equation is equivalent to 2 0 −1 1 1 2 0 0 0 c1 c2 c3 = 0 0 0 .
Performing elementary row operations, 2 0 −1 | 0 1 1 2 | 0 0 0 0 | 0 → 1 0 −1/2 | 0 0 1 5/2 | 0 0 0 0 | 0 .
We obtain the system of equations c1 −c3/2 = 0 c2 + 5c3/2 = 0 .
Setting c3 = 2 , it follows that c1 = 1 and c3 = −5 . Hence x(1) −5x(2) + 2x(3) = 0 .
10. The matrix containing the given vectors as columns is X = 1 2 −1 3 2 3 0 −1 −1 1 2 1 0 −1 2 3 .
We find that det(X) = −70. Hence the given vectors are linearly independent.
268 Chapter 7. Systems of First Order Linear Equations 11. Write the given vectors as columns of the matrix X = 1 3 2 4 2 1 −1 3 −2 0 1 −2 .
The four vectors are necessarily linearly dependent. Hence there are nonzero scalars such that c1x(1) + c2x(2) + c3x(3) + c4x(4) = 0 . The latter equation is equivalent to 1 3 2 4 2 1 −1 3 −2 0 1 −2 c1 c2 c3 c4 = 0 0 0 .
Performing elementary row operations, 1 3 2 4 | 0 2 1 −1 3 | 0 −2 0 1 −2 | 0 → 1 0 0 1 | 0 0 1 0 1 | 0 0 0 1 0 | 0 .
We end up with an equivalent linear system c1 + c4 = 0 c2 + c4 = 0 c3 = 0 .
Let c4 = −1 . Then c1 = 1 and c2 = 1 . Therefore we find that x(1) + x(2) −x(4) = 0 .
12. The matrix containing the given vectors as columns, X, is of size n × m . Since n < m, we can augment the matrix with m −n rows of zeros. The resulting matrix, ˜ X, is of size m × m . Since ˜ X is a square matrix, with at least one row of zeros, it follows that det( ˜ X) = 0. Hence the column vectors of ˜ X are linearly dependent.
That is, there is a nonzero vector, c, such that ˜ Xc= 0m×1 . If we write only the first n rows of the latter equation, we have Xc= 0n×1 . Therefore the column vectors of X are linearly dependent.
13. By inspection, we find that x(1)(t) −2x(2)(t) = −e−t 0 .
Hence 3 x(1)(t) −6 x(2)(t)+x(3)(t) = 0 , and the vectors are linearly dependent.
17. The eigenvalues λ and eigenvectors x satisfy the equation 3 −λ −2 4 −1 −λ x1 x2 = 0 0 .
For a nonzero solution, we must have (3 −λ)(−1 −λ) + 8 = 0 , that is, λ2 −2λ + 5 = 0 .
7.3 269 The eigenvalues are λ1 = 1 −2 i and λ2 = 1 + 2 i . The components of the eigen-vector x(1) are solutions of the system 2 + 2 i −2 4 −2 + 2 i x1 x2 = 0 0 .
The two equations reduce to (1 + i)x1 = x2 . Hence x(1) = (1 , 1 + i)T . Now setting λ = λ2 = 1 + 2 i , we have 2 −2 i −2 4 −2 −2 i x1 x2 = 0 0 , with solution given by x(2) = (1 , 1 −i)T .
18. The eigenvalues λ and eigenvectors x satisfy the equation −2 −λ 1 1 −2 −λ x1 x2 = 0 0 .
For a nonzero solution, we must have (−2 −λ)(−2 −λ) −1 = 0 , that is, λ2 + 4λ + 3 = 0 .
The eigenvalues are λ1 = −3 and λ2 = −1 . For λ1 = −3 , the system of equations becomes 1 1 1 1 x1 x2 = 0 0 , which reduces to x1 + x2 = 0 . A solution vector is given by x(1) = (1 , −1)T . Sub-stituting λ = λ2 = −1 , we have −1 1 1 −1 x1 x2 = 0 0 .
The equations reduce to x1 = x2 . Hence a solution vector is given by x(2) = (1 , 1)T .
20. The eigensystem is obtained from analysis of the equation 1 −λ √ 3 √ 3 −1 −λ x1 x2 = 0 0 .
For a nonzero solution, the determinant of the coefficient matrix must be zero.
That is, λ2 −4 = 0 .
Hence the eigenvalues are λ1 = −2 and λ2 = 2 . Substituting the first eigenvalue, λ = −2 , yields 3 √ 3 √ 3 1 x1 x2 = 0 0 .
The system is equivalent to the equation √ 3 x1 + x2 = 0 . A solution vector is given by x(1) = (1 , − √ 3 )T . Substitution of λ = 2 results in −1 √ 3 √ 3 −3 x1 x2 = 0 0 , 270 Chapter 7. Systems of First Order Linear Equations which reduces to x1 = √ 3 x2 . A corresponding solution vector is x(2) = ( √ 3 , 1)T .
21. The eigenvalues λ and eigenvectors x satisfy the equation −3 −λ 3/4 −5 1 −λ x1 x2 = 0 0 .
For a nonzero solution, we must have (−3 −λ)(1 −λ) + 15/4 = 0 , that is, λ2 + 2λ + 3/4 = 0 .
Hence the eigenvalues are λ1 = −3/2 and λ2 = −1/2 . In order to determine the eigenvector corresponding to λ1 , set λ = −3/2 . The system of equations becomes −3/2 3/4 −5 5/2 x1 x2 = 0 0 , which reduces to −2 x1 + x2 = 0 .
A solution vector is given by x(1) = (1 , 2)T .
Substitution of λ = λ2 = −1/2 results in −5/2 3/4 −5 3/2 x1 x2 = 0 0 , which reduces to 10 x1 = 3 x2 . A corresponding solution vector is x(2) = (3 , 10)T .
23. The eigensystem is obtained from analysis of the equation 3 −λ 2 2 1 4 −λ 1 −2 −4 −1 −λ x1 x2 x3 = 0 0 0 .
The characteristic equation of the coefficient matrix is λ3 −6λ2 + 11λ −6 = 0 , with roots λ1 = 1 , λ2 = 2 and λ3 = 3 . Setting λ = λ1 = 1 , we have 2 2 2 1 3 1 −2 −4 −2 x1 x2 x3 = 0 0 0 .
This system is reduces to the equations x1 + x3 = 0 x2 = 0 .
A corresponding solution vector is given by x(1) = (1 , 0 , −1)T . Setting λ = λ2 = 2 , the reduced system of equations is x1 + 2 x2 = 0 x3 = 0 .
A corresponding solution vector is given by x(2) = (−2 , 1 , 0)T .
Finally, setting λ = λ3 = 3 , the reduced system of equations is x1 = 0 x2 + x3 = 0 .
A corresponding solution vector is given by x(3) = (0 , 1 , −1)T .
7.3 271 24. For computational purposes, note that if λ is an eigenvalue of B, then c λ is an eigenvalue of the matrix A= c B . Eigenvectors are unaffected, since they are only determined up to a scalar multiple. So with B = 11 −2 8 −2 2 10 8 10 5 , the associated characteristic equation is µ3 −18µ2 −81µ + 1458 = 0 , with roots µ1 = −9 , µ2 = 9 and µ3 = 18 . Hence the eigenvalues of the given matrix, A, are λ1 = −1 , λ2 = 1 and λ3 = 2 . Setting λ = λ1 = −1 , (which corresponds to using µ1 = −9 in the modified problem) the reduced system of equations is 2 x1 + x3 = 0 x2 + x3 = 0 .
A corresponding solution vector is given by x(1) = (1 , 2 , −2)T . Setting λ = λ2 = 1 , the reduced system of equations is x1 + 2 x3 = 0 x2 −2 x3 = 0 .
A corresponding solution vector is given by x(2) = (2 , −2 , −1)T . Finally, setting λ = λ2 = 1 , the reduced system of equations is x1 −x3 = 0 2 x2 −x3 = 0 .
A corresponding solution vector is given by x(3) = (2 , 1 , 2)T .
26.(b) By definition, (Ax , y) = n X i = 0 (Ax)i yi = n X i = 0 n X j = 0 aij xj yi .
Let bij = aji , so that aij = bji . Now interchanging the order or summation, (Ax , y) = n X j = 0 xj n X i = 0 aij yi = n X j = 0 xj n X i = 0 bji yi .
Now note that n X i = 0 bji yi = n X i = 0 bji yi = (A∗y)j .
Therefore (Ax , y) = n X j = 0 xj (A∗y)j = (x , A∗y) .
(c) By definition of a Hermitian matrix, A=A∗.
272 Chapter 7. Systems of First Order Linear Equations 27. Suppose that Ax= 0 , but that x̸= 0 . Let A= (aij). Using elementary row operations, it is possible to transform the matrix into one that is not upper trian-gular. If it were upper triangular, backsubstitution would imply that x= 0 . Hence a linear combination of all the rows results in a row containing only zeros. That is, there are n scalars, βi , one for each row and not all zero, such that for each for column j , n X i = 1 βi aij = 0 .
Now consider A∗= (bij). By definition, bij = aji , or aij = bji . It follows that for each j , n X i = 1 βi bji = n X k = 1 bjk βk = n X k = 1 bjk βk = 0 .
Let y= (β1, β2, · · · , βn)T . Hence we have a nonzero vector, y, such that A∗y= 0 .
29. By linearity, A(x(0) + α ξ) = Ax(0) + α Aξ = b + 0 = b .
30. Let cij = aji . By the hypothesis, there is a nonzero vector, y, such that n X j = 1 cij yj = n X j = 1 aji yj = 0 , i = 1, 2, · · · , n .
Taking the conjugate of both sides, and interchanging the indices, we have n X i = 1 aij yi = 0 .
This implies that a linear combination of each row of A is equal to zero. Now consider the augmented matrix [A |B]. Replace the last row by n X i = 1 yi [ai1 , ai2 , · · · , ain , bi] = " 0 , 0 , · · · , 0 , n X i = 1 yi bi # .
We find that if (B , y) = 0, then the last row of the augmented matrix contains only zeros. Hence there are n −1 remaining equations. We can now set xn = α , some parameter, and solve for the other variables in terms of α . Therefore the system of equations Ax=b has a solution.
31. If λ = 0 is an eigenvalue of A, then there is a nonzero vector, x, such that Ax = λ x = 0 .
That is, Ax= 0 has a nonzero solution. This implies that the mapping defined by A is not 1-to-1, and hence not invertible. On the other hand, if A is singular, then det(A) = 0. Thus, Ax= 0 has a nonzero solution. The latter equation can be written as Ax= 0 x.
7.4 273 32.(a) Based on Problem 26, (Ax , x) = (x , Ax) .
(b) Let x be an eigenvector corresponding to an eigenvalue λ . It then follows that (Ax , x) = (λx , x) and (x , Ax) = (x , λx) . Based on the properties of the inner product, (λx , x) = λ(x , x) and (x , λx) = λ(x , x) . Then from part (a), λ(x , x) = λ(x , x) .
(c) From part (b), (λ −λ)(x , x) = 0 .
Based on the definition of an eigenvector, (x , x) = ∥x∥2 > 0 . Hence we must have λ −λ = 0 , which implies that λ is real.
33. From Problem 26(c), (Ax(1) , x(2)) = (x(1) , Ax(2)) .
Hence λ1(x(1) , x(2)) = λ2(x(1) , x(2)) = λ2(x(1) , x(2)) , since the eigenvalues are real. Therefore (λ1 −λ2)(x(1) , x(2)) = 0 .
Given that λ1 ̸= λ2 , we must have (x(1) , x(2)) = 0 .
7.4 3. Equation (14) states that the Wronskian satisfies the first order linear ODE dW dt = (p11 + p22 + · · · + pnn)W.
The general solution of this is given by Equation (15): W(t) = C e R (p11+p22+···+pnn)dt , in which C is an arbitrary constant. Let X1 and X2 be matrices representing two sets of fundamental solutions. It follows that det(X1) = W1(t) = C1e R (p11+p22+···+pnn)dt det(X2) = W2(t) = C2e R (p11+p22+···+pnn)dt .
Hence det(X1)/det(X2) = C1/C2 . Note that C2 ̸= 0.
4. First note that p11 + p22 = −p(t). As shown in Problem 3, W h x(1) , x(2)i = c e− R p(t)dt.
274 Chapter 7. Systems of First Order Linear Equations For second order linear ODE, the Wronskian (as defined in Chapter 3) satisfies the first order differential equation W ′ + p(t)W = 0 . It follows that W h y(1) , y(2)i = c1 e− R p(t)dt.
Alternatively, based on the hypothesis, y(1) = α11 x11 + α12 x12 y(2) = α21 x11 + α22 x12 .
Direct calculation shows that W h y(1) , y(2)i = α11 x11 + α12 x12 α21 x11 + α22 x12 α11 x ′ 11 + α12 x ′ 12 α21 x ′ 11 + α22 x ′ 12 = (α11α22 −α12α21)x11x ′ 12 −(α11α22 −α12α21)x12x ′ 11 = (α11α22 −α12α21)x11x22 −(α11α22 −α12α21)x12x21 .
Here we used the fact that x′ 1 = x2 . Hence W h y(1) , y(2)i = (α11α22 −α12α21)W h x(1) , x(2)i .
5. The particular solution satisfies the ODE (x(p))′ =P(t)x(p)+g(t) . Now let x be any solution of the homogeneous equation, x′ =P(t)x . We know that x=x(c), in which x(c) is a linear combination of some fundamental solution. By linearity of the differential equation, it follows that x =x(p)+x(c) is a solution of the ODE.
Based on the uniqueness theorem, all solutions must have this form.
7.(a) By definition, W h x(1) , x(2)i = t2 et 2t et = (t2 −2t)et.
(b) The Wronskian vanishes at t0 = 0 and t0 = 2 . Hence the vectors are linearly independent on D = (−∞, 0) ∪(0 , 2) ∪(2 , ∞).
(c) It follows from Theorem 7.4.3 that one or more of the coefficients of the ODE must be discontinuous at t0 = 0 and t0 = 2 .
If not, the Wronskian would not vanish.
(d) Let x = c1 t2 2t + c2 et et .
Then x ′ = c1 2t 2 + c2 et et .
7.4 275 On the other hand, p11 p12 p21 p22 x = c1 p11 p12 p21 p22 t2 2t + c2 p11 p12 p21 p22 et et = c1 p11t2 + 2p12t + c2 [p11 + p12] et c1 [p21t2 + 2p22t] + c2 [p21 + p22] et .
Comparing coefficients, we find that p11t2 + 2p12t = 2t p11 + p12 = 1 p21t2 + 2p22t = 2 p21 + p22 = 1 .
Solution of this system of equations results in p11(t) = 0, p12(t) = 1, p21(t) = 2 −2t t2 −2t, p22(t) = t2 −2 t2 −2t .
Hence the vectors are solutions of the ODE x ′ = 1 t2 −2t 0 t2 −2t 2 −2t t2 −2 x .
8. Suppose that the solutions x(1), x(2),· · · , x(m) are linearly dependent at t = t0 .
Then there are constants c1 , c2, · · · , cm (not all zero) such that c1x(1)(t0) + c2x(2)(t0) + · · · + cmx(m)(t0) = 0 .
Now let z(t) = c1x(1)(t) + c2x(2)(t) + · · · + cmx(m)(t) . Then clearly, z(t) is a solu-tion of x ′ =P(t)x, with z(t0) = 0 . Furthermore, y(t) ≡0 is also a solution, with y(t0) = 0 . By the uniqueness theorem, z(t) =y(t) = 0 . Hence c1x(1)(t) + c2x(2)(t) + · · · + cmx(m)(t) = 0 on the entire interval α < t < β . Going in the other direction is trivial.
9.(a) Let y(t) be any solution of x ′ =P(t)x. It follows that z(t) + y(t) = c1x(1)(t) + c2x(2)(t) + · · · + cnx(n)(t) + y(t) is also a solution. Now let t0 ∈(α , β) . Then the collection of vectors x(1)(t0), x(2)(t0), . . . , x(n)(t0), y(t0) constitutes n + 1 vectors, each with n components.
Based on the assertion in Problem 12, Section 7.3, these vectors are necessarily linearly dependent. That is, there are n + 1 constants b1, b2, . . . , bn, bn+1 (not all zero) such that b1x(1)(t0) + b2x(2)(t0) + · · · + bnx(n)(t0) + bn+1 y(t0) = 0 .
From Problem 8, we have b1x(1)(t) + b2x(2)(t) + · · · + bnx(n)(t) + bn+1 y(t) = 0 276 Chapter 7. Systems of First Order Linear Equations for all t ∈(α , β). Now bn+1 ̸= 0 , otherwise that would contradict the fact that the first n vectors are linearly independent. Hence y(t) = − 1 bn+1 (b1x(1)(t) + b2x(2)(t) + · · · + bnx(n)(t)), and the assertion is true.
(b) Consider z(t) = c1x(1)(t) + c2 x(2)(t) + · · · + cn x(n)(t), and suppose that we also have z(t) = k1x(1)(t) + k2x(2)(t) + · · · + knx(n)(t) .
Based on the assumption, (k1 −c1)x(1)(t) + (k2 −c2)x(2)(t) + · · · + (kn −cn)x(n)(t) = 0 .
The collection of vectors x(1)(t), x(2)(t), . . . , x(n)(t) is linearly independent on α < t < β . It follows that ki −ci = 0 , for i = 1, 2, · · · , n .
7.5 2.(a) Setting x= ξ ert, and substituting into the ODE, we obtain the algebraic equations 1 −r −2 3 −4 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we must have det(A −rI) = r2 + 3r + 2 = 0 . The roots of the characteristic equation are r1 = −1 and r2 = −2. For r = −1, the two equations reduce to ξ1 = ξ2. The corresponding eigenvector is ξ(1) = (1, 1)T . Substitution of r = −2 results in the single equation 3ξ1 = 2ξ2.
A corresponding eigenvector is ξ(2) = (2, 3)T . Since the eigenvalues are distinct, the general solution is x = c1 1 1 e−t + c2 2 3 e−2t.
(b) 7.5 277 3.(a) Setting x= ξ ert results in the algebraic equations 2 −r −1 3 −2 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we must have det(A −rI) = r2 −1 = 0. The roots of the characteristic equation are r1 = 1 and r2 = −1. For r = 1, the system of equations reduces to ξ1 = ξ2. The corresponding eigenvector is ξ(1) = (1, 1)T . Substitution of r = −1 results in the single equation 3ξ1 = ξ2. A corresponding eigenvector is ξ(2) = (1, 3)T . Since the eigenvalues are distinct, the general solution is x = c1 1 1 et + c2 1 3 e−t.
(b) The system has an unstable eigendirection along ξ(1) = (1, 1)T . Unless c1 = 0, all solutions will diverge.
4.(a) Solution of the ODE requires analysis of the algebraic equations 1 −r 1 4 −2 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we must have det(A −rI) = r2 + r −6 = 0.
The roots of the characteristic equation are r1 = 2 and r2 = −3. For r = 2, the system of equations reduces to ξ1 = ξ2. The corresponding eigenvector is ξ(1) = (1, 1)T . Sub-stitution of r = −3 results in the single equation 4ξ1 + ξ2 = 0. A corresponding eigenvector is ξ(2) = (1, −4)T . Since the eigenvalues are distinct, the general solu-tion is x = c1 1 1 e2t + c2 1 −4 e−3t.
278 Chapter 7. Systems of First Order Linear Equations (b) The system has an unstable eigendirection along ξ(1) = (1, 1)T . Unless c1 = 0, all solutions will diverge.
8.(a) Setting x= ξ ert results in the algebraic equations 3 −r 6 −1 −2 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we must have det(A −rI) = r2 −r = 0. The roots of the characteristic equation are r1 = 1 and r2 = 0. With r = 1, the system of equations reduces to ξ1 + 3ξ2 = 0. The corresponding eigenvector is ξ(1) = (3, −1)T . For the case r = 0, the system is equivalent to the equation ξ1 + 2ξ2 = 0. An eigenvector is ξ(2) = (2, −1)T . Since the eigenvalues are distinct, the general solution is x = c1 3 −1 et + c2 2 −1 .
(b) The entire line along the eigendirection ξ(2) = (2, −1)T consists of equilibrium points.
All other solutions diverge.
The direction field changes across the line x1 + 2x2 = 0. Eliminating the exponential terms in the solution, the trajectories are given by x1 + 3 x2 = −c2.
7.5 279 10. The characteristic equation is given by 2 −r 2 + i −1 −1 −i −r = r2 −(1 −i)r −i = 0 .
The equation has complex roots r1 = 1 and r2 = −i. For r = 1, the components of the solution vector must satisfy ξ1 + (2 + i)ξ2 = 0 . Thus the corresponding eigen-vector is ξ(1) = (2 + i , −1)T . Substitution of r = −i results in the single equation ξ1 + ξ2 = 0. A corresponding eigenvector is ξ(2) = (1 , −1)T . Since the eigenvalues are distinct, the general solution is x = c1 2 + i −1 et + c2 1 −1 e−it.
11. Setting x= ξ ert results in the algebraic equations 1 −r 1 2 1 2 −r 1 2 1 1 −r ξ1 ξ2 ξ3 = 0 0 0 .
For a nonzero solution, we must have det(A −rI) = r3 −4r2 −r + 4 = 0 .
The roots of the characteristic equation are r1 = 4 , r2 = 1 and r3 = −1 . Setting r = 4 , we have −3 1 2 1 −2 1 2 1 −3 ξ1 ξ2 ξ3 = 0 0 0 .
This system is reduces to the equations ξ1 −ξ3 = 0 ξ2 −ξ3 = 0 .
A corresponding solution vector is given by ξ(1) = (1 , 1 , 1)T . Setting λ = 1 , the reduced system of equations is ξ1 −ξ3 = 0 ξ2 + 2 ξ3 = 0 .
A corresponding solution vector is given by ξ(2) = (1 , −2 , 1)T .
Finally, setting λ = −1 , the reduced system of equations is ξ1 + ξ3 = 0 ξ2 = 0 .
A corresponding solution vector is given by ξ(3) = (1 , 0 , −1)T . Since the eigenval-ues are distinct, the general solution is x = c1 1 1 1 e4t + c2 1 −2 1 et + c3 1 0 −1 e−t.
280 Chapter 7. Systems of First Order Linear Equations 12. The eigensystem is obtained from analysis of the equation 3 −r 2 4 2 −r 2 4 2 3 −r ξ1 ξ2 ξ3 = 0 0 0 .
The characteristic equation of the coefficient matrix is r3 −6r2 −15r −8 = 0 , with roots r1 = 8 , r2 = −1 and r3 = −1 . Setting r = r1 = 8 , we have −5 2 4 2 −8 2 4 2 −5 ξ1 ξ2 ξ3 = 0 0 0 .
This system is reduced to the equations ξ1 −ξ3 = 0 2 ξ2 −ξ3 = 0 .
A corresponding solution vector is given by ξ(1) = (2 , 1 , 2)T . Setting r = −1 , the system of equations is reduced to the single equation 2 ξ1 + ξ2 + 2 ξ3 = 0 .
Two independent solutions are obtained as ξ(2) = (1 , −2 , 0)T and ξ(3) = (0 , −2 , 1)T .
Hence the general solution is x = c1 2 1 2 e8t + c2 1 −2 0 e−t + c3 0 −2 1 e−t.
13. Setting x= ξ ert results in the algebraic equations 1 −r 1 1 2 1 −r −1 −8 −5 −3 −r ξ1 ξ2 ξ3 = 0 0 0 .
For a nonzero solution, we must have det(A −rI) = r3 + r2 −4r −4 = 0 .
The roots of the characteristic equation are r1 = 2 , r2 = −2 and r3 = −1 . Setting r = 2 , we have −1 1 1 2 −1 −1 −8 −5 −5 ξ1 ξ2 ξ3 = 0 0 0 .
This system is reduces to the equations ξ1 = 0 ξ2 + ξ3 = 0 .
A corresponding solution vector is given by ξ(1) = (0 , 1 , −1)T . Setting λ = −1 , the reduced system of equations is 2 ξ1 + 3 ξ3 = 0 ξ2 −2 ξ3 = 0 .
7.5 281 A corresponding solution vector is given by ξ(2) = (3 , −4 , −2)T . Finally, setting λ = −2 , the reduced system of equations is 7 ξ1 + 4 ξ3 = 0 7ξ2 −5 ξ3 = 0 .
A corresponding solution vector is given by ξ(3) = (4 , −5 , −7)T . Since the eigen-values are distinct, the general solution is x = c1 0 1 −1 e2t + c2 3 −4 −2 e−t + c3 4 −5 −7 e−2t.
15. Setting x= ξ ert results in the algebraic equations 5 −r −1 3 1 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we must have det(A −rI) = r2 −6r + 8 = 0 . The roots of the characteristic equation are r1 = 4 and r2 = 2 . With r = 4 , the system of equations reduces to ξ1 −ξ2 = 0. The corresponding eigenvector is ξ(1) = (1 , 1)T .
For the case r = 2 , the system is equivalent to the equation 3 ξ1 −ξ2 = 0 . An eigenvector is ξ(2) = (1 , 3)T . Since the eigenvalues are distinct, the general solution is x = c1 1 1 e4t + c2 1 3 e2t.
Invoking the initial conditions, we obtain the system of equations c1 + c2 = 2 c1 + 3 c2 = −1 .
Hence c1 = 7/2 and c2 = −3/2 , and the solution of the IVP is x = 7 2 1 1 e4t −3 2 1 3 e2t.
17. Setting x= ξ ert results in the algebraic equations 1 −r 1 2 0 2 −r 2 −1 1 3 −r ξ1 ξ2 ξ3 = 0 0 0 .
For a nonzero solution, we must have det(A −rI) = r3 −6r2 + 11r −6 = 0 . The roots of the characteristic equation are r1 = 1 , r2 = 2 and r3 = 3 . Setting r = 1 , we have 0 1 2 0 1 2 −1 1 2 ξ1 ξ2 ξ3 = 0 0 0 .
This system reduces to the equations ξ1 = 0 ξ2 + 2 ξ3 = 0 .
282 Chapter 7. Systems of First Order Linear Equations A corresponding solution vector is given by ξ(1) = (0 , −2 , 1)T . Setting λ = 2 , the reduced system of equations is ξ1 −ξ2 = 0 ξ3 = 0 .
A corresponding solution vector is given by ξ(2) = (1 , 1 , 0)T . Finally, upon setting λ = 3 , the reduced system of equations is ξ1 −2 ξ3 = 0 ξ2 −2 ξ3 = 0 .
A corresponding solution vector is given by ξ(3) = (2 , 2 , 1)T . Since the eigenvalues are distinct, the general solution is x = c1 0 −2 1 et + c2 1 1 0 e2t + c3 2 2 1 e3t.
Invoking the initial conditions, the coefficients must satisfy the equations c2 + 2 c3 = 2 −2 c1 + c2 + 2 c3 = 0 c1 + c3 = 1 .
It follows that c1 = 1 , c2 = 2 and c3 = 0 . Hence the solution of the IVP is x = 0 −2 1 et + 2 1 1 0 e2t.
18. The eigensystem is obtained from analysis of the equation −r 0 −1 2 −r 0 −1 2 4 −r ξ1 ξ2 ξ3 = 0 0 0 .
The characteristic equation of the coefficient matrix is r3 −4r2 −r + 4 = 0 , with roots r1 = −1 , r2 = 1 and r3 = 4 . Setting r = r1 = −1 , we have −1 0 −1 2 −1 0 −1 2 3 ξ1 ξ2 ξ3 = 0 0 0 .
This system is reduced to the equations ξ1 −ξ3 = 0 ξ2 + 2 ξ3 = 0 .
A corresponding solution vector is given by ξ(1) = (1 , −2 , 1)T . Setting r = 1 , the system reduces to the equations ξ1 + ξ3 = 0 ξ2 + 2 ξ3 = 0 .
7.5 283 The corresponding eigenvector is ξ(2) = (1 , 2 , −1)T . Finally, upon setting r = 4 , the system is equivalent to the equations 4 ξ1 + ξ3 = 0 8 ξ2 + ξ3 = 0 .
The corresponding eigenvector is ξ(3) = (2 , 1 , −8)T . Hence the general solution is x = c1 1 −2 1 e−t + c2 1 2 −1 et + c3 2 1 −8 e4t.
Invoking the initial conditions, c1 + c2 + 2 c3 = 7 −2 c1 + 2 c2 + c3 = 5 c1 −c2 −8 c3 = 5 .
It follows that c1 = 3 , c2 = 6 and c3 = −1 . Hence the solution of the IVP is x = 3 1 −2 1 e−t + 6 1 2 −1 et − 2 1 −8 e4t.
19. Set x= ξ tr. Substitution into the system of differential equations results in t · rtr−1ξ = A ξ tr, which upon simplification yields is, A ξ −rξ = 0 . Hence the vector ξ and constant r must satisfy (A −rI)ξ = 0 .
21. Setting x= ξ tr results in the algebraic equations 5 −r −1 3 1 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we must have det(A −rI) = r2 −6r + 8 = 0 . The roots of the characteristic equation are r1 = 4 and r2 = 2 . With r = 4 , the system of equations reduces to ξ1 −ξ2 = 0. The corresponding eigenvector is ξ(1) = (1 , 1)T .
For the case r = 2 , the system is equivalent to the equation 3 ξ1 −ξ2 = 0 . An eigenvector is ξ(2) = (1 , 3)T . It follows that x(1) = 1 1 t4 and x(2) = 1 3 t2.
The Wronskian of this solution set is W x(1), x(2) = 2t6. Thus the solutions are linearly independent for t > 0 . Hence the general solution is x = c1 1 1 t4 + c2 1 3 t2.
22. As shown in Problem 19, solution of the ODE requires analysis of the equations 4 −r −3 8 −6 −r ξ1 ξ2 = 0 0 .
284 Chapter 7. Systems of First Order Linear Equations For a nonzero solution, we must have det(A −rI) = r2 + 2r = 0 .
The roots of the characteristic equation are r1 = 0 and r2 = −2 . For r = 0 , the system of equations reduces to 4 ξ1 = 3 ξ2 . The corresponding eigenvector is ξ(1) = (3 , 4)T .
Setting r = −2 results in the single equation 2 ξ1 −ξ2 = 0 . A corresponding eigen-vector is ξ(2) = (1 , 2)T . It follows that x(1) = 3 4 and x(2) = 1 2 t−2.
The Wronskian of this solution set is W x(1), x(2) = 2 t−2. These solutions are linearly independent for t > 0 . Hence the general solution is x = c1 3 4 + c2 1 2 t−2.
23. Setting x= ξ tr results in the algebraic equations 3 −r −2 2 −2 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we must have det(A −rI) = r2 −r −2 = 0. The roots of the characteristic equation are r1 = 2 and r2 = −1. Setting r = 2, the system of equations reduces to ξ1 −2ξ2 = 0. The corresponding eigenvector is ξ(1) = (2, 1)T .
With r = −1, the system is equivalent to the equation 2ξ1 −ξ2 = 0. An eigenvector is ξ(2) = (1, 2)T . It follows that x(1) = 2 1 t2 and x(2) = 1 2 t−1.
The Wronskian of this solution set is W x(1), x(2) = 3 t. Thus the solutions are linearly independent for t > 0 . Hence the general solution is x = c1 2 1 t2 + c2 1 2 t−1.
24.(a) The general solution is x = c1 −1 2 e−t + c2 1 2 e−2t.
7.5 285 (b) (c) 26.(a) The general solution is x = c1 −1 2 e−t + c2 1 2 e2t.
286 Chapter 7. Systems of First Order Linear Equations (b) (c) 28.(a) We note that (A −riI)ξ(i) = 0 , for i = 1, 2 .
(b) It follows that (A −r2I)ξ(1) =A ξ(1) −r2ξ(1) = r1ξ(1) −r2ξ(1) .
(c) Suppose that ξ(1) and ξ(2) are linearly dependent. Then there exist constants c1 and c2 , not both zero, such that c1ξ(1) + c2ξ(2) = 0 . Assume that c1 ̸= 0 . It is clear that (A −r2I)(c1ξ(1) + c2 ξ(2)) = 0 . On the other hand, (A −r2I)(c1ξ(1) + c2 ξ(2)) = c1(r1 −r2)ξ(1) + 0 = c1(r1 −r2)ξ(1).
Since r1 ̸= r2 , we must have c1 = 0 , which leads to a contradiction.
(d) Note that (A −r1I)ξ(2) = (r2 −r1)ξ(2).
(e) Let n = 3, with r1 ̸= r2 ̸= r3. Suppose that ξ(1), ξ(2) and ξ(3) are indeed linearly dependent. Then there exist constants c1, c2 and c3, not all zero, such that c1ξ(1) + c2ξ(2) + c3ξ(3) = 0 .
Assume that c1 ̸= 0 . It is clear that (A −r2I)(c1ξ(1) + c2 ξ(2) + c3ξ(3)) = 0 . On the other hand, (A −r2I)(c1ξ(1) + c2 ξ(2) + c3ξ(3)) = c1(r1 −r2)ξ(1) + c3(r3 −r2)ξ(3).
7.5 287 It follows that c1(r1 −r2)ξ(1) + c3(r3 −r2)ξ(3) = 0 . Based on the result of part (a), which is actually not dependent on the value of n , the vectors ξ(1)and ξ(3) are linearly independent.
Hence we must have c1(r1 −r2) = c3(r3 −r2) = 0 , which leads to a contradiction.
29.(a) Let x1 = y and x2 = y ′. It follows that x ′ 1 = x2 and x ′ 2 = y ′′ = −1 a(c y + b y ′).
In terms of the new variables, we obtain the system of two first order ODEs x ′ 1 = x2 x ′ 2 = −1 a(c x1 + b x2) .
(b) The coefficient matrix is given by A = 0 1 −c a −b a .
Setting x= ξ ert results in the algebraic equations −r 1 −c a −b a −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we must have det(A −rI) = r2 + b ar + c a = 0 .
Multiplying both sides of the equation by a , we obtain a r2 + b r + c = 0 .
30.(a) Solution of the ODE requires analysis of the algebraic equations −1/10 −r 3/40 1/10 −1/5 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we must have det(A −rI) = 0 . The characteristic equation is 80 r2 + 24 r + 1 = 0 , with roots r1 = −1/4 and r2 = −1/20 . With r = −1/4 , the system of equations reduces to 2 ξ1 + ξ2 = 0 . The corresponding eigenvector is ξ(1) = (1 , −2)T . Substitution of r = −1/20 results in the equation 2 ξ1 −3 ξ2 = 0 .
A corresponding eigenvector is ξ(2) = (3 , 2)T . Since the eigenvalues are distinct, the general solution is x = c1 1 −2 e−t/4 + c2 3 2 e−t/20.
Invoking the initial conditions, we obtain the system of equations c1 + 3 c2 = −17 −2 c1 + 2 c2 = −21 .
288 Chapter 7. Systems of First Order Linear Equations Hence c1 = 29/8 and c2 = −55/8 , and the solution of the IVP is x = 29 8 1 −2 e−t/4 −55 8 3 2 e−t/20.
(b) (c) Both functions are monotone increasing. It is easy to show that −0.5 ≤x1(t) < 0 and −0.5 ≤x2(t) < 0 provided that t > T ≈74.39 .
32.(a) The system of differential equations is d dt I V = −1/2 −1/2 3/2 −5/2 I V .
Solution of the system requires analysis of the eigenvalue problem −1/2 −r −1/2 3/2 −5/2 −r ξ1 ξ2 = 0 0 .
The characteristic equation is r2 + 3r + 2 = 0, with roots r1 = −1 and r2 = −2.
With r = −1, the equations reduce to ξ1 −ξ2 = 0. A corresponding eigenvector is given by ξ(1) = (1, 1)T .
Setting r = −2, the system reduces to the equation 3ξ1 −ξ2 = 0. An eigenvector is ξ(2) = (1, 3)T . Hence the general solution is I V = c1 1 1 e−t + c2 1 3 e−2t.
(b) The eigenvalues are distinct and both negative. We find that the equilibrium point (0, 0) is a stable node. Hence all solutions converge to (0, 0).
33.(a) Solution of the ODE requires analysis of the algebraic equations −R1 L −r −1 L 1 C − 1 CR2 −r ξ1 ξ2 = 0 0 .
The characteristic equation is r2 + (L + CR1R2 LCR2 )r + R1 + R2 LCR2 = 0 .
7.6 289 The eigenvectors are real and distinct, provided that the discriminant is positive.
That is, (L + CR1R2 LCR2 )2 −4(R1 + R2 LCR2 ) > 0, which simplifies to the condition ( 1 CR2 −R1 L )2 − 4 LC > 0 .
(b) The parameters in the ODE are all positive. Observe that the sum of the roots is −L + CR1R2 LCR2 < 0 .
Also, the product of the roots is R1 + R2 LCR2 > 0 .
It follows that both roots are negative. Hence the equilibrium solution I = 0, V = 0 represents a stable node, which attracts all solutions.
(c) If the condition in part (a) is not satisfied, that is, ( 1 CR2 −R1 L )2 − 4 LC ≤0 , then the real part of the eigenvalues is Re(r1,2) = −L + CR1R2 2 LCR2 .
As long as the parameters are all positive, then the solutions will still converge to the equilibrium point (0 , 0).
7.6 2.(a) Setting x= ξ ert results in the algebraic equations −1 −r −4 1 −1 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we require that det(A −rI) = r2 + 2r + 5 = 0. The roots of the characteristic equation are r = −1 ± 2i . Substituting r = −1 −2i , the two equations reduce to ξ1 + 2i ξ2 = 0 . The two eigenvectors are ξ(1) = (−2i , 1)T and ξ(2) = (2i , 1)T . Hence one of the complex-valued solutions is given by x(1) = −2i 1 e−(1+2i)t = −2i 1 e−t(cos 2t −i sin 2t) = = e−t −2 sin 2t cos 2t + i e−t −2 cos 2t −sin 2t .
290 Chapter 7. Systems of First Order Linear Equations Based on the real and imaginary parts of this solution, the general solution is x = c1 e−t −2 sin 2t cos 2t + c2 e−t 2 cos 2t sin 2t .
(b) 3.(a) Solution of the ODEs is based on the analysis of the algebraic equations 2 −r −5 1 −2 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we require that det(A −rI) = r2 + 1 = 0. The roots of the characteristic equation are r = ±i . Setting r = i , the equations are equivalent to ξ1 −(2 + i)ξ2 = 0 . The eigenvectors are ξ(1) = (2 + i , 1)T and ξ(2) = (2 −i , 1)T .
Hence one of the complex-valued solutions is given by x(1) = 2 + i 1 eit = 2 + i 1 (cos t + i sin t) = = 2 cos t −sin t cos t + i cos t + 2 sin t sin t .
Therefore the general solution is x = c1 2 cos t −sin t cos t + c2 cos t + 2 sin t sin t .
The solution may also be written as x = c1 5 cos t 2 cos t + sin t + c2 5 sin t −cos t + 2 sin t .
7.6 291 (b) 4.(a) Setting x= ξ ert results in the algebraic equations 2 −r −5/2 9/5 −1 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we require that det(A −rI) = r2 −r + 5 2 = 0. The roots of the characteristic equation are r = (1 ± 3i)/2 . With r = (1 + 3 i)/2 , the equa-tions reduce to the single equation (3 −3i)ξ1 −5 ξ2 = 0 . The corresponding eigen-vector is given by ξ(1) = (5 , 3 −3 i)T . Hence one of the complex-valued solutions is x(1) = 5 3 −3i e(1+3i)t/2 = 2 + i 1 et/2(cos 3 2t + i sin 3 2t) = = et/2 2 cos 3 2t −sin 3 2t cos 3 2t + iet/2 cos 3 2t + 2 sin 3 2t sin 3 2t .
The general solution is x = c1 et/2 2 cos 3 2t −sin 3 2t cos 3 2t + c2 et/2 cos 3 2t + 2 sin 3 2t sin 3 2t .
The solution may also be written as x = c1 et/2 5 cos 3 2t 3 cos 3 2t + 3 sin 3 2t + c2 et/2 5 sin 3 2t −3 cos 3 2t + 3 sin 3 2t .
292 Chapter 7. Systems of First Order Linear Equations (b) 5.(a) Setting x= ξ tr results in the algebraic equations 1 −r −1 5 −3 −r ξ1 ξ2 = 0 0 .
The characteristic equation is r2 + 2 r + 2 = 0 , with roots r = −1 ± i . Substitut-ing r = −1 −i reduces the system of equations to (2 + i)ξ1 −ξ2 = 0 . The eigenvec-tors are ξ(1) = (1 , 2 + i)T and ξ(2) = (1 , 2 −i)T . Hence one of the complex-valued solutions is given by x(1) = 1 2 + i e−(1+i)t = 1 2 + i e−t(cos t −i sin t) = = e−t cos t 2 cos t + sin t + ie−t −sin t cos t −2 sin t .
The general solution is x = c1 e−t cos t 2 cos t + sin t + c2 e−t sin t −cos t + 2 sin t .
(b) 7.6 293 6.(a) Solution of the ODEs is based on the analysis of the algebraic equations 1 −r 2 −5 −1 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we require that det(A −rI) = r2 + 9 = 0. The roots of the characteristic equation are r = ± 3 i . Setting r = 3 i , the two equations reduce to (1 −3 i)ξ1 + 2ξ2 = 0 .
The corresponding eigenvector is ξ(1) = (−2 , 1 −3i)T .
Hence one of the complex-valued solutions is given by x(1) = −2 1 −3i e3it = −2 1 −3i (cos 3t + i sin 3t) = = −2 cos 3t cos 3t + 3 sin 3t + i −2 sin 3t −3 cos 3t + sin 3t .
The general solution is x = c1 −2 cos 3t cos 3t + 3 sin 3t + c2 2 sin 3t 3 cos 3t −sin 3t .
(b) 8. The eigensystem is obtained from analysis of the equation −3 −r 0 2 1 −1 −r 0 −2 −1 −r ξ1 ξ2 ξ3 = 0 0 0 .
The characteristic equation of the coefficient matrix is r3 + 4r2 + 7r + 6 = 0 , with roots r1 = −2 , r2 = −1 − √ 2 i and r3 = −1 + √ 2 i . Setting r = −2 , the equa-tions reduce to −ξ1 + 2 ξ3 = 0 ξ1 + ξ2 = 0 .
The corresponding eigenvector is ξ(1) = (2 , −2 , 1)T .
With r = −1 − √ 2 i , the system of equations is equivalent to (2 −i √ 2 )ξ1 −2 ξ3 = 0 ξ1 + i √ 2 ξ2 = 0 .
294 Chapter 7. Systems of First Order Linear Equations An eigenvector is given by ξ(2) = (−i √ 2 , 1 , −1 −i √ 2 )T . Hence one of the complex-valued solutions is given by x(2) = −i √ 2 1 −1 −i √ 2 e−(1+i √ 2)t = −i √ 2 1 −1 −i √ 2 e−t(cos √ 2 t −i sin √ 2 t) = = e−t − √ 2 sin √ 2 t cos √ 2 t −cos √ 2 t − √ 2 sin √ 2 t + ie−t − √ 2 cos √ 2 t −sin √ 2 t − √ 2 cos √ 2 t + sin √ 2 t .
The other complex-valued solution is x(3) = ξ(2) er3t. The general solution is x = c1 2 −2 1 e−2t+ + c2 e−t √ 2 sin √ 2 t −cos √ 2 t cos √ 2 t + √ 2 sin √ 2 t + c3e−t √ 2 cos √ 2 t sin √ 2 t √ 2 cos √ 2 t −sin √ 2 t .
It is easy to see that all solutions converge to the equilibrium point (0 , 0 , 0) .
10. Solution of the system of ODEs requires that −3 −r 2 −1 −1 −r ξ1 ξ2 = 0 0 .
The characteristic equation is r2 + 4 r + 5 = 0 , with roots r = −2 ± i . Substi-tuting r = −2 + i , the equations are equivalent to ξ1 −(1 −i)ξ2 = 0 . The corre-sponding eigenvector is ξ(1) = (1 −i , 1)T . One of the complex-valued solutions is given by x(1) = 1 −i 1 e(−2+i)t = 1 −i 1 e−2t(cos t + i sin t) = = e−2t cos t + sin t cos t + ie−2t −cos t + sin t sin t .
Hence the general solution is x = c1 e−2t cos t + sin t cos t + c2 e−2t −cos t + sin t sin t .
Invoking the initial conditions, we obtain the system of equations c1 −c2 = 1 c1 = −2 .
Solving for the coefficients, the solution of the initial value problem is x = −2 e−2t cos t + sin t cos t −3 e−2t −cos t + sin t sin t = e−2t cos t −5 sin t −2 cos t −3 sin t .
7.6 295 The solution converges to (0, 0) as t →∞.
12. Solution of the ODEs is based on the analysis of the algebraic equations −4 5 −r 2 −1 6 5 −r ξ1 ξ2 = 0 0 .
The characteristic equation is 25 r2 −10 r + 26 = 0 , with roots r = 1/5 ± i . Set-ting r = 1/5 + i , the two equations reduce to ξ1 −(1 −i)ξ2 = 0 . The correspond-ing eigenvector is ξ(1) = (1 −i , 1)T . One of the complex-valued solutions is given by x(1) = 1 −i 1 e( 1 5 +i)t = 1 −i 1 et/5(cos t + i sin t) = = et/5 cos t + sin t cos t + iet/5 −cos t + sin t sin t .
Hence the general solution is x = c1et/5 cos t + sin t cos t + c2et/5 −cos t + sin t sin t .
(b) Let x(0) = (x0 1 , x0 2)T . The solution of the initial value problem is x = x0 2 et/5 cos t + sin t cos t + (x0 2 −x0 1)et/5 −cos t + sin t sin t = et/5 x0 1 cos t + (2 x0 2 −x0 1) sin t x0 2 cos t + (x0 2 −x0 1) sin t .
With x(0) = (1 , 2)T , the solution is x = et/5 cos t + 3 sin t 2 cos t + sin t .
296 Chapter 7. Systems of First Order Linear Equations (c) (d) 13.(a) The characteristic equation is r2 −2αr + 1 + α2 = 0, with roots r = α ± i .
(b) When α < 0 and α > 0 , the equilibrium point (0 , 0) is a stable spiral and an unstable spiral, respectively. The equilibrium point is a center when α = 0 .
(c) (a) α = −1/8 (b) α = 1/8 7.6 297 14.(a) The roots of the characteristic equation, r2 −α r + 5 = 0 , are r1,2 = α 2 ± 1 2 p α2 −20 .
(b) Note that the roots are complex when − √ 20 < α < √ 20 . For the case when α ∈(− √ 20 , 0), the equilibrium point (0 , 0) is a stable spiral. On the other hand, when α ∈(0 , √ 20 ), the equilibrium point is an unstable spiral. For the case α = 0, the roots are purely imaginary, so the equilibrium point is a center. When α2 > 20 , the roots are real and distinct. The equilibrium point becomes a node, with its stability dependent on the sign of α . Finally, the case α2 = 20 marks the transition from spirals to nodes.
(c) (a) α = −5 (b) α = −3 (c) α = −1/2 (d) α = 1/2 17. The characteristic equation of the coefficient matrix is r2 + 2r + 1 + α = 0 , with roots given formally as r1,2 = −1 ± √−α . The roots are real provided that α ≤0 . First note that the sum of the roots is −2 and the product of the roots is 1 + α . For negative values of α , the roots are distinct, with one always negative.
When α < −1, the roots have opposite signs.
Hence the equilibrium point is a saddle. For the case −1 < α < 0 , the roots are both negative, and the equilibrium point is a stable node. α = −1 represents a transition from saddle to node. When α = 0 , both roots are equal. For the case α > 0 , the roots are complex conjugates, with negative real part. Hence the equilibrium point is a stable spiral.
(a) α = −3/2 (b) α = −1/2 (c) α = 1/2 298 Chapter 7. Systems of First Order Linear Equations 19. The characteristic equation for the system is given by r2 + (4 −α)r + 10 −4α = 0.
The roots are r1,2 = −2 + α 2 ± p α2 + 8α −24 .
First note that the roots are complex when −4 −2 √ 10 < α < −4 + 2 √ 10 . We also find that when −4 −2 √ 10 < α < −4 + 2 √ 10 , the equilibrium point is a stable spiral.
For α > −4 + 2 √ 10 , the roots are real.
When α > 2.5 , the roots have opposite signs, with the equilibrium point being a saddle.
For the case −4 + 2 √ 10 < α < 2.5 , the roots are both negative, and the equilibrium point is a stable node. Finally, when α < −4 −2 √ 10 , both roots are negative, with the equilibrium point being a stable node.
(a) α = −11 (b) α = −8 (c) α = 2 (d) α = 2.4 (e) α = 4 20. The characteristic equation is r2 + 2 r −(24 + 8α) = 0 , with roots r1,2 = −1 ± √ 25 + 8α .
The roots are complex when α < −25/8 . Since the real part is negative, the origin is a stable spiral. Otherwise the roots are real. When −25/8 < α < −3 , both roots are negative, and hence the equilibrium point is a stable node. For α > −3 , the roots are of opposite sign and the origin is a saddle.
7.6 299 (a) α = −4 (b) α = −3.05 (c) α = −2 22. Based on the method in Problem 19 of Section 7.5, setting x= ξ tr results in the algebraic equations 2 −r −5 1 −2 −r ξ1 ξ2 = 0 0 .
The characteristic equation for the system is r2 + 1 = 0 , with roots r1,2 = ± i .
With r = i , the equations reduce to the single equation ξ1 −(2 + i)ξ2 = 0. A cor-responding eigenvector is ξ(1) = (2 + i , 1)T . One complex-valued solution is x(1) = 2 + i 1 ti .
We can write ti = ei ln t. Hence x(1) = 2 + i 1 ei ln t = 2 + i 1 [cos(ln t) + i sin(ln t)] = = 2 cos(ln t) −sin(ln t) cos(ln t) + i cos(ln t) + 2 sin(ln t) sin(ln t) .
Therefore the general solution is x = c1 2 cos(ln t) −sin(ln t) cos(ln t) + c2 cos(ln t) + 2 sin(ln t) sin(ln t) .
Other combinations are also possible.
24.(a) The characteristic equation of the system is r3 + 2 5r2 + 81 80r −17 160 = 0 , with eigenvalues r1 = 1/10, and r2,3 = −1/4 ± i. For r = 1/10, simple calculations reveal that a corresponding eigenvector is ξ(1) = (0, 0, 1)T . Setting r = −1/4 −i, we obtain the system of equations ξ1 −i ξ2 = 0 ξ3 = 0 .
300 Chapter 7. Systems of First Order Linear Equations A corresponding eigenvector is ξ(2) = (i , 1 , 0)T . Hence one solution is x(1) = 0 0 1 et/10.
Another solution, which is complex-valued, is given by x(2) = i 1 0 e−( 1 4 +i)t = i 1 0 e−t/4(cos t −i sin t) = = e−t/4 sin t cos t 0 + ie−t/4 cos t −sin t 0 .
Using the real and imaginary parts of x(2), the general solution is constructed as x = c1 0 0 1 et/10 + c2 e−t/4 sin t cos t 0 + c3 e−t/4 cos t −sin t 0 .
(b) Let x(0) = (x0 1 , x0 2 , x0 3) . The solution can be written as x = 0 0 x0 3 et/10 + e−t/4 x0 2 sin t + x0 1 cos t x0 2 cos t −x0 1 sin t 0 .
With x(0) = (1 , 1 , 1), the solution of the initial value problem is x = 0 0 et/10 + e−t/4 sin t + cos t cos t −sin t 0 .
(a) x1 −x2 (b) x1 −x3 (c) x2 −x3 7.6 301 (c) 25.(a) Based on Problems 19-21 of Section 7.1, the system of differential equations is d dt I V = −R1 L −1 L 1 C − 1 CR2 I V = −1 2 −1 8 2 −1 2 I V , since R1 = R2 = 4 ohms, C = 1/2 farads and L = 8 henrys.
(b) The eigenvalue problem is −1 2 −r −1 8 2 −1 2 −r ξ1 ξ2 = 0 0 .
The characteristic equation of the system is r2 + r + 1 2 = 0 , with eigenvalues r1,2 = −1 2 ± 1 2i .
Setting r = −1/2 + i/2 , the algebraic equations reduce to 4iξ1 + ξ2 = 0 . It follows that ξ(1) = (1 , −4i)T . Hence one complex-valued solution is I V (1) = 1 −4i e(−1+i)t/2 = 1 −4i e−t/2 [cos(t/2) + i sin(t/2)] = = e−t/2 cos(t/2) 4 sin(t/2) + ie−t/2 sin(t/2) −4 cos(t/2) .
Therefore the general solution is I V = c1e−t/2 cos(t/2) 4 sin(t/2) + c2e−t/2 sin(t/2) −4 cos(t/2) .
(c) Imposing the initial conditions, we arrive at the equations c1 = 2 and c2 = −3/4 , and I V = e−t/2 2 cos(t/2) −3 4 sin(t/2) 8 sin(t/2) + 3 cos(t/2) .
(d) Since the eigenvalues have negative real parts, all solutions converge to the origin.
302 Chapter 7. Systems of First Order Linear Equations 26.(a) The characteristic equation of the system is r2 + 1 RC r + 1 CL = 0 , with eigenvalues r1,2 = − 1 2RC ± 1 2RC r 1 −4R2C L .
The eigenvalues are real and different provided that 1 −4R2C L > 0 .
The eigenvalues are complex conjugates as long as 1 −4R2C L < 0 .
(b) With the specified values, the eigenvalues are r1,2 = −1 ± i . The eigenvec-tor corresponding to r = −1 + i is ξ(1) = (1 , −4i)T . Hence one complex-valued solution is I V (1) = 1 −1 + i e(−1+i)t = 1 −1 + i e−t(cos t + i sin t) = = e−t cos t −cos t −sin t + ie−t sin t cos t −sin t .
Therefore the general solution is I V = c1e−t cos t −cos t −sin t + c2e−t sin t cos t −sin t .
(c) Imposing the initial conditions, we arrive at the equations c1 = 2 −c1 + c2 = 1 , with c1 = 2 and c2 = 3 . Therefore the solution of the IVP is I V = e−t 2 cos t + 3 sin t cos t −5 sin t .
(d) Since Re(r1,2) = −1 , all solutions converge to the origin.
27.(a) Suppose that c1a+c2b= 0 . Since a and b are the real and imaginary parts of the vector ξ(1) , respectively, a= (ξ(1) + ξ(1))/2 and b= (ξ(1) −ξ(1))/2i . Hence c1(ξ(1) + ξ(1)) −ic2(ξ(1) −ξ(1)) = 0 , which leads to (c1 −ic2)ξ(1) + (c1 + ic2)ξ(1) = 0 .
7.6 303 (b) Now since ξ(1) and ξ(1) are linearly independent, we must have c1 −ic2 = 0 c1 + ic2 = 0 .
It follows that c1 = c2 = 0 .
(c) Recall that u(t) = eλt(a cos µt −b sin µt) v(t) = eλt(a cos µt + b sin µt) .
Consider the equation c1u(t0) + c2v(t0) = 0 , for some t0. We can then write c1eλt0(a cos µt0 −b sin µt0) + c2eλt0(a cos µt0 + b sin µt0) = 0 . (∗) Rearranging the terms, and dividing by the exponential, (c1 + c2) cos µt0 a + (c2 −c1) sin µt0 b = 0 .
From part (b), since a and b are linearly independent, it follows that (c1 + c2) cos µt0 = (c2 −c1) sin µt0 = 0 .
Without loss of generality, assume that the trigonometric factors are nonzero. Oth-erwise proceed again from Equation (∗), above. We then conclude that c1 + c2 = 0 and c2 −c1 = 0 , which leads to c1 = c2 = 0. Thus u(t0) and v(t0) are linearly independent for some t0, and hence the functions are linearly independent at every point.
28.(a) Let x1 = u and x2 = u ′. It follows that x ′ 1 = x2 and x ′ 2 = u ′′ = −k m u .
In terms of the new variables, we obtain the system of two first order ODEs x ′ 1 = x2 x ′ 2 = −k m x1 .
(b) The associated eigenvalue problem is −r 1 −k/m −r ξ1 ξ2 = 0 0 .
The characteristic equation is r2 + k/m = 0 , with roots r1,2 = ± i p k/m .
(c) Since the eigenvalues are purely imaginary, the origin is a center. Hence the phase curves are ellipses, with a clockwise flow. For computational purposes, let k = 1 and m = 2 .
304 Chapter 7. Systems of First Order Linear Equations (a) k = 1, m = 2 (b) x1 −x2 (c) x1, x2 vs t (d) The general solution of the second order equation is u(t) = c1 cos r k m t + c2 sin r k m t .
The general solution of the system of ODEs is given by x = c1 p m k sin q k m t cos q k m t + c2 p m k cos q k m t −sin q k m t .
It is evident that the natural frequency of the system is equal to |r1| = |r2|.
29.(a) Set x = (x1 , x2 )T . We can rewrite Equation (22) in the form 2 0 0 9/4 d2x1 dt2 d2x2 dt2 !
= −4 3 3 −27 4 x1 x2 .
Multiplying both sides of this equation by the inverse of the diagonal matrix, we obtain d2x1 dt2 d2x2 dt2 !
= −2 3/2 4/3 −3 x1 x2 .
(b) Substituting x = ξ ert , r2 ξ1 ξ2 ert = −2 3/2 4/3 −3 ξ1 ξ2 ert, which can be written as (A −r2I)ξ = 0 .
(c) The eigenvalues are r2 1 = −1 and r2 2 = −4 , with corresponding eigenvectors ξ(1) = 3 2 and ξ(2) = 3 −4 .
(d) The linearly independent solutions are x(1) = ˜ C1 3 2 eit and x(2) = ˜ C2 3 −4 e2it .
7.6 305 in which ˜ C1 and ˜ C2 are arbitrary complex coefficients. In scalar form, x1 = 3c1 cos t + 3c2 sin t + 3c3 cos 2t + 3c4 sin 2t x2 = 2c1 cos t + 2c2 sin t −4c3 cos 2t −4c4 sin 2t (e) Differentiating the above expressions, x ′ 1 = −3c1 sin t + 3c2 cos t −6c3 sin 2t + 6c4 cos 2t x ′ 2 = −2c1 sin t + 2c2 cos t + 8c3 sin 2t −8c4 cos 2t It is evident that y= (x1, x2, x ′ 1, x ′ 2)T as in Equation (31).
31.(a) The second order system is given by d2x1 dt2 = −2x1 + x2 d2x2 dt2 = x1 −2x2 Let y1 = x1, y2 = x2, y3 = x ′ 1 and y4 = x ′ 2. In terms of the new variables, we have y ′ 1 = y3 y ′ 2 = y4 y ′ 3 = −2y1 + y2 y ′ 4 = y1 −2y2 hence the coefficient matrix is A = 0 0 1 0 0 0 0 1 −2 1 0 0 1 −2 0 0 .
(b) The eigenvalues and corresponding eigenvectors of A are: r1 = i , ξ(1) = (1, 1, i, i)T r2 = −i , ξ(2) = (1, 1, −i, −i)T r3 = √ 3 i , ξ(3) = (1, −1, √ 3 i, − √ 3 i)T r4 = − √ 3 i , ξ(4) = (1, −1, − √ 3 i, √ 3 i)T (c) Note that ξ(1)eit = 1 1 i i (cos t + i sin t) and ξ(3)e √ 3 it = 1 −1 √ 3 i − √ 3 i (cos √ 3 t + i sin √ 3 t).
306 Chapter 7. Systems of First Order Linear Equations Hence the general solution is y = c1 cos t cos t −sin t −sin t + c2 sin t sin t cos t cos t + c3 cos √ 3 t −cos √ 3 t − √ 3 sin √ 3 t √ 3 sin √ 3 t + c4 sin √ 3 t −sin √ 3 t √ 3 cos √ 3 t − √ 3 cos √ 3 t .
(d) The two modes have natural frequencies of ω1 = 1 rad/sec and ω2 = √ 3 rad/sec.
(e) For the initial condition y(0) = (−1, 3, 0, 0)T , it is necessary that −1 3 0 0 = c1 1 1 0 0 + c2 0 0 1 1 + c3 1 −1 0 0 + c4 0 0 √ 3 − √ 3 , resulting in the coefficients c1 = 1, c2 = 0, c3 = −2 and c4 = 0.
The solutions are not periodic, since the two natural frequencies are incommensu-rate.
7.7 307 7.7 1.(a) The eigenvalues and eigenvectors were found in Problem 1, Section 7.5.
r1 = −1, ξ(1) = 1 2 ; r2 = 2, ξ(2) = 2 1 .
The general solution is x = c1 e−t 2 e−t + c2 2 e2t e2t .
Hence a fundamental matrix is given by Ψ(t) = e−t 2 e2t 2 e−t e2t .
(b) We now have Ψ(0) = 1 2 2 1 and Ψ−1(0) = 1 3 −1 2 2 −1 , So that Φ(t) = Ψ(t)Ψ−1(0) = 1 3 −e−t + 4e2t 2e−t −2e2t −2e−t + 2e2t 4e−t −e2t .
3.(a) The eigenvalues and eigenvectors were found in Problem 3, Section 7.5. The general solution of the system is x = c1 et et + c2 e−t 3e−t .
Hence a fundamental matrix is given by Ψ(t) = et e−t et 3e−t .
(b) Given the initial conditions x(0) =e(1), we solve the equations c1 + c2 = 1 c1 + 3c2 = 0 , to obtain c1 = 3/2 , c2 = −1/2 . The corresponding solution is x = 3 2et −1 2e−t 3 2et −3 2e−t .
Given the initial conditions x(0) =e(2), we solve the equations c1 + c2 = 0 c1 + 3c2 = 1 , 308 Chapter 7. Systems of First Order Linear Equations to obtain c1 = −1/2 , c2 = 1/2 . The corresponding solution is x = −1 2et + 1 2e−t −1 2et + 3 2e−t .
Therefore the fundamental matrix is Φ(t) = 1 2 3et −e−t −et + e−t 3et −3e−t −et + 3e−t .
5.(a) The general solution, found in Problem 3, Section 7.6, is given by x = c1 5 cos t 2 cos t + sin t + c2 5 sin t −cos t + 2 sin t .
Hence a fundamental matrix is given by Ψ(t) = 5 cos t 5 sin t 2 cos t + sin t −cos t + 2 sin t .
(b) Given the initial conditions x(0) =e(1), we solve the equations 5c1 = 1 2c1 −c2 = 0 , resulting in c1 = 1/5 , c2 = 2/5 . The corresponding solution is x = cos t + 2 sin t sin t .
Given the initial conditions x(0) =e(2), we solve the equations 5c1 = 0 2c1 −c2 = 1 , resulting in c1 = 0 , c2 = −1 . The corresponding solution is x = −5 sin t cos t −2 sin t .
Therefore the fundamental matrix is Φ(t) = cos t + 2 sin t −5 sin t sin t cos t −2 sin t .
7.(a) The general solution, found in Problem 15, Section 7.5, is given by x = c1 e2t 3e2t + c2 e4t e4t .
Hence a fundamental matrix is given by Ψ(t) = e2t e4t 3e2t e4t .
7.7 309 (b) Given the initial conditions x(0) =e(1), we solve the equations c1 + c2 = 1 3c1 + c2 = 0 , resulting in c1 = −1/2 , c2 = 3/2 . The corresponding solution is x = 1 2 −e2t + 3e4t −3e2t + 3e4t .
The initial conditions x(0) =e(2) require that c1 + c2 = 0 3c1 + c2 = 1 , resulting in c1 = 1/2 , c2 = −1/2 . The corresponding solution is x = 1 2 e2t −e4t 3e2t −e4t .
Therefore the fundamental matrix is Φ(t) = 1 2 −e2t + 3e4t e2t −e4t −3e2t + 3e4t 3e2t −e4t .
8.(a) The general solution, found in Problem 5, Section 7.6, is given by x = c1e−t cos t 2 cos t + sin t + c2e−t sin t −cos t + 2 sin t .
Hence a fundamental matrix is given by Ψ(t) = e−t cos t e−t sin t 2e−t cos t + e−t sin t −e−t cos t + 2e−t sin t .
(b) The specific solution corresponding to the initial conditions x(0) =e(1) is x = e−t cos t + 2 sin t 5 sin t .
For the initial conditions x(0) =e(2), the solution is x = e−t −sin t cos t −2 sin t .
Therefore the fundamental matrix is Φ(t) = e−t cos t + 2 sin t −sin t 5 sin t cos t −2 sin t .
9.(a) The general solution, found in Problem 13, Section 7.5, is given by x = c1 4e−2t −5e−2t −7e−2t + c2 3e−t −4e−t −2e−t + c3 0 e2t −e2t .
310 Chapter 7. Systems of First Order Linear Equations Hence a fundamental matrix is given by Ψ(t) = 4e−2t 3e−t 0 −5e−2t −4e−t e2t −7e−2t −2e−t −e2t .
(b) Given the initial conditions x(0) =e(1), we solve the equations 4c1 + 3c2 = 1 −5c1 −4c2 + c3 = 0 −7c1 −2c2 −c3 = 0 , resulting in c1 = −1/2 , c2 = 1 , c3 = 3/2 . The corresponding solution is x = −2e−2t + 3e−t 5e−2t/2 −4e−t + 3e2t/2 7e−2t/2 −2e−t −3e2t/2 .
The initial conditions x(0) =e(2), we solve the equations 4c1 + 3c2 = 0 −5c1 −4c2 + c3 = 1 −7c1 −2c2 −c3 = 0 , resulting in c1 = −1/4 , c2 = 1/3 , c3 = 13/12 . The corresponding solution is x = −e−2t + e−t 5e−2t/4 −4e−t/3 + 13e2t/12 7e−2t/4 −2e−t/3 −13e2t/12 .
The initial conditions x(0) =e(3), we solve the equations 4c1 + 3c2 = 0 −5c1 −4c2 + c3 = 0 −7c1 −2c2 −c3 = 1 , resulting in c1 = −1/4 , c2 = 1/3 , c3 = 1/12 . The corresponding solution is x = −e−2t + e−t 5e−2t/4 −4e−t/3 + e2t/12 7e−2t/4 −2e−t/3 −e2t/12 .
Therefore the fundamental matrix is Φ(t) = 1 12 −24e−2t + 36e−t −12e−2t + 12e−t −12e−2t + 12e−t 30e−2t −48e−t + 18e2t 15e−2t −16e−t + 13e2t 15e−2t −16e−t + e2t 42e−2t −24e−t −18e2t 21e−2t −8e−t −13e2t 21e−2t −8e−t −e2t .
12. The solution of the initial value problem is given by x = Φ(t)x(0) = e−t cos 2t −2e−t sin 2t 1 2e−t sin 2t e−t cos 2t 3 1 = = e−t 3 cos 2t −2 sin 2t 3 2 sin 2t + cos 2t .
7.7 311 13. Let Ψ(t) = x(1) 1 (t) · · · x(n) 1 (t) .
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x(1) n (t) · · · x(n) n (t) .
It follows that Ψ(t0) = x(1) 1 (t0) · · · x(n) 1 (t0) .
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x(1) n (t0) · · · x(n) n (t0) is a scalar matrix, which is invertible, since the solutions are linearly independent.
Let Ψ−1(t0) = (cij). Then Ψ(t)Ψ−1(t0) = x(1) 1 (t) · · · x(n) 1 (t) .
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x(1) n (t) · · · x(n) n (t) c11 · · · c1n .
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cn1 · · · cnn .
The j-th column of the product matrix is Ψ(t)Ψ−1(t0) (j) = n X k = 1 ckj x(k), which is a solution vector, since it is a linear combination of solutions. Furthermore, the columns are all linearly independent, since the vectors x(k) are.
Hence the product is a fundamental matrix. Finally, setting t = t0 , Ψ(t0)Ψ−1(t0) =I . This is precisely the definition of Φ(t).
14. The fundamental matrix Φ(t) for the system x ′ = 1 1 4 1 x is given by Φ(t) = 1 4 2e3t + 2e−t e3t −e−t 4e3t −4e−t 2e3t + 2e−t .
Direct multiplication results in Φ(t)Φ(s) = 1 16 2e3t + 2e−t e3t −e−t 4e3t −4e−t 2e3t + 2e−t 2e3s + 2e−s e3s −e−s 4e3s −4e−s 2e3s + 2e−s = 1 16 8(e3t+3s + e−t−s) 4(e3t+3s −e−t−s) 16(e3t+3s −e−t−s) 8(e3t+3s + e−t−s) .
Hence Φ(t)Φ(s) = 1 4 2e3(t+s) + 2e−(t+s) e3(t+s) −e−(t+s) 4e3(t+s) −4e−(t+s) 2e3(t+s) + 2e−(t+s) = Φ(t + s).
312 Chapter 7. Systems of First Order Linear Equations 15.(a) Let s be arbitrary, but fixed, and t variable. Similar to the argument in Problem 13, the columns of the matrix Φ(t)Φ(s) are linear combinations of funda-mental solutions. Hence the columns of Φ(t)Φ(s) are also solution of the system of equations. Further, setting t = 0 , Φ(0)Φ(s) =I Φ(s) = Φ(s) . That is, Φ(t)Φ(s) is a solution of the initial value problem Z ′ =AZ, with Z(0) = Φ(s) . Now consider the change of variable τ = t + s . Let W(τ) =Z(τ −s). The given initial value problem can be reformulated as d dτ W = AW , with W(s) = Φ(s) .
Since Φ(t) is a fundamental matrix satisfying Φ ′ =AΦ , with Φ(0) =I, it follows that W(τ) = Φ(τ)Φ−1(s) Φ(s) = Φ(τ).
That is, Φ(t + s) = Φ(τ) =W(τ) =Z(t) = Φ(t)Φ(s) .
(b) Based on part (a), Φ(t)Φ(−t) = Φ(t + (−t)) = Φ(0) =I. Hence Φ(−t) = Φ−1(t).
(c) It also follows that Φ(t −s) = Φ(t + (−s)) = Φ(t)Φ(−s) = Φ(t)Φ−1(s).
16. Let A be a diagonal matrix, with A= a1e(1), a2e(2), · · · , ane(n) . Note that for any positive integer k, Ak = h ak 1 e(1), ak 2 e(2), · · · , ak n e(n)i .
It follows, from basic matrix algebra, that I + m X k = 1 Ak tk k! = Pm k= 0 ak 1 tk k!
0 · · · 0 0 Pm k= 0 ak 2 tk k!
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0 0 · · · Pm k= 0 ak n tk k!
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It can be shown that the partial sums on the left hand side converge for all t .
Taking the limit as m →∞on both sides of the equation, we obtain eAt = ea1t 0 · · · 0 0 ea2t · · · 0 .
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0 0 · · · eant .
Alternatively, consider the system x ′ =Ax . Since the ODEs are uncoupled, the vectors x(j) = eajt e(j), j = 1, 2, · · · n , are a set of linearly independent solutions.
Hence the matrix x = h ea1t e(1), ea2t e(2), · · · , eant e(n)i is a fundamental matrix. Finally, since X(0) =I, it follows that h ea1t e(1), ea2t e(2), · · · , eant e(n)i = Φ(t) = eAt .
7.7 313 17.(a) Let x1 = u and x2 = u ′; then u ′′ = x ′ 2 . In terms of the new variables, we have x ′ 2 + ω2 x1 = 0 with the initial conditions x1(0) = u0 and x2(0) = v0 . The equivalent first order system is x ′ 1 = x2 x ′ 2 = −ω2 x1 which can be expressed in the form x1 x2 ′ = 0 1 −ω2 0 x1 x2 ; x1(0) x2(0) = u0 v0 .
(b) Setting A = 0 1 −ω2 0 , it is easy to show that A2 = −ω2 I, A3 = −ω2 A and A4 = ω4 I.
It follows inductively that A2k = (−1)kω2k I and A2k+1 = (−1)kω2k A .
Hence eAt = ∞ X k= 0 (−1)k ω2k t2k (2k)! I + (−1)k ω2k t2k+1 (2k + 1)! A = " ∞ X k= 0 (−1)k ω2k t2k (2k)!
I + 1 ω " ∞ X k= 0 (−1)k ω2k+1 t2k+1 (2k + 1)!
A and therefore eAt = cos ω tI + 1 ω sin ω t A.
(c) From Equation (28), x1 x2 = cos ω tI + 1 ω sin ω t A u0 v0 = cos ωt u0 v0 + 1 ω sin ω t v0 −ω2 u0 .
18.(a) Assuming that x= φ(t) is a solution, then φ ′ =Aφ, with φ(0) =x0. Integrate both sides of the equation to obtain φ(t) −φ(0) = Z t 0 Aφ(s)ds .
314 Chapter 7. Systems of First Order Linear Equations Hence φ(t) = x0 + Z t 0 Aφ(s)ds .
(b) Proceed with the iteration φ(i+1)(t) = x0 + Z t 0 Aφ(i)(s)ds .
With φ(0)(t) =x0, and noting that A is a constant matrix, φ(1)(t) = x0 + Z t 0 Ax0ds = x0 + Ax0t .
That is, φ(1)(t) = (I + At)x0.
(c) We then have φ(2)(t) = x0 + Z t 0 A(I + At)x0ds = x0 + Ax0t + A2x0 t2 2 = (I + At + A2 t2 2 )x0.
Now suppose that φ(n)(t) = (I + At + A2 t2 2 + · · · + An tn n!)x0.
It follows that Z t 0 A(I + At + A2 t2 2 + · · · + An tn n!)x0ds = = A(It + At2 2 + A2 t3 3! + · · · + An tn+1 (n + 1)!)x0 = (At + A2 t2 2 + A3 t3 3! + · · · + An+1 tn n!)x0.
Therefore φ(n+1)(t) = (I + At + A2 t2 2 + · · · + An+1 tn+1 (n + 1)!)x0.
By induction, the asserted form of φ(n)(t) is valid for all n ≥0 .
(d) Define φ(∞)(t) = limn →∞φ(n)(t). It can be shown that the limit does exist. In fact, φ(∞)(t) = eAtx0.
Term-by-term differentiation results in d dtφ(∞)(t) = d dt(I + At + A2 t2 2 + · · · + An tn n! + · · · )x0 = (A + A2t + · · · + An tn−1 (n −1)! + · · · )x0 = A(I + At + A2 t2 2 + · · · + An−1 tn−1 (n −1)! + · · · )x0.
7.8 315 That is, d dtφ(∞)(t) = Aφ(∞)(t).
Furthermore, φ(∞)(0) =x0. Based on uniqueness of solutions, φ(t) = φ(∞)(t) .
7.8 2.(a) (b) All of the points on the line x2 = 2x1 are equilibrium points. Solutions starting at all other points become unbounded.
(c) Setting x= ξ tr results in the algebraic equations 4 −r −2 8 −4 −r ξ1 ξ2 = 0 0 .
The characteristic equation is r2 = 0, with the single root r = 0. Substituting r = 0 reduces the system of equations to 2ξ1 −ξ2 = 0 . Therefore the only eigenvector is ξ = (1 , 2)T . One solution is x(1) = 1 2 , which is a constant vector.
In order to generate a second linearly independent solution, we must search for a generalized eigenvector. This leads to the system of equations 4 −2 8 −4 η1 η2 = 1 2 .
This system also reduces to a single equation, 2η1 −η2 = 1/2 . Setting η1 = k , some arbitrary constant, we obtain η2 = 2k −1/2 . A second solution is x(2) = 1 2 t + k 2k −1/2 = 1 2 t + 0 −1/2 + k 1 2 .
316 Chapter 7. Systems of First Order Linear Equations Note that the last term is a multiple of x(1) and may be dropped. Hence x(2) = 1 2 t + 0 −1/2 .
The general solution is x = c1 1 2 + c2 1 2 t + 0 −1/2 .
4.(a) (b) All trajectories converge to the origin.
(c) Solution of the ODE requires analysis of the algebraic equations −3 −r 5 2 −5 2 2 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we must have det(A −rI) = r2 + r + 1/4 = 0. The only root is r = −1/2 , which is an eigenvalue of multiplicity two. Setting r = −1/2 is the coefficient matrix reduces the system to the single equation −ξ1 + ξ2 = 0 .
Hence the corresponding eigenvector is ξ = (1 , 1)T . One solution is x(1) = 1 1 e−t/2.
In order to obtain a second linearly independent solution, we find a solution of the system −5/2 5/2 −5/2 5/2 η1 η2 = 1 1 .
There equations reduce to −5η1 + 5η2 = 2 . Set η1 = k , some arbitrary constant.
Then η2 = k + 2/5 . A second solution is x(2) = 1 1 te−t/2 + k k + 2/5 e−t/2 = 1 1 te−t/2 + 0 2/5 e−t/2 + k 1 1 e−t/2.
Dropping the last term, the general solution is x = c1 1 1 e−t/2 + c2 1 1 te−t/2 + 0 2/5 e−t/2 .
7.8 317 6. The eigensystem is obtained from analysis of the equation −r 1 1 1 −r 1 1 1 −r ξ1 ξ2 ξ3 = 0 0 0 .
The characteristic equation of the coefficient matrix is r3 −3r −2 = 0 , with roots r1 = 2 and r2,3 = −1 . Setting r = 2 , we have −2 1 1 1 −2 1 1 1 −2 ξ1 ξ2 ξ3 = 0 0 0 .
This system is reduced to the equations ξ1 −ξ3 = 0 ξ2 −ξ3 = 0 .
A corresponding eigenvector is given by ξ(1) = (1 , 1 , 1)T .
Setting r = −1 , the system of equations is reduced to the single equation ξ1 + ξ2 + ξ3 = 0 .
An eigenvector vector is given by ξ(2) = (1 , 0 , −1)T . Since the last equation has two free variables, a third linearly independent eigenvector (associated with r = −1) is ξ(3) = (0 , 1 , −1)T . Therefore the general solution may be written as x = c1 1 1 1 e2t + c2 1 0 −1 e−t + c3 0 1 −1 e−t.
7.(a) Solution of the ODE requires analysis of the algebraic equations 1 −r −4 4 −7 −r ξ1 ξ2 = 0 0 .
For a nonzero solution, we must have det(A −rI) = r2 + 6r + 9 = 0 .
The only root is r = −3 , which is an eigenvalue of multiplicity two. Substituting r = −3 into the coefficient matrix, the system reduces to the single equation ξ1 −ξ2 = 0 .
Hence the corresponding eigenvector is ξ = (1 , 1)T . One solution is x(1) = 1 1 e−3t.
For a second linearly independent solution, we search for a generalized eigenvector.
Its components satisfy 4 −4 4 −4 η1 η2 = 1 1 , that is, 4η1 −4η2 = 1 . Let η2 = k , some arbitrary constant. Then η1 = k + 1/4 .
It follows that a second solution is given by x(2) = 1 1 te−3t + k + 1/4 k e−3t = 1 1 te−3t + 1/4 0 e−3t + k 1 1 e−3t.
318 Chapter 7. Systems of First Order Linear Equations Dropping the last term, the general solution is x = c1 1 1 e−3t + c2 1 1 te−3t + 1/4 0 e−3t .
Imposing the initial conditions, we require that c1 + c2/4 = 3, c1 = 2, which results in c1 = 2 and c2 = 4 . Therefore the solution of the IVP is x = 3 2 e−3t + 4 4 te−3t.
(b) 8.(a) Solution of the ODEs is based on the analysis of the algebraic equations −5 2 −r 3 2 −3 2 1 2 −r ξ1 ξ2 = 0 0 .
The characteristic equation is r2 + 2 r + 1 = 0 , with a single root r = −1 . Setting r = −1, the two equations reduce to −ξ1 + ξ2 = 0. The corresponding eigenvector is ξ = (1 , 1)T . One solution is x(1) = 1 1 e−t.
A second linearly independent solution is obtained by solving the system −3/2 3/2 −3/2 3/2 η1 η2 = 1 1 .
The equations reduce to the single equation −3η1 + 3η2 = 2. Let η1 = k. We obtain η2 = 2/3 + k , and a second linearly independent solution is x(2) = 1 1 te−t + k 2/3 + k e−t = 1 1 te−t + 0 2/3 e−t + k 1 1 e−t.
Dropping the last term, the general solution is x = c1 1 1 e−t + c2 1 1 te−t + 0 2/3 e−t .
7.8 319 Imposing the initial conditions, we find that c1 = 3, c1 + 2c2/3 = −1, so that c1 = 3 and c2 = −6 . Therefore the solution of the IVP is x = 3 −1 e−t − 6 6 te−t.
(b) 10.(a) The eigensystem is obtained from analysis of the equation 3 −r 9 −1 −3 −r ξ1 ξ2 = 0 0 .
The characteristic equation is r2 = 0 , with a single root r = 0 . Setting r = 0 , the two equations reduce to ξ1 + 3ξ2 = 0. The corresponding eigenvector is ξ = (−3 , 1)T . Hence one solution is x(1) = −3 1 , which is a constant vector. A second linearly independent solution is obtained from the system 3 9 −1 −3 η1 η2 = −3 1 .
The equations reduce to the single equation η1 + 3η2 = −1 . Let η2 = k. We obtain η1 = −1 −3k , and a second linearly independent solution is x(2) = −3 1 t + −1 −3k k = −3 1 t + −1 0 + k −3 1 .
Dropping the last term, the general solution is x = c1 −3 1 + c2 −3 1 t + −1 0 .
Imposing the initial conditions, we require that −3c1 −c2 = 2, c1 = 4, which results in c1 = 4 and c2 = −14 . Therefore the solution of the IVP is x = 2 4 −14 −3 1 t.
320 Chapter 7. Systems of First Order Linear Equations (b) 13. Setting x= ξ tr results in the algebraic equations 3 −r −4 1 −1 −r ξ1 ξ2 = 0 0 .
The characteristic equation is r2 −2r + 1 = 0 , with a single root of r1,2 = 1 . With r = 1 , the system reduces to a single equation ξ1 −2 ξ2 = 0. An eigenvector is given by ξ = (2 , 1)T . Hence one solution is x(1) = 2 1 t .
In order to find a second linearly independent solution, we search for a generalized eigenvector whose components satisfy 2 −4 1 −2 η1 η2 = 2 1 .
These equations reduce to η1 −2 η2 = 1 . Let η2 = k , some arbitrary constant.
Then η1 = 1 + 2k . (Before proceeding, note that if we set u = ln t, the original equation is transformed into a constant coefficient equation with independent vari-able u.
Recall that a second solution is obtained by multiplication of the first solution by the factor u. This implies that we must multiply first solution by a factor of ln t.) Hence a second linearly independent solution is x(2) = 2 1 t ln t + 1 + 2k k t = 2 1 t ln t + 1 0 t + k 2 1 t.
Dropping the last term, the general solution is x = c1 2 1 t + c2 2 1 t ln t + 1 0 t .
16.(a) Using the result in Problem 15, the eigenvalues are r1,2 = − 1 2RC ± √ L2 −4R2CL 2RCL .
The discriminant vanishes when L = 4R2C.
7.8 321 (b) The system of differential equations is d dt I V = 0 1 4 −1 −1 I V .
The associated eigenvalue problem is −r 1 4 −1 −1 −r ξ1 ξ2 = 0 0 .
The characteristic equation is r2 + r + 1/4 = 0 , with a single root of r1,2 = −1/2 .
Setting r = −1/2 , the algebraic equations reduce to 2ξ1 + ξ2 = 0. An eigenvector is given by ξ = (1 , −2)T . Hence one solution is I V (1) = 1 −2 e−t/2 .
A second solution is obtained from a generalized eigenvector whose components satisfy 1 2 1 4 −1 −1 2 η1 η2 = 1 −2 .
It follows that η1 = k and η2 = 4 −2k . A second linearly independent solution is I V (2) = 1 −2 t e−t/2 + k 4 −2k e−t/2 = 1 −2 t e−t/2 + 0 4 e−t/2 + k 1 −2 e−t/2.
Dropping the last term, the general solution is I V = c1 1 −2 e−t/2 + c2 1 −2 t e−t/2 + 0 4 e−t/2 .
Imposing the initial conditions, we require that c1 = 1, −2c1 + 4c2 = 2, which re-sults in c1 = 1 and c2 = 1 . Therefore the solution of the IVP is I V = 1 2 e−t/2 + 1 −2 te−t/2.
19.(a) The eigensystem is obtained from analysis of the equation 5 −r −3 −2 8 −5 −r −4 −4 3 3 −r ξ1 ξ2 ξ3 = 0 0 0 .
The characteristic equation of the coefficient matrix is r3 −3r2 + 3r −1 = 0 , with a single root of multiplicity three, r = 1 . Setting r = 1 , we have 4 −3 −2 8 −6 −4 −4 3 2 ξ1 ξ2 ξ3 = 0 0 0 .
The system of algebraic equations reduces to a single equation 4ξ1 −3ξ2 −2ξ3 = 0 .
322 Chapter 7. Systems of First Order Linear Equations An eigenvector vector is given by ξ(1) = (1 , 0 , 2)T . Since the last equation has two free variables, a second linearly independent eigenvector (associated with r = 1) is ξ(2) = (0 , 2 , −3)T . Therefore two solutions are obtained as x(1) = 1 0 2 et and x(2) = 0 2 −3 et.
(b) It follows directly that x ′ = ξtet + ξet + ηet .
Hence the coefficient vectors must satisfy ξtet + ξet + ηet =Aξtet+Aηet. Rearranging the terms, we have ξet = (A −I)ξtet + (A −I)ηet.
Given an eigenvector ξ , it follows that (A −I)ξ = 0 and (A −I)η = ξ .
(c) Clearly, (A −I)2η = (A −I)(A −I)η = (A −I)ξ = 0 . Also, 4 −3 −2 8 −6 −4 −4 3 2 4 −3 −2 8 −6 −4 −4 3 2 = 0 0 0 0 0 0 0 0 0 (d) We get that ξ = (A −I)η = 4 −3 −2 8 −6 −4 −4 3 2 0 0 1 = −2 −4 2 .
This is an eigenvector: 5 −3 −2 8 −5 −4 −4 3 3 −2 −4 2 = −2 −4 2 .
(e) Given the three linearly independent solutions, a fundamental matrix is given by Ψ(t) = et 0 −2t et 0 2et −4t et 2et −3et 2t et + et .
(f) We construct the transformation matrix T = 1 −2 0 0 −4 0 2 2 1 , with inverse T−1 = 1 −1/2 0 0 −1/4 0 −2 3/2 1 .
The Jordan form of the matrix A is J = T−1AT = 1 0 0 0 1 1 0 0 1 .
7.8 323 21.(a) Direct multiplication results in J2 = λ2 0 0 0 λ2 2λ 0 0 λ2 , J3 = λ3 0 0 0 λ3 3λ2 0 0 λ3 , J4 = λ4 0 0 0 λ4 4λ3 0 0 λ4 .
(b) Suppose that Jn = λn 0 0 0 λn nλn−1 0 0 λn .
Then Jn+1 = λn 0 0 0 λn nλn−1 0 0 λn λ 0 0 0 λ 1 0 0 λ = λ · λn 0 0 0 λ · λn λn + nλ · λn−1 0 0 λ · λn .
Hence the result follows by mathematical induction.
(c) Note that J is block diagonal. Hence each block may be exponentiated. Using the result in Problem 20, eJt = eλt 0 0 0 eλt teλt 0 0 eλt .
(d) Setting λ = 1 , and using the transformation matrix T in Problem 19, TeJt = 1 2 0 0 4 0 2 −2 −1 et 0 0 0 et tet 0 0 et = et 2et 2t et 0 4et 4t et 2et −2et −2t et −et .
Based on the form of J, eJt is the fundamental matrix associated with the solutions y(1) = ξ(1)et, y(2) = (2ξ(1) + 2ξ(2))et and y(3) = (2ξ(1) + 2ξ(2))tet + ηet.
Hence the resulting matrix is the fundamental matrix associated with the solution set n ξ(1)et , (2ξ(1) + 2ξ(2))et, (2ξ(1) + 2ξ(2))tet + ηeto , as opposed to the solution set in Problem 19, given by n ξ(1)et , ξ(2)et, (2ξ(1) + 2ξ(2))tet + ηeto .
22.(a) Direct multiplication results in J2 = λ2 2λ 1 0 λ2 2λ 0 0 λ2 , J3 = λ3 3λ2 3λ 0 λ3 3λ2 0 0 λ3 , J4 = λ4 4λ3 6λ2 0 λ4 4λ3 0 0 λ4 .
324 Chapter 7. Systems of First Order Linear Equations (b) Suppose that Jn = λn nλn−1 n(n−1) 2 λn−2 0 λn nλn−1 0 0 λn .
Then Jn+1 = λn nλn−1 n(n−1) 2 λn−2 0 λn nλn−1 0 0 λn λ 1 0 0 λ 1 0 0 λ = λ · λn λn + nλ · λn−1 nλn−1 + n(n−1) 2 λ · λn−2 0 λ · λn λn + nλ · λn−1 0 0 λ · λn .
The result follows by noting that nλn−1 + n(n −1) 2 λ · λn−2 = n + n(n −1) 2 λn−1 = n2 + n 2 λn−1.
(c) We first observe that ∞ X n = 0 λn tn n! = eλt ∞ X n = 0 nλn−1 tn n! = t ∞ X n = 1 λn−1 tn−1 (n −1)! = t eλt ∞ X n = 0 n(n −1) 2 λn−2 tn n! = t2 2 ∞ X n = 2 λn−2 tn−2 (n −2)! = t2 2 eλt.
Therefore eJt = eλt teλt t2 2 eλt 0 eλt teλt 0 0 eλt .
(d) Setting λ = 2 , and using the transformation matrix T in Problem 18, TeJt = 0 1 2 1 1 0 −1 0 3 e2t te2t t2 2 e2t 0 e2t te2t 0 0 e2t = 0 e2t te2t + 2e2t e2t te2t + e2t t2 2 e2t + te2t −e2t −te2t −t2 2 e2t + 3e2t .
7.9 5. As shown in Problem 2, Section 7.8, the general solution of the homogeneous equation is xc = c1 1 2 + c2 t 2t −1 2 .
7.9 325 An associated fundamental matrix is Ψ(t) = 1 t 2 2t −1 2 .
The inverse of the fundamental matrix is easily determined as Ψ−1(t) = 4t −3 −2t + 2 8t −8 −4t + 5 .
We can now compute Ψ−1(t)g(t) = −1 t3 2t2 + 4t −1 −2t −4 , and Z Ψ−1(t)g(t) dt = −1 2t−2 + 4t−1 −2 ln t −2t−2 −2t−1 .
Finally, v(t) = Ψ(t) Z Ψ−1(t)g(t) dt, where v1(t) = −1 2t−2 + 2t−1 −2 ln t −2, v2(t) = 5t−1 −4 ln t −4 .
Note that the vector (2 , 4)T is a multiple of one of the fundamental solutions.
Hence we can write the general solution as x = c1 1 2 + c2 t 2t −1 2 −1 t2 1/2 0 + 1 t 2 5 −2 ln t 1 2 .
6. The eigenvalues of the coefficient matrix are r1 = 0 and r2 = −5 . It follows that the solution of the homogeneous equation is xc = c1 1 2 + c2 −2e−5t e−5t .
The coefficient matrix is symmetric. Hence the system is diagonalizable. Using the normalized eigenvectors as columns, the transformation matrix, and its inverse, are T = 1 √ 5 1 −2 2 1 T−1 = 1 √ 5 1 2 −2 1 .
Setting x=Ty, and h(t) =T−1g(t), the transformed system is given, in scalar form, as y ′ 1 = 5 + 8t √ 5 t y ′ 2 = −5y2 + 4 √ 5 .
The solutions are readily obtained as y1(t) = √ 5 ln t + 8 √ 5 t + c1 and y2(t) = c2 e−5t + 4 5 √ 5 .
326 Chapter 7. Systems of First Order Linear Equations Transforming back to the original variables, we have x=Ty, with x = 1 √ 5 1 −2 2 1 y1(t) y2(t) = 1 √ 5 1 2 y1(t) + 1 √ 5 −2 1 y2(t).
Hence the general solution is x = k1 1 2 + k 2 −2e−5t e−5t + 1 2 ln t + 8 5 1 2 t + 4 25 −2 1 .
7. The solution of the homogeneous equation is xc = c1 e−t −2e−t + c2 e3t 2e3t .
Based on the simple form of the right hand side, we use the method of undetermined coefficients. Set v=a et. Substitution into the ODE yields a1 a2 et = 1 1 4 1 a1 a2 et + 2 −1 et.
In scalar form, after canceling the exponential, we have a1 = a1 + a2 + 2 a2 = 4a1 + a2 −1 , with a1 = 1/4 and a2 = −2 . Hence the particular solution is v = 1/4 −2 et, so that the general solution is x = c1 e−t −2e−t + c2 e3t 2e3t + 1 4 et −8et .
9. Note that the coefficient matrix is symmetric. Hence the system is diagonalizable.
The eigenvalues and eigenvectors are given by r1 = −1 2 , ξ(1) = 1 1 and r2 = −2 , ξ(2) = 1 −1 .
Using the normalized eigenvectors as columns, the transformation matrix, and its inverse, are T = 1 √ 2 1 1 1 −1 T−1 = 1 √ 2 1 1 1 −1 .
Setting x=Ty, and h(t) =T−1g(t) , the transformed system is given, in scalar form, as y ′ 1 = −1 2y1 + √ 2 t + 1 √ 2 et y ′ 2 = −2y2 + √ 2 t − 1 √ 2 et .
7.9 327 Using any elementary method for first order linear equations, the solutions are y1(t) = k1e−t/2 + √ 2 3 et −4 √ 2 + 2 √ 2 t y2(t) = k 2e−2t − 1 3 √ 2 et − 1 2 √ 2 + 1 √ 2 t .
Transforming back to the original variables, x=Ty, the general solution is x = c1 1 1 e−t/2 + c2 1 −1 e−2t −1 4 17 15 + 1 2 5 3 t + 1 6 1 3 et.
10.
Since the coefficient matrix is symmetric, the differential equations can be decoupled. The eigenvalues and eigenvectors are given by r1 = −4 , ξ(1) = √ 2 −1 and r2 = −1 , ξ(2) = 1 √ 2 .
Using the normalized eigenvectors as columns, the transformation matrix, and its inverse, are T = 1 √ 3 √ 2 1 −1 √ 2 T−1 = 1 √ 3 √ 2 −1 1 √ 2 .
Setting x=Ty, and h(t) =T−1g(t) , the transformed system is given, in scalar form, as y ′ 1 = −4y1 + 1 √ 3(1 + √ 2)e−t y ′ 2 = −y2 + 1 √ 3(1 − √ 2) e−t .
The solutions are easily obtained as y1(t) = k1e−4t + 1 3 √ 3(1 + √ 2)e−t, y2(t) = k 2e−t + 1 √ 3(1 − √ 2 )te−t.
Transforming back to the original variables, the general solution is x = c1 √ 2 −1 e−4t + c2 1 √ 2 e−t + 1 9 2 + √ 2 + 3 √ 3 3 √ 6 − √ 2 −1 e−t + 1 3 1 − √ 2 √ 2 −2 te−t.
Note that 2 + √ 2 + 3 √ 3 3 √ 6 − √ 2 −1 = 2 + √ 2 − √ 2 −1 + 3 √ 3 1 √ 2 .
The second vector is an eigenvector, hence the solution may be written as x = c1 √ 2 −1 e−4t + c2 1 √ 2 e−t + 1 9 2 + √ 2 − √ 2 −1 e−t + 1 3 1 − √ 2 √ 2 −2 te−t.
11. Based on the solution of Problem 3 of Section 7.6, a fundamental matrix is given by Ψ(t) = 5 cos t 5 sin t 2 cos t + sin t −cos t + 2 sin t .
328 Chapter 7. Systems of First Order Linear Equations The inverse of the fundamental matrix is easily determined as Ψ−1(t) = 1 5 cos t −2 sin t 5 sin t 2 cos t + sin t −5 cos t .
It follows that Ψ−1(t)g(t) = cos t sin t −cos2 t , and Z Ψ−1(t)g(t) dt = 1 2 sin2 t −1 2 cos t sin t −1 2t .
A particular solution is constructed as v(t) = Ψ(t) Z Ψ−1(t)g(t) dt, where v1(t) = 5 2 cos t sin t −cos2 t + 5 2t + 1, v2(t) = cos t sin t −1 2 cos2 t + t + 1 2 .
Hence the general solution is x = c1 5 cos t 2 cos t + sin t + c2 5 sin t −cos t + 2 sin t − −t sin t 5/2 1 + t cos t 0 1/2 −cos t 5/2 1 .
13.(a) As shown in Problem 25 of Section 7.6, the solution of the homogeneous system is x(c) 1 x(c) 2 = c1e−t/2 cos(t/2) 4 sin(t/2) + c2e−t/2 sin(t/2) −4 cos(t/2) .
Therefore the associated fundamental matrix is given by Ψ(t) = e−t/2 cos(t/2) sin(t/2) 4 sin(t/2) −4 cos(t/2) .
(b) The inverse of the fundamental matrix is Ψ−1(t) = et/2 4 4 cos(t/2) sin(t/2) 4 sin(t/2) −cos(t/2) .
It follows that Ψ−1(t)g(t) = 1 2 cos(t/2) sin(t/2) , and Z Ψ−1(t)g(t) dt = sin(t/2) −cos(t/2) .
A particular solution is constructed as v(t) = Ψ(t) Z Ψ−1(t)g(t) dt, 7.9 329 where v1(t) = 0, v2(t) = 4 e−t/2. Hence the general solution is x = c1e−t/2 cos(t/2) 4 sin(t/2) + c2e−t/2 sin(t/2) −4 cos(t/2) + 4 e−t/2 0 1 .
Imposing the initial conditions, we require that c1 = 0, −4c2 + 4 = 0, which results in c1 = 0 and c2 = 1 . Therefore the solution of the IVP is x = e−t/2 sin(t/2) 4 −4 cos(t/2) .
15. The general solution of the homogeneous problem is x(c) 1 x(c) 2 = c1 1 2 t−1 + c2 2 1 t2, which can be verified by substitution into the system of ODEs. Since the vectors are linearly independent, a fundamental matrix is given by Ψ(t) = t−1 2t2 2t−1 t2 .
The inverse of the fundamental matrix is Ψ−1(t) = 1 3 −t 2t 2t−2 −t−2 .
Dividing both equations by t, we obtain g(t) = −2 t3 −t−1 .
Proceeding with the method of variation of parameters, Ψ−1(t)g(t) = 2 3t4 + 2 3t −2 3 −1 3t −4 3t−2 + 1 3t−3 , and Z Ψ−1(t)g(t) dt = 2 15t5 + 1 3t2 −2 3t −1 6t2 + 4 3t−1 −1 6t−2 .
Hence a particular solution is obtained as v = −1 5t4 + 3t −1 1 10t4 + 2t −3 2 .
The general solution is x = c1 1 2 t−1 + c2 2 1 t2 + 1 10 −2 1 t4 + 3 2 t − 1 3/2 .
16. Based on the hypotheses, φ ′(t) = P(t)φ(t) + g(t) and v ′(t) = P(t)v(t) + g(t) .
Subtracting the two equations results in φ ′(t) −v ′(t) = P(t)φ(t) −P(t)v(t) , 330 Chapter 7. Systems of First Order Linear Equations that is, [φ(t) −v(t)] ′ = P(t) [φ(t) −v(t)] .
It follows that φ(t)−v(t) is a solution of the homogeneous equation. According to Theorem 7.4.2, φ(t) −v(t) = c1x(1)(t) + c2x(2)(t) + · · · + cnx(n)(t).
Hence φ(t) = u(t) + v(t), in which u(t) is the general solution of the homogeneous problem.
17.(a) Setting t0 = 0 in Equation (34), x = Φ(t)x0 + Φ(t) Z t 0 Φ−1(s)g(s)ds = Φ(t)x0 + Z t 0 Φ(t)Φ−1(s)g(s)ds .
It was shown in Problem 15(c) in Section 7.7 that Φ(t)Φ−1(s) = Φ(t −s). There-fore x = Φ(t)x0 + Z t 0 Φ(t −s)g(s)ds .
(b) The principal fundamental matrix is identified as Φ(t) = eAt. Hence x = eAtx0 + Z t 0 eA(t−s)g(s)ds .
In Problem 27 of Section 3.6, the particular solution is given as y(t) = Z t t0 K(t −s)g(s)ds , in which the kernel K(t) depends on the nature of the fundamental solutions.
18. Similarly to Eq.(43), here (sI −A)X(s) = G(s) + α1 α2 , where G(s) = 2/(s + 1) 3/s2 and sI −A = s + 2 −1 −1 s + 2 .
The transfer matrix is given by Eq.(46): (sI −A)−1 = 1 (s + 1)(s + 3) s + 2 1 1 s + 2 .
From these equations we obtain that X(s) = 2(s+2) (s+1)2(s+3) + 3 s2(s+1)(s+3) + α1(s+2) (s+1)(s+3) + α2 (s+1)(s+3) 2 (s+1)2(s+3) + 3(s+2) s2(s+1)(s+3) + α1 (s+1)(s+3) + α2(s+2) (s+1)(s+3) !
.
7.9 331 The inverse Laplace transform gives us that x(t) = 4+α1+α2 2 e−t + −4+3α1−3α2 6 e−3t + t + te−t −4 3 2+α1+α2 2 e−t + 4−3α1+3α2 6 e−3t + 2t + te−t −5 3 , so α1 and α2 should be chosen so that 4 + α1 + α2 2 = c2 + 1 2 and −4 + 3α1 −3α2 6 = c1.
This gives us α1 = (−5 + 6c1 + 6c2)/6 and α2 = −c1 + c2 −13/6.
332 Chapter 7. Systems of First Order Linear Equations |
845 | https://fiveable.me/ap-calc/unit-4/intro-related-rates/study-guide/WxyKc3lpYx3sCEzkH9y2 | Intro to Related Rates - AP Calc Study Guide | Fiveable | Fiveable
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♾️AP Calculus AB/BC Unit 4 Review
4.4 Intro to Related Rates
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AP Calculus AB/BC
Unit 4 – Contextual Applications of Differentiation
Topic: 4.4
♾️AP Calculus AB/BC Unit 4 Review
4.4 Intro to Related Rates
Written by the Fiveable Content Team • Last updated September 2025
Verified for the 2026 exam
Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025
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♾️AP Calculus AB/BC
Unit & Topic Study Guides
AP Calculus AB/BC Exams
Unit 1 – Limits and Continuity
Unit 2 – Fundamentals of Differentiation
Unit 3 – Composite, Implicit, and Inverse Functions
Unit 4 – Contextual Applications of Differentiation
Unit 4 Overview: Contextual Applications of Differentiation
4.1 Interpreting the Meaning of the Derivative in Context
4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration
4.3 Rates of Change in Applied Contexts other than Motion
4.4 Intro to Related Rates
4.5 Solving Related Rates Problems
4.6 Approximating Values of a Function Using Local Linearity and Linearization
4.7 Using L'Hopitals Rule for Determining Limits in Indeterminate Forms
Unit 5 – Analytical Applications of Differentiation
Unit 6 – Integration and Accumulation of Change
Unit 7 – Differential Equations
Unit 8 – Applications of Integration
Unit 9 – Parametric Equations, Polar Coordinates, and Vector–Valued Functions (BC Only)
Unit 10 – Infinite Sequences and Series (BC Only)
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4.4 Introduction to Related Rates
Using the concept of the rate of change, we can use derivatives to solve real-world problems particularly when dealing with how much something has changed in relation to another quantity. This is known as related rates in AP Calculus AB. 🧠
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🤔 Using Derivative Rules
Related rates describe the process of using derivatives to calculate unknown quantities in relation to other quantities and their rates of change. Typically, problems regarding this topic are calculated concerning the quantity of time. ⏰
To solve related rate problems, several derivative rules must be applied, depending on the problem and the known and unknown quantities. This includes the chain rule, the Product Rule, and the Quotient Rule. For in-depth explanations, check out our previous study guides for explanations on these rules! They are all found in unit two, linked here.
✏️ How to Solve Related Rate Problems
Here is roughly a step-by-step process of solving related rates problems:
⛳️ Determine quantities from the problem
This includes known quantities where the amounts are explicitly mentioned in the problem and unknown quantities that are not mentioned in the problem.
📈Model the relationship between quantities
This is done through a formula. It is important to note that if some quantities mentioned in the problem are not explicitly related to the unknown quantity that is trying to be solved, it should be replaced to simplify the formula when computing its derivatives.
👍Apply derivative rules with respect to the unknown quantity
After coming up with a formula that encapsulates all of the quantities mentioned in the problem, use the derivative rules to derive the equation, its quantities, and their rates of change. Note that the variables in the formula must be derived WITH RESPECT to the same independent variable that is being solved (typically this is usually with respect to a variable of time d x d t\frac{dx}{dt}d t d x.
🔢 Substitute known quantities into the derived equation & solve for the remaining quantities
Doing this will simplify the equation to the point where algebra can be applied to determine the value of unknown quantities.
💡 Another essential piece of advice when solving related rates problems is drawing/modeling out the problem and labeling the quantities provided in the question. This can help not only provide a visual understanding of the problem but also lead to finding a formula that will be needed to determine the unknown quantities.
This may be a lot to digest, so let’s go through a couple of examples to put this into action!
📝Related Rates Practice
Here are two practice problems you can use to get familiar with related rates.
Practice Example #1
Let’s start with an easy problem and work our way up!
❓ Suppose an oil spill occurs and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 3 m/s, how fast is the area of the spill increasing when the radius is 30 m?
To approach this problem, we can draw a picture or model of how this might look:
Model of presented question. Image Courtesy of Sonia Sohail
Notice how the radius and rate of change of the radius are labeled in the image above. Labeling your quantities in relation to the picture is an excellent way to figure out the known and unknown quantities of the problem. Here, we see that the known quantities are the radius, which is 30 meters and the rate of change of radius with respect to time, denoted as d r d t\frac{dr}{dt}d t d r, is 3 meters per second (m/s).
Because we are asked to find the area given the radius and its rate of change, we can model the relationship between the radius and the area of the oil spill using the formula for the area of a circle:
A=π r 2 A = \pi r^2 A=π r 2
Since the problem explicitly states that the unknown quantity to solve for is how fast the area is increasing at a particular radius (r = 30), we can translate this as “the rate of change in the area with respect to time”, otherwise denoted as d A d t\frac{dA}{dt}d t d A.
Now that we know what quantity we are looking for, we can begin differentiating the formula with respect to time. In finding the derivative of this formula, we use implicit differentiation to get:
d d t(A)=d d t(π r 2)=π d d t(r 2)=2 π r d r d t\frac{d}{dt}(A) =\frac{d}{dt}(\pi r^2) = \pi \frac{d}{dt}(r^2) = 2\pi r\frac{dr}{dt}d t d(A)=d t d(π r 2)=π d t d(r 2)=2 π r d t d r
For simplification, this can be written as
d A d t=2 π r d r d t\frac{dA}{dt} = 2\pi r\frac{dr}{dt}d t d A=2 π r d t d r
By this point, you may have noticed that we have most of the quantities that are present in this derivative, so we can substitute our values and solve for the unknown quantity. Make sure to include the units!
d A d t=2 π∗30 m∗3 m/s=180 π m 2/s\frac{dA}{dt} = 2\pi 30 m3m/s = 180\pi m^2/s d t d A=2 π∗30 m∗3 m/s=180 π m 2/s
Therefore, our final answer is 180 π m 2/s 180\pi m^2/s 180 π m 2/s ✅
Practice Example #2
Here is another problem:
❓ Suppose a 15ft tall ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 5 ft/s, how fast is the top of the ladder sliding down if the bottom of the ladder is 9 ft away from the wall?
At first glance, this one may not be as clear as the previous one. This is what is known as the “ladder” problem. To understand what we are looking for, it is best, once again, to draw or model the problem, using the quantities provided in the question:
Model of presented question. Image courtesy of Sonia Sohail
Here we draw the ladder resting against the vertical y-axis with the bottom part of the ladder reaching the horizontal x-axis. In doing so, we have created a right triangle, and because we are given the length of the ladder (which could be considered the hypotenuse) and the length of one of the sides, we can use the Pythagorean Theorem to model the relationship between the lengths of the wall, the ground, and the ladder. 📐
x 2+y 2=225 x^2+y^2= 225 x 2+y 2=225
Now that we have a formula that models our problem, we can see that our unknown quantity is the rate of change in y because as the ladder slides away from the wall (increasing the x-value), the length between the ground and the ladder is shrinking. This is denoted as d y d t\frac{dy}{dt}d t d y. We can perform implicit differentiation on the formula to find the rate of change of all quantities with respect to time:
d d t(x 2+y 2)=d d t(225)\frac{d}{dt}(x^2+y^2)=\frac{d}{dt}(225)d t d(x 2+y 2)=d t d(225)
2 x d x d t+2 y d y d t=0 2x\frac{dx}{dt}+2y\frac{dy}{dt} = 0 2 x d t d x+2 y d t d y=0
At this point, you may have noticed that while we have values for most of the quantities in this derivative, we still don't know what y is. You may think you’ve made a mistake somewhere, but all you need to do is solve for y using the original formula.
Given that x=9 x= 9 x=9, we can find the value of y by:
9 2+y 2=15 2 81+y 2=225 y 2=225−81 y=144=12 y=12 9^2+y^2=15^2 \ 81 + y^2 = 225 \y^2 = 225 -81 \ y = \sqrt{144} = 12 \ y= 12 9 2+y 2=1 5 2 81+y 2=225 y 2=225−81 y=144=12 y=12
Now that we have the value of y, we can go back to our derivative and substitute the other values to solve for d y d t\frac{dy}{dt}d t d y:
2 x d x d t+2 y d y d t=0 2x\frac{dx}{dt}+2y\frac{dy}{dt} = 0 2 x d t d x+2 y d t d y=0
(2∗9∗5)+(2∗12∗d y d t)=0(295)+ (212\frac{dy}{dt}) = 0(2∗9∗5)+(2∗12∗d t d y)=0
90+24 d y d t=0 90 + 24\frac{dy}{dt} = 0 90+24 d t d y=0
24 d y d t=−90 24\frac{dy}{dt} = -90 24 d t d y=−90
d y d t=−90 24=−15 4 f t/s\frac{dy}{dt} = \frac{-90}{24} = \frac{-15}{4} ft/s d t d y=24−90=4−15f t/s
By substituting the values found for each of the quantities, we can use algebra to solve for d y d t\frac{dy}{dt}d t d y, which is −15 4 f t/s\frac{-15}{4}ft/s 4−15f t/s!
💫 Closing
Using the rules of differentiation, related rates is just one way to apply calculus to solving real-world problems. In the next guide, we’ll provide a different set of steps for you and additional practice questions. That way, you can decide what technique is best for you!
🔗 AP Calculus 4.5 Solving Related Rates Problems
Frequently Asked Questions
How do I solve related rates problems step by step?
Step-by-step you can always follow this checklist for Topic 4.4 related-rates problems: 1. Read context and assign variables with units (e.g., x(t), y(t), r(t)). Identify which rate is given and which you need (dx/dt, dy/dt). 2. Write an equation relating the variables (geometry, similar triangles, Pythagorean, volume formulas). 3. Differentiate both sides wrt time t using the chain rule (and product/quotient/implicit as needed). That gives an equation in rates (e.g., 2x dx/dt + 2y dy/dt = 0). 4. Substitute the known variable values at the instant and the known rates (use correct units). 5. Solve algebraically for the unknown rate and state it with units and sign (positive = increasing). 6. Check reasonableness (limits, sign, magnitude). Key AP tips: always label dx/dt, dy/dt, use implicit differentiation (chain rule is the foundation per CED CHA-3.D), and show work—this is exactly what AP free-response expects. For extra examples and practice, see the Topic 4.4 study guide ( the Unit 4 overview ( and lots of practice problems (
What's the formula for related rates in calculus?
Related-rates problems use the chain rule: you differentiate an equation relating several changing quantities with respect to the same independent variable (usually time t), then solve for the desired rate (like dy/dt). Two useful templates: - If F(x, y, …) = 0, implicitly differentiate: F_x dx/dt + F_y dy/dt + … = 0 (sum of partials times the corresponding rates). - If z = f(x,y), use the multivariable chain rule: dz/dt = f_x(x,y)·dx/dt + f_y(x,y)·dy/dt. Concrete example (CED keywords: chain rule, implicit differentiation): for a cylinder V = πr^2h, dV/dt = 2πr h·dr/dt + πr^2·dh/dt so you plug known r, h, and one or two known rates to find the unknown. This is exactly what Topic 4.4 tests (CHA-3.D in the CED). For more worked examples and practice, see the Topic 4.4 study guide ( and the unit page (
When do I use the chain rule vs product rule in related rates?
Use the chain rule any time you’re differentiating a quantity that depends on a variable which itself depends on time—in related rates you always differentiate every variable with respect to t (that’s CHA-3.D). Use the product (or quotient) rule when the equation you’re differentiating has a product (or quotient) of two or more time-varying quantities. Quick recipe: - Step 1: Write the relation (e.g., x^2 + y^2 = 25 or V = πr^2h). - Step 2: Differentiate both sides with respect to t. For each variable use the chain rule: d/dt[f(x)] = f′(x)·dx/dt. - Step 3: If a term is a product (like r^2·h), apply the product rule, remembering to use chain rule inside: d/dt(r^2·h) = (2r dr/dt)·h + r^2·dh/dt. Examples: - Implicit (circle): d/dt(x^2 + y^2) = 2x dx/dt + 2y dy/dt (chain rule). - Cylinder V=πr^2h: dV/dt = π[2r dr/dt · h + r^2 dh/dt] (product + chain). This is exactly what the CED expects for related rates (CHA-3.D and CHA-3.D.2). For more worked examples and practice, see the Topic 4.4 study guide ( and lots of practice problems (
I don't understand how to find dy/dt when I only know dx/dt - can someone explain?
You use the chain rule / implicit differentiation: write an equation relating x and y, differentiate every term with respect to time t (treat dx/dt and dy/dt as derivatives), then solve for dy/dt by plugging the known dx/dt and the point values. Quick example (common on the AP): x^2 + y^2 = 25. Differentiate w.r.t. t: 2x(dx/dt) + 2y(dy/dt) = 0. Solve for dy/dt: dy/dt = −[x(dx/dt)] / y. So if dx/dt = 3 cm/s and the point is (4, 3), dy/dt = −(4·3)/3 = −4 cm/s. Steps to follow on any problem: 1) Write the geometric/physical relation (Pythagorean, similar triangles, volume formula...). 2) Differentiate implicitly w.r.t. t using chain/product/quotient rules as needed (CHA-3.D). 3) Plug in x, y, dx/dt values at the instant given and solve for dy/dt. This is exactly what the AP CED expects for related rates (Topic 4.4). For more worked examples and practice, see the Topic 4.4 study guide ( and try problems at (
How do I know which variables are related in a word problem?
Look for any quantity in the story that depends on another quantity—those are your related variables. Steps you can use every time: - Identify what’s changing (rates asked for: e.g., dV/dt, dx/dt). Write variables for those quantities with units. - Find a connecting equation from geometry/physics (Pythagorean for ladders, V = (1/3)πr^2h for cones, similar triangles for shadow problems, area = πr^2 for balloons). That equation shows how the variables relate. - Decide which variables are independent/dependent in context (time t is the common independent variable in related rates: use chain rule and write dx/dt, dr/dt, dh/dt). - Implicitly differentiate the connecting equation with respect to t, using product/quotient/chain rules as needed (CED CHA-3.D.1 & CHA-3.D.2). - Plug known values and solve for the requested rate. If you’re unsure which equation to use, scan for geometric relations or conservation laws in the text. For examples and guided practice, see the Topic 4.4 study guide ( and try practice problems (
What's the difference between implicit differentiation and related rates?
Implicit differentiation is a differentiation technique: when an equation defines y implicitly (like x^2 + y^2 = 25), you differentiate both sides with respect to x and use dy/dx for every y-term (and product/quotient rules as needed). Related rates is an application: you have several quantities that change with time, so you differentiate an equation that ties them together with respect to the same independent variable (usually t). In practice, related rates uses implicit differentiation + the chain rule (replace dy/dt for y′, dx/dt for x′, etc.) to get equations relating rates (e.g., dx/dt, dV/dt). Key AP points: Topic 4.4 expects you to differentiate with respect to time, use chain/product/quotient rules, and interpret units (CHA-3.D). Common contexts: ladder, balloon, conical tank, shadow problems. For a focused review see the Topic 4.4 study guide ( and practice problems (
Can someone walk me through a balloon expanding problem with related rates?
Pick variables, write relation, differentiate, solve. Example: A spherical balloon’s volume V = (4/3)πr^3. Let r = r(t), V = V(t). If the volume is increasing at dV/dt = 8 cm^3/s when r = 2 cm, find dr/dt. 1. Differentiate both sides with respect to time t (use chain rule / implicit differentiation): dV/dt = 4πr^2 · dr/dt. 2. Solve for dr/dt: dr/dt = (dV/dt) / (4πr^2) = 8 / (4π·2^2) = 8 / (16π) = 1/(2π) cm/s. So the radius is increasing at 1/(2π) cm/s at that instant. Always state units and what each rate means (dV/dt in cm^3/s, dr/dt in cm/s). On the AP exam show: variable definitions, the equation, differentiation step (chain rule), substitution, and units—that’s aligned with CHA-3.D (CED). For more worked examples and practice, see the Topic 4.4 study guide ( and try problems at Fiveable’s practice page (
Why do I need to use the chain rule for related rates problems?
You use the chain rule in related-rates problems because you’re differentiating quantities that depend on the same hidden independent variable (usually time). If an equation ties two or more changing quantities together (for example, V = πr^2h or x^2 + y^2 = 25), each variable is really a function of t. The chain rule tells you how to take d/dt of those expressions: differentiate with respect to the variable in the equation, then multiply by that variable’s rate (dx/dt, dy/dt, etc.). That’s exactly CHA-3.D from the CED: “The chain rule is the basis for differentiating variables in a related rates problem with respect to the same independent variable.” Often you combine chain rule with implicit differentiation and other rules (product/quotient) to get an equation involving the desired rates, then plug in known values. For more examples and step-by-step practice, see the Topic 4.4 study guide ( and try problems at (
How do I set up the equation before I start differentiating in related rates?
Before you differentiate, build one clear equation that relates the changing quantities (use one equation only). Steps: 1. Identify all variables and what rates you’re given/asked for (use dx/dt notation and include units). 2. Draw a diagram and label the variables—geometry often helps (Pythagorean, similar triangles, volume formulas, area, etc.). 3. Write one relationship connecting the variables (e.g., x^2 + y^2 = 25 for a ladder, V = (1/3)πr^2h for a cone, or similar-triangle ratios for shadows). 4. If needed, eliminate extra variables using a second algebraic relation so the main equation contains only the variables you’ll differentiate. 5. Check units and which variable is implicitly a function of time (t). Only then differentiate implicitly with respect to t, using the chain rule—plus product or quotient rules if they appear—and substitute the known values to solve for the requested rate. This follows CED ideas (CHA-3.D: chain rule, implicit differentiation; use product/quotient as needed). For examples and guided practice, see the Topic 4.4 study guide ( and try problems from Fiveable’s practice set (
I'm confused about when to substitute values in related rates - before or after differentiating?
Short answer: almost always differentiate first, then substitute the known numerical values. Why: related-rates problems use implicit differentiation with respect to the same independent variable (usually t). Treat every letter as a function of t, apply the chain rule (and product/quotient rules if needed) to get an equation relating the rates (dx/dt, dy/dt, etc.). Only after you have that equation do you plug in the specific values of the variables and any known rates to solve for the unknown rate. Quick steps: 1) Write the equation linking the quantities (e.g., x^2 + y^2 = r^2). 2) Differentiate implicitly with respect to t: 2x dx/dt + 2y dy/dt = 2r dr/dt. 3) Substitute the given numbers (x, y, known rates) and solve for the unknown rate. Exception: if a letter really is a fixed constant (not changing with t), you can substitute it beforehand—but check the problem carefully. This approach matches CED CHA-3.D (chain rule, implicit differentiation). For more examples and practice, see the Topic 4.4 study guide ( and Unit 4 overview ( For lots of practice problems, go to (
What does it mean when they ask for the rate of change with respect to time?
When a problem asks for a "rate of change with respect to time," it simply means take the derivative of a quantity using time (usually t) as the independent variable—e.g., dx/dt, dV/dt, dr/dt. Practically, that derivative tells you how fast that quantity is changing per one unit of time and includes units (meters/second, liters/minute). The sign matters: positive means increasing, negative means decreasing. In related-rates problems (Topic 4.4 on the CED) you usually have two or more variables related by an equation (Pythagorean theorem, volume formula, similar triangles, etc.). Differentiate that equation implicitly with respect to t (use chain rule, product/quotient rules as needed) to get an equation in the time-derivatives (dx/dt, dy/dt, dV/dt). Then plug in known values to solve for the unknown rate. For review and worked examples, check the Topic 4.4 study guide ( and do lots of practice problems ( free-response often tests this skill, so practice implicit differentiation and interpreting units.
How do I solve ladder sliding down wall problems using related rates?
Start by labeling: let x(t) = distance from wall to ladder base, y(t) = ladder top height, and L = ladder length (constant). Use the Pythagorean relation x^2 + y^2 = L^2. Differentiate implicitly with respect to time t (chain rule / implicit differentiation): 2x dx/dt + 2y dy/dt = 0. Solve for the unknown rate (usually dy/dt): dy/dt = −(x/y) · dx/dt. Steps to solve any ladder problem 1. Draw and label x, y, L; write the geometric equation. 2. Differentiate both sides w.r.t. t using chain rule and other rules if needed (CHA-3.D). 3. Plug in the given x, y (or compute y from L and x) and the known rate (dx/dt or dy/dt). 4. Compute sign and units—a negative dy/dt means the top is sliding down. Quick numeric example: L = 10 ft, base moves away at dx/dt = 3 ft/s when x = 6 ft. Then y = 8 ft, so dy/dt = −(6/8)(3) = −2.25 ft/s (top moving down at 2.25 ft/s). For more worked examples and AP-aligned practice on related rates, see the Topic 4.4 study guide ( and lots of practice problems (
Do I always need to use d/dt notation in related rates or can I just use regular derivatives?
Short answer: you don’t have to literally write d/dt every time, but you must show you’re differentiating with respect to the same independent variable (usually time). AP CED Topic 4.4 expects you to use the chain rule/implicit differentiation to turn relationships among variables into relationships among their rates (CHA-3.D.1/CHA-3.D.2). Why d/dt is helpful: writing dy/dt, dx/dt (or dV/dt) makes it explicit that x, y, V are functions of t and keeps units clear. Using y′ or x′ is fine too—as long as you state what the prime means (e.g., x′ = dx/dt). The key error exam graders watch for is treating dx and dy as independent or forgetting the common independent variable. Example: if x^2 + y^2 = 25 and x = x(t), y = y(t), differentiate with respect to t: 2x dx/dt + 2y dy/dt = 0 (showing you used d/dt). That’s the AP-style approach. For practice and more worked examples, see the Topic 4.4 study guide ( and try problems at (
I keep getting confused about which rate they're asking for in the problem - how do I identify what I'm solving for?
Start by reading the scenario and asking: what physical quantity do they want as a rate (e.g., how fast is the radius changing—dr/dt—or how fast is the volume changing—dV/dt)? Steps that make that clear: 1. Pick symbols: write each quantity as a variable (r, h, V, x, y, s, L, etc.) and label the one you’re asked for with d(·)/dt (dx/dt, dV/dt). 2. Find the relation: write an equation linking the variables (Pythagorean, V = πr^2h, similar triangles, area formula, etc.). This uses CHA-3.D knowledge (chain rule + implicit differentiation). 3. Differentiate both sides with respect to time t (use chain rule, product/quotient as needed) so every derivative is a rate like dr/dt, dh/dt. 4. Solve algebraically for the requested rate and plug in the given numbers (including units!). Check units and sign (positive vs negative). 5. State the answer with units and a short sentence interpreting it. Common traps: don’t confuse “how fast x is changing” with x itself; if they give dx/dt but ask for dy/dt you must use the relation. For extra practice and worked examples (ladder, cone, balloon), see the Topic 4.4 study guide ( and try problems from Fiveable’s practice set (
When do I need to use the quotient rule in related rates problems?
Use the quotient rule whenever the equation that relates the quantities is a quotient of two functions that both depend on time and you can’t (or don’t want to) simplify it into a product first. In related rates you always differentiate implicitly with respect to t (chain rule/CED CHA-3.D). If your relation is F(t) = u(t)/v(t), apply d/dt[u/v] = (v du/dt − u dv/dt)/v^2 so every occurrence of u and v gets a corresponding du/dt and dv/dt. Examples where this shows up: a density = mass/area where both mass and area change, or a rate expressed as a ratio of two changing lengths. Quick tips: - Try algebraic simplification first (rewrite u/v as u·v⁻¹ and use product + chain rule)—same result but sometimes cleaner. - Always label which variable depends on t and include units (AP free-response expects clear notation like dx/dt). For a quick refresher on solving related-rates problems and examples, see the Topic 4.4 study guide ( For lots of practice problems, check Fiveable’s practice page (
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0 |
846 | https://mathworld.wolfram.com/BinomialSeries.html | Binomial Series
Download
Wolfram Notebook
There are several related series that are known as the binomial series.
The most general is
| | |
--- |
| | (1) |
where
is a binomial coefficient and is a real number. This series converges for an integer, or (Graham et al. 1994, p. 162). When is a positive
integer ,
the series terminates at and can be written in the form
| | |
--- |
| | (2) |
The theorem that any one of these (or several other related forms) holds is known
as the binomial theorem.
Special cases give the Taylor series
| | | | |
--- --- |
| | | | (3) |
| | | | (4) |
where
is a Pochhammer symbol and . Similarly,
| | | | |
--- --- |
| | | | (5) |
| | | | (6) |
which is the so-called negative binomial series.
In particular, the case gives
| | | | |
--- --- |
| | | | (7) |
| | | | (8) |
| | | | (9) |
(OEIS A001790 and A046161), where
is a double factorial and is a binomial coefficient.
The binomial series has the continued fraction
representation
| | |
--- |
| | (10) |
(Wall 1948, p. 343).
See also
Binomial, Binomial Identity, Binomial Theorem, Multinomial
Series, Negative Binomial Series
Explore with Wolfram|Alpha
More things to try:
binomial series
arccot x
feigenbaum alpha
References
Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing.
New York: Dover, pp. 14-15, 1972.Graham, R. L.; Knuth, D. E.;
and Patashnik, O. Concrete
Mathematics: A Foundation for Computer Science, 2nd ed. Reading, MA: Addison-Wesley,
1994.Pappas, T. "Pascal's Triangle, the Fibonacci Sequence &
Binomial Formula." The
Joy of Mathematics. San Carlos, CA: Wide World Publ./Tetra, pp. 40-41,
1989.Sloane, N. J. A. Sequences A001790/M2508
and A046161 in "The On-Line Encyclopedia
of Integer Sequences."
Referenced on Wolfram|Alpha
Binomial Series
Cite this as:
Weisstein, Eric W. "Binomial Series."
From MathWorld--A Wolfram Resource.
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847 | https://www.tiger-algebra.com/drill/m~2-3m_2=0/ | Copyright Ⓒ 2013-2025
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Solution - Quadratic equations
Other Ways to Solve
Step by Step Solution
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring m2-3m+2
The first term is, m2 its coefficient is 1 .
The middle term is, -3m its coefficient is -3 .
The last term, "the constant", is +2
Step-1 : Multiply the coefficient of the first term by the constant 1 • 2 = 2
Step-2 : Find two factors of 2 whose sum equals the coefficient of the middle term, which is -3 .
| | | | | | | |
--- --- ---
| | -2 | + | -1 | = | -3 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -2 and -1
m2 - 2m - 1m - 2
Step-4 : Add up the first 2 terms, pulling out like factors :
m • (m-2)
Add up the last 2 terms, pulling out common factors :
1 • (m-2)
Step-5 : Add up the four terms of step 4 :
(m-1) • (m-2)
Which is the desired factorization
Equation at the end of step 1 :
Step 2 :
Theory - Roots of a product :
2.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation :
2.2 Solve : m-1 = 0
Add 1 to both sides of the equation :
m = 1
Solving a Single Variable Equation :
2.3 Solve : m-2 = 0
Add 2 to both sides of the equation :
m = 2
Supplement : Solving Quadratic Equation Directly
Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula
Parabola, Finding the Vertex :
3.1 Find the Vertex of y = m2-3m+2
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero).
Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.
Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.
For any parabola,Am2+Bm+C,the m -coordinate of the vertex is given by -B/(2A) . In our case the m coordinate is 1.5000
Plugging into the parabola formula 1.5000 for m we can calculate the y -coordinate :
y = 1.0 1.50 1.50 - 3.0 1.50 + 2.0
or y = -0.250
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = m2-3m+2
Axis of Symmetry (dashed) {m}={ 1.50}
Vertex at {m,y} = { 1.50,-0.25}
m -Intercepts (Roots) :
Root 1 at {m,y} = { 1.00, 0.00}
Root 2 at {m,y} = { 2.00, 0.00}
Solve Quadratic Equation by Completing The Square
3.2 Solving m2-3m+2 = 0 by Completing The Square .
Subtract 2 from both side of the equation :
m2-3m = -2
Now the clever bit: Take the coefficient of m , which is 3 , divide by two, giving 3/2 , and finally square it giving 9/4
Add 9/4 to both sides of the equation :
On the right hand side we have :
-2 + 9/4 or, (-2/1)+(9/4)
The common denominator of the two fractions is 4 Adding (-8/4)+(9/4) gives 1/4
So adding to both sides we finally get :
m2-3m+(9/4) = 1/4
Adding 9/4 has completed the left hand side into a perfect square :
m2-3m+(9/4) =
(m-(3/2)) • (m-(3/2)) =
(m-(3/2))2
Things which are equal to the same thing are also equal to one another. Since
m2-3m+(9/4) = 1/4 and
m2-3m+(9/4) = (m-(3/2))2
then, according to the law of transitivity,
(m-(3/2))2 = 1/4
We'll refer to this Equation as Eq. #3.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(m-(3/2))2 is
(m-(3/2))2/2 =
(m-(3/2))1 =
m-(3/2)
Now, applying the Square Root Principle to Eq. #3.2.1 we get:
m-(3/2) = √ 1/4
Add 3/2 to both sides to obtain:
m = 3/2 + √ 1/4
Since a square root has two values, one positive and the other negative
m2 - 3m + 2 = 0
has two solutions:
m = 3/2 + √ 1/4
or
m = 3/2 - √ 1/4
Note that √ 1/4 can be written as
√ 1 / √ 4 which is 1 / 2
Solve Quadratic Equation using the Quadratic Formula
3.3 Solving m2-3m+2 = 0 by the Quadratic Formula .
According to the Quadratic Formula, m , the solution for Am2+Bm+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
m = ————————
2A
In our case, A = 1
B = -3
C = 2
Accordingly, B2 - 4AC =
9 - 8 =
1
Applying the quadratic formula :
3 ± √ 1
m = ————
2
So now we are looking at:
m = ( 3 ± 1) / 2
Two real solutions:
m =(3+√1)/2= 2.000
or:
m =(3-√1)/2= 1.000
Two solutions were found :
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848 | https://www.youtube.com/watch?v=VX_gBuZ6byA | C23--Construct a Right Trapezoid Given Bases and Height
Glenn Olson
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Posted: 2 Apr 2014
Transcript:
this is video c23 we're going to construct a right trapezoid with base one base 2 and the height provided uh so step one I'll just put the uh I like to put the longer base on the bottom so we'll start with that so just anywhere on the bottom here we'll make a longer base make sure it's longer than needed then we will measure the space we will transfer that over to here so I now have base one copied what I need to do is construct a perpendicular so what I'll do is I'll extend for this to be um and I'll do the perpendicular bis sector so now I will extend in the other direction and we can do the perpendicular bis sector trick again so what I'm going to do is force this to be the midpoint of a segment so I'll just go Arch here Arc there and then from each of these end points go more than halfway Arc there Arc there uh I now have my two points I'm good to go I'm living on the edge there but it's okay so I now know that that's a right angle I want to know how high up to go I'm going to go the height up so I know that's a Vertex Point that's a Vertex point and that's a Vertex point point you want to make sure you don't get confused with all these points and arcs now what I want to do is uh have this go parallel again what I can do is um if I'm perpendicular here I know I'll be parallel there's a thorm in Geometry that states that so I'm going to do another perpendicular bis sector thing so I'm going to force this to be the midpoint I'll just use the same distance we're got it set so now this segment here you're the midpoint so I'll just go more than halfway I'll Arc it from here I'll Arc it from here and now this is the perpendicular bis sector it's going sideways [Music] now this base so just so you know this is base one this is the height this is base two base two is that length and I'm done I've got all four vertex points I'll just connect them in green so it stands out a little bit [Music] better and there you have it a right trapezoid constructed from base one one base 2 and a given height |
849 | https://blog.csdn.net/u013541048/article/details/81584594 | 数学：ln函数的链式法则导数解析-CSDN博客
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【Math】ln链式求导
最新推荐文章于 2024-12-26 10:20:39 发布
原创于 2018-08-11 11:11:39 发布·3.4k 阅读
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版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。
本文介绍了ln函数求导的两种方法，包括直接应用链式法则和转换为加法形式后再求导，详细阐述了每种方法的步骤与思路。
对求导的两种方法：
直接链式求导：
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微积分-导数5（链式 法则）
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待续。。。。。
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Detecting a subsequence that's an arithmetic progression, in a sorted sequence
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I have following problem: I have a sorted sequence of N integers (assume they are monotonically increasing). I want to check whether there is any subsequence of length ≥N/4, such that consecutive elements of the subsequence all differ by the same value.
For example, in the sequence [3,4,5,8,12] there are two such subsequences: [3,4,5] (the difference is 1) and [4,8,12] (the difference is 4). Thus, the length of longest such subsequence is 3 for this example. Since 3≥5/4, the answer is yes, there is a subsequence of length ≥N/4 with the desired property.
In my real-life situation, the sequence is of length N≈106, and the elements are all 9-digit numbers. Is there an efficient algorithm to solve this problem?
My naive approach was to create Cartesian product with absolute differences between numbers:
⎛⎝⎜⎜⎜⎜⎜⎜0125910148210375430498740⎞⎠⎟⎟⎟⎟⎟⎟
And then focus on top-right part and compute number of occurrences of each difference, so:
||diff-by-1||=2=>3 numbers diff by 1||diff-by-4||=2=>3 numbers diff by 4
This is very simple and very ineffective. It requires lot of comparisons and does not scale (at all): its running time is Θ(N2). In my real life scenario my sequence is ~10^6 long, so this is too slow.
To give you wider picture as maybe there is much better (probabilistic) approach to this problem: after largest sub-sequence is found I want to compute simple ratio:
r:=largest sub-sequence lengthsequence length
and if r is greater then some fixed value I want to raise alarm (or do whatever I have to do ;-)).
Thanks for any help, references, pointers, etc.
BTW: here are things that I was/am looking at:
Update: was thinking a little bit more about it and started from the end, so instead of computing all differences between numbers (top-right corner of the matrix) I can derive small k value from "fixed value" I mentioned at the end of original question. For instance if I am going to raise the alarm when 25% of all numbers are in some sequence I need to focus on small "triangles" in matrix and number of computations required is smaller (much smaller).
When I add some sampling then it should be simple enough to implement at scale.
Update 2 - Implemented @D.W. algorithm, sample run below:
```
11:51:06 ~$ time nodejs progression.js
L: 694000000,694000002,694000006,694000007,694000009,694000010,
694000013,694000015,694000018,694000019,694000021,694000022,694000023,
694000026,694000028,694000030,694000034,694000036,694000038,694000040,
694000043,694000045,694000046,694000048,694000051,694000053,694000055,
694000057,694000060,694000061,694000063,694000067,694000069,694000072,
694000074,694000076,694000077,694000079,694000080,694000082,694000083,
694000084,694000086,694000090,694000091,694000093,694000095,694000099,
694000102,694000103,694000105,694000108,694000109,694000113,694000116,
694000118,694000122,694000125,694000128,694000131,694000134,694000137,
694000141,694000143,694000145,694000148,694000152,694000153,694000154,
694000157,694000160,694000162,694000163,694000166,694000170,694000173,
694000174,694000177,694000179,694000180,694000181,694000184,694000185,
694000187,694000189,694000193,694000194,694000198,694000200,694000203,
694000207,694000211,694000215,694000219,694000222,694000226,694000228,
694000232,694000235,694000236
N: 100
P: 0.1
L: 10 (min)
D: 26 (max)
[ 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 ]
Found progression of 10 elements, difference: 16 starts: 694000045, ends: 694000189.
real 0m0.065s
user 0m0.052s
sys 0m0.004s
```
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edited Dec 16, 2013 at 10:54
Maciej Łopaciński
asked Dec 13, 2013 at 13:35
Maciej ŁopacińskiMaciej Łopaciński
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What kind of value of r were you thinking of? For example, testing whether there exists an arithmetic progression of length ≥1000 might be easier than finding the longest arithmetic progression (maybe; I'm not sure). Also, what's the distribution on the inputs? Can we assume the numbers in the input sequence are random, i.e., we draw N integers uniformly and independently at random from the range [108,109−1] and then sort them to get the input sequence?
– D.W.
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Dec 14, 2013 at 6:45
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You recently clarified that you want to solve the following problem:
Given an input sequence L[0..N−1], check whether an arithmetic progression of length ≥N/4 appears as a subsequence of L.
This is a significantly easier problem, and I can suggest a very efficient algorithm for you. Let me build up to it by first explaining a few key ideas underlying the algorithm.
Idea 1. Given any two elements of L, it is easy to extend them to the maximum-length arithmetic progression containing those two elements. In particular, suppose we have x,y∈L with x<y. Set d=y−x. Then we can extend this sequence on the right as the longest prefix of y+d,y+2d,y+3d,… that is wholly contained in L, and we can extend it on the left with the longest prefix of x−d,x−2d,x−3d,… that is wholly contained in L.
Idea 2. Say that an arithmetic progression has gap d if the difference between consequence elements of the arithmetic progression is d. If L contains an arithmetic progression of length ≥N/4, then its gap d must satisfy d≤4(L[N−1]−L)/N.
Idea 3. Suppose there is an arithmetic progression of length ≥N/4, and let α,β denote the start and end index (in L) of the subsequence. Since the subsequence has to be of length ≥N/4, the interval [α,β] must contain one or more of the following numbers: 0.2N, 0.4N, 0.6N, 0.8N (rounding all of them to an integer).
To give you some intuition for this: Imagine you're playing Battleship on a N×1 board, and your opponent has a single ship: a super-long battleship, that is N/4 cells long. Could you find his battleship? You sure could. It's only gonna take you 4 shots to find his battleship. You can shoot at cells 0.2N, 0.4N, 0.6N, 0.8N, and it's guaranteed that at least one of them will be a hit (that's Idea 3). Moreover, once you get a single hit, it's not going to be hard to sink the rest of his battleship (that's basically Ideas 1 and 2).
With these ideas, I'm ready to propose an algorithm:
Set dmax←4(L[N−1]−L)/N.
For each i0∈{0.2N,0.4N,0.6N,0.8N}, do:
For each i such that i≥i0 and L[i]≤L[i0]+dmax, do:
For each j such that j>i and L[j]≤L[i]+dmax, do:
Extend the arithmetic progression L[i],L[j] on the left and right as far as possible (using Idea 1).
Look at the longest arithmetic progression found at any point above. If it has length ≥N/4, then yes, there exists an arithmetic progression of length ≥N/4. If its length is <N/4, then no, there is no arithmetic progression of length ≥N/4.
Hopefully you can see why this works. Idea 3 means that, if there is an arithmetic progression of length ≥N/4, then at least one of the indices i that are visited in the above algorithm must be the index of one element of that progression. Moreover, the index of the next element of that progression must be one of the values taken on by j in the innermost loop. Therefore, at some point the pair L[i],L[j] will be a pair of consecutive elements of the arithmetic progression -- which is enough to discover that long arithmetic progression.
I would expect that this algorithm will typically be very efficient, on real-world data. If all of the integers in L are in the range [1,M], then we know dmax≤4M/N. Moreover, the loop over i typically involves about dmaxN/M≤4 iterations (on average), and similarly for the loop over j. So, we expect to perform maybe 5×4×4=80 iterations of the innermost statement, regardless of N, if the data is randomly chosen (and in particular, not adversarily chosen). Consequently, this should be extremely efficient if the data doesn't conspire against you.
It is possible to further improve this algorithm, for instance, to improve its worst-case running time and reduce the opportunity for worst-case data to cause the running time to blow up, at the cost of making the code more complicated. However, I suspect that this simple algorithm might be sufficient for your needs.
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edited Dec 15, 2013 at 17:35
answered Dec 15, 2013 at 5:46
D.W.♦D.W.
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Thanks for this. Will look on Monday. And yes, starting from the end of the problem makes it simpler and subject for optimizations.
– Maciej Łopaciński
Commented
Dec 15, 2013 at 11:00
@D.W.- Looking good, implemented in nodejs and works very well.
– Maciej Łopaciński
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Dec 16, 2013 at 10:39
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Subsequences of the form a,a+d,a+2d,a+3d,… are called arithmetic progressions. Your problem is to find the longest arithmetic progression that appears as a subsequence of your input (a sorted sequence).
Unfortunately, the method you describe in your question doesn't quite work to find the longest arithmetic progression, but I'll describe one way to fix it up in this answer.
You suggested enumerating all pairwise differences and counting how many times each difference occurs. However, that isn't enough to find the longest subsequence, in general. There might be 3 pairs of numbers that differ by 4, but that doesn't necessarily mean that there exists an arithmetic progression of length 3. Consider, e.g., [1,2,5,6,20,24], or [10,14,100,104,1000,1004].
The following algorithm is slightly better (i.e., it is correct, and often not too much slower):
For each i,j with i<j:
Consider the arithmetic progression L[i],L[j]. It has length 2, and its difference is d←L[j]−L[i].
Extend this arithmetic progression rightward as far as possible. In other words, set z←L[j]+d; while z∈L, append z to the arithmetic progression and set z←z+d.
If the input is randomly distributed, each iteration of the loop takes O(1) expected time, so the expected running time is O(N2). (You can ensure that the test z∈L can be done in O(1) time by precomputing an appropriate data structure, e.g., a hash table.)
Warning: If the input is not randomly distributed, this approach could take be much slower in the worst case. However, this approach might be a pragmatic method, as it's plausible that you might not run into those anomalies in practice if the inputs don't have any structure or nasty patterns.
Incidentally, for your parameters, if the numbers are all random, it's very likely that the longest arithmetic progression that appears as a subsequence has length 5 or 6. (There's maybe a 1/1000 chance that there will be an arithmetic progression of length 7 contained in your list.) I don't know how to find the length-5 or -6 progression efficiently, though, if efficiently means "much faster than Θ(N2) time".
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edited Dec 14, 2013 at 6:42
answered Dec 14, 2013 at 6:36
D.W.♦D.W.
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Thank you for your answer. You are right re algorithm being incorrect. I will try your approach. One thing I did not mention is that I will have to run it on database and anything like iteration through rows is bad idea in terms of performance. I think I will try to mix approaches and focus on smaller samples and check if there is arithmetic progression in each of them. Will try to update question on Monday. Cheers.
– Maciej Łopaciński
Commented
Dec 14, 2013 at 10:51
@MaciejŁopaciński, glad this was helpful! It seems like the database issue is a distraction. If accessing the data via the database is slow, then export the entire input sequence from the database and store it in an array in memory. Once you've done that, you can access elements of the input sequence efficiently. That said... I have an even better algorithm to propose, given your recent comments. See my newest answer.
– D.W.
♦
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Dec 15, 2013 at 5:48
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851 | http://carmengeometry.weebly.com/uploads/2/6/3/3/26333517/worksheet_7.3.pdf | Pythagorean Triples Worksheet I NAME_________ I. Use Pythagorean Theorem to find the missing dimension of each right triangle. Then complete the chart. 1. 2. Short Leg Long Leg Hypotenuse 1. 2. 3. 4. 5. 6. Given the triples above that you put in the table, use the factors in the table below to compute additional triples. (Use the chart above, and fill in the chart below.) Triple Factor New Triple Factor New Triple 7. 2 10 8. 2 10 9. 2 10 10. 2 10 11. 2 10 12 2 10 13. 6, 8, 10 2 10 14. 3, 4, 5 3 5 15. 7 5 , 7 4 , 7 3 2 10 Looking at #15 above, if all three numbers of a triple have the same radical ( , 2 , 7 . . .) – would you get a similar result as you did in #15? 4 3 13 5 25 7 17 15 9 40 61 60 3 . 4 .
. 5 .
. 6 .
. Using the Pythagorean Common Triples, find the missing side (triangles are rarely to scale). 16. 17. 18. 19. 20. 21. . 22. 23. 24. 25. 26. 27. 50 130 ? 10 ? 24 60 ? 80 ? 17 8 ? 65 25 8 10 ? 4 ? 3 40 ? 9 10 26 ? 5 ? 12 30 ? 40 ? 3 25 3 24 |
852 | https://www.mas.ncl.ac.uk/ask/numeracy-maths-statistics/core-mathematics/calculus/integrating-factor.html | Numeracy, Maths and Statistics - Academic Skills Kit
Integrating Factor
Contents- [x] Toggle Main Menu1 Definition2 Method3 Worked Examples4 Video Example5 Workbook6 See Also7 External Resources
Definition
The integrating factor method is a technique for solving first order ordinary differential equations of the form
d y d x+f(x)y=g(x),d y d x+f(x)y=g(x),
where f(x)f(x) and g(x)g(x) are any two arbitrary functions of x x only. Equations of this form are not separable, however we can combine the two terms on the left-hand side into a single derivative by using an integrating factor.
The integrating factor IF is given by integratingf(x)f(x) and then exponentiating:
I F=e F(x),I F=e F(x),
where F(x)F(x) is defined by F(x)=∫f(x)d x F(x)=∫f(x)d x.
It is important to note that the constant of integration is not included; this method requires that the derivative of F(x)F(x) is f(x)f(x), i.e. F′(x)=f(x)F′(x)=f(x), and the constant of integration is not necessary to meet this condition.
Method
Multiplying the original differential equation by the integrating factor I F=e F(x)I F=e F(x) gives
e F(x)d y d x+f(x)e F(x)y=g(x)e F(x).e F(x)d y d x+f(x)e F(x)y=g(x)e F(x).
Now note that the left hand side is the derivative of the integrating factor multiplied by y y:
d d x[e F(x)y]=e F(x)d y d x+f(x)e F(x)y.d d x[e F(x)y]=e F(x)d y d x+f(x)e F(x)y.
This result comes from the product rule, with the first term on the right-hand side containing the derivative of y y and the second term containing the derivate of e F(x)e F(x).
The original equation can now be written in the form
d d x[e F(x)y]=g(x)e F(x).d d x[e F(x)y]=g(x)e F(x).
This equation can be solved by integrating both sides:
∫d d x[e F(x)y]d x=∫g(x)e F(x)d x,∫d d x[e F(x)y]d x=∫g(x)e F(x)d x,
As integration is the opposite of differentiation, the left-hand side can be simplified:
∫d d x[e F(x)y]d x=e F(x)y.∫d d x[e F(x)y]d x=e F(x)y.
Hence the equation becomes:
e F(x)y=∫g(x)e F(x)d x.e F(x)y=∫g(x)e F(x)d x.
Provided that the integral on the right-hand side is simple enough to compute, this will lead to a solution.
Worked Examples
Example 1
Solve d y d x+3 y=x.d y d x+3 y=x.
Solution
The general form of a first order linear ordinary differential equation is:
d y d x+f(x)y=g(x).d y d x+f(x)y=g(x).
In the given equation, f(x)=3 f(x)=3 and g(x)=x g(x)=x.
Recall that the integrating factor is given by I F=e F(x)I F=e F(x), where F(x)=∫f(x)d x F(x)=∫f(x)d x. For this equation, the function F(x)F(x) is:
F(x)=∫3 d x=3 x.F(x)=∫3 d x=3 x.
As mentioned above, it is not necessary to include a constant of integration.
The integrating factor for this equation is therefore given by:
I F=e 3 x.I F=e 3 x.
Multiplying both sides of the original equation by the integrating factor gives:
e 3 x d y d x+3 e 3 x y=x e 3 x.e 3 x d y d x+3 e 3 x y=x e 3 x.
Note: At this point it is advisable to check that the left-hand side is indeed the derivative of the integrating factor multiplied by y y:
[\frac{\mathrm{d {\mathrm{d} x} \left[ e^{3x}y \right] = e^{3x}\frac{\mathrm{d} y}{\mathrm{d} x} + 3 e^{3x}y.]
The simplified form of the equation is therefore:
d d x[e 3 x y]=x e 3 x.d d x[e 3 x y]=x e 3 x.
Integrating both sides gives:
e 3 x y=∫x e 3 x d x.e 3 x y=∫x e 3 x d x.
To compute the integral on the right-hand side it is necessary to use integration by parts.
Recall the formula for integration by parts:
∫u v′d x=u v−∫u′v d x.∫u v′d x=u v−∫u′v d x.
Choose u=x u=x and v′=e 3 x v′=e 3 x. Then:
u=x,v=1 3 e 3 x,u′=1,v′=e 3 x,u=x,v=1 3 e 3 x,u′=1,v′=e 3 x,
and the integral becomes:
∫x e 3 x d x=x⋅1 3 e 3 x−∫1⋅1 3 e 3 x d x,=x 3 e 3 x−1 9 e 3 x+C.∫x e 3 x d x=x⋅1 3 e 3 x−∫1⋅1 3 e 3 x d x,=x 3 e 3 x−1 9 e 3 x+C.
Note: At this point it is necessary to include the constant of integration.
Substituting this result for the integral back into the equation gives:
e 3 x y=x 3 e 3 x−1 9 e 3 x+C.e 3 x y=x 3 e 3 x−1 9 e 3 x+C.
Multiplying both sides by e−3 x e−3 x gives the solution for y(x)y(x):
y=x 3−1 9+C e−3 x.y=x 3−1 9+C e−3 x.
}}
Video Example 1
Newcastle University Maths-Aid uses the integrating factor method to find the general solution of d y d x+3 y=x d y d x+3 y=x.
Example 1
Solve d y d x−1 x y=x 2.d y d x−1 x y=x 2.
Solution
The general form of a first order linear ordinary differential equation is:
d y d x+f(x)y=g(x).d y d x+f(x)y=g(x).
In the given equation, f(x)=−1 x f(x)=−1 x and g(x)=x 2 g(x)=x 2.
Note: If the function f(x)f(x) includes a minus sign it is essential to include this minus sign when computing the integrating factor.
Recall that the integrating factor is given by I F=e F(x)I F=e F(x), where F(x)=∫f(x)d x F(x)=∫f(x)d x. For this equation, the function F(x)F(x) is:
F(x)=∫−1 x d x=−ln(x)=ln(x−1).F(x)=∫−1 x d x=−ln(x)=ln(x−1).
As mentioned above, it is not necessary to include a constant of integration.
The integrating factor for this equation is therefore given by:
I F=e ln(x−1).I F=e ln(x−1).
By the laws of logarithms this simplifies to become:
I F=1 x.I F=1 x.
Multiplying both sides of the original equation by the integrating factor gives:
1 x d y d x−1 x 2 y=x.1 x d y d x−1 x 2 y=x.
Note: At this point it is advisable to check that the left-hand side is indeed the derivative of the integrating factor multiplied by y y:
[\frac{\mathrm{d {\mathrm{d} x} \left[ \frac{1}{x}y \right] = \frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x} - \frac{1}{x^2}y.]
The simplified from of the equation is therefore:
d d x[1 x y]=x.d d x[1 x y]=x.
Integrating both sides gives:
1 x y=∫x d x=1 2 x 2+C 1 x y=∫x d x=1 2 x 2+C
Multiplying both sides by x x gives the solution for y y:
y=1 2 x 3+C x.y=1 2 x 3+C x.
}}
Video Example 2
Newcastle University Maths-Aid uses the integrating factor method to find the general solution of d y d x−1 x y=x 2 d y d x−1 x y=x 2.
Video Example
Prof. Robin Johnson uses the integrating factor method to find the general solution of x d y d x+2 y=x 2 x d y d x+2 y=x 2.
Workbook
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.
First order differential equations
See Also
Separable ODEs
Homogeneous First Order Differential Equations
External Resources
Solving differential equations with integrating factors workbook at mathcentre.
Integrating factors videos at Khan Academy.
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You can get one-to-one support from Maths-Aid. |
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Home Anatomy of an mRNA
Anatomy of an mRNA
CAP
Eukaryotic mRNA has a 7-methyl-G cap (mCAP) at the 5' end. The cap protects the mRNA from nucleolytic degradation and recruits the ribosomal initiation complex to the start of the mRNA. Translation is dramatically enhanced by the presence of a cap. A cap can be incorporated in a transcription by including a mixture of cap analog and GTP (usually at a 4:1 ratio). 50% of the time, mCAP is inserted in the correct orientation to enhance translation. The other 50% of molecules are not substrates for efficient translation, reducing the specific activity of the transcript. More recently, Anti Reverse Cap Analog (ARCA) was introduced (1). ARCA can only insert in the proper orientation, leading to 100% activity.
5' Untranslated Region (5' UTR)
The 5' UTR does not encode for protein. However, these generally short regions can influence translational efficiency. The 5' UTR contains a sequence called the Kozak sequence which strongly influences translation efficiency. The optimal consensus Kozak sequence is GCCGCCRCCAUGG, where R is adenine or guanine. The AUG sequence is the site of translation initiation.
Open Reading Frame
The open reading frame of the mRNA is usually determined by the first AUG codon in the mRNA. Translation proceeds to the first in frame stop codon.
3' Untranslated Region (3' UTR) and Poly(A) Tail
The 3' UTR of an mRNA contains one or more polyadenylation sequences. In cells, the mRNA is cleaved at a polyadenylation sequence and a poly(A) tail is added to the end of the mRNA by poly(A) polymerase. The length of the poly(A) tail is an important determinant of translational efficiency and mRNA stability. When making mRNA in vitro by transcription, researchers commonly bypass the cleavage and polyadenylation step. Instead, a string of Ts is incorporated into the 3' end of the template and upon transcription a poly(A) tail is added opposite the Ts. It is desirable to have a long poly(A) tail, preferably 200 nucleotides or longer. Some natural 3' UTR sequences contain RNA instability elements that direct accelerated degradation of the mRNA. 3' UTRs are the main target for repression by microRNA. For applications where it is important to maintain normal regulation of the mRNA, it is desirable to utilize the endogenous 3' UTR of the transcript. For applications where maximal translation and duration of expression is desired, researchers commonly make chimeric mRNA that contains the 3' UTR of very stable transcripts such as alpha-globin or beta-globin. |
854 | https://ctools.ece.utah.edu/Probability/BasicProbability/LawTotalProb/ProbBasicLawTotProb.htm | | | | |
---
| | By: Neil E. Cotter | Probability |
| | | Basic probability |
| | | Law of total probability |
| | | definition |
| | | |
Tool: The Law of Total Probability states that, given a partition A1, A2, A3, ..., An of sample space S, the probability of any event B is given by the sum of the probabilities of B intersected with each of the Ai:
The Law of Total Probability is often used to find one unknown probability of the intersection of events when all of the other terms (including P(B)) are known.
A Venn diagram illustrates the law of total probability in an intuitively obvious way.
Here, n = 6. It is easy to see that the area of B, which represents P(B), is equal to the areas of the overlaps of B with each of the Ai. |
855 | https://www.geeksforgeeks.org/chemistry/bromine-water-test/ | Bromine Water Test
Last Updated :
23 Jul, 2025
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Bromine Water Test is an important test in Practical Organic Chemistry that deals with detecting the presence of unsaturation, such as carbon-carbon (C-C) double or triple bonds in organic compounds, i.e. it is used to identify the presence of alkane or alkene in an organic compound.
In this article, we will discuss this test in detail, including the preparation of bromine water, the principle of the test, and various substances that undergo this test with their reactions.
Table of Content
What is Bromine Water?
Preparation of Bromine Water
Principles of Bromine Water Test
Bromine Water Test Reaction
Bromine Water Test for Different Substances
Application of Bromine Water Test
What is Bromine Water?
Bromine Water is a chemical solution comprising diatomic bromine (Br2) dissolved in water(H2O). It is also known as bromide bromate solution or bromine solution. The molecular weight of bromine water is 159.808, with a density of 2.91 g/cm3, melting point of -7.2°C, boiling point of 58.8°C at 760 mm Hg, solubility of 35 g/L (20°C), and vapor pressure of 190 mmHg at 25°C with high oxidizing properties. Recognizable by its characteristic reddish-brown color, bromine water serves as a versatile reagent in chemistry, notably in the Bromine Water Test.
Bromine Water Formula
Bromine water, also called as Bromide Bromate solution or Bromine solution. The chemical formula of Bromine Water is Br2
Bromine Water Color
The color of Bromine water is red-brown due to presence of Bromine in it.
Preparation of Bromine Water
Bromine water can be prepared in the laboratory by direct mixing of fumes of diatomic bromine (Br2) in water (H2O) , but this isn't a safe method and alternatively it is prepared safely by breaking sodium bromide (NaBr) in the presence of bleach (NaClO) and hydrochloric acid (HCl).
2NaBr + NaClO + 2HCl → Br2 + 3NaCl + H2O
Principles of Bromine Water Test
Bromine Water Test is based on the principle of bromine's ability to decolorize in the presence of substances with unsaturated bonds, such as carbon-carbon (C-C) double or triple bonds. The key principles underlying the Bromine Water Test are as follows .
Unsaturation Detection: Bromine (Br2) is a reddish-brown liquid, but when it reacts with a substance containing unsaturated bonds (like alkenes or alkynes), it undergoes addition reactions and becomes colorless. This change in color from reddish-brown to colorless is a visual indicator of the presence of unsaturation.
Electrophilic Addition: Bromine is an electrophilic reagent, meaning it seeks electrons. Unsaturated compounds, such as alkenes or alkynes, readily provide electrons to bromine, leading to the formation of bromonium ions. This electrophilic addition results in the breaking of the bromine molecule and the decolorization of bromine water.
Selective Reaction: The Bromine Water Test is selective for unsaturated compounds. Saturated hydrocarbons, which lack carbon-carbon double or triple bonds, do not undergo this reaction, and therefore, bromine water remains in its characteristic reddish-brown color when added to saturated compounds.
Qualitative Analysis: The Bromine Water Test is primarily qualitative, indicating the presence or absence of unsaturation in a given compound. It does not provide quantitative information about the number or nature of unsaturated bonds.
Bromine Water Test Reaction
When Bromine Water is made to react with an unsaturated compound, it undergoes addition reaction and it turns from reddish-brown to a colorless liquid, whereas saturated compounds doesn't react with it, so there isn't any color changes observed. Hence it is used for Saturation Test.
Bromine Water Test for Different Substances
The bromine water test is a chemical analysis technique used to identify the presence of unsaturation in organic compounds. It involves the addition of bromine water ( Br2 + H2O ) to various substances, with the reaction revealing the presence of carbon-carbon double or triple bonds. This test is particularly useful in POC for distinguishing between saturated and unsaturated hydrocarbons. However this test is also useful to identify some other functional groups such as enols and ketones.
Bromine Water Test for Alkanes
Alkanes CnH2n+2 (such as methane, ethane, propane) does not react with the bromine water solution, so there isn't any color changes observed. This indicates presence of saturation.
CH4 (Methane) + Br2 → No reaction
C2H6 (Ethane) +Br2 → No reaction
C3H8( Propane ) +Br2 → No reaction
Bromine Water Test for Alkenes
Alkenes CnH2n ( such as ethene , propene ) undergoes an addition reaction with the bromine water solution. For example, ethene reacts with bromine water to give 1,2 dibromo ethane. This reaction takes place at room temperature if the reactants are in the gaseous state .Bromine water change from reddish-brown to a colorless liquid. This indicates presence of unsaturation.
Br2 + C2H4 → Br-CH2-CH2-Br
Bromine Water Test for Alkynes
Alkynes CnH2n-2 (such as Ethyne, Propyne) undergoes an addition reaction similar to the alkenes .For example, ethyne reacts with bromine water to give 1,1,2,2 tetrabromo ethane. Bromine water change from reddish-brown to a colorless liquid. This indicates presence of unsaturation .
2Br2 + C2H2 → Br2-CH-CH-Br2
Bromine Water Test for Phenol
Phenol undergoes substitution reactions with the bromine water solution giving a white precipitate named as 2, 4, 6-tribromophenol .This reaction is also known as bromination of phenol . Bromine water change from reddish-brown to a colorless liquid .This indicates presence of unsaturation.
Bromine Water Test for Aniline
Aniline or phenylamine undergoes substitution reactions with the bromine water solution giving a white precipitate named as 2, 4, 6-tribromoaniline. Bromine water change from reddish-brown to a colorless liquid. This indicates presence of unsaturation.
Bromine Water Test for Benzene
Benzene is an unsaturated compound but it doesn't react with bromine water . This is because benzene is an aromatic compound and if it undergoes anti addition with bromine, it will lose its aromaticity, which is not possible. Hence we observe no color changes.
C6H6 (Benzene) + Br2 → No reaction
Bromine Water Test for Enols
Enols are characterized by an alkene (C=C) and an adjacent alcohol group (-OH). It undergoes the bromine water test, and brominated ketone is formed. This reaction, which is highly selective for enols, causes the bromine water to change from reddish-brown to colorless indicating the presence of unsaturation.
Bromine Water Test for Glucose and Fructose
Bromine Water test is a simple test to distinguish between Glucose and Fructose. Glucose undergoes an oxidation reaction to give gluconic acid in reaction with the bromine water solution. Since bromine water is a mild oxidizing agent, Fructose fails to undergo an oxidation reaction with bromine water. Hence bromine water change from reddish-brown to a colorless liquid when it reacts with Glucose and there is no color changes observed when it reacts with Fructose.
Bromine Water Test for Ketones and Aldehydes
A reaction between the ketone and bromine water is an electrophilic alpha substitution reaction adjacent to the carbonyl group, and gives the colorless solution of brominated compounds. Similarly when an aldehyde interacts with bromine water, an oxidation effect proceeds, resulting in the formation of a colorless solution. Hence bromine water change from reddish-brown to a colorless liquid for both ketones and aldehydes.
Application of Bromine Water Test
Testing for the presence of bromine in water systems is a popular use for bromine water tests, especially in situations where bromine is used as a biocide or disinfectant. In order to maintain the water's ability to regulate microbial development while averting possible health problems, monitoring bromine levels is essential. A few uses for bromine water tests are as follows:
Swimming Pools and Spas: To preserve water quality and guarantee the efficacy of the disinfection procedure, it is important to conduct routine testing of bromine levels in swimming pools and spas. In order to keep the water clean and safe for swimmers, pool managers use bromine water test kits to monitor and regulate bromine concentrations.
Systems for Cooling Towers: Tests on bromine water in industrial cooling tower systems are carried out to confirm the bromine content used as a biocide. By preventing microbial fouling, corrosion, and scaling, maintaining the proper bromine levels contributes to the cooling system's effective functioning.
Plants that Process Poultry: In poultry processing facilities, bromine water tests are utilized to verify the amount of bromine utilized for disinfection. By doing this, the processing environment is kept free of dangerous germs, preserving the safety and quality of chicken products.
Breweries: Bromine water tests are used by breweries to confirm the level of bromine in their sanitation procedures. In order to keep brewing equipment hygienic and avoid contamination, monitoring bromine levels is essential for retaining the beer's quality.
Facilities for Treating Wastewater: Tests using bromine water are used in wastewater treatment to determine how effective bromine is as a disinfectant. In order to assure effective pathogen elimination and minimize any potential negative environmental repercussions, wastewater treatment plants benefit from routine monitoring that helps them adjust bromine dosages.
Limitation of Bromine Water Test
Although bromine water tests are useful for tracking bromine levels in a variety of applications, it's critical to understand their limitations. The following are some typical restrictions connected to testing using bromine water:
Sensitivity to Temperature: Temperature fluctuations may affect the results of bromine water tests.
Shelf Life and Storage: Test kit expiration or incorrect storage can have an impact on the accuracy of bromine water tests.
Sensitivity to pH: The water's pH can have an impact on how accurate bromine water testing are.
Other Chemicals' Interference: Tests using bromine water may be sensitive to specific compounds or impurities in the water, which could cause interference and provide false findings.
Restricted Quantitative Accuracy: Tests on bromine-contaminated water are frequently more qualitative than quantitative.
Also, Check
Chemical Properties of Caron Compounds
Bromine Formula
Chemical Reaction of Amine
Sample Questions on Bromine Water Test
Question 1: What is the principle involved in the bromine water test?
Answer:
Principle involved in the bromine water test is based on the reaction between bromine and unsaturated compounds, specifically alkenes or alkynes. Bromine adds across carbon-carbon double or triple bonds, causing the characteristic reddish-brown color of bromine water to disappear. Saturated compounds, lacking double or triple bonds, do not undergo this reaction and do not show a color change.
Question 2: What is the function of sunlight in the bromine water and iodine solution tests?
Answer:
Sunlight is not typically a factor in the bromine water test. In contrast, the iodine solution test, which assesses the presence of starch, can benefit from sunlight.
Question 3: Which hydrocarbon is most easily oxidized by KMnO4 and which is most reactive towards bromine water?
Answer:
Alkenes are readily oxidized by potassium permanganate (KMnO4), leading to the decolorization of the purple KMnO4 solution. This reaction involves oxidative cleavage of the carbon-carbon double bond, forming diols. On the other hand, alkynes are highly reactive towards bromine water, causing the rapid disappearance of its reddish-brown color.
Question 4: What happens when D-glucose is treated with bromine water?
Answer:
When D-glucose is treated with bromine water, D- gluconic acid is produced and bromine water changes from reddish - brown to a colorless solution indicating the presence of unsaturation .
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856 | https://examine.com/glossary/dose-response/?srsltid=AfmBOoobS7eCUZKbnjEfMDtAR2Eysq19K0zyT1NxwWiBzZRaP4LCNWq- | Dose-response
The term dose-response describes the relationship between the dose of a substance or stimulus and its effect. A dose-response study explores the dose-response relationship between an intervention and an outcome in terms of the efficacy and/or safety of a substance or stimulus.
Last Updated:April 29, 2025
Summary
What is dose-response?
Dose-response refers to the relationship between the dose of a substance or stimulus and its effect on an outcome of interest in a specific organism. Dose-response relationships can be simple and linear, but are often complex and nonlinear. For example, a dose-response relationship often means that a response tends to get larger as the dose increases. However, there is typically also a maximal response reached at a particular dose.
What is a dose-response study?
A dose-response study is a useful way to determine the efficacy (pharmacology) and/or safety (toxicology) of a substance or stimulus that humans or other organisms are exposed to. Such substances and stimuli can include drugs, foods, supplements, pollutants, pathogens, temperature, light, and more.
For example, to determine the dose of supplement A that has the greatest efficacy for increasing HDL cholesterol (HDL-C), a group of people with metabolic syndrome would be randomized to receive 0, 25, 50, 100, 200, or 400 mg of supplement A per day for 6 weeks, and the researchers would measure HDL-C before and after 6 weeks of treatment. To understand the dose-response of supplement A on HDL-C, the investigators would construct a dose-response curve like the one below (Figure 1), which displays the change in HDL-C after 6 weeks of supplementation with the different doses of supplement A. The dose-response curve in Figure 1 indicates that the lowest dose of the substance (25 mg/day) is insufficient to cause a response, whereas increasingly higher doses eventually lead to a maximal response. Specifically, 0 and 25 mg/day of supplement A have no effect on HDL-C, while 50, 100, 200, and 400 mg/day increase HDL-C after 6 weeks of treatment. However, there is no further increase in efficacy beyond 200 mg/day of supplement A, suggesting that there is no reason for people with metabolic syndrome to take a dose greater than this level.
Note that this is an example of a simple dose-response relationship. A real dose-response study would also be designed to determine the toxicity of supplement A at various doses and subsequently identify the lowest dose that causes the greatest efficacy with the lowest toxicity.
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1988 USAMO Problems/Problem 2
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
3 Solution 3
4 See Also
Problem
The cubic polynomial has real coefficients and three real roots . Show that and that .
Solution
Solution 1
By Vieta's Formulas, , , and . Now we know ; in terms of r, s, and t, then, Now notice that we can multiply both sides by 2, and rearrange terms to get . But since , the three terms of the RHS are all non-negative (as the square of a real number is always non-negative), and therefore their sum is also non-negative -- that is, .
Now, we will show that . We can square both sides, and the inequality will hold since they are both non-negative (it is given that , therefore ). This gives . Now we already have , so substituting this for k gives Note that this is a quadratic. Since its leading coefficient is positive, its value is less than 0 when s is between the two roots. Using the quadratic formula: The quadratic is 0 when s is equal to r or t, and the inequality holds when its value is less than or equal to 0 -- that is, . (Its value is less than or equal to 0 when s is between the roots, since the graph of the quadratic opens upward.) In fact, the problem tells us this is true. Q.E.D.
Solution 2
From Vieta's Formula (which tells us that and ), we have that clearly non-negative. To prove , it suffices to prove the square of this relation, or This in turn simplifies to or which is clearly true as . This completes the proof.
Solution 3
By Vieta's Formulas, and . .
To show that , simply note that by the trivial inequality, all three squares are greater than as they are the squares of real numbers.
To show that , since both are positive, it is sufficient to show that . implies that . . Let and . We then have , which is clearly true as both and are positive.
See Also
1988 USAMO (Problems • Resources)
Preceded by
Problem 1Followed by
Problem 3
1•2•3•4•5
All USAMO Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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858 | https://promova.com/confusing-words/ton-tonne | Ton and Tonne | Meaning, Examples & Difference | Promova
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Ton vs Tonne
Type your word here
Try:Affect vs Effect
What’s the difference between them?
Ton
Meaning:
Ton is a unit of measure equal to 2,000 pounds (907.18 kilograms) in North America and also known as a 'short ton.' There is also an 'imperial' ton (also known as British ton), which is 1.016.047 kg or 2.240 lb.
Examples:
The ton of bricks that was shipped to the building site was a lot heavier than expected.
Our car was able to tow a ton of weight with ease.
His voice had a soothing ton that made everyone calm.
Tonne
Meaning:
A tonne, also referred to as a metric ton, is a unit of mass in the metric system, equal to 1,000 kilograms or 2,204.6 pounds.
Examples:
The truck was carrying five tonnes of coal.
The recycling centre had to transport twenty tonnes of plastic to the factory.
We managed to move the tonne of gravel with a forklift.
Learn similar and opposite words to spot the difference
Synonyms
Antonyms
Ton
There are no direct synonyms for this word.
There are no direct antonyms for this word.
Tonne
Metric ton
There are no direct antonyms for this word.
Tricks for mastery
Useful tips to understand the difference between confusing words "Ton", "Tonne".
Ton is a unit of weight/mass used in the US and is equal to 2000 pounds (907 kg).
Tonne is a metric unit of mass and is equal to 1000 kg.
Use the phrase 'metric tonne or US ton' to easily remember the difference between the two.
Practice English with the Promova app and avoid misusing confusing words
Download the app
Frequently asked questions
When to use the first word 'Ton'?
The word ton is a unit of measure for weight. It is often used to talk about the weight of large items, such as vehicles and cargo. It is equal to 2,000 pounds (907.185 kilograms). For example, a large truck might be said to weigh 10 tons. In the United States, the ton is sometimes called a short ton, while in the United Kingdom, it is called a long ton.
When to use the second word 'Tonne'?
The word tonne is also a unit of measure for weight. It is the metric equivalent of the ton, and is equal to 1,000 kilograms. It is mostly used in scientific contexts and for measuring the weight of large objects, such as vehicles and ships. For example, a large truck might be said to weigh 10 tonnes.
Do the words have the same pronunciation?
No, the words have different pronunciations. The word ton is pronounced with a short 'o' sound (as in 'don') while the word tonne is pronounced with a long 'o' sound (as in 'bone').
What are common mistakes associated with the words 'ton' and 'tonne'?
One common mistake is using the wrong unit of measure for a particular context. For example, it is common for people to use the word tonne when they should be using the word ton. Another common mistake is confusing the metric tonne with the British long ton, which is slightly different. Finally, people often forget about the different pronunciations for the two words.
Fill in the gaps to check yourself
Test Questions Correct Answers
The elephant weighs nearly 6 ______.
In the UK, a ______ is equivalent to 1,000 kilograms.
The American ______ is commonly used in the United States and is equal to 2,000 pounds.
The shipment weighs approximately 10 ______.
In British English, colour is spelled with a u, similarly to how ______ is spelled with an e.
The vehicle can carry up to 2 ______ of cargo.
Answer: tons or tonnes
Explanation: There is no specific clue in the sentence to indicate whether American or British English should be used, so both ton and tonne are acceptable answers.
Answer: tonne
Explanation: The sentence provides a clue by mentioning in the UK and providing the metric equivalent, which indicates that the British English spelling, tonne, should be used.
Answer: ton
Explanation: The sentence provides a clue by mentioning the American ton and the United States, which indicates that the American English spelling, ton, should be used.
Answer: tons or tonnes
Explanation: There is no specific clue in the sentence to indicate whether American or British English should be used, so both ton and tonne are acceptable answers.
Answer: tonne
Explanation: The sentence provides a clue by discussing British English spelling and giving an example (colour), indicating that tonne, which is British English spelling, should be used.
Answer: tons or tonnes
Explanation: There is no specific clue in the sentence to indicate whether American or British English should be used, so both ton and tonne are acceptable answers.
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List of Commonly Confused Words
Finding your way around the English language can be hard, especially since there are so many confusing words and rules. So, a list of the most confusing words in English is an extremely useful tool for improving language accuracy and sharing the ideas clearly.
Affect vs. EffectAnalysis vs. AnalysesGreatful or GratefulCenter vs. CentreLoss vs. LostSeat vs. Sit
How to pronounce
How to pronounce wonHow to pronounce sonHow to pronounce ownHow to pronounce oneHow to pronounce ion
Grammar Articles
Past Perfect ContinuousReflexive PronounsActive VoicePresent PerfectAdjective Phrase
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What part of speech is camaraderieWhat part of speech is pronounWhat part of speech is teacherWhat part of speech is chooseWhat part of speech is proud
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859 | https://www.youtube.com/watch?v=0WSzz0k-mAI | Khan Academy Tutorial: tangents of circles problems
West Explains Best
4960 subscribers
13 likes
Description
595 views
Posted: 7 Jul 2021
Special shout out to George: I figured out the setting to the microphone and it is much better now!
Circles, tangents, quadrilaterals, radii, triangle angle sum, Pythagorean theorem, chords, central angles
I hope you enjoyed the video! Please leave a comment if you'd like to see a topic covered or have any mathematics related question. Happiness begins with formation. Formation happens with education. Education starts with you.
Transcript:
hi everyone this is mr west today we're doing tangents of circles problems in khan academy this is a tutorial to help you with tangent circles a geometry concept let's get started angle a is circumscribed about circle o what is the length of ac a couple things one because this is tangents of circles problems ac is assumed to be a tangent so let me identify ac here ac is this guy so we're going to have to find his length eventually but what is a tangent a tangent is a line that touches a circle at only one place so right there now the cool thing about a tangent line is that when it connects with the radius drawn in purple here it creates a 90 degree angle with that tangent line so how is this helpful for this problem for this problem it's useful because we can draw an additional line right through here and now you can see that we have a triangle being formed two triangles actually so we have this triangle here and we have this triangle here so two triangles one of which has a side the green side is what we're looking for but i'm actually not going to start there because only i don't have any information on this triangle i have no sides so that's not good so i'm gonna start with this triangle right here because it has two sides given out of three let's take a look at this so if i'm going to redraw this notice i have this side here that's a question mark we're not sure what that is then we have this side here that's three so i'm gonna go ahead and copy that down that's three and then i have this bottom side which i know is six as we said earlier this is our tangent line okay it's assumed that that is our tangent line so we know this creates a 90 degree angle right there well if we have a 90 degree angle in this triangle that makes it a right triangle with the right triangle we can use pythagorean theorem to solve for that missing side so we know that 6 squared which is a leg plus 3 squared which is another leg is going to be equal to the hypotenuse squared so instead of calling this question mark let's call it c because that's the pythagorean theorem a squared plus b squared equals c squared now we have 36 plus 9 equals c squared so what is that that's 45 summer so been out of math for a while let's see 45 equals c squared and so c equals the square root of 45 and we're just going to leave it the square root of 45 for now uh technically this could be simplified even in radical form or it could be in decimal form but for now it just we since this is the distance we're looking for we can just leave it square to 45 okay so now if we're looking back at this original triangle we have one side now the other thing we know is because this is a tangent line and this is a radius okay so that line's a radius so i drew that very dark that works we know that this is 90 degrees because the tangent and its radius make a 90 degree angle well this is a radius then we also know it's three so we have square root of 45 we have three so we can find the last leg here by using the pythagorean theorem again this time it would be almost the same except instead of c we would have uh square root of 45 we have uh 3 there and would have 6 would be unknown now if you're thinking okay well we know it's 6 because we just did 6 squared plus 3 squared equals square root of 45 then square root of 45 minus 9 squared is going to excuse me yeah minus 9 is going to give us 6. so this is actually going to be 6 when we solve again so that is the distance of 6 right there so we just used the principles of a right triangle based on the tangent lines and then pythagorean theorem from there to find that missing side so you got it onward next question ac is tangent to circle o at point c what is the measure of angle o all right so here we go again now let's first identify our tangent line our tangent line as it says is ac so we have ac is our tangent line circle o now whenever there's a center that is always designated by that point which circle so obviously o is that circle right there usually pretty obvious but i thought i would mention that what is the measure of angle oh so in this case angle o is being referred to as that guy right there so trying to do a better job that angle right there and actually i'm not super satisfied with that that is the angle we're looking for okay now how do we do this well we know that's a tangent line and this is a radius we don't know the length of that radius but we don't really care because all we're interested in in this example is the angle so we have 90 degree angles there we have 23 degree angle there we know that in a triangle it adds up to 180 degrees so simply we just need to do 180 minus 90 well that's 90 minus 23 because that was there and plus this angle x so we can find angle x by doing 90 minus 23 or 180 minus 90 minus 23 same deal and we get what is that that is 57 no that's 67 67 degrees yes that's correct okay so 67 degrees so we know angle 0 is 67 because 67 plus 23 plus 90 equals 180 so that is 67 degrees check it all right next question all sides of quadrilateral abcd are tangent to circle p what is the perimeter of the quadrilateral below okay so we have all these tangent lines being drawn here we have 3.3 we have 12.2 okay and we have a really interesting quadrilateral in place here so how are we going to go about doing this hmm well we know a couple things well we could draw some tangent lines or to the radius so we know that these angles are all 90 degrees okay we know that that this is 90 that this is 90 this is 90 and that this is 90. now this doesn't help us that much um unless that we use well if we know that is 90 and that is 90 then we know that's a little square right there and then we have a little kite so hold on let me highlight this so you know what i'm talking about this would be a square because that's 90 and 90. no that wouldn't be that's just these are all just different forms of a kite however because this is a kite okay so this is a kite because we have 90 degree angle here and here and unknown lengths here in here a kite by definition has these opposite angles right here that are the same and the two other angles that are not the same so because all these um have opposite angles the same these are technically all different types of kites here so we have four different kites and we'll go to my last color i guess i'll go red because all their angles are the same here here and here and here okay so they have opposite angles so these are all different kinds of kites that's a little too loaded of stuff so i'm going to erase some of it and concentrate on this yellow kite to start now one of the definitions of kites is because these lengths are the same they're both radius this length is also the same so that is going to be 3.3 right there those lengths are going to be the same so likewise here we go there's the other part of my kite we have our kite here those angles are the same here and here so we know that these angles this bit is also going to be the same this blue and this blue are also going to be the same length now we don't know what this is yet because we don't know this length right here so we're going to have to come back to this blue one actually let's go to a green one i'm just going to make this green this one should be easy to solve because we have one of the distances there again here's my radius which forms the other part of the kite same distance there so the green needs to be the same distance likewise because this is 5.1 this part is also going to be 5.1 now from here we can get a rough tally we have 3.3 plus 5 point uh let me use red actually because i'm going to use purple for the next one so we have 3.3 plus 5.1 that is going to give us 8.4 we don't know this side we need to come back to it we already know this whole side is 12.1 that's a good news okay but and we know this whole side is 12.2 so we're only missing that side but we need to kind of go around the whole kite to keep figuring it out because this is 5.1 and this whole section is 12 point let me get my highlighter because this whole section is 12.1 we can just simply subtract 12.1 sorry it's 5.1 from 12.1 to find that purple distance there and that purple distance is going to be again 12.1 minus 5.1 and that gives us what is that 7. so we know this is seven and if that's seven then this is also seven now we have one last piece of the puzzle consider this like it's just like a puzzle it's just like sudoku so now we have 12.2 minus seven to find out that blue so we have 12.2 minus 7 gives us the blue and that's going to be 5.2 so this is going to be 5.2 but we still have to add that 3.3 so that total distance right there is going to be 8.5 so now we have our final distances 8.5 8.4 12.1 and 12.2 so i don't have a calculator on me so we're just going to have to do this old school so we have 12.2 plus 12.1 that's going to give us 24.3 we're going to add 8.4 to that and we're going to get 32.7 and then we have 8.5 to that 2 carry the 1. that takes us to 11. carry the 1. we have 4. 41.2 hopefully i did that right otherwise this might be embarrassing 41.2 let's check it out got it whoo close call all right angle a is circumscribed about circle o what is the measure of a okay great problem to end on i'm happy you got this we got all the hard problems in this particular khan academy was the angle of uh measure of angle a angle a is right here okay again why do i keep using that one here's what i want okay angle a so here we have 65 65 is this right here and this is a cord so notice how it doesn't go through the center so any line that doesn't go through the well i mean it can be any line that goes across the circle these are called chords okay so that's a chord now one of the properties of a chord is that if it's 65 degrees here okay then that means this angle is going to be 2 times 65 degrees because they end on the same place okay that's important they end in the same place so this is going to be 2 times 65 degrees and i think i put 66.5 i meant 65 degrees so that angle right there this red angle is going to be 2 times 65 that's 130. so this is 130 degrees okay so that's 130 degrees now we have almost all the information we need so here we have a tangent here's a tangent that means this is 90 degrees 90 degrees we're going to focus here on this quadrilateral right here that quadrilateral needs to add up to 360. so so far we have two 90s 90 90 we have a 130 and then we have that missing angle a angle a which i made in red so i'm going to add up these three first and that's going to be 180 plus 130 that's 3 10. now i know that these all four need to add up to 360. so angle a needs to help us get to 360. so that's got to be 50 degrees so 50 plus 130 plus 90 plus 90 equals 360. so angle a equals 50 degrees and that's how we can kind of go backwards to figure that out so 50 degrees the chord here is probably the the toughest part of understanding this this chord is half the distance or half the measure of that central angle here or that's that arc uh measure right there now arc length that arc measure so that's another thing to consider i'm going to make another video on that uh later on so if you want to check that out make sure to check west explains how to do those circles and that's it that's all there is for khan academy with changes of circle problems hope you enjoyed this make sure to check out more and i'll be right here waiting for you on west explains best take care |
860 | https://www.gettyimages.com/photos/louella-o-parsons | 61 Louella O Parsons Stock Photos, High-Res Pictures, and Images - Getty Images
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Director Busby Berkeley shows how he wants a scene played in the Warner Bros. 1938 film Hollywood Hotel. Actors and actresses Glenda Farrell, Lola...
Director and Cast on the Set of Hollywood Hotel
Beverly Hills, Calif.: Actress Loretta Young is given the "Louella O. Parsons Award" by last years recipient, Jimmy Stewart as Young is honored by...
Loretta Young and Jimmy Stewart Holding Award
Noted Newspaper Columnist Miss Louella O. Parsons.
Portrait of Miss Louella O. Parsons
"Hollywood Hotel", palatial in its construction is dedicated. Not an actual place of residence is his hotel but a radio program of one-hour duration...
Colbert with Louella Parsons and Warren Williams
At the Cosmopolitan Ball for Disabled War Vets. New York, New York: Many notable persons were present at the Cosmopolitan Ball and midnight frolic at...
Louella Parsons and Friends
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Debbie Reynolds Greeting Eddie Fisher
When Eddie Fisher first looked at Debbie Reynolds, the girl of his choice, writes Louella O. Parsons in her story of the Eddie-Debbie romance, he...
Portrait of Debbie Reynolds
When Debbie Reynolds is around, says Louella O. Parsons, Eddie Fisher can't look anywhere else. And when Eddie's nearby, Miss Parsons relates in her...
Debbie Reynolds Holding Hands With Eddie Fisher
"That's my girl!" says Louella O. Parsons of Debbie Reynolds with whom Hollywood's most famous columnist is shown here. Miss Parsons, who tells the...
Louella Parsons Sitting Next to Debbie Reynolds
Filmland made a deep bow to its acknowledged queen tonight in an hour-long telecast of the dramatic and glamorous life of Louella O. Parsons,...
Celebrities Gathering to Watch Television Show
Host and her Guest Stars. Los Angeles, California: Louella O. Parsons, famed movie columnist, is shown with her guest stars, Bruce Cabot and Marlene...
Bruce Cabot with Louella Parsons and Marlene Dietrich
New York. Eleanor Parker , star of Above and Beyond which had its premiere tonight at Brandt's Mayfair Theatre, is pictured with Jinx Falkenburg...
Eleanor Parker Smiling
Parsons' Showcase Feature for Sunday, March 31. Hollywood, California: Editors: these pictures of actress Lauren Bacall are for suggested use with...
Lauren Bacall Reading a Magazine
Hollywood, California: "The King" Talks To "The Queen." In Hollywood, they call Clark Gable, "The King," and Louella O. Parsons, Motion Picture...
Actor Clark Gable Conversing with Editor Louella Parsons
George McManus luncheon, 16 January 1952. Richard A Carrington, Junior;George Jessel;George McManus;David Hearst;Louella O Parsons;Walt Disney..
George McManus luncheon, 1952
George McManus luncheon, 16 January 1952. Richard A Carrington, Junior;George Jessel;George McManus;David Hearst;Louella O Parsons;Walt Disney..
George McManus luncheon, 1952
George McManus luncheon, 16 January 1952. Richard A Carrington, Junior;George Jessel;George McManus;David Hearst;Louella O Parsons;Walt Disney..
George McManus luncheon, 1952
George McManus luncheon, 16 January 1952. Richard A Carrington, Junior;George Jessel;George McManus;David Hearst;Louella O Parsons;Walt Disney..
George McManus luncheon, 1952
Hearst party, 23 June 1961. J.D Gortatowsky;Ewen Dingwall;Louella Parsons;William R Hearst, Jr.;Kingsbury Smith;Fred Archibald;Mrs Lucille...
Hearst party, 1961
Hearst party, 23 June 1961. J.D Gortatowsky;Ewen Dingwall;Louella Parsons;William R Hearst, Jr.;Kingsbury Smith;Fred Archibald;Mrs Lucille...
Hearst party, 1961
Hearst party, 23 June 1961. J.D Gortatowsky;Ewen Dingwall;Louella Parsons;William R Hearst, Jr.;Kingsbury Smith;Fred Archibald;Mrs Lucille...
Hearst party, 1961
Hearst party, 23 June 1961. J.D Gortatowsky;Ewen Dingwall;Louella Parsons;William R Hearst, Jr.;Kingsbury Smith;Fred Archibald;Mrs Lucille...
Hearst party, 1961
Hearst party, 23 June 1961. J.D Gortatowsky;Ewen Dingwall;Louella Parsons;William R Hearst, Jr.;Kingsbury Smith;Fred Archibald;Mrs Lucille...
Hearst party, 1961
Hearst party, 23 June 1961. J.D Gortatowsky;Ewen Dingwall;Louella Parsons;William R Hearst, Jr.;Kingsbury Smith;Fred Archibald;Mrs Lucille...
Hearst party, 1961
Hearst party, 23 June 1961. J.D Gortatowsky;Ewen Dingwall;Louella Parsons;William R Hearst, Jr.;Kingsbury Smith;Fred Archibald;Mrs Lucille...
Hearst party, 1961
Hearst party, 23 June 1961. J.D Gortatowsky;Ewen Dingwall;Louella Parsons;William R Hearst, Jr.;Kingsbury Smith;Fred Archibald;Mrs Lucille...
Hearst party, 1961
Hearst party, 23 June 1961. J.D Gortatowsky;Ewen Dingwall;Louella Parsons;William R Hearst, Jr.;Kingsbury Smith;Fred Archibald;Mrs Lucille...
Hearst party, 1961
Fiesta of Stars, April 29, 1951. Loretta Young;Van Johnson;Father Patrick Peyton;Margaret O'Brien;Clare Boothe Luce;Irene Dunn;Louella Parsons;George...
Fiesta of Stars, 1951
Fiesta of Stars, April 29, 1951. Loretta Young;Van Johnson;Father Patrick Peyton;Margaret O'Brien;Clare Boothe Luce;Irene Dunn;Louella Parsons;George...
Fiesta of Stars, 1951
Fiesta of Stars, April 29, 1951. Loretta Young;Van Johnson;Father Patrick Peyton;Margaret O'Brien;Clare Boothe Luce;Irene Dunn;Louella Parsons;George...
Fiesta of Stars, 1951
Fiesta of Stars, April 29, 1951. Loretta Young;Van Johnson;Father Patrick Peyton;Margaret O'Brien;Clare Boothe Luce;Irene Dunn;Louella Parsons;George...
Fiesta of Stars, 1951
Fiesta of Stars, April 29, 1951. Loretta Young;Van Johnson;Father Patrick Peyton;Margaret O'Brien;Clare Boothe Luce;Irene Dunn;Louella Parsons;George...
Fiesta of Stars, 1951
Fiesta of Stars, April 29, 1951. Loretta Young;Van Johnson;Father Patrick Peyton;Margaret O'Brien;Clare Boothe Luce;Irene Dunn;Louella Parsons;George...
Fiesta of Stars, 1951
Fiesta of Stars, April 29, 1951. Loretta Young;Van Johnson;Father Patrick Peyton;Margaret O'Brien;Clare Boothe Luce;Irene Dunn;Louella Parsons;George...
Fiesta of Stars, 1951
Fiesta of Stars, April 29, 1951. Loretta Young;Van Johnson;Father Patrick Peyton;Margaret O'Brien;Clare Boothe Luce;Irene Dunn;Louella Parsons;George...
Fiesta of Stars, 1951
Fiesta of Stars, April 29, 1951. Loretta Young;Van Johnson;Father Patrick Peyton;Margaret O'Brien;Clare Boothe Luce;Irene Dunn;Louella Parsons;George...
Fiesta of Stars, 1951
Fiesta of Stars, April 29, 1951. Loretta Young;Van Johnson;Father Patrick Peyton;Margaret O'Brien;Clare Boothe Luce;Irene Dunn;Louella Parsons;George...
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Fiesta of Stars, 1951
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861 | https://allen.in/dn/qna/644369617 | Can you explain why the column of mercury first descends and then rises when a mercury in glass thermometers put in a flame?
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Explain the construction of a mercury thermometer.
Explain why an iron nail floats on mercury, but it sinks in water.
On a Celsius thermometer the distance between the readings 0^@C and 100^@C is 30cm and the area of cross section of the narrow tube containing mercury is 15xx10^-4cm^2 . Find the total volume of mercuty in the thermometer at 0^@C . alpha of glass =9xx10^-6//K and the coefficient of real expansion of mercury =18xx10^(-5)//K .
In a mercury-glass thermometer the cross-section of the capillary portion is A_(0) and the volume of the bulb is V_(0) at 273K . If alpha and gamma are the coefficients of linear and cubical expansion coefficients of glass and mercury respectively then length of mercury in the capillary at temperature t^(@)C is (Ignore the increase in cross-sectional area of capillary)
Explain why : Mercury loses its meniscus in contact with ozone.
In a constant volume gas thermometer, the pressure of the working gas is measured by the difference in the levels of mercury in the two arms of a U-tube connected to the gas at one end. When the bulb is placed at the room temperature 27.0^0 C , the mercury column in the arm open to atmosphere stands 5.00 cm above the level of mercury in the other arm. When the bulb is placed in a hot liquid, the difference of mercury levels becomes 45.0 cm . Calculate the temperature of the liquid. (Atmospheric pressure = 75.0cm of mercury).
In a constant volume gas thermometer, the pressure of the working gas is measured by the differenced in the levels of mercury in the two arms of a U-tube connected to the gas at one end. When the bulb is placed at the room temperature 27.0^0 C , the mercury column in the arm open to atmosphere stands 5.00 cm above the level of mercury in the other arm. When the bulb is placed in a hot liquid, the difference of mercury levels becomes 45..0 Cm . Calculate the temperature of the liquid. (Atmospheric pressure = 75.0cm of mercury).
Explain why does the mercury level in a barometer go down when atmospheric pressure decreases?
The coefficient of linear expansion of glass is alpha_g per .^(@)C and the cubical expansion of mercury is gamma_m per .^(@)C . The volume of the bulb of a mercury thermometer at 0^@C is V_0 and cross section of the capillary is A_0 . What is the length of mercury column in capillary at T^@C , if the mercury just fills the bulb at 0^@C ?
A glass tube 80 cm long and open ends is half immersed in mercury. Then the top of the tube is closed and it is taken out of the mercury. A column of murcury 20 cm long then remains in the tube. The atmospheric (in cm of Hg) is (assume temperature to be constant )
AAKASH INSTITUTE ENGLISH-THERMAL PROPERTIES OF MATTER-Assignment (Section-J) Akash Challengers Questions
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862 | https://www.physicsforums.com/threads/mechanics-of-materials-by-russell-hibbeler.667415/ | Mechanics of Materials by Russell Hibbeler • Physics Forums
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Mechanics of Materials by Russell Hibbeler
Thread starter rdjon
Start date May 26, 2003
TagsMaterialsMechanicsMechanics of materials
AI Thread Summary
The discussion highlights several recommended geology books, emphasizing their significance and appeal. "The Fossil Trail" by Ian Tattersall provides an insightful overview of human evolution, though it may be somewhat outdated. Richard Fortey's "Trilobite" captivates readers with its engaging exploration of these ancient arthropods. "Global Tectonics" by Keary and Vine serves as a comprehensive yet accessible textbook on plate tectonics. "Potential Theory in Gravity & Magnetic Applications" by Blakely is praised for its clarity in explaining complex geophysical concepts, making it essential for students in the field. Peter D. Ward's work on mass extinctions is noted for its poetic depth and relevance to current environmental issues, urging even those outside geology to engage with its themes. The conversation invites further opinions on these titles, particularly Ward's book, which raises awareness about humanity's impact on the planet.
For those who have used this book
Strongly Recommend
Votes: 1 50.0%
### Lightly Recommend
Votes: 1 50.0%
### Lightly don't Recommend
Votes: 0 0.0%
### Strongly don't Recommend
Votes: 0 0.0%
Total voters 2
May 26, 2003
1
rdjon
Seeing as no-one has posted any good geology books, I'll recommend some of my favourites...
The Fossil Trail, Ian Tattersall - an excellent summation of Human evolution, a little bit out of date now.
Trilobite, Richard Fortey - how he makes a whole book about small arthropods interesting, I have no idea, but it is a fascinating read.
Global Tectonics, Keary and Vine - everything you ever wanted to know about plate tectonics. More of a textbook than a popular science book, but very readable.
Potential Theory in Gravity & Magnetic Applications, Blakely - this book got me through some of the trickier bits of my Geophysics degree. Explains complicated ideas very well.
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New perspectives on light-matter interaction: How virtual charges influence material responses
Nov 17, 2003
2
Mentat
3,935 3
Anyone who hasn't read this book is missing out, big time!
The author, Peter D. Ward, uses amazing (practically poetic) profundity in explaining the previous three major "mass" extinctions, and then reasons on the current state of the Earth, to show that we are at the beginning of the fourth such major "mass" extinction - and that it's our fault.
Seriously, even those of you who are not normally interested in Biology or Geology (like me) should check this book out, because it is awesome.
Has anyone here read it? What did you think of it?
Jan 27, 2013
3
Astronuc
Staff Emeritus
Science Advisor
Gold Member
22,355 7,174
Author: Russell C. Hibbeler
Title: Mechanics of Materials
Amazon Link:
Prerequisities: Prior or concurrent experience in Calculus, Introductory Physics
Level: Undergraduate
Table of Contents:
Code:
```
Chapter 1: Stress
1.1 Introduction
1.2 Equilibrium of a Deformable Body
1.3 Stress
1.4 Average Normal Stress in an Axially Loaded Bar
1.5 Average Shear Stress
1.6 Allowable Stress
1.7 Design of Simple Connections
Chapter 2: Strain
2.1 Deformation
2.2 Strain
Chapter 3: Mechanical Properties of Materials
3.1 The Tension and Compression Test
3.2 The Stress–Strain Diagram
3.3 Stress–Strain Behavior of Ductile and Brittle Materials
3.4 Hooke’s Law
3.5 Strain Energy
3.6 Poisson’s Ratio
3.7 The Shear Stress–Strain Diagram
3.8 Failure of Materials Due to Creep and Fatigue
Chapter 4: Axial Load
4.1 Saint-Venant’s Principle
4.2 Elastic Deformation of an Axially Loaded Member
4.3 Principle of Superposition
4.4 Statically Indeterminate Axially Loaded Member
4.5 The Force Method of Analysis for Axially Loaded Members
4.6 Thermal Stress
4.7 Stress Concentrations
4.8 Inelastic Axial Deformation
4.9 Residual Stress
Chapter 5: Torsion
5.1 Torsional Deformation of a Circular Shaft
5.2 The Torsion Formula
5.3 Power Transmission
5.4 Angle of Twist
5.5 Statically Indeterminate Torque-Loaded Members
5.6 Solid Noncircular Shafts
5.7 Thin-Walled Tubes Having Closed Cross Sections
5.8 Stress Concentration
5.9 Inelastic Torsion
5.10 Residual Stress
Chapter 6: Bending
6.1 Shear and Moment Diagrams
6.2 Graphical Method for Constructing Shear and Moment Diagrams
6.3 Bending Deformation of a Straight Member
6.4 The Flexure Formula
6.5 Unsymmetric Bending
6.6 Composite Beams
6.7 Reinforced Concrete Beams
6.8 Curved Beams
6.9 Stress Concentrations
6.10 Inelastic Bending
Chapter 7: Transverse Shear
7.1 Shear in Straight Members
7.2 The Shear Formula
7.3 Shear Flow in Built-Up Members
7.4 Shear Flow in Thin-Walled Members
7.5 Shear Center for Open Thin-Walled Members
Chapter 8: Combined Loadings
8.1 Thin-Walled Pressure Vessels
8.2 State of Stress Caused by Combined Loadings
Chapter 9: Stress Transformation
9.1 Plane-Stress Transformation
9.2 General Equations of Plane-Stress Transformation
9.3 Principal Stresses and Maximum In-Plane Shear Stress
9.4 Mohr’s Circle—Plane Stress
9.5 Absolute Maximum Shear Stress
Chapter 10: Strain Transformation
10.1 Plane Strain
10.2 General Equations of Plane-Strain Transformation
10.3 Mohr’s Circle—Plane Strain
10.4 Absolute Maximum Shear Strain
10.5 Strain Rosettes
10.6 Material-Property Relationships
10.7 Theories of Failure
Chapter 11: Design of Beams and Shafts
11.1 Basis for Beam Design
11.2 Prismatic Beam Design
11.3 Fully Stressed Beams
11.4 Shaft Design
Chapter 12: Deflection of Beams and Shafts
12.1 The Elastic Curve
12.2 Slope and Displacement 12 by Integration
12.3 Discontinuity Functions
12.4 Slope and Displacement by the Moment-Area Method
12.5 Method of Superposition
12.6 Statically Indeterminate Beams and Shafts
12.7 Statically Indeterminate Beams and Shafts—Method of Integration
12.8 Statically Indeterminate Beams and Shafts—Moment-Area Method
12.9 Statically Indeterminate Beams and Shafts—Method of Superposition
Chapter 13: Buckling of Columns
13.1 Critical Load
13.2 Ideal Column with Pin Supports
13.3 Columns Having Various Types of Supports
13.4 The Secant Formula
13.5 Inelastic Buckling
13.6 Design of Columns for Concentric Loading
13.7 Design of Columns for Eccentric Loading
Chapter 14: Energy Methods
14.1 External Work and Strain Energy
14.2 Elastic Strain Energy for Various Types of Loading
14.3 Conservation of Energy
14.4 Impact Loading
14.5 Principle of Virtual Work
14.6 Method of Virtual Forces Applied to Trusses
14.7 Method of Virtual Forces Applied to Beams
14.8 Castigliano’s Theorem
14.9 Castigliano’s Theorem Applied to Trusses
14.10 Castigliano’s Theorem Applied to Beams
Appendix A: Geometric Properties of An Area
A.1 Centroid of an Area
A.2 Moment of Inertia for an Area
A.3 Product of Inertia for an Area
A.4 Moments of Inertia for an Area about Inclined Axes
A.5 Mohr’s Circle for Moments of Inertia
Appendix B: Geometric Properties of Structural Shapes
Appendix C: Slopes and Deflections of Beams
```
Last edited by a moderator: May 6, 2017
Mar 17, 2013
4
Astronuc
Staff Emeritus
Science Advisor
Gold Member
22,355 7,174
Author: M. L. Wilson, S. H. Kosmatka
Title: Design and Control of Concrete Mixtures
Amazon Link:
Prerequisities: First two years of Civil Engineering program
Level: Undergraduate, intermediate
Table of Contents
Chapter 1 Introduction to Concrete
Chapter 2 Sustainability
Chapter 3 Portland, Blended and Other Hydraulic Cement
Chapter 4 Supplementary Cementitious Materials
Chapter 5 Mixing Water for Concrete
Chapter 7 Chemical Admixtures for Concrete
Chapter 8 Reinforcement
Chapter 9 Properties of Concrete
Chapter 10 Volume Changes of Concrete
Chapter 11 Durability
Chapter 12 Designing and Proportioning Concrete Mixtures
Chapter 13 Batching, Mixing, Transporting, and Handling Concrete
Chapter 14 Placing and Finishing Concrete
Chapter 15 Curing Concrete
Chapter 16 Hot Weather Concreting
Chapter 17 Cold Weather Concreting
Chapter 18 Test Methods
Chapter 19 High-Performance Concrete
Chapter 20 Special Types of Concrete
Publisher's product page -
Portland Cement Association said:
Design and Control of Concrete Mixtures--the guide to applications, methods, and materials, has been the industry’s primary reference on concrete technology for over 85 years This fully revised 15th edition is a concise, current reference on concrete, that includes the many advances that occurred since the last edition was published in 2002. The text is backed by over 95 years of research by the Portland Cement Association. It reflects the latest information on standards, specifications, test methods and guides of ASTM International (ASTM), the American Association of State Highway and Transportation Officials (AASHTO), the American Concrete Institute (ACI), and the National Ready Mixed Concrete Association (NRMCA).
This book presents the properties of concrete as needed in concrete construction, including strength and durability. All concrete ingredients (cementing materials, water, aggregates, chemical admixtures, and fibers) are reviewed for their optimal use in designing and proportioning concrete mixtures. The use of concrete from design to batching, mixing, transporting, placing, consolidating, finishing, and curing is addressed. Besides presenting a 30% increase in new information over the prior edition within the previous chapters, this edition has added four new chapters on concrete sustainability, reinforcement, properties of concrete, and durability.
This comprehensive manual benefits personnel from ready-mixed concrete producers, concrete contractors, and material suppliers and is a reference publication for many national and state certification programs. Design and Control of Concrete Mixtures also meets the growing needs of architects, engineers, builders, and students. The 15th edition also includes a foreward by William F. Baker, PE, SE, FIStructE Partner, Skidmore, Owings & Merrill LLP. This book is a “must have” for anyone involved with concrete.
14th Edition -
Last edited by a moderator: May 6, 2017
Aug 3, 2013
5
Astronuc
Staff Emeritus
Science Advisor
Gold Member
22,355 7,174
Authors: John Conrad Jaeger, Neville G. W. Cook, Robert Zimmerman
Title: Fundamentals of Rock Mechanics, 4th Ed
Amazon Link:
Prerequisities: Calculus, Introductory Physics/Engineering, Mechanics
Level: Graduate, Introductory; Undergraduate, Advanced
Table of Contents
Code:
```
1. Rock as a Material.
1.1 Introduction.
1.2 Joints and faults.
1.3 Rock-forming minerals.
1.4 The fabric of rocks.
1.5 The mechanical nature of rock.
Analysis of Stress and Strain.
2.1 Introduction.
2.2 Definition of traction and stress.
2.3 Analysis of stress in two dimensions.
2.4 Graphical representations of stress in two dimensions.
2.5 Stresses in three dimensions.
2.6 Stress transformations in three dimensions.
2.7 Mohr’s representation of stress in three dimensions.
2.8 Stress invariants and stress deviation.
2.9 Displacement and strain.
2.10 Infinitesimal strain in two dimensions.
2.11 Infinitesimal strain in three dimensions.
2.12 Determination of principle stresses or strains from measurements.
2.13 Compatibility equations.
2.14 Stress and strain in polar and cylindrical coordinates.
2.15 Finite strain.
Friction on Rock Surfaces.
3.1 Introduction.
3.2 Amonton’s law.
3.3 Friction on rock surfaces.
3.4 Stick-slip oscillations.
3.5 Sliding on a plane of weakness.
3.6 Effects of time and velocity.
Deformation and Failure of Rock.
4.1 Introduction.
4.2 The stress-strain curve.
4.3 Effects of confining stress and temperature.
4.4 Types of fracture.
4.5 Coulomb failure criterion.
4.6 Mohr’s hypothesis.
4.7 Effects of pore fluids.
4.8 Failure under true-triaxial conditions.
4.9 The effect of anisotropy on strength.
Linear Elasticity.
5.1 Introduction.
5.2 Stress-strain relations for an isotropic linear elastic solid.
5.3 Special cases.
5.4 Hooke’s law in terms of deviatoric stresses and strains.
5.5 Equations of stress equilibrium.
5.6 Equations of stress equilibrium in cylindrical and spherical coordinates.
5.7 Airy stress functions.
5.8 Elastic strain energy and related principles.
5.9 Uniqueness theorem for elasticity problems.
5.10 Stress-strain relations for anisotropic materials.
Laboratory Testing of Rocks.
6.1 Introduction.
6.2 Hydrostatic tests.
6.3 Uniaxial compression.
6.4 Triaxial tests.
6.5 Stability and stiff testing machines.
6.6 True-triaxial tests.
6.7 Diametral compression of cylinders.
6.8 Torsion of circular cylinders.
6.9 Bending tests.
6.10 Hollow cylinders.
Poroelasticity and Thermoelasticity.
7.1 Introduction.
7.2 Hydrostatic poroelasticity.
7.3 Undrained compression.
7.4 Constitutive equations of poroelasticity.
7.5 Equations of stress equilibrium and fluid flow.
7.6 One-dimensional consolidation.
7.7 Applications of poroelasticity.
7.8 Thermoelasticity.
Stresses around Cavities and Excavations.
8.1 Introduction.
8.2 Complex variable method for two-dimensional elasticity problems.
8.3 Homogeneous state of stress.
8.4 Pressurised hollow cylinder.
8.5 Circular hole in a rock mass with given far-field principal stresses.
8.6 Stresses applied to a circular hole in an infinite rock mass.
8.7 Stresses applied to the surface of a solid cylinder.
8.8 Inclusions in an infinite region.
8.9 Elliptical hole in an infinite rock mass.
8.10 Stresses near a crack tip.
8.11 Nearly rectangular hole.
8.12 Spherical cavities.
8.13 Penny-shaped cracks.
8.14 Interactions between nearby cavities.
Inelastic Behavior.
9.1 Introduction.
9.2 Plasticity and yield.
9.3 Elastic-plastic hollow cylinder.
9.4 Circular hole in an elastic-brittle-plastic rock mass.
9.5 Perfectly plastic behavior.
9.6 Flow between flat surfaces.
9.7 Flow rules and hardening.
9.8 Creep.
9.9 Simple rheological models.
9.10 Theory of viscoelasticity.
9.11 Some simple viscoelastic problems.
Micromechanical Models.
10.1 Introduction.
10.2 Effective moduli of heterogeneous rocks.
10.3 Effect of pores on compressibility.
10.4 Crack closure and elastic nonlinearity.
10.5 Effective medium theories.
10.6 Sliding crack friction and hysteresis.
10.7 Griffith cracks and the Griffith locus.
10.8 Griffith theory of failure.
10.9 Linear elastic fracture mechanics.
Wave Propagation in Rocks.
11.1 Introduction.
11.2 One-dimensional elastic wave propagation.
11.3 Harmonic waves and group velocity.
11.4 Elastic waves in unbounded media.
11.5 Reflection and refraction of waves at an interface.
11.6 Surface and interface waves.
11.7 Transient waves.
11.8 Effects of fluid saturation.
11.9 Attenuation.
11.10 Inelastic waves.
Hydromechanical Behavior of Fractures.
12.1 Introduction.
12.2 Geometry of rock fractures.
12.3 Normal stiffness of rock fractures.
12.4 Behaviour of rock fractures under shear.
12.5 Hydraulic transmissivity of rock fractures.
12.6 Coupled hydro-mechanical behavior.
12.7 Seismic response of rock fractures.
12.8 Fractured rock masses.
State of Stress Underground.
13.1 Introduction.
13.2 Simple models for the state of stress in the subsurface.
13.3 Measured values of subsurface stresses.
13.4 Surface loads on a half-space: two-dimensional theory.
13.5 Surface loads on a half-space: three-dimensional theory.
13.6 Hydraulic fracturing.
13.7 Other stress measurement methods.
Geological Applications.
14.1 Introduction.
14.2 Stresses and faulting.
14.3 Overthrust faulting and sliding under gravity.
14.4 Stresses around faults.
14.5 Mechanics of intrusion.
14.6 Beam models for crustal folding.
14.7 Earthquake mechanics.
References
```
Last edited by a moderator: May 6, 2017
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50+ Lateral Thinking Puzzles That’ll Stretch Your Mind in a Whole New Way
Discover mind-blowing puzzles to keep your brain sharp.
ByMaryn Liles
Lifestyle Writer, Parade
Canva/Parade
Jul 9, 2025 9:27 AM EDT
Are you looking for a fun and exciting way to stretch out your creative thinking muscles? Are you obsessed with solving mysteries before everyone else? If you answered ‘yes’ to any of these questions, then solving lateral thinking puzzles may be your new favorite hobby! We’ve rounded up 50+ of the best lateral thinking puzzles around! Enjoy!
Related: 33 Best Online Games To Play With Friends for Fun
Table of Contents
What Is Lateral Thinking?
How Is Lateral Thinking Different From Vertical Thinking?
Hard Lateral Thinking Puzzles
Challenging Lateral Thinking Puzzles
Lateral Thinking Puzzles for Kids and Adults
Show More
What Is Lateral Thinking?
Lateral thinking refers to a person’s capacity to problem solve by imagining creative solutions that cannot be figured out by deductive or logical solutions. Basically, it’s thinking outside of the box.
Coined by Maltese physician and psychologist, Edward de Bono, lateral thinking is the essence of creativity. This gives you a chance to put on a detective’s hat while solving fun mysteries!
How Is Lateral Thinking Different From Vertical Thinking?
Vertical thinking is when a person goes straight to the obvious and logical answer. With lateral thinking, someone might come up with a solution to a riddle or scenario that requires you to piece facts together to come up with a creative answer.
You can stretch your lateral thinking by taking on a lateral thinking puzzle. All you have to do is look at the scenario presented and find context clues. Sometimes, the most correct answer to the most lateral thinking puzzles is actually the most obvious, which is why you don’t get there immediately. Other times, the correct answer is really difficult to find because you have to come up with the rest of the scenario yourself to get there.
If this sounds a little confusing, that’s because it’s supposed to be! Have a little fun and try out these lateral thinking puzzles below to test your super sleuthing skills.
Related: Give Your Mind a Workout! 101 Brain Teasers That’ll Improve Your Memory
Hard Lateral Thinking Puzzles
A couple enters a grand ballroom only to find the occupants are dead. There has been no crime committed. The couple is not concerned by what they have found. How is this possible?
Answer: The couple is scuba diving on a sunken cruise ship. The occupants in the ballroom had drowned when the ship sank suddenly in a terrible disaster at sea.
A carrot, a scarf and five pieces of coal are found lying on your neighbor’s lawn. Nobody put them on the lawn, but there is a simple, logical reason why they are there. What is it?
Answer: The items were used by children to build a snowman that has now melted.
An elderly woman lives on the 35th floor and hates taking the stairs. Every day, she takes the elevator down to the lobby floor to go to work. When she comes home from work, she takes the elevator to the 25th floor and walks the rest of the way up, except on days when it rains. Those days, she takes the elevator all the way home. Why does she walk the last ten flights of stairs if she hates it so much?
Answer: The elderly woman is too petite to reach the button for the 35th floor. She can only reach the 25th-floor button. On days when it rains, she uses her umbrella to hit the button for the 35th floor.
You are driving a school bus. The bus is empty when you begin your route. At the first stop, 4 people get on. At the second stop, 8 people get on and 2 get off. At the third stop, 3 people get off and 4 get on. What color are the bus driver’s eyes?
Answer: The color of your eyes.
A man walks up to a woman, who is standing behind a counter. After handing her a book, the woman looks at it and says, “That will be $4.00.” He promptly pays her and walks out without it. Why?
Answer:He was returning an overdue library book.
Related: 25 Logic Puzzles That Will Totally Blow Your Mind, But Also Prove You’re Kind of a Genius
Source: Canva/Parade
A baby falls out of a window of a 20-story building, yet survives. How?
Answer: The window was on the ground floor.
A man pushes his car until he reaches a hotel. When he arrives, he goes bankrupt. What happened?
Answer:He’s playing Monopoly, and his piece is the car. He lands on a space with a hotel and doesn’t have the money to pay the fee.
Rachael and Monica have moved into a shared house with several other occupants. Monica doesn’t like their new accommodation, though, because of the angry and aggressive cat that greeted them when they first walked in. Rachael realizes that there isn’t really a problem. Why?
Answer: There is a mirror near the entrance. Monica is a cat, saw her reflection, and thought that it was another cat challenging her for territory. Rachael can cover up the mirror until Monica becomes settled, but Monica will figure it out eventually even if Rachael does nothing.
A family has a chicken coop for their one dozen hens. Late one night, a terrible storm came through their area and killed all but eight chickens. How many chickens did the family have the next morning?
Answer: The family still had 12 hens: four dead and eight alive.
A woman with her barking dog enters a room and presses a button. Within seconds, she instantly loses 20 pounds, and the dog stops barking. How did she lose the weight?
Answer: The room is actually an elevator. When she gets inside to go down to her room, the elevator accelerates downward, making her weight temporarily lower.
Related: Give These 110 Totally Twisted Tongue Twisters a Try To Test Your Tongue’s Talents
Source: Canva/Parade
The sound of Brian’s snoring is usually just annoying, but today it causes terror and panic. Why?
Answer: Brian is a bus driver.
A woman who is in a department store fills her basket to the top and leaves the store without paying. Although she is seen, no one calls the police or attempts to stop her. Why?
Answer: The woman works there. She is filling the basket with trash and taking it out to the dumpster.
A man is sentenced to death. He has to choose from three rooms to receive his punishment. The first room has a firing squad with guns loaded. The second room has a blazing fire. The third room is full of tigers that haven’t eaten for six months. Which room should he choose?
Answer: The room of tigers. If the tigers haven’t eaten in six months, they’re dead.
A man opened a door, screamed and then was found dead a few minutes later. No gunshots were heard in the area. What happened to him?
Answer: He was on an airplane and opened the door while the plane was in flight, falling to his death.
While walking across a large open field, you come across a woman who has died very suddenly, unexpectedly and only a very short time ago. You look around, but there is no one and nothing except for an unopened package next to the woman. How did she die?
Answer: The unopened package was the woman’s parachute.
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16.Why is it better to have round manhole covers than square ones?
Answer: A square manhole cover can be turned and dropped down the diagonal of the manhole. A round manhole cover cannot be dropped down the manhole. So, for safety reasons, all manhole covers should be round
Jason’s dad couldn’t keep his son from playing video games. To keep Jason from playing video games all the time, the dad grabbed a hammer and solved the problem. Now he can play video games, but Jason cannot. What did the dad do?
Answer: The dad built a shelf out of Jason’s reach and put the video game console up there. The dad can still reach it to play, but Jason cannot.
A man and his wife raced through the streets. They stopped, and the husband got out of the car. When he returned, his wife was dead, and there was a stranger in the car. What happened?
Answer: The wife was about to have a baby. The husband drove them to the hospital. The husband left to get a wheelchair, but the baby was born in the meantime. The wife died in childbirth.
A woman enters a large metal tube. She is overcome with fear and clenches her boyfriend’s hand tightly. She is visibly shaken. Her boyfriend gently rubs her arms and speaks softly to her, but is unaffected himself. A few hours later, the woman’s boyfriend tells her that it’s time to go, and her torment ends. What was happening to the woman?
Answer: The woman is terrified of flying but must travel to see a dying relative, and this adds to her stress level. Her boyfriend does his best to comfort her during the few hours of the flight.
After a night of partying with her friends, a sorority girl arrives home and finds that she cannot enter her house. She’s sure that she’s at the right home, but she cannot get inside. What happened that makes entering her home impossible?
Answer: Her friends took her keys away from her the night before because she had too much to drink. They ordered her an Uber to get home and kept her keys, which meant she didn’t have her house key.
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21.One of Michael’s dearest loved ones binds him to a chair, but Michael doesn’t mind. Why?
Answer: Michael is a kid. The chair is a car seat. One of Michael’s parents put his seatbelt on for him.
At the bottom of a large hill, more than 50 cars are involved in an accident. Some of the cars are overturned, and others are resting on top of other cars. The pile-up is so large that a few military vehicles and a fire truck are also involved in the crash. What happened to cause this big accident?
Answer: The cars, trucks and other vehicles are toys being played with by a toddler.
Every two weeks, a woman sits down and writes two words on 60 sheets of paper. Why does she do this?
Answer: The woman runs her own small business with 60 employees. Every week, she signs her name on their paychecks.
The man finds himself commuting to work in his car every day. Upon arrival to work each morning, he proceeds to drive his car in a circle four times before finally parking it and entering his office building. Why does the man drive in circles every day?
Answer: The man works in a crowded part of downtown and is required to park on the fifth floor of a parking structure every day.
A woman is driving on a sunny day. She makes a turn, and water suddenly starts to pelt her car very hard for about five minutes. Then the car is buffeted by extremely hard-blowing air. The car even begins to shake. The blowing stops, and the woman makes another turn onto the main road and heads home safely. What happened?
Answer: The woman drove her car into a drive-through car wash. After about 10 minutes, she leaves and drives her clean car home.
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26.A man ran into a fire and lived. A man stayed where there was no fire and died. What caused this?
Answer: The two men were working in a small room protected by a carbon dioxide gas fire extinguisher system when a fire broke out in the next room. One of the men ran through the fire and escaped with only minor burns. The other one stayed in the room until the fire extinguishers kicked in and died of oxygen starvation.
A woman hears her name and is then taken away by two men. Later, the woman dies while under the care of many. What happened to this poor woman?
Answer: The woman had committed a murder and was determined to be guilty after a fair trial. A judge called her name, gave her the death penalty and then two bailiffs took her out of the courtroom. Years later, she was executed while under the care and supervision of a doctor and detention officers.
Three kids enter a room, but only two walk out. The room is empty. Where is the third kid?
Answer: The third kid uses a wheelchair, so they roll out of the room instead of walking out.
A woman died and went to Heaven. There were thousands of other people there. They were all naked and all looked young. He looked around to see if there was anyone he recognized. He saw a couple, and he knew immediately that they were Adam and Eve. How did he know it was them?
Answer: He recognized Adam and Eve as the only people without navels. Because they were not born from a woman, they had never had umbilical cords, and therefore, they never had a belly button.
Max’s Trucking Company has been delivering goods to and from local businesses for years. Max’s business is steady, and his trucks drive the same routes and roads every day without any trouble. This year, Max decided to start delivering and receiving goods on Christmas Day as a way to earn additional money. When Christmas Day finally came, Max sent a few drivers out to do business with those companies that chose to remain open. Later that day, Max received a call that one of his delivery drivers had struck a bridge and severely damaged the truck. How could this happen if the drivers have always used the same route without any previous problems?
Answer: On Christmas Day, many of the local businesses that contract with Max’s company had shut down for the day. Having made a final delivery, and not receiving any new goods to transport, this particular truck was empty. When returning to headquarters the empty truck had a higher profile and struck the bridge on the way back.
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A woman grabbed a guy’s ring, pulled on it and dropped it, thereby saving his life. What happened?
Answer: They were skydiving. He broke his arm as he jumped from the plane by hitting it on the plane door, and he couldn’t reach his ripcord with his other arm to pull out the parachute. She pulled the ripcord for him.
If you were alone in a dark room, with only one match and an oil lamp, a fireplace, and a candle to choose from, which would you light first?
Answer: The match.
33.On a hot Saturday afternoon, a woman was walking slowly through the savannah when she spotted a lion in the distance. Instead of turning around, hiding or seeking help, the woman began to run towards where the lion was. Why wasn’t she afraid?
Answer: The woman was visiting a zoo.
Fiona entered a restaurant where a crowd of people was enjoying a meal. When the people saw Fiona they dropped their forks and fled the restaurant. What happened?
Answer: Fiona was the name of a reticulated python who escaped from the local zoo and found herself entering a crowded restaurant.
In a large, completely empty, wooden barn is Old Man Jenkins, hanging dead from the middle of the central rafter. He is hanging three feet off the ground by a rope ten feet long. The closest wall is 20 feet away from the hanging man. It would not be possible to climb up the walls or along the rafters. Below him is a puddle of water. How did Old Man Jenkins hang himself?
Answer:Old Man Jenkins stood on a block of ice and hung when it melted.
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Mind-Bending Lateral Thinking Puzzles
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A girl celebrated her birthday. Two days later, her older twin celebrated his. How?
Answer: When she went into labor the mother of the twins was traveling by boat. The older twin brother was born first early on March 1st. The boat then crossed the time zone line and the younger twin girl was born on February the 28th. In a leap year, the younger twin celebrates her birthday two days before her older brother.
A man was wanted for burglarizing dozens of businesses over a period of almost a year. Surveillance footage from the businesses clearly showed this man’s face, and the local news station showed this footage to viewers on many occasions. The local police department even placed “wanted” posters around the community to help catch the burglar. However, when the man was spotted by two police officers familiar with the burglaries, he was not arrested. What happened?
Answer: The man was currently in jail, and the two police officers came to get a statement from him for the upcoming trial.
A woman has indisputable proof in court that her husband was murdered by her sister. The judge declares, “This is the strangest case I’ve ever seen. Though it’s a clear case, this woman cannot be punished.” How can this be?
Answer: The sisters are Siamese twins.
One sunny day, you’re cruising on your boat in the middle of the Pacific Ocean. In the middle of the ocean, you spot a yacht and see several corpses floating in the water nearby. What happened?
Answer: A group of people was vacationing on a yacht. One day, they decided to go swimming, put on their swimsuits, and dove off the side. They discovered belatedly that they had forgotten to put a ladder down the side of the yacht and were unable to climb back in, so they drowned.
Several voices were heard coming from the other room. There was a lot of commotion, but when the room was checked, only one person was inside. What’s going on?
Answer: The other voices were coming from a TV.
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There are a dozen eggs in a carton. Twelve people each take a single egg, but there is one egg left in the carton. How?
Answer: The 12th person takes the egg and the carton, leaving the egg inside.
Two men ordered dinner in a restaurant. They both ordered the same items from the menu. After they both tasted it, one of the men went outside the restaurant and shot himself. Why did he do so?
Answer:The food the two men ordered was swordfish. Many years earlier, they had both been stranded on a desert island after a plane crash. When one of the men tasted the swordfish, he realized he had never tasted it before. This meant that the meat he had been given on the island was actually not swordfish, as he had been told. He realized that he had eaten the flesh of his son, who died when the accident happened. The thought drove him to insanity, and he killed himself.
A body is discovered in a park in New York in the middle of summer. It has a fractured skull and many other broken bones, but the cause of death was hypothermia. What happened?
Answer:The body was a poor refugee from somewhere in the Middle East who desperately wants to come to the United States. Lacking money for a ticket, he stowed away in the landing gear compartment of an airplane jet. He dies of hypothermia in mid-flight and falls out when the compartment opens as the plane makes its final approach.
One day, a woman received a parcel in the mail. She found a carefully packed human arm inside. She examined it, repacked it and sent it on to another woman. The second woman also carefully examined the arm and then took it to the woods and buried it. Why did they both do these things?
Answer: Years earlier, three women had been stranded on a desert island. When they became desperate for food, they agreed to amputate their left arms in order to eat them. One of them, a doctor, cut off the arms of the other two women. Because they swore an oath to all have their left arm cut off after they were rescued, the doctor had her arm amputated and sent to the other two women.
A hunter aimed his gun carefully and fired. Seconds later, he realized his mistake. Minutes later, he was dead. What happened?
Answer: It was winter. He fired the gun near a snowy cliff, which started an avalanche.
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The music stops and a woman dies. What’s going on?
Answer: The woman performs as a tightrope walker in a circus. Her act consists of walking the rope blindfolded, accompanied by music, without a net. The conductor is supposed to stop the music when she reaches the end of the rope, signaling that it’s safe to step off onto the platform. That day, the usual conductor came down with the flu. The substitute stopped the music early.
A man is lying dead, face down in the desert. There’s a match near his outstretched hand. What happened?
Answer: He was with several others in a hot air balloon, crossing the Sahara desert. The balloon was punctured and they began to lose altitude. They tossed all their non-essentials overboard and then their clothing and food, but they were still sinking too fast. They drew matches to see who would jump over the side and save the others. This man lost.
Hans and Fritz are German spies during World War II. They try to enter America, posing as returning tourists. Hans is immediately arrested. What happened to give him away?
Answer: Crossing the border, Hans and Fritz were required to fill out a personal information form, which asked about their birthdays. The German date ordering is day/month/year, rather than the American way, month/day/year. Fritz was born on July 7th, so he wrote down 7/7/15—no problem. Hans was born on, say, July 20th, so he wrote down 20/7/15 instead of the American way, 7/20/15. Since Hans had claimed to be a returning American, he was found out by the border patrol.
A woman had two sons. They were born at the same hour on the same day of the same month in the same year. However, they were not twins. How could this be?
Answer: Her two sons were part of a set of triplets (or quadruplets!).
50.A man walks into a bar and asks the bartender for a glass of water. The bartender pulls out a gun and points it at the man. The man says, “Thank you” and walks out. What happened?
Answer: The man had hiccups and the bartender’s gun scared them out of him, to which he said, “Thank you.”
51.A man is sitting in a dark room, having a great time. Suddenly, he stops breathing and can’t speak. In a matter of moments, his breathing and ability to speak come back to him. What happened?
Answer: The man is in a movie theater enjoying his new favorite movie. He takes a drink of his favorite soda and begins to choke on an ice cube. This prevents him from breathing or speaking momentarily. Soon, the ice melts and the man is able to breathe and speak again.
A two-year-old named Leia goes missing, and her parents are absolutely devastated by the news. In their search to find her, they begin putting up flyers, asking for her safe return. However, the local police and FBI refuse to help. Why?
Answer: Leia is a dog.
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Coprime
Coprime (or Relatively Prime or Mutually Prime) numbers
have no common factors other than 1
And no fractions! Just integers, and positive factors.
Example: 21 and 22 are coprime
The positive factors of 21 are 1, 3, 7 and 21
The positive factors of 22 are 1, 2, 11 and 22
The only common factor is 1, so they are coprime.
Example: 21 and 24 are NOT coprime
The factors of 21 are 1, 3, 7 and 21
The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24
The common factors are 1 AND 3, so they are NOT coprime.
With no common factors other than 1 their greatest common factor is also 1
And 1 is not prime, so coprimes don't share any prime factors either.
Coprime Table 1 to 12
Consecutive (following each other) numbers like 3,4 or 126,127 are always coprime: imagine the first number n has a prime factor p, then n+1 cannot have the factor p as n+1 divided by p will have a remainder of 1.
Useful for Fractions
35
When we simplify a fraction as much as possible, then the top and bottom numbers (the numerator and denominator) are coprime.
610 = 35
If the top and bottom numbers of a fraction are not coprime (like 6 and 10 that share the factor 2), then we can simplify the fraction.
Useful in Cryptography
Coprime numbers are needed in the field of cryptography, such as for RSA Cryptography that helps ensure secure communication and data protection.
𝄞 Music
In music, when two notes have coprime ratios such as 1:2, 2:3, 3:4, etc, they create a harmony that is pleasing to the ear.
Example: the ratios between the pitches of the scale (C, D, E, F, G, A, B, C) all have coprime ratios.
By using coprime intervals within a chord or melody composers can achieve a sense of complexity and interest in the music.
And just like musical harmony, coprimes have a special relationship that makes them mathematically "harmonious".
Testing
We can test if two numbers are coprime by using the Euclidean Algorithm.
Grid
When two numbers are coprime (like 7 and 5) they can "see" each other through the grid (no other grid points get in the way):
Mathopolis:Q1)Q2)Q3)Q4)Q5)Q6)Q7)Q8)Q9)Q10)
Coprime CalculatorPrime Factors of a NumberAll Factors of a NumberIntroduction to Number TheoryNumbers Index
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Second-Generation Antipsychotics and Extrapyramidal Adverse Effects
Nevena Divac
Milica Prostran
Igor Jakovcevski
Natasa Cerovac
Nevena Divac: ndivac@med.bg.ac.rs
Academic Editor: Nevena Radonjic
Received 2013 Dec 30; Accepted 2014 May 6; Issue date 2014.
This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
Antipsychotic-induced extrapyramidal adverse effects are well recognized in the context of first-generation antipsychotic drugs. However, the introduction of second-generation antipsychotics, with atypical mechanism of action, especially lower dopamine receptors affinity, was met with great expectations among clinicians regarding their potentially lower propensity to cause extrapyramidal syndrome. This review gives a brief summary of the recent literature relevant to second-generation antipsychotics and extrapyramidal syndrome. Numerous studies have examined the incidence and severity of extrapyramidal syndrome with first- and second-generation antipsychotics. The majority of these studies clearly indicate that extrapyramidal syndrome does occur with second-generation agents, though in lower rates in comparison with first generation. Risk factors are the choice of a particular second-generation agent (with clozapine carrying the lowest risk and risperidone the highest), high doses, history of previous extrapyramidal symptoms, and comorbidity. Also, in comparative studies, the choice of a first-generation comparator significantly influences the results. Extrapyramidal syndrome remains clinically important even in the era of second-generation antipsychotics. The incidence and severity of extrapyramidal syndrome differ amongst these antipsychotics, but the fact is that these drugs have not lived up to the expectation regarding their tolerability.
Background
Antipsychotic drugs are the cornerstone of the pharmacological treatment of schizophrenia. The introduction of the first antipsychotic chlorpromazine in 1952 marked the new era in psychopharmacology . However, those early antipsychotics, now referred to as first-generation antipsychotics (FGAs), such as chlorpromazine, haloperidol, or fluphenazine, though effective in relieving positive symptoms of the disease, have some serious limitations. Lack of efficacy regarding negative symptoms and the adverse effects, especially extrapyramidal symptoms (EPS), are serious drawbacks of these drugs. The development of newer antipsychotics (risperidone, olanzapine, quetiapine, etc.) since 1990s was met with great expectations. These novel antipsychotics, now referred to as second-generation antipsychotics (SGAs), have been modeled on the prototype drug clozapine .
Clozapine was the first antipsychotic that proved to be efficacious in treatment-refractory schizophrenia , but it was also the first antipsychotic devoid of EPS. However, the ability of clozapine to cause agranulocytosis as a serious adverse effect led to voluntary withdrawal of the drug by the manufacturer, with subsequent reintroduction in 1989, followed by strict regulation regarding indications and white blood cells count followup . The efficacy of clozapine and its inability to produce EPS were motives for the development of similar antipsychotics, but with the safer profile. Second-generation antipsychotics such as olanzapine, risperidone, quetiapine, and more recently ziprasidone and aripiprazole soon became the mainstay of the treatment of schizophrenia, despite their higher costs and inconsistency of the data showing their superior efficacy versus FGAs [5, 6].
Clozapine, as the first SGA, actually discredited the theory that EPS are an unavoidable accompaniment of antipsychotic efficacy. Previously, EPS were considered as an essential component of antipsychotic “neuroleptic” effect. The association of antidopaminergic (D2) potency, antipsychotic effect, and EPS (due to loss of dopamine in the extrapyramidal part of the central nervous system) was the foundation for the dopamine hypothesis of schizophrenia [7, 8]. The ability of a substance to induce EPS experimentally was considered as proof of its antipsychotic potential. However, dopamine hypothesis of schizophrenia became obsolete with the introduction of clozapine and other SGAs.
All antipsychotic agents have some degree of antagonistic affinity for dopaminergic D2 receptors. It was shown that first-generation antipsychotics, though known to block other receptors, not only exert their antipsychotic, but also their extrapyramidal effects, primarily by binding to D2 receptors in the central nervous system. First-generation antipsychotics produce their therapeutic (antipsychotic) effect at 60–80% of D2 occupancy, while the 75–80% of D2 receptor occupancy leads to the acute EPS [9–11]. Therefore, the overlap between desired and adverse D2 receptor occupancy is mostly unavoidable with FGAs. On the other hand, the therapeutic effects of SGAs are attributable also to some degree to D2 antagonism, but more to blockade of certain serotonin (mostly 5HT2A) receptors. Surprisingly, clozapine, as the most effective antipsychotic so far, has the lowest D2 affinity (Table 1). It was also suggested and shown in animal models that SGAs actually bind to and dissociate from D2 receptors in an atypical manner (Kapur, 2001). Loose binding to and fast dissociation of SGAs from D2 receptors may be the cause of their lower EPS propensity . The affinity of antipsychotic drugs for D2 receptors is shown in Table 1. While the antipsychotic effect of FGAs correlates with D2 affinity, that is not the case with SGAs.
Table 1.
First- and second-generation antipsychotics and D2 antagonism.
Antagonistic D2 effect | First-generation antipsychotics | Second-generation antipsychotics
Low | Chlorpromazine LevomepromazineThioridazine | ClozapineQuetiapine
Intermediate | TrifluoperazinePerphenazine | Olanzapine
High | HaloperidolFluphenazineFlupentixol | RisperidoneZiprasidoneAripiprazole (possible D2 agonism)
The efficacy of a pharmacological treatment cannot be interpreted independently from its adverse effects profile. Better tolerability of SGAs was considered as one of their major advantages as a class . The idea of treating schizophrenia without producing EPS was very attractive for psychiatric care professionals, as well as for the patients. However, post-clozapine SGAs have not fully lived up to these expectations and intolerability due to the fact that EPS remain a considerable problem in the treatment of schizophrenia [7, 13]. It is now evident that all SGAs, apart from clozapine, have propensity to cause certain degree of EPS. The results of recent clinical trials and meta-analyses have shown that there is no advantage of SGAs regarding tolerability and effectiveness compared with FGAs [13, 14]. Also, postmarketing followup of SGAs surfaced other adverse effects such as weight gain and metabolic side effects. However, notable metabolic side effects are also caused by FGAs and the higher cardiometabolic risk of SGAs versus FGAs has not been confirmed . Therefore, the oversimplified distinction of antipsychotic drugs classes, in which FGAs are responsible for EPS and SGAs for metabolic side effects, though ingrained in clinical practice, is actually not supported by recent findings [1, 16].
This review summarizes the recent reported results regarding the risk of EPS development in patients treated with different classes of antipsychotic drugs.
Extrapyramidal Symptoms
EPS include acute dystonias, akathisia, Parkinsonism, and tardive dyskinesia (TD). EPS are serious, sometimes debilitating and stigmatizing adverse effects, and require additional pharmacotherapy. EPS develop into two phases. Early, acute EPS most often develop upon the beginning of treatment with antipsychotics or when the dose is increased. The later-onset EPS usually occur after prolonged treatment and present as tardive dyskinesia (TD). The motor manifestations include akathisia (restlessness and pacing), acute dystonia (sustained abnormal postures and muscle spasms, especially of the head or neck), and Parkinsonism (tremor, skeletal muscle rigidity, and/or bradykinesia) [13, 17]. TD is characterized by involuntary, repetitive facial movements such as grimacing, tongue protruding, oculogyric crisis, and lips puckering, as well as torso and limb movements. Acute EPS are one of the main causes of poor adherence to antipsychotic treatment due to the reversibility of symptoms, while late-onset TD has the most serious impact on patients and caregivers with respect to quality of life [18, 19]. TD may persist after the discontinuation of treatment or even be irreversible. It is estimated that approximately 50% of patients treated with high-potency FGAs (such as haloperidol) develop acute EPS within the first several days of treatment. The prevalence of TD is somewhat less known due to differences in design and methodologies among studies that have investigated this problem [13, 20, 21]. Prevalence of TD has been reported to be 0.5% to 70% of patients receiving FGAs, with the average rate being between 24% and 30% [22, 23].
Acute EPS usually respond to dose reduction of the antipsychotic agent or require additional pharmacological treatment.
Acute dystonia occurs within first few days after the initiation of the antipsychotic treatment and can be effectively prevented or reversed with anticholinergic drugs such as biperiden [24–26]. Risk factors for acute dystonia are young age and male gender, history of substance abuse, and family history of dystonia [27, 28]. Acute dystonia is common with FGAs such as haloperidol and less common with SGAs. It is reported that approximately 7.2% treated with long-acting parenteral risperidone develop acute dystonic reactions . Also, case reports regarding acute dystonia after initiation of antipsychotic treatment with aripiprazole and ziprasidone have been published [31, 32].
Akathisia is very common (about one half of all cases of EPS), poorly understood, and difficult to treat. It occurs mostly within the first three months of treatment. Akathisia does not respond to anticholinergic medication, but antipsychotic dose reduction, liposoluble beta adrenergic blockers, and benzodiazepines have proved effective [24, 25]. The rough estimation is that about 25% of patients treated with FGAs develop akathisia, but it is also common with SGAs. Some researchers suggest that akathisia rates do not differ between FGAs and SGAs . It was previously suggested that SGAs clozapine and quetiapine carry the lowest risk for akathisia, yet it was not confirmed in some blinded reviews . Also, the Clinical Antipsychotic Trials of Intervention Effectiveness (CATIE) study as a randomized, partially open-label study in which efficacy and side effects of multiple SGAs with an FGA perphenazine showed that akathisia remains a problem with SGAs, though at lower rates compared to FGAs [24, 34]. Based on CATIE study, it appeared that risperidone and perphenazine, for example, both cause akathisia in 7% of patients. Further analysis of the CATIE study data revealed no difference between any of the antipsychotics tested in this study regarding incidence of akathisia and other EPS in patients with chronic schizophrenia during maintenance of antipsychotic treatment for up to 18 months . However, the well-known limitations of the CATIE (the choice of an intermediate-potency FGA perphenazine, the nonrandomized allocation of patients with the tardive dyskinesia to a SGA treatment) should be considered when interpreting these results.
Parkinsonism induced by antipsychotics occurs between few days and up to several months after the initiation of the treatment. Risk factors for this type of Parkinsonism are age (elderly), gender (females), cognitive deficit, and early onset EPS . Antipsychotic-induced Parkinsonism is considered a reversible condition although its duration is variable. The treatment of choice is not established, but dose reduction and anticholinergic drugs may be useful. However, anticholinergics should be avoided in the elderly patients due to their side effects such as cognitive deterioration, urinary retention, dry mouth, and risk of glaucoma exacerbation. Although switching to SGAs is often recommended in cases of Parkinsonism, the rates of Parkinsonism induced by SGAs (e.g., 26% with olanzapine) are lower than those with the FGAs (55% with haloperidol), but not negligible . Other evidence shows virtually no advantages of SGAs compared to FGAs in relation to Parkinsonism as an adverse effect, especially when the potency and dose are considered. It was shown that high doses of SGAs (such as olanzapine, risperidone, or quetiapine) caused Parkinsonism in high doses at a similar rate as low-potency FGA (chlorpromazine), but the risk was 50% higher in high-potency FGA group . In CATIE study, the results regarding Parkinsonism were also conflicting. CATIE study includes patients with previous tardive dyskinesia, who at baseline were excluded from perphenazine branch. This could lead to potential bias, meaning that patients with previous vulnerability to EPS were allocated exclusively to SGA branch. In order to avoid this potential bias, only patients without previous TD were included in comparisons for Parkinsonism. The proportion of patients showing no evidence of Parkinsonism at baseline who met at least one of the three criteria for Parkinsonism during the subsequent follow-up period revealed no substantial differences between treatment groups. At the 12-month followup, covariate-adjusted rates of Parkinsonism were 37%–44% for SGAs and 37% for perphenazine . However, the choice of an intermediate-potency FGA (perphenazine) as a comparator in modest doses in CATIE could probably be responsible for the lack of significant difference between FGAs and SGAs regarding incidence of Parkinsonism. The Cost Utility of the Latest Antipsychotics in Schizophrenia Study Band 1 (CUtLASS-1) as a randomized controlled trial (RCT) that tested the hypothesis that the clinical and cost-effectiveness of SGAs is superior in individuals whose antipsychotic treatment is changed due to insufficient efficacy or side-effects of previous treatment. This study also did not show statistically significant difference between the treatment groups in terms of Parkinsonism between SGA and FGA patients between SGA and FGA patients. The results were similar regarding akathisia. As in CATIE study, the main limitation of this study is the choice of FGA comparator. Haloperidol as the high-potency FGA was a rare choice at baseline, while sulpiride was the most common. Sulpiride is considered as an FGA with atypical properties and its low propensity for EPS is well established .
Tardive dyskinesia occurs after months or years of antipsychotic therapy. The risk of TD development is highest in the first five years of treatment with FGAs . Leading risk factors for TD are increased age, non-Caucasian race, female gender, a history of diabetes, organic brain damage, and the presence of negative symptoms of schizophrenia . TD can also occur spontaneously in patients diagnosed with schizophrenia at the rate of 0.5% per year . Management of TD is different than the management of acute EPS. Anticholinergic drugs are not recommended (actually, these drugs have been shown to exacerbate TD). The primary step is, according to guidelines, switching from the causative agent to an SGA followed by, if necessary, additional pharmacological treatment. An empirical treatment algorithm from Margolese et al. suggests tapering of anticholinergic drugs, switching to an SGA and, if necessary addition of tetrabenazine. Finally adding experimental therapy including donepezil/melatonin/vitamin E/vitamin B6/branched-chain amino acids (BCAAs) should be considered if previous steps do not provide relief . Clozapine is considered the safest, even beneficial, SGA regarding TD due its ability to improve involuntary symptoms . A recent prospective cohort study on TD incidence amongst outpatients on antipsychotic maintenance therapy showed some disappointing results regarding SGAs and TD incidence. While most of the previously conducted studies showed that the risk of TD with SGAs is one-quarter that of FGAs, the results of this study suggest that the risk with SGAs is more than half that of FGAs (excluding clozapine patients) or more than two-thirds of the risk (including clozapine patients) . The finding of surprisingly high rate of TD among clozapine patients in this study was attributed to certain confounding factors, such as confounding by indication (prescribing of clozapine to patients with TD or at-risk for TD), and should be interpreted with caution. In CATIE study, patients with TD were excluded from being randomized to perphenazine treatment. There were no statistically significant differences in the rate of new onset TD across the group of antipsychotic drugs. The rates ranged from 13% (quetiapine) to 17% (perphenazine) . Since patients in the FGA (perphenazine) group were free from previous TD, CATIE study does not enable true comparison between FGAs and SGAs regarding TD, but it offers some valuable insight into predisposing factors for TD registered as baseline. These factors are older age, previous exposure to FGA and anticholinergic medication, previous longer antipsychotic treatment, and acute EPS [13, 24]. The CUtLASS-1 study showed unexpectedly the increase of TD incidence in the SGA group of patients during the 12th week of treatment, but this was probably due to switch of treatment (withdrawal of D2 blocking drug and the initiation of an SGA with more anticholinergic effects). This difference in the TD incidence was diminished by 52nd week of the followup .
Recent studies on the propensity of FGAs and SGAs to cause EPS yielded conflicting results [35, 37, 39, 45]. When interpreting these studies, it is of utmost importance to consider methodological issues and limitations, some of which are doses of antipsychotics, choice of an FGA comparator, duration of the study, inclusion and exclusion criteria, baseline patients' characteristics, and sensitivity of the criteria for EPS.
EPS remain the most serious problem among patients affected with schizophrenia, even in the era of new antipsychotics with less affinity towards D2 receptors. Upon the introduction of second-generation antipsychotics, these agents were defined as atypical based on their mechanism of action. Atypical antipsychotics expressed less affinity for striatal D2 receptors than typical, FGAs, and different levels of 5-HT2A antagonism, alpha-1 antagonism, or cholinergic antagonism. However, all SGAs still affect D2 receptors to some degree, with clozapine having the least affinity [7, 46] and therefore have some nonnegligible EPS liabilities.
Conclusion
SGAs have not completely fulfilled the expectation of being EPS-free antipsychotic drugs. Though recommended by current guidelines as the first-line therapy in the treatment of schizophrenia , the superiority of these drugs in terms of better efficacy and tolerability is not clear. Recent studies showed that SGAs do not significantly differ from FGAs in terms of efficacy (with the exception of clozapine for treatment-resistant patients) and have in general lower liability to cause EPS than FGAs, but with great variations within the class .
The likelihood of causing EPS with an SGA exists and depends on many factors. The patient's characteristics (age, gender, and concomitant conditions), history of the disease, previous treatment, the choice of a particular antipsychotic, its dose, and duration of treatment and adjuvant therapy should be taken into consideration in the order to minimize the risk of EPS and provide the best quality of care. At this moment, the trial-and-error approach is recommended, since the therapeutic outcome and adverse effects are not easily predictable. Hopefully, the recent, promising advances in pharmacogenomics and neurobiology could provide predictive markers of antipsychotic response and adverse effects and lead towards personalized therapy .
Acknowledgment
This work was supported by Ministry of Education, Science and Technological Development of Serbia (Grant no. 175023).
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Authors' Contribution
All authors have read and approved the final paper.
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867 | https://www.chilimath.com/lessons/introductory-algebra/order-of-operations-practice-problems/ | Skip to content
Order of Operations Exercises
Order of Operations Practice Problems with Answers
There are nine (9) problems below that can help you practice your skills in applying the order of operations to simplify numerical expressions. The exercises have varying levels of difficulty which are designed to challenge you to be more extra careful in every step while you apply the rules of the Order of Operations. Good luck!
Part 1: Order of Operations problems involving addition, subtraction, multiplication, and division
Problem 1: Simplify the numerical expression below.
Answer
Problem 2: Simplify the numerical expression below.
Answer
Problem 3: Simplify the numerical expression below.
Answer
Part 2: Order of Operations problems involving the four arithmetic operations and parenthesis (or nested grouping symbols)
Problem 4: Simplify the numerical expression below.
Answer
Problem 5: Simplify the numerical expression below.
Answer
Problem 6: Simplify the numerical expression below.
Answer
Part 3: Order of Operations problems involving the four arithmetic operations, parentheses and exponents
Problem 7: Simplify using the Order of Operations.
Answer
Problem 8: Simplify using the Order of Operations.
Answer
Problem 9: Simplify using the Order of Operations.
Answer
Take a Quiz
Order of Operations Quiz
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Order of Operations
PEMDAS Rule
Tags: Introductory Algebra, Lessons |
868 | https://math.stackexchange.com/questions/1676331/apply-the-maximum-and-minimum-modulus-theorems-why-how | complex analysis - Apply the maximum and minimum modulus theorems - Why/How? - Mathematics Stack Exchange
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Apply the maximum and minimum modulus theorems - Why/How?
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Here is the question I have to answer for context, please don't solve the whole problem for me!
Suppose |a k|<R|a k|<R for k=1,2,…,n k=1,2,…,n. Prove that (unless |a k|=0|a k|=0 for all k k), |z−a 1|⋅|z−a 2|⋯|z−a n|−−−−−−−−−−−−−−−−−−−−−−√n|z−a 1|⋅|z−a 2|⋯|z−a n|n assumes a maximum value greater than R R, and a minimum value less than R R, at some points z z on |z|=R|z|=R. Hint: Apply the maximum and minimum modulus theorems to ∏n k=1(R 2−a k¯¯¯¯¯z)∏k=1 n(R 2−a k¯z).
So the maximum for |z|≤R|z|≤R is where |z|=R|z|=R, so we set z=R e i θ z=R e i θ and rewrite the "Hint" function as ∏R|e i θ−R a k¯¯¯¯¯|∏R|e i θ−R a k¯|, but I dont see how we can show the maximum is always greater than R! Surely its possible that |e i θ−R a k¯¯¯¯¯|=R−1|e i θ−R a k¯|=R−1 or smaller!
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edited Feb 29, 2016 at 0:01
HunterHunter
asked Feb 28, 2016 at 22:31
HunterHunter
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the maximum modulus theorem states that if f(z)f(z) is holomorphic on a closed path-connected subset U U then its maximum (modulus) lies on the boundary of U U. the minimum modulus theorem is the same applied to 1 f(z)1 f(z) if f(z)≠0 f(z)≠0 on U U (hint : U U can have weird shapes)reuns –reuns 2016-02-28 22:43:57 +00:00 Commented Feb 28, 2016 at 22:43
I simply dont understand the use of this. Our functions are holomorphic everywhere from what I can tell, so if you consider it with |z|≤R|z|≤R, then the max must be where |z|=R|z|=R, but I don't see it helpful to show that the maximum value is always greater than R. I've considered the "Hint" function with z=R e i θ z=R e i θ, but its not clear to me then that you cant select some {a k}{a k} such that the maximum is less than R.Hunter –Hunter 2016-02-28 22:47:56 +00:00 Commented Feb 28, 2016 at 22:47
try and see what you get reuns –reuns 2016-02-28 22:52:05 +00:00 Commented Feb 28, 2016 at 22:52
Try selecting {a k}{a k}? With z=R e i θ z=R e i θ you get ∏R(R−a k¯¯¯¯¯e i θ)=R n∏(R−a k¯¯¯¯¯e i θ)∏R(R−a k¯e i θ)=R n∏(R−a k¯e i θ), but I'm not convinced you cant select some {a k}{a k} such that |∏(R−a k¯¯¯¯¯e i θ)|<R−n+1|∏(R−a k¯e i θ)|<R−n+1 Hunter –Hunter 2016-02-28 22:54:31 +00:00 Commented Feb 28, 2016 at 22:54
@user1952009 Is this the wrong approach?Hunter –Hunter 2016-02-28 23:03:31 +00:00 Commented Feb 28, 2016 at 23:03
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I think the answer you are looking for is here: Show that |z−a 1|⋯|z−a k|−−−−−−−−−−−−−−−√k|z−a 1|⋯|z−a k|k has a max greater than R R, and a min less than R R
You don't actually need to convert z=e θ i z=e θ i form.
Let f(z)=|z−a 1|⋅|z−a 2|⋯|z−a n|f(z)=|z−a 1|⋅|z−a 2|⋯|z−a n|. You want to show that there exists a point in |z|≤R|z|≤R where f(z)=R f(z)=R. For instance, f(0)f(0). If you plug z=0 z=0 into ∏n k=1(R 2−a k¯¯¯¯¯z)∏k=1 n(R 2−a k¯z) since they are equivalent expressions, you will find that |R n|f(0)=∏n k=1 R 2=R 2 n⟹f(0)=R n|R n|f(0)=∏k=1 n R 2=R 2 n⟹f(0)=R n. So the function you care about is g(z)=f(z)1 n g(z)=f(z)1 n, which means that g(0)=R g(0)=R. z=0 z=0, however is not the boundary. So by Min/Max Mod Theorem, there must be a minimum or a maximum on the boundary such that it is less than/greater than R R and we are done.
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answered Nov 1, 2018 at 7:43
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3Show that |z−a 1|⋯|z−a k|−−−−−−−−−−−−−−−√k|z−a 1|⋯|z−a k|k has a max greater than R R, and a min less than R R
Related
5Equality of modulus at the boundary of a bounded domain
3Show that |z−a 1|⋯|z−a k|−−−−−−−−−−−−−−−√k|z−a 1|⋯|z−a k|k has a max greater than R R, and a min less than R R
1Maximum modulus principle exercise.
3Maximum /Minimum Modulus theorem for Harmonic Function ( Corollary 6.16 )
1Show that ∣∣∣R(z−a)R 2−a¯¯¯z∣∣∣|R(z−a)R 2−a¯z| is analytic for |z|≤R|z|≤R, and maps the circle |z|=R|z|=R into the unit circle.
0how to find the maximum number
1Looking for the maximum and minimum of a modulus in the unit disk.
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869 | https://pmc.ncbi.nlm.nih.gov/articles/PMC2606773/ | The evolution of photosynthesis…again? - PMC
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The evolution of photosynthesis…again?
Lynn J Rothschild
Lynn J Rothschild
1 NASA Ames Research Center, Mail Stop 239-20, Moffett Field, CA 94035-1000, USA
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1 NASA Ames Research Center, Mail Stop 239-20, Moffett Field, CA 94035-1000, USA
Author for correspondence (lynn.j.rothschild@nasa.gov)
Issue date 2008 Aug 27.
© 2008 The Royal Society
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PMCID: PMC2606773 PMID: 18487134
Abstract
‘Replaying the tape’ is an intriguing ‘would it happen again?’ exercise. With respect to broad evolutionary innovations, such as photosynthesis, the answers are central to our search for life elsewhere. Photosynthesis permits a large planetary biomass on Earth. Specifically, oxygenic photosynthesis has allowed an oxygenated atmosphere and the evolution of large metabolically demanding creatures, including ourselves. There are at least six prerequisites for the evolution of biological carbon fixation: a carbon-based life form; the presence of inorganic carbon; the availability of reductants; the presence of light; a light-harvesting mechanism to convert the light energy into chemical energy; and carboxylating enzymes. All were present on the early Earth. To provide the evolutionary pressure, organic carbon must be a scarce resource in contrast to inorganic carbon. The probability of evolving a carboxylase is approached by creating an inventory of carbon-fixation enzymes and comparing them, leading to the conclusion that carbon fixation in general is basic to life and has arisen multiple times. Certainly, the evolutionary pressure to evolve new pathways for carbon fixation would have been present early in evolution. From knowledge about planetary systems and extraterrestrial chemistry, if organic carbon-based life occurs elsewhere, photosynthesis—although perhaps not oxygenic photosynthesis—would also have evolved.
Keywords: photosynthesis, carbon fixation, carboxylase, evolution, rubisco, enzyme evolution
1. Introduction
To an economist, solar power may seem like a boutique product next to oil, but to life on Earth it is by far the most important source of energy. Even fossil fuels are traced to solar energy. Today, an average of 342 W m−2 of solar radiation reaches our atmosphere and 198 W m−2 of that reaches the Earth's surface (Nielsen 2005). Thus, it is clear that for organisms inhabiting planets with a source of light such as their Sun, solar power is a functionally limitless source of energy.
While the first life was most probably heterotrophic (reviewed in Fenchel et al. 1998), early on, life on Earth evolved to exploit solar energy. Autotrophy—the biotic production of organic carbon from inorganic sources—allowed life to go beyond a reliance on abiotically produced organic carbon. Using solar radiation as the means to produce the energy for carbon fixation, i.e. photosynthesis, remains a very powerful combination. With the use of H 2 O as a reductant, the raw materials were available to support aerobic metabolism and energy-intensive species such as ourselves. Today, there is an estimated 600–1000 Gt of living biomass on Earth (Falkowski et al. 2000) supported largely by photosynthesis. Photosynthesis, and possibly oxygenic photosynthesis more specifically, appears critical for maintaining large planetary populations.
It is clear from our vantage point that photosynthesis, especially oxygenic photosynthesis, was the key to the history of life on Earth. But was that the result of contingency—an unpredictable and lucky accident (e.g. Gould 1989)—or, rather, a predictable outcome given the physical, chemical and biological conditions on the early Earth, as might be suggested by Conway Morris (1998)? This question is important to evolutionary biology because it asks whether the field has predictive power; in this case, could we have predicted the evolution of photosynthesis?
The question of whether life exists elsewhere is particularly intriguing, and is thus one of the three fundamental foci of astrobiology. Owing to its importance to life on Earth, the question of whether photosynthesis may—or is likely—to evolve again or elsewhere is fundamental. This question is the focus of this study.
To assess the possibility of photosynthesis arising again requires an inventory of the necessary prerequisites including the following: a carbon-based life form; the presence of inorganic carbon; the availability of reductants; the presence of light; a light-harvesting mechanism to convert the light energy into chemical energy; and carboxylating enzymes. Second, evolutionary pressures should be present, which provide a clear advantage to organisms with this function. Third, there should be, if not data, at least suggestive evidence. This comes in the form of convergence.
2. Autotrophy
Autotrophs are organisms that can derive some or all of their organic carbon needs from inorganic sources. Autotrophy is widespread among all three domains, including plants, algae, cyanobacteria and even a significant fraction of marine archaea (Ingalls et al. 2006). There are several major pathways for autotrophy (reviewed in Shively & Barton 1991), but the best known and most prevalent one involves the fixation of CO 2 by the enzyme ribulose-1,5-bisphosphate carboxylase/oxygenase (rubisco; EC 4.1.1.39; figure 1).
Figure 1.
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Fixation of carbon dioxide by ribulose-1,5-bisphosphate carboxylase/oxygenase (rubisco). This step is the first crucial step in the CBB cycle, which is shown in the abridged form.
(a) The origin of autotrophy
When did autotrophy arise? The ‘Great Oxidation Event’ provides a minimum age constraint for oxygenic photosynthesis. This Great Oxidation Event occurred ca 2.4 billion years (Ga), when atmospheric oxygen concentrations rose from less than 10−5 present atmospheric level (PAL) to more than 0.01 PAL or, possibly, more than 0.1 PAL. Geological and theoretical arguments suggest that oxygenic photosynthesis arose at least 300 Myr before the Great Oxidation Event (e.g. Goldblatt et al. 2006; Anbar et al. 2007), although others suggest a greatly compressed interval with cyanobacteria evolving only a million years or so earlier (Kopp et al. 2005). Suggestive of autotrophy arising even earlier, the Archaean fossil record contains cellularly preserved microfossils that are morphologically similar to extant autotrophic organisms, and some are oriented in a way that implies that they were photosynthetic (Walter 1983; Schopf 1992). Stromatolite evidence of erect filaments in shallow isolated basins with insignificant sulphate concentrations in the Tumbiana Formation, Australia, suggests that oxygenic photosynthesis evolved by 2700 Myr ago (Buick 1992). Similarly, the presence of the biological lipids 2α-methylhopanes, characteristic of cyanobacteria, in the Pilbara Craton, Australia, also indicates that oxygenic photosynthesis was well established by 2.7 Ga (Brocks et al. 1999). Filamentous fossils suggesting that either cyanobacteria or other bacteria date to 3.5 Ga (Walsh & Lowe 1985). In the 3.416 Ga Buck Reef Chert of South Africa, there are probably remains of photosynthetic bacteria that used H 2 as a reductant (Tice & Lowe 2006). Stable carbon isotope data suggest that carbon fixation evolved by 3.5–3.8 Ga (Schidlowski 1988), and, more recently, carbon isotope data suggesting production by form I rubisco in oxic environments by ca 2900 Myr ago (Nisbet et al. 2007). Rosing & Frei (2004) found organic carbon in a more than 3.7-Ga-old shale from Isua, West Greenland, with a carbon isotopic signature suggesting oxygenic photosynthesis coupled with the presence of oxidized ocean water.
Some have argued that autotrophy is the fundamental process around which life evolved, and thus dates from the origin of life. Wachtershauser (1990) proposed that the primordial metabolism was an autocatalytic cycle that can be derived from the reductive tricarboxylic acid (rTCA) cycle (figure 2) by replacing thioesters by thioacids and by assuming that the required reducing power was obtained from the oxidative formation of pyrite (FeS 2). This proposal is appealing because it circumvents the issue of a dilute solution of prebiotic precursors having sufficient concentration to organize into a life form. Smith & Morowitz (2004) pointed out that the rTCA cycle is a network-autocatalytic cycle, which provides the organic building blocks for the synthesis of all major classes of biomolecules. Furthermore, they proposed that the rTCA cycle is favoured relative to other redox reaction pathways under early Earth conditions and that this feature drove its emergence and also accounts for its evolutionary robustness and universality. Thus, subsequent layers of biochemical complexity were built around the rTCA cycle.
Figure 2.
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Reverse (reductive) TCA cycle. The rTCA cycle is a carbon dioxide fixation pathway found in autotrophic eubacteria and archaea. It is considered to be a primordial pathway for the production of starting organic molecules for the biosynthesis of sugars, lipids, amino acids, pyrimidines and pyrroles. The rTCA cycle is largely the oxidative, catabolic TCA cycle operating in reverse. Most of the enzymes of the TCA cycle work reversibly and could catalyse both directions. Only three steps are thought to be non-reversible, and thus determine the oxidative or reductive direction of the cycle: the conversion of citrate to oxaloacetate and acetyl-CoA; the conversion of fumarate to succinate; and the conversion of 2-ketoglutarate to isocitrate. The presence of such enzyme activities in autotrophically grown bacteria and archaea is considered indicative of the presence of the rTCA cycle in these organisms.
Persuasive though these proposals might be, only the heterotrophic theory of the origin of life has experimental support because it has been shown that the reduction of CO 2 under prebiotic conditions results in the synthesis of organic compounds (Lazcano & Miller 1999). Others (e.g. Orgel 2000) argue that the organization of the rTCA cycle on mineral surfaces makes unreasonable assumptions about the catalytic properties of minerals and the ability of minerals to organize sequences of dissimilar reactions. Clearly, the origin of life on Earth remains highly controversial.
In summary, the origin of oxygenic photosynthesis must have occurred prior to 2.4 Ga, and possibly as far back as the earliest fossil record at 3.7 Ga. Autotrophy has been proposed to be as old as life itself.
(b) Pathways for autotrophy
There are four main pathways for autotrophic carbon fixation: the Calvin–Benson–Bassham (CBB) cycle; the rTCA cycle; the 3-hydroxypropionate cycle; and the reductive acetyl-CoA pathway. In addition, there are two pathways for the assimilation of one-carbon organic compounds: the ribulose monophosphate cycle and the serine pathway. The eukaryotes use only the CBB cycle, whereas, at the other extreme, the Crenarchaeota use three or possibly four autotrophic pathways (Hügler et al. 2003 a,b). This diversity of pathways demonstrates that autotrophy has arisen multiple times on the Earth, increasing the likelihood that it would arise elsewhere.
(i) Calvin–Benson–Bassham cycle
The CBB cycle is the most widespread route for autotrophic carbon fixation on Earth. It is the cycle used by all photosynthetic eukaryotes, cyanobacteria, chemoautotrophs such as purple sulphur and non-sulphur bacteria (Shively et al. 1998), and most anoxygenic photolithotrophs. The incorporation of CO 2 into cellular materials requires a carboxylase, as well as energy and a reductant, in the case of the CBB cycle, ATP and NADPH, which are obtained from the ‘light reactions’.
The CBB cycle contains three unique enzymes. Ribulose-1,5-bisphosphate carboxylase/oxygenase (rubisco) fixes CO 2 onto the five-carbon sugar, ribulose-1,5-bisphosphate, yielding two molecules of 3-phosophoglycerate through an unstable six-carbon intermediate (figure 1). While the CO 2 usually comes from the environment, it can also come from methanol, formaldehyde (e.g. Beijerinckia mobilis: Dedysh et al. 2005) or CO 2 produced by CO oxidation. Functional rubisco has been found in all the three domains of life. The other unique enzymes in the CBB cycle are sedoheptulose bisphosphatase (SBPase), which catalyses the dephosphorylation of sedoheptulose-1,7-bisphosphate, and phosphoribulokinase (PRK), which phosphorylates ribulose-5-phosphate (RuMP).
A key to the evolution of rubisco may be found by studying the form IV subfamily of the rubisco large subunit. Although similar in sequence to forms I, II and III, form IV lacks multiple conserved active-site residues, and thus none of these enzymes has been demonstrated to have robust rubisco activity, even when recombinant rubisco-like protein (RLP) was synthesized in E. coli (Hanson & Tabita 2001). Such RLPs have been found in the green sulphur bacteria Chlorobium tepidum and Chlorobium limicola, organisms that use the rTCA cycle for carbon fixation, as well as in Bacillus subtilis and Archaeoglobus fulgidus. Although the RLPs do not appear to fix carbon, the loss of the gene in C. tepidum resulted in a series of effects including a decreased production of bacteriochlorophyll c, a decrease in photoautotrophic growth rate, a decrease in carbon fixation, the accumulation of sulphur grains and the accumulation of two oxidative stress proteins. Hanson & Tabita (2001) hypothesized that these effects were attributable to a defect in the oxidation of sulphur compounds that provide the reducing power for carbon fixation in C. tepidum.
Ashida et al. (2005) studied the RLP of the non-photosynthetic bacterium B. subtilis and found that it was the 2,3-diketo-5-methylthiopentyl-1-phosphate enolase in the methionine salvage pathway. They suggested that photosynthetic rubiscos evolved from RLPs. Ashida et al. (2003) rescued a RLP-deficient B. subtilis with the gene for the photosynthetic rubisco large subunit from Rhodospirillum rubrum. Yet, the question remains: are RLPs precursors to bona fide functional rubisco or do they represent descendant states (Hanson & Tabita 2001)?
(ii) Reductive TCA cycle
The rTCA cycle (reverse Krebs cycle or reverse citric acid cycle) is used by some bacteria to produce carbon compounds from carbon dioxide and water, sometimes using hydrogen or sulphates as electron donors. The reaction is approximately the citric acid cycle run in reverse (figure 2). The reaction is one of the possible candidates for prebiotic early Earth reactions and so is of interest in the origin of life research. Some of the steps can be catalysed by minerals. Among the bacteria, the rTCA cycle is known for at least some Aquificales, Chlorobiales and Proteobacteria including magnetotactic proteobacteria (Williams et al. 2006). The rTCA cycle may predominate over the CBB cycle among autotrophs in hydrothermal vents (Campbell & Cary 2004) and possibly other microaerophilic environments, suggesting that the rTCA cycle is a key route for carbon fixation on the Earth even today.
(iii) 3-Hydroxypropionate cycle
The 3-hydroxypropionate cycle (figure 3) is proposed as the pathway for autotrophic carbon fixation in Chloroflexus aurantiacus (Ivanovsky et al. 1993) and for some acidophilic archaebacteria of the phylum Crenarchaeota, such as Acidianus brierleyi (Ishii et al. 1997) and Sulfolobus metallicus (Hügler et al. 2003 a,b). For each turn of the 3-hydroxypropionate cycle, two molecules of are fixed into one molecule of glyoxylate (Alber & Fuchs 2002). The thermophilic acidophile Metallosphaera sedula uses a modified 3-hydroxypropionate cycle. In M. sedula, there are ATP-dependent carboxylations of acetyl-CoA and propionyl-CoA, both of which are catalysed by one large enzyme, acetyl-CoA/propionyl-CoA carboxylase (Hügler et al. 2003 a,b). The enzyme contains biotin carboxylase and carboxytransferase, and a small biotin carrier protein (Hügler et al. 2003 a,b).
Figure 3.
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Proposed 3-hydroxypropionate cycle of autotrophic CO 2 fixation in the phototrophic green non-sulphur bacterium Chloroflexus aurantiacus. Carboxylating enzymes in the 3-hydroxypropionate cycle include acetyl-CoA carboxylase and propionyl-CoA carboxylase. After Alber & Fuchs (2002).
(iv) Reductive acetyl-CoA
The reductive acetyl-CoA pathway is found among many anaerobes in the archaea and bacteria including acetogens, such as Clostridium thermoaceticum, methanogens and most autotrophic sulphate reducers. In the reductive acetyl-CoA pathway, two CO 2 are ultimately combined and reduced to form acetyl-CoA (figure 4). In this pathway, one CO 2 is captured on tetrahydrofolate and reduced to a methyl group, while the other CO 2 is reduced to a carbonyl group (C=O), and then added to the methyl group (Fuchs 1986).
Figure 4.
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Reductive acetyl-CoA pathway (Ljungdahl–Wood pathway). Note that there are three carboxylation steps, including the final one where carbon monoxide is added to form acetyl-CoA by acetyl-CoA carboxylase, also known as acetyl-CoA synthase complex. Pathway after
(v) Ribulose monophosphate (RuMP) cycle and serine pathway
There are two known pathways that are used by bacteria for the assimilation of formaldehyde or methane: the serine pathway (figure 5) and the RuMP cycle. Since these C1 compounds are not oxidized to CO 2 before assimilation, their fixation is not strictly autotrophic. However, as both are also produced abiotically and even found in space in hot molecular cores and comets (Ehrenfreund & Charnley 2000), these pathways could be considered a way to get abiotically produced organic carbon into the biosphere. In the RuMP cycle, fixation begins with the condensation of formaldehyde and RuMP to form hexulose-6-phosphate (HuMP), which in turn is converted to fructose-6-phosphate (FMP). The FMP is then cleaved to form two glyceraldehyde-3-phosphates. This overall pattern is certainly reminiscent of the CBB cycle using rubisco. 3-Hexulose-6-phosphate synthase (HPS) and 6-phospho-3-hexuloisomerase (PHI) are the only enzymes that are unique to this cycle. Overall, three molecules of formaldehyde are assimilated forming a three-carbon intermediate of central metabolism. The RuMP cycle has been found in chemolithoautotrophs including several methanotrophs.
Figure 5.
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Serine pathway (assimilation of formaldehyde) is used by some chemolithoautotrophs, for example several methanotrophs. Methanotrophic bacteria oxidize methane and methanol to formaldehyde that condenses with glycine to form serine that can be assimilated to form intermediates of the central metabolic pathways. During this stage, phosphoenolpyruvate is carboxylated by phosphoenolpyruvate carboxylase to form oxaloacetate. The intermediate compounds are then used for biosynthesis. The net balance of this cycle is the fixation of 2 mol of formaldehyde and 1 mol of CO 2 into 1 mol of 3-phosphoglycerate, which is used for biosynthesis, at the expense of 2 mol of ATP and the oxidation of 2 mol of NAD(P)H.
The serine pathway is similar to the RuMP pathway in that the initial substrate is formaldehyde, which itself may be derived from methane or methanol in methanotrophic bacteria. It also contains a carboxylation step. Some of the 2-phosphoglycerate produced is converted to phosphoenolpyruvate that is carboxylated by phosphoenolpyruvate carboxylase to oxaloacetate. Similar to the RuMP cycle, the serine pathway has been found in chemolithoautotrophs including several methanotrophs.
Both the RuMP and serine pathways oxidize CO or organic compounds including methanol and formate to CO 2 prior to fixation. This suggests that CO 2 fixation is the primitive substrate and the oxidation of the initial substrate was added later in evolution.
(vi) Variations on a theme
There are organisms known that have variations on these well-known pathways. For example, the thermophilic hydrogen-oxidizing bacterium Hydrogenobacter thermophilus fixes CO 2 by the rTCA cycle (Shiba et al. 1985). However, pyruvic carboxylase catalyses the carboxylation of pyruvic to oxaloacetate instead of phosphoenolpyruvate catalysing the carboxylation of phosphoenolpyruvate to oxaloacetate as is often done in the rTCA cycle.
(vii) Which came first?
The most direct pathway imaginable for an autotroph is to condense two molecules of CO 2, as acetogens do using the acetyl-CoA pathway and the rTCA cycle. This mode of autotrophy is used by anaerobes, which evolved prior to aerobes. In fact, formate dehydrogenase, which catalyses the reversible reduction of CO 2 or , is extremely oxygen sensitive. H 2, CO 2, CO and metals such as cobalt, nickel, iron, tungsten, molybdenum and selenium, which are involved in the catalysis of the pathways, were present on the early Earth. For these reasons, Wood & Ljungdahl (1991) suggested that these might be considered as earliest autotrophic pathways. Some bacteria and archaea, such as Chlorobium tepidum, use the rTCA cycle but their genome encodes an orthologue of the large subunit of rubisco, although the purified protein has no detectable carbon-fixation activity (Hanson & Tabita 2001; Eisen et al. 2002).
Type III rubiscos, found only in the archaea, have rubisco activity although a functional CBB pathway has not been identified in that domain (Sato et al. 2007).
(c) Why is rubisco dominant today?
Why did rubisco-based carbon fixation come to dominate the biosphere? The enzyme itself is not particularly efficient (e.g. Tcherkez et al. 2006). One possibility is that the cyanobacteria used the CBB pathway, and thus this pathway was ‘fixed’ in all eukaryotes since carbon fixation is the result of the chloroplast, an evolved cyanobacterian. As eukaryotes came to dominate the biosphere, so did rubisco. In turn, cyanobacteria use the CBB pathway because they are oxygenic, and the acetyl-CoA and rTCA cycles are oxygen sensitive, as discussed above. Thus, even with its oxygenase activity, rubisco is superior in an oxygenated environment. Similarly, the 3-hydroxypropionate cycle has a narrow taxonomic distribution, and is found in a few thermophiles and acidophiles. Again, this suggests that, in cooler and aerobic environments, rubisco is superior.
Using a different approach, McFadden (1973) postulated that the acquisition of rubisco and PRK by ancient heterotrophic bacteria would have given huge selective advantage if hexoses or intermediates of the CBB cycle were limiting in the environment, assuming that all other enzymes of the CBB cycles were already present. He hypothesized that rubisco could have evolved from phosphofructokinase, noting that fructose-1,6-disphosphate is a good competitive inhibitor for rubisco.
3. Photosynthesis, a specialized case of autotrophy
Photosynthesis is a special case of autotrophy in that it exploits light as its source of energy. Photosynthesis can be aerobic (oxygenic), where water is used as a source of electrons and thus oxygen is produced as a by-product. Alternatively, it may be anaerobic (anoxygenic), i.e. a reductant other than water is used. Reductants for anaerobic photosynthesis are listed in table 1. Photosynthesis is found among the eukaryotes, notably the plants and algae, and six phyla of bacteria including cyanobacteria, Chlorobi, Proteobacteria, Chloroflexi, Firmicutes and Acidobacteria (Bryant et al. 2007).
Table 1.
Examples of reductants used by autotrophs.
| reductants for CO 2 | type of photosynthesis |
:--- |
| H 2 O | aerobic (oxygenic) |
| H 2 | anaerobic (anoxygenic) |
| | anaerobic (anoxygenic) |
| | anaerobic (anoxygenic) |
| S° | anaerobic (anoxygenic) |
| H 2 S | anaerobic (anoxygenic) |
| CO | anaerobic (anoxygenic) |
| CH 4 | anaerobic (anoxygenic) |
| Mn 2+ | anaerobic (anoxygenic) |
| Fe 2+ | anaerobic (anoxygenic) |
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How probable is photosynthesis, specifically, to arise again? Once again, the diversity of reductants and even pathways, some of which have more than one step where carbon fixation occurs, increases the probability of it arising again.
Some cyanobacteria and algae have carbon-concentrating mechanisms ranging from active influx of CO 2 or , passive influx of CO 2, to prefixation of the inorganic carbon on a three-carbon sugar prior to release and re-fixation (Giordano et al. 2005). Among the plants, there are three major variations in photosynthetic pathways: the C3 pathway (no carbon concentration); the C4 pathway; and Crassulacean acid metabolism (CAM). Over 95% of the plants on Earth use the C3 pathway, in which inorganic carbon is incorporated directly into the CBB cycle. In C4 and CAM, there is an initial fixation of inorganic carbon with a subsequent decarboxylation prior to incorporation into the CBB cycle, an adaptation that aids in water conservation. In C4 plants, a distinct leaf anatomy keeps the two steps physically separated. Inorganic carbon is fixed to a three-carbon sugar, phosphoenolpyruvate, in the mesophyll cells to form oxaloacetate, which is converted to malate and then transported to the bundle-sheath cell where it is decarboxylated thus delivering CO 2 to the CBB cycle enzymes. CAM is an adaptation for arid conditions. Similar to C4, there is an initial fixation of CO 2 and a subsequent release to the CBB cycle, but in this case the two processes are temporally separated with CO 2 entering the stomata during the night, and decarboxylation occurring during the day when the stomata are closed, thus minimizing water loss.
4. Could photosynthesis evolve again?
In order to estimate the probability of photosynthesis evolving again on Earth or elsewhere requires that the prerequisites for photosynthesis be met and selective pressures to act. When these components have arisen multiple times, this increases the likelihood of photosynthesis arising again.
(a) Prerequisites for the evolution of biological carbon fixation
There are at least six prerequisites for the evolution of biological carbon fixation as stated earlier. Each are now explored in order to assess the probability of the evolution of carbon fixation.
(i) Carbon-based life form
Organic carbon forms the basis for life on Earth, and is likely to do so if life arose again on Earth or elsewhere. Carbon is the fourth most abundant element in the Universe and forms the basis for a complex and varied chemistry. It may be twice as common relative to hydrogen in our Solar System than elsewhere (Snow & Witt 1995), but even at half the amount of our Solar System, it is still abundant. While silicon is also common (though not nearly as common as carbon in the Universe as a whole) and can form interesting polymers (Benner et al. 2004), its flexibility pales in comparison with organic chemistry, particularly in the ability of carbon to form polymers. Most stunning of all, organic compounds—including amino acids and nucleotide bases—have been detected in the interstellar medium (Irvine 1998; Ehrenfreund & Charnley 2000; Allamandola & Hudgins 2003). Even Earth, which is composed of a substantial quantity of silicates and thus should be biased towards silicon-based life, harbours carbon-based life.
(ii) The presence of inorganic carbon
While organic carbon is quite common in the Universe, the predominant form of carbon is inorganic. Carbon is formed by the triple collision of alpha particles (helium nuclei) within the core of a giant or supergiant star, and is made available for life when it disperses from its parent star during a supernova. Today's Earth is estimated to contain approximately 38 000 Gt inorganic carbon (Falkowski et al. 2000).
(iii) Availability of reductants
Much of the photosynthesis on Earth today is oxygenic; that is, the reducing power is derived from H 2 O. Yet many reductants, from elemental sulphur to hydrogen, are used by prokaryotes that do not produce oxygen (table 1). Some can use multiple electron donors for photosynthesis, such as Chlorobium tepidum, which can use both sulphide and thiosulphate (Eisen et al. 2002).
(iv) The presence of light
Any planet, comet or asteroid near enough to a parent star will receive sufficient radiation for photosynthesis on its surface. ‘Near enough’ is surprisingly low, as only a small photon flux is needed for photosynthesis, with an absolute minimum thought to be 0.01 μmol of photons m−2 s−1 (Raven & Cockell 2006). By contrast, midday sunlight in summer in the continental USA is approximately 2000 μmol of photons m−2 s−1 (L. J. Rothschild 2008, unpublished). We have shown that interstitial microbial mats found just below the sand surface where they receive a tiny fraction of the surface solar flux can still fix approximately 41 mg C m−2 d−1 (Rothschild & Giver 2003). Nisbet et al. (1995) proposed that even the presence of solar radiation may not be required for the evolution of the light-harvesting pigment chlorophyll as it could have arisen for thermal detection in order for organisms to properly orient with respect to hydrothermal vents.
Which wavelengths are necessary? On Earth, most organisms use solar radiation between 400 and 700 nm. Much below, ultraviolet radiation damage becomes of increasing concern, a problem with any organic carbon-based life form. At the other extreme, longer wavelengths are more benign but are also of much less energy, hence may not be energetic enough to drive photosynthesis. However, there are prokaryotes that can access slightly longer wavelengths. For example, the cyanobacterium Acaryochloris marina is unique among phototrophs in that it uses chlorophyll d as its main light-harvesting pigment instead of chlorophyll a (Kühl et al. 2005). The peak absorption for chlorophyll d is 700–720 nm (Kühl et al. 2005). Some anoxygenic prokaryotes use even higher wavelengths into the near infrared (IR) because bacteriochlorophylls a and b have secondary absorption peaks at 770 and 795 nm, respectively. The purple bacteria are able to access solar radiation even further into the IR, with the LH1 complex absorbing as high as 963 nm (Permentier et al. 2001).
IR radiation contains less energy than visible radiation, so it is difficult to understand why bacteriochlorophyll should use this portion of the spectrum. Nisbet et al. (1995) have suggested that bacteriochlorophylls arose as heat (i.e. near IR) sensing devices for early organisms that lived near hydrothermal vents. In practice, niche partitioning of different species with different suites of primary and secondary pigments occurs so that each organism can access as much solar radiation as possible (e.g. Stomp et al. 2004).
(v) A light-harvesting mechanism to convert the light energy into chemical energy
The synthesis of chlorophyll/bacteriochlorophyll arose only once in evolution (Xiong & Bauer 2002). But chlorophyll biosynthesis is closely related to the widespread haem, the former being a magnesium porphyrin and the latter an iron porphyrin. Haem is extremely widespread as it is the prosthetic group for such proteins as myoglobin, haemoglobin, catalase, peroxidase and cytochrome c. Thus, it seems plausible biochemically that a chlorophyll-like molecule could have arisen multiple times in multiple taxa.
In addition, several types of photoreceptor molecules are known, some of which convert light energy into chemical energy, as in photosynthesis, or use the light energy for signal transduction. Béjá et al. (2000) presented evidence that bacterial rhodopsin could be the photoreceptor driving photoheterotrophy or even photosynthesis in a variety of marine bacteria.
Photopigments include carotenoids, bilins, chlorophylls, flavins and pterins (table 2). There may be a few other photopigments such as parahydroxycinnamic acid that may act as the photoactive pigment in association with green fluorescent protein (Sancar 2000). A common theme in the molecular structure of light-harvesting compounds is the network of alternating single and double bonds called polyenes that makes them particularly effective photoreceptors. Chlorophyll is a substituted tetrapyrrole, with the four nitrogen atoms of the pyrroles coordinated to a magnesium atom. Carotenoids, used as secondary compounds in photosynthesis, also contain polyenes (figure 6). Rhodopsin, the photosensitive molecule in rods, contains opsin (a protein) and 11-cis-retinal, a prosthetic group that contains alternating polyenes.
Table 2.
Photoreactive pigments (see Sancar 2000 for more details). (The actual absorption depends on both the pigment and the chemical environment in which it is found.)
| class of pigment | absorption range | description | examples |
:--- :--- |
| carotenoids | ∼400–550 | photoantenna (accessory) pigments in the photosystems and the catalytic pigments in animal and bacterial rhodopsins | xanthophylls, carotenes such as β-carotene, retinal |
| bilins | 400–500 nm, 600–700 nm | photoantenna pigments in the photosystems and the chromophore of the plant photoreceptor phytochrome | |
| chlorophylls | 350–450 nm, 600–700 nm, >700 for chlorophyll d | photoantenna pigments in the photosystems and the primary electron donors in the reaction centre of photosystems | chlorophylls a, b, c and d |
| flavins | 360 nm in the two-electron reduced form; 370 and 440 nm in the two-electron oxidized form; 380, 480, 580, 625 nm in the one-electron reduced (blue neutral radical) form | photoactive cofactor in the photolyase/blue-light photoreceptor family | flavin adenine dinucleotide (photoactive cofactor in photolyase/blue-light photoreceptor family), deazoriboflavin |
| pterins | 360–420 nm | photoantenna in the majority of photolyase/cryptochrome blue-light photoreceptors | 5,10-methenyltetrahydrofolate (MTHF) |
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Figure 6.
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β-Carotene, a light-harvesting pigment with peak absorptions of 455 and 480 nm. It is an effective photoreceptor owing to the alternating single and double bonds.
Not all of these molecules originated as light-harvesting compounds. For example, it is thought that the carotenoids arose in archaea as lipids reinforcing cell membranes (Vershinin 1999). However, their structure contains a number of properties that pre-adapted them to the various functions they serve today ranging from light harvesting to quenching singlet oxygen, while still maintaining their ability to reinforce membrane structure as they do today in mycoplasms, some fungi and some animals such as molluscs. Cryptochromes may have evolved from photolyases, but no longer possess the photolyase function (Lin & Shalitin 2003).
Often, such as in the case of the bilins, the photoreceptor is covalently attached to a protein. Grossman et al. (1995) hypothesized that the phycobiliprotein subunits were derived from a common precursor protein that already had the amino acid residues needed for photoreceptor attachment and stability. The integral membrane proteins called opsins bind retinal (vitamin A aldehyde) forming rhodopsins. Rhodopsins are capable of generating a chemiosmotic membrane potential in response to light. Rhodopsins belong to two distinct protein families: the visual rhodopsins and archaeal and bacterial rhodopsins (Béjá et al. 2000). Although they have no significant sequence similarity and may have different origins, they have identical topologies (Béjá et al. 2000).
(vi) Carbon-fixation enzymes
Carbon fixation, in the broadest sense, is a process where inorganic carbon is converted or incorporated into organic compounds. Carbon dioxide is a ‘featureless molecule’ (Tcherkez et al. 2006) as is oxygen, leading to confusion in catalysis with at least rubisco, resulting in both a carboxylase and oxygenase function. Carboxylases are usually associated with autotrophs only, particularly phototrophs, but, in fact, multiple enzymes found in all taxa have evolved to fix carbon (table 3). Different types of compounds can be carboxylated from sugars to proteins to assorted other compounds.
Table 3.
Carboxylating enzymes.
| name | alternate name | number | reaction catalysed | cofactor |
:--- :---
| pyruvic-malic carboxylase | malate dehydrogenase (oxaloacetate-decarboxylating) | 1.1.1.38 | (S)-malate+NAD+↔pyruvate+CO 2+NADH | |
| pyruvic-malic carboxylase | malate dehydrogenase (decarboxylating) | 1.1.1.39 | (S)-malate+NAD+↔pyruvate+CO 2+NADH | |
| pyruvic-malic carboxylase | malate dehydrogenase (oxaloacetate-decarboxylating) (NADP+) | 1.1.1.40 | (S)-malate+NADP+↔pyruvate+CO 2+NADPH | |
| β-ketoglutaric-isocitric carboxylase | isocitrate dehydrogenase (NAD+) | 1.1.1.41 | isocitrate+NAD+↔2-oxoglutarate+CO 2+NADH | manganese or magnesium |
| triphosphopyridine nucleotide-linked isocitrate dehydrogenase-oxalosuccinate carboxylase | isocitrate dehydrogenase (NADP+) | 1.1.1.42 | isocitrate+NADP+↔2-oxoglutarate+CO 2+NADPH | |
| 6-phosphogluconic carboxylase | phosphogluconate dehydrogenase (decarboxylating) | 1.1.1.44 | 6-phospho-d-gluconate+NADP+↔d-ribulose-5-phosphate+CO 2+NADPH | |
| 2-oxopropyl-CoM reductase (carboxylating) | 2-oxopropyl-CoM reductase (carboxylating) | 1.8.1.5 | 2-mercaptoethanesulphonate+acetoacetate+NADP+↔2-(2-oxopropylthio)ethanesulphonate+CO 2+NADPH | |
| α-ketoacid carboxylase | pyruvate decarboxylase | 4.1.1.1 | a 2-oxo acid↔an aldehyde+CO 2 | thiamine diphosphate |
| phosphoribosylaminoimidazole carboxylase | | 4.1.1.21 | 5-amino-1-(5-phospho-d-ribosyl)imidazole-4-carboxylate →5-amino-1-(5-phospho-d-ribosyl)imidazole+CO 2 | |
| phosphoenolpyruvate carboxylase | phosphoenolpyruvate carboxykinase (GTP) | 4.1.1.32 | GTP+oxaloacetate→GDP+phosphoenolpyruvate+CO 2 | |
| phosphopyruvate carboxylase | phosphoenolpyruvate carboxykinase (diphosphate) | 4.1.1.38 | diphosphate+oxaloacetate→phosphate+phosphoenolpyruvate+CO 2 | |
| ribulose-1,5-bisphosphate carboxylase/oxygenase | carboxydismutase | 4.1.1.39 | 2 3-phospho-d-glycerate+2H+→d-ribulose-1,5-bisphosphate+CO 2+H 2 O | |
| phosphoenolpyruvate carboxylase | methylmalonyl-CoA decarboxylase | 4.1.1.41 | ATP+oxaloacetate→ADP+phosphoenolpyruvate+CO 2 | |
| urea carboxylase | urea amidolyase | 6.3.4.6 | ATP+urea+→ADP+phosphate+urea-1-carboxylate→→ 2NH 3+2CO 2 | biotin |
| biotin carboxylase | | 6.3.4.14 | ATP+biotin-carboxyl-carrier protein+CO 2→ADP+phosphate+carboxybiotin-carboxyl-carrier protein | |
| pyruvate carboxylase | | 6.4.1.1 | ATP+pyruvate+→ ADP+phosphate+oxaloacetate | biotin, manganese or zinc |
| acetyl-CoA carboxylase | | 6.4.1.2 | ATP+acetyl-CoA+→ ADP+phosphate+malonyl-CoA | biotin |
| propionyl-CoA carboxylase | | 6.4.1.3 | ATP+propionyl-CoA+→ ADP+phosphate+(S)-methylmalonyl-CoA | biotin |
| methylcrotonyl-CoA carboxylase | | 6.4.1.4 | ATP+3-methylcrotonyl-CoA+→ ADP+phosphate+3-methylglutaconyl-CoA | biotin |
| geranoyl-CoA carboxylase | | 6.4.1.5 | ATP+geranoyl-CoA+→ ADP+phosphate+3-(4-methylpent-3-en-1-yl)pent-2-enedioyl-CoA | biotin |
| acetone carboxylase | | 6.4.1.6 | Acetone+CO 2+ATP+2H 2 O→ acetoacetate+AMP+2 phosphate | magnesium |
| 2-oxoglutarate carboxylase | oxalosuccinate synthetase | 6.4.1.7 | ATP+2-oxoglutarate+→ ADP+phosphate+oxalosuccinate | biotin, magnesium |
| γ-glutamyl carboxylase | | | carboxylation of specific glutamate residues in vitamin K-dependent proteins to Gla in the presence of carbon dioxide, oxygen and reduced vitamin K; in the process, vitamin K is converted to vitamin K epoxide, which is subsequently converted to vitamin K by vitamin K epoxide reductase and used in the carboxylation reaction | |
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Examples of a carboxylating enzyme that is not used in autotrophy are the γ-carboxylation enzymes. These enzymes are important in blood clotting in vertebrates, but have recently been found in some invertebrates including Drosophila and the marine cone snail Conus, where they have a role in the production of venom peptides (Bandyopadhyay et al. 2002). γ-Carboxylation enzymes are integral proteins of the endoplasmic reticulum, which, in the presence of carbon dioxide, oxygen and reduced vitamin K, results in the carboxylation of specific glutamate residues in vitamin K-dependent proteins. The general topology of the enzyme from all the three taxa are similar, as was the sequence, especially in the transmembrane portion of the carboxylase and the putative substrate-binding pocket (Bandyopadhyay et al. 2002).
Degradation of urea is widely accomplished by the action of urease, but in algae, yeast and some bacteria, the degradation occurs through the action of the enzyme complex urea amidolyase (Kanamori et al. 2004). Bicarbonate is phosphorylated by ATP, with the subsequent transfer of the CO 2 to biotin. Through the urea carboxylase activity, the urea is carboxylated leading to its degradation to two molecules each of NH 3 and CO 2.
Biotin, a water-soluble member of the B complex of vitamins, is a mobile carrier of activated CO 2 (figure 7). Biotin is a cofactor for many carboxylating, decarboxylating and transcarboxylating enzymes, for example pyruvate carboxylase (table 2; Nikolau et al. 2003). For all carboxylase reactions, the carboxyl donor is bicarbonate/ATP (Visser & Kellogg 1978). Biotin-dependent carboxylases share a catalytic mechanism and a structure that consists of biotin carboxylase, biotin-carboxyl-carrier protein and carboxyl transferase. The carboxyl terminus of biotin is linked to the amino group of a specified lysine residue by an amide bond. The biotin itself is carboxylated by biotin carboxylase (EC 6.3.4.14) during the following reaction:
Most coenzymes are ribonucleotides, or at least have a cyclic nitrogenous base in common, except biotin and lipoic acid. Coenzymes and prosthetic groups and tRNA can be seen as relics from the RNA world, except biotin and lipoic acid. This suggests that biotin and lipoic acid arose after the RNA world (Visser & Kellogg 1978).
Figure 7.
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Biotin (vitamin B7 or vitamin H) is a cofactor for CO 2 transfer. (a) The carboxyl terminus of biotin is linked to the amino group of a specified lysine residue by an amide bond. Carboxylating enzymes using biotin are: acetyl-CoA carboxylase alpha; acetyl-CoA carboxylase beta; methylcrotonyl-CoA carboxylase; propionyl-CoA carboxylase; and pyruvate carboxylase. (b) An example of the use of biotin in the carboxylation of pyruvate by pyruvate carboxylase.
Some carboxylating enzymes are used for autotrophic carbon fixation in some cases and not others. Acetyl-CoA (figure 8) and propionyl-CoA carboxylation are normally used in fatty acid synthesis; however, they are also found in archaea that do not have fatty acids. It is thought that, in these archaea, the enzymes function in autotrophic carbon fixation. The carboxylase enzyme from the archaeans Metallosphaera sedula (Hügler et al. 2003 b) and Acidianus brierleyi (Chuakrut et al. 2003) catalyses equally effectively both acetyl-CoA and propionyl-CoA carboxylation. The holoenzyme contains subunits of biotin carboxylase, carboxytransferase and a biotin-carrier protein.
Figure 8.
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Acetyl-CoA carboxylase catalyses the first step in fatty acid synthesis. This carboxylation is irreversible. Photo of a wombat by L.J.R.
Acetone degradation in the actinomycete Rhodococcus rhodochrous is accomplished with an initial carboxylation by acetone carboxylase. Carboxylation is increased in the presence of a nucleotide triphosphate, with GTP and IPT particularly effective (Clark & Ensign 1999).
How probable is the evolution of a carboxylase? The wide diversity of carboxylases outlined above and their substrates suggests multiple origins of carboxylases, implying a relative ease of evolutionary origin. Other putative examples of convergence in enzyme function have been identified, including the d- and l-specific 6-hydroxynicotine oxidases, both found in the soil bacterium Arthrobacter nicotinovorans, but unrelated at the DNA and protein levels (Schenk & Decker 1999). Sugar kinases, enzymes that catalyse the phosphorylation of sugars, arose at least three times leading to distinct non-homologous families (Bork et al. 1993), and the lactate dehydrogenase from the amitochondriate protists Trichomonas vaginalis appears to have evolved from the cytosolic malate dehydrogenase of the same species rather than from another lactate dehydrogenase (Wu et al. 1999).
As Pain (1982) pointed out, natural selection should act on the active site of an enzyme rather than its amino acid sequence, and with the featureless nature of the substrate, CO 2, again, one should predict relative ease in the evolution of carboxylation activity although, as in the case of rubisco, it may lack in specificity. But, if there are similarities in the active site, is this the result of common ancestry or of a limited number of ways to construct an active site with a particular catalytic activity? Conversely, the subtilisin group of the microbial serine proteases is similar to the mammalian type in their active sites but not conformation, suggesting that this catalytic activity was solved at least twice in evolution with a resulting convergence on the active site (Pain 1982). Furthermore, there is modularity in the evolution of many carboxylases in that the incorporation of biotin allows the rapid evolution of carboxylation.
But biomolecules do not exist in either physical or functional isolation. When an enzyme is part of a metabolic network, changes in components may need compensation in order to keep the entire network functional (Vitkup et al. 2006). Not surprisingly, analyses of the photosynthetic machinery of cyanobacterial gene products that need to interact show genetic linkage suggesting coordinated evolution (Shi et al. 2005). However, rubisco may not be one of them, but rather a ‘frozen accident’ because it allows oxygen to act as a competitive inhibitor of carboxylase activity (Shi et al. 2005). Rather than evolution acting on removing the oxygenase activity during geological time as atmospheric levels of oxygen rose, organisms have responded by evolving carbon-concentrating mechanisms and increasing the production of rubisco.
(b) Selection pressure for photosynthesis
In addition to the prerequisites for the evolution of biological carbon fixation, a selective pressure should be present. An excess of inorganic carbon (e.g. CO 2, CO, carbonates, cyanides, cyanates, carbides and thiocyanates) over organic carbon, in most cases a vast surfeit, should provide the selective pressure. Even on today's life-infested Earth, there are approximately 38 000 Gt carbon in contrast to 1000 Gt organic carbon in the oceans (Falkowski et al. 2000).
(c) Proposed evolutionary scenario for the origin of photosynthesis
Given these conditions, it is easy to envision an evolutionary scenario that includes strong selective pressure for organisms that evolve autotrophy, particularly photosynthesis. The evolutionary background includes the following:
Life on Earth is based on organic carbon.
Most of the carbon in the Universe is inorganic.
Life is lazy, or, to put it more scientifically, one must be efficient to maintain the competitive edge.
Evolution uses what is available; ‘evolution as tinkerer’ rather than an engineer (Jacob 1977). Parts are co-opted from within or from others rather than designed, as Ashida et al. (2005) have suggested for the origin of rubisco from RLPs. Another source of novelty are units—from genes to teeth—which replicate and diversify.
Of course, contingency does occur. Asteroids, floods, atmospheric catastrophes and so on may change the rules substantially.
From these rules, I hypothesize a planet-independent scenario for the evolution of life in the Universe.
The first living organisms were built from abiotically produced organic carbon, as the heterotrophic model for the origin of life postulates (e.g. Lazcano & Miller 1999). The struggle for existence began over the remains of the first ‘free lunch’, and life was forced to make its own organic carbon in order to persist. Perhaps this was doubly beneficial because it was more efficient; for example, organic carbon production could be combined with the generation of energy. In either case, the first autotroph was born.
But life conserves energy whenever possible for other uses. Why synthesize organic carbon when it can be delivered in an integrated package? In other words, why be autotrophic when organic carbon is not limiting?
There is experimental evidence to support this point. For example, a mixed community of the red alga Cyanidium caldarium, a green alga described as ‘Chlorella-like’ or ‘Chlorella’ protothecoides var. acidicola and a minor component of the fungus Dactylaria (Belly et al. 1973; Ferris et al. 2005) was tested for gross carbon fixation in the presence and absence of added fixed carbon (L. J. Rothschild 2008, unpublished). Acetate and glucose were added to mat samples, which were then incubated in situ with supplemental 14 CO 2. While, for the most part, there was little difference with the addition of glucose, the presence of acetate had a powerful inhibitory effect on carbon fixation (figure 9). This suggests that these photosynthetic algae were able to take up a source of fixed carbon; they downregulated their own carbon fixation. Similarly, oceanic photoheterotrophic bacteria metabolize organic carbon from the environment when available, but photosynthesize when organic carbon is sparse (Kolber et al. 2001).
Figure 9.
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Effect of exogenous fixed carbon on carbon-fixation rates in Nymph Creek (N 44°45.178′, W 110°43.439′), a natural red and green algal ecosystem in Yellowstone National Park, WY, USA. Carbon fixation was measured by the acid-stable incorporation of in situ, as described by Rothschild & Mancinelli (1990) and Rothschild (1991). The incubation mix was stream water (control: filled circle, solid line) or stream water supplemented with 25 mM acetate (open circle, solid line) or 25 mM glucose (open circle, dashed line) and 1 μCi ml−1 14 C-bicarbonate (New England Nuclear, Wilmington, DE, USA, catalogue number NEC 086H). Algal mat cores (surface area=17 cm 2) were placed in Whirlpack bags, incubation mix was added and the bags were incubated in Nymph Creek. Experiments were conducted in duplicate, and all data points were averaged. Dark controls were performed by wrapping the incubation vessel in foil. The dissolved inorganic carbon of Nymph Creek water was determined using a CO 2 probe (Lazar Research Laboratories, Los Angeles, CA, USA) calibrated with aqueous solutions of . Incubation times were 15 min in order to minimize the impact of remineralization of the fixed carbon and to be sensitive to light levels that are constantly changing in nature.
And thus, herbivory arose. Of course, this required various modifications such as sufficient size and structure. Organic ‘slurp’ was still available for those unable to feed on autotrophs, but unlike the first life forms, the soup course was now biologically generated; a satisfying broth for countless members of all the three domains of life. Even today, it is fashionable to acquire an in-house source of organic carbon through symbiosis, a strategy that has been brilliantly exploited on Earth by a variety of protists and animals.
Assimilating fixed carbon is metabolically less expensive than fixing carbon, and if the fixed carbon comes with proteins, nucleic acids and other nutrients, so much the better. The autotrophs provided a new biological source of organic carbon for heterotrophs (more specifically saprotrophs) by excreting organic compounds. More significantly, a new type of heterotrophy that could not be supported previously arose: herbivory. Many microbes have remained mixotrophic (e.g. Kolber et al. 2001). Raven's (1997) excellent analysis of the costs and benefits of phagotrophy points out that for all the versatility of mixotrophy, that is, retaining both the ability to photosynthesize and consume, there are costs. The most obvious cost is the necessity of maintaining two sets of machinery, which he estimates as approximately 50% for the photosynthetic apparatus and approximately 10% to maintain phagotrophic capability, although there can be some regulation in the amount of machinery based on environmental conditions. A possible secondary problem is that a phagotrophic cell must regulate volume. A photosynthetic eukaryote must make sure that all of the components for the photosynthetic apparatus including genetic instructions are inherited properly.
(d) The danger of oxygenic photosynthesis
Oxygenic photosynthesis is tremendously advantageous owing to the wide availability of water. But oxygenic photosynthesis creates hazards in the form of reactive oxygen species that can damage nucleic acids, proteins and lipids. Oxygen, water, superoxide and hydrogen peroxide are not particularly reactive, but hydrogen peroxide can be reduced to the hydroxyl radical that is extremely reactive.
During the course of the production of O 2 from H 2 O in photosystem II, oxygen of the ground (triplet) state (3 O 2) may be excited to singlet state (1 O 2) as a side or back reaction that generates triplet chlorophyll. Fortunately, 1 O 2 is rapidly quenched by water (Asada 2006). Photoreduction of oxygen to hydrogen peroxide (H 2 O 2) occurs in photosystem I. The primary reduced product is superoxide anion (), and its enzymatically catalysed disproportionation products are H 2 O 2 and O 2 (Asada 2006). In isolated thylakoids, H 2 O 2 accumulates, but not in intact chloroplasts owing to enzymatic degradation. However, the prerequisite for oxygenic photosynthesis is the ability to tolerate oxygen, and aerobic metabolism also generates reactive oxygen species during the reduction of O 2 to H 2 O. Thus, oxygenic photosynthesis is fraught with biochemical risks that must be addressed with antioxidants.
5. Photosynthesis, a universal phenomenon?
The preceding argument is that given certain prerequisites—a carbon-based life form, the presence of inorganic carbon, the availability of reductants, the presence of light, a light-harvesting mechanism to convert the light energy into chemical energy and carboxylating enzymes—and evolutionary pressure, photosynthesis is likely to evolve again, whether on Earth or elsewhere.
Oxygenic photosynthesis, however, carries with it the danger of reactive oxygen species but the advantage of, at least on Earth, a widely available source of reducing power relative to other reductants. It is unclear what the relative costs to benefits are, but these must always be accompanied by a cautionary note. By contrast, Lovelock (1975) assumed that oxygenic photosynthesis would evolve multiple times, and thus the detection of O 3 on other planets would be an indication of life. A thorough and thoughtful analysis of the likelihood of oxygenic photosynthesis arising elsewhere has been presented by Wolstencroft & Raven (2002).
Leaving oxygenic photosynthesis aside, are the prerequisites for the evolution of any form of photosynthesis available elsewhere? The argument was advanced above that life anywhere in the Universe is likely to be based on organic carbon, and that the inorganic and organic carbon is widely available in the Universe. Even in our Solar System, Europa and Enceladus are thought to contain substantial quantities of organic carbon at least on their surface, and the Huygens probe confirmed that Titan has large quantities of organics. Meteorites, comets and dust particles rain organic carbon through the Solar System, so it is unclear why we have yet to detect it on Mars unless it is being oxidized. Mars does have a large inventory of CO 2. All these bodies have water and other compounds that could be used as reductants.
Light becomes increasingly scarce with distance from the Sun, decreasing with the inverse of the square of the distance. Mars at 1.6 AU has approximately 44% of the terrestrial solar radiation flux. Of the bodies mentioned, Saturn's moons Titan and Enceladus are the farthest from the Sun at approximately 9.2 AU, which means they should receive approximately 1.1% of the terrestrial solar radiation flux, a flux sufficient to support photosynthesis on Earth. Even beyond our Solar System, Tarter et al. (2007) showed that an M dwarf emits enough light for photosynthesis. From a star of T eff=4000 K (slightly hotter than a M0 star), photosynthetically active radiation (400–700 nm) incident at the top of the atmosphere would be roughly a third that of the Earth, and the range that penetrates clear ocean water (450–550 nm) would be about a quarter that incident on the Earth. And even if the average wavelength is shifted towards the IR, there should be enough useful radiation for photosynthesis (Raven 2007). But note that if a planet is tidally locked, then any stationary organism, such as terrestrial plants, would always see light from the same angle, so less of the canopy would be able to access it.
The wide diversity of photoreceptors and carboxylation enzymes on Earth and their multiple origin argue for a relative ease in the evolution of these mechanisms. If so, and the selection pressure arose with regard to a dearth of organic carbon relative to biotic demands, photosynthesis would arise.
Acknowledgments
I am indebted to Euan and Ellen Nisbet for the invitation to participate in this meeting, and especially to Euan for his ever-stimulating suggestions and perspectives. This manuscript has benefited by the insightful suggestions of John Ellis, Robert Tabita and two anonymous reviewers, and the sharp eyes of Derek Bendall. This work was supported by grants from NASA's Astrobiology Program, including the NASA Astrobiology Institute.
Footnotes
One contribution of 15 to a Discussion Meeting Issue ‘Photosynthetic and atmospheric evolution’.
References
Alber B.E, Fuchs G. Propionyl-coenzyme A synthase from Chloroflexus aurantiacus, a key enzyme of the 3-hydroypropionate cycle for autotrophic CO2 fixation. J. Biol. Chem. 2002;277:12 137–12 143. doi: 10.1074/jbc.M110802200. doi:10.1074/jbc.M110802200 [DOI] [PubMed] [Google Scholar]
Allamandola L, Hudgins D. From interstellar polycyclic aromatic hydrocarbons and ice to astrobiology. In: Pirronello V, Krelowski J, editors. Proc. NATO ASI entitled “Solid State Astrochemistry”. Kluwer; Dordrecht, The Netherlands: 2003. pp. 251–316. [Google Scholar]
Anbar A.D, et al. A whiff of oxygen before the great oxidation event? Science. 2007;317:1903–1906. doi: 10.1126/science.1140325. doi:10.1126/science.1140325 [DOI] [PubMed] [Google Scholar]
Asada K. Production and scavenging of reactive oxygen species in chloroplasts and their functions. Plant Physiol. 2006;141:391–396. doi: 10.1104/pp.106.082040. doi:10.1104/pp.106.082040 [DOI] [PMC free article] [PubMed] [Google Scholar]
Ashida H, Salto Y, Kojima C, Kobayashi K, Ogasawara N, Yokota A. A functional link between rubisco-like protein of Bacillus and photosynthetic rubisco. Science. 2003;302:286–290. doi: 10.1126/science.1086997. doi:10.1126/science.1086997 [DOI] [PubMed] [Google Scholar]
Ashida H, Danchin A, Yokota A. Was photosynthetic rubisco recruited by acquisitive evolution from rubisco-like proteins involved in sulfur metabolism? Res. Microbiol. 2005;156:611–618. doi: 10.1016/j.resmic.2005.01.014. doi:10.1016/j.resmic.2005.01.014 [DOI] [PubMed] [Google Scholar]
Bandyopadhyay P.K, Garrett J.E, Shetty R.P, Keate T, Walker C.S, Olivera B.M. γ-Glutamyl carboxylation: an extracellular posttranslational modification that antedates the divergence of mollusks, arthropods, and chordates. Proc. Natl Acad. Sci. USA. 2002;99:1264–1269. doi: 10.1073/pnas.022637099. doi:10.1073/pnas.022637099 [DOI] [PMC free article] [PubMed] [Google Scholar]
Béjá O, et al. Bacterial rhodopsin: evidence for a new type of phototrophy in the sea. Science. 2000;289:1902–1906. doi: 10.1126/science.289.5486.1902. doi:10.1126/science.289.5486.1902 [DOI] [PubMed] [Google Scholar]
Belly R.T, Tansey M.R, Brock T.D. Algal excretion of 14C-labeled compounds and microbial interactions in Cyanidium caldarium mats. J. Phycol. 1973;9:123–127. doi:10.1111/j.1529-8817.1973.tb04067.x [Google Scholar]
Benner S.A, Ricardo A, Carrigan M.A. Is there a common chemical model for life in the universe? Curr. Opin. Chem. Biol. 2004;8:672–689. doi: 10.1016/j.cbpa.2004.10.003. doi:10.1016/j.cbpa.2004.10.003 [DOI] [PubMed] [Google Scholar]
Bork P, Sander C, Valencia A. Convergent evolution of similar enzymatic function on different protein folds: the hexokinase, ribokinase, and galactokinase families of sugar kinases. Protein Sci. 1993;2:31–40. doi: 10.1002/pro.5560020104. [DOI] [PMC free article] [PubMed] [Google Scholar]
Brocks J.J, Logan G.A, Buick R, Summons R.E. Archean molecular fossils and the early rise of eukaryotes. Science. 1999;285:1033–1036. doi: 10.1126/science.285.5430.1033. doi:10.1126/science.285.5430.1033 [DOI] [PubMed] [Google Scholar]
Bryant D.A, et al. Candidatus Chloracidobacterium thermophilum: an aerobic phototrophic acidobacterium. Science. 2007;317:523–526. doi: 10.1126/science.1143236. doi:10.1126/science.1143236 [DOI] [PubMed] [Google Scholar]
Buick R. The antiquity of oxygenic photosynthesis: evidence from stromatolites in sulphate-deficient Archaean lakes. Science. 1992;255:74–77. doi: 10.1126/science.11536492. doi:10.1126/science.11536492 [DOI] [PubMed] [Google Scholar]
Campbell B.J, Cary S.C. Abundance of reverse tricarboxylic acid cycle genes in free-living microorganisms at deep-sea hydrothermal vents. Appl. Environ. Microbiol. 2004;70:6282–6289. doi: 10.1128/AEM.70.10.6282-6289.2004. doi:10.1128/AEM.70.10.6282-6289.2004 [DOI] [PMC free article] [PubMed] [Google Scholar]
Chuakrut S, Arai H, Ishii M, Igarashi Y. Characterization of a bifunctional archaeal acyl coenzyme A carboxylase. J. Bacteriol. 2003;185:938–947. doi: 10.1128/JB.185.3.938-947.2003. doi:10.1128/JB.185.3.938-947.2003 [DOI] [PMC free article] [PubMed] [Google Scholar]
Clark D.D, Ensign S.A. Evidence for an inducible nuceotide-dependent acetone carboxylase in Rhodococcus rhodochrous B276. J. Bacteriol. 1999;181:2752–2758. doi: 10.1128/jb.181.9.2752-2758.1999. [DOI] [PMC free article] [PubMed] [Google Scholar]
Conway Morris S. Oxford University Press; Oxford, UK: 1998. The crucible of creation: the Burgess Shale and the rise of animals. p. 272. [Google Scholar]
Dedysh S.N, Smirnova K.V, Khmelenina V.N, Suzina N.E, Liesack W, Trotsenko Y.A. Methylotrophic autotrophy in Beijerinckia mobilis. J. Bacteriol. 2005;187:3884–3888. doi: 10.1128/JB.187.11.3884-3888.2005. doi:10.1128/JB.187.11.3884-3888.2005 [DOI] [PMC free article] [PubMed] [Google Scholar]
Ehrenfreund P, Charnley S.B. Organic molecules in the interstellar medium, comets, and meteorites: a voyage from dark clouds to the early Earth. Annu. Rev. Astron. Astrophys. 2000;38:427–483. doi:10.1146/annurev.astro.38.1.427 [Google Scholar]
Eisen J.A, et al. The complete genome sequence of Chlorobium tepidum TLS, a photosynthetic, anaerobic, green-sulfur bacterium. Proc. Natl Acad. Sci. USA. 2002;99:9509–9514. doi: 10.1073/pnas.132181499. doi:10.1073/pnas.132181499 [DOI] [PMC free article] [PubMed] [Google Scholar]
Falkowski P, et al. The global carbon cycle: a test of our knowledge of Earth as a system. Science. 2000;290:291–296. doi: 10.1126/science.290.5490.291. doi:10.1126/science.290.5490.291 [DOI] [PubMed] [Google Scholar]
Fenchel T, King G.M, Blackburn T.H. 2nd edn. Academic Press; San Diego, CA: 1998. Bacterial biogeochemistry: the ecophysiology of mineral cycling. [Google Scholar]
Ferris M.J, Sheehan K.B, Kühl M, Cooksey K, Wigglesworth-Cooksey B, Harvey R, Henson J.M. Algal species and light microenvironment in a low-pH, geothermal microbial mat community. Appl. Environ. Microbiol. 2005;71:7164–7171. doi: 10.1128/AEM.71.11.7164-7171.2005. doi:10.1128/AEM.71.11.7164-7171.2005 [DOI] [PMC free article] [PubMed] [Google Scholar]
Fuchs G. CO2 fixation in acetogenic bacteria: variations on a theme. FEMS Microbiol. Lett. 1986;39:181–213. doi:10.1111/j.1574-6968.1986.tb01859.x [Google Scholar]
Giordano M, Beardall J, Raven J.A. CO2 concentrating mechanisms in algae: mechanisms, environmental modulation, and evolution. Annu. Rev. Plant Biol. 2005;56:99–131. doi: 10.1146/annurev.arplant.56.032604.144052. doi:10.1146/annurev.arplant.56.032604.144052 [DOI] [PubMed] [Google Scholar]
Goldblatt C, Lenton T.M, Watson A.J. Bistability of atmospheric oxygen and the great oxidation. Nature. 2006;443:683–686. doi: 10.1038/nature05169. doi:10.1038/nature05169 [DOI] [PubMed] [Google Scholar]
Gould S.J. W. W. Norton; New York, NY: 1989. Wonderful life: the Burgess Shale and the nature of history. p. 323. [Google Scholar]
Grossman A.R, Bhaya D, Apt K.E, Kehoe D.M. Light-harvesting complexes in oxygenic photosynthesis: diversity, control, and evolution. Annu. Rev. Genet. 1995;29:231–288. doi: 10.1146/annurev.ge.29.120195.001311. doi:10.1146/annurev.ge.29.120195.001311 [DOI] [PubMed] [Google Scholar]
Hanson T.E, Tabita F.R. A ribulose-1,5-bisphosphate carboxylase/oxygenase (rubisco)-like protein from Chlorobium tepidum that is involved with sulfur metabolism and the response to oxidative stress. Proc. Natl Acad. Sci. USA. 2001;98:4397–4402. doi: 10.1073/pnas.081610398. doi:10.1073/pnas.081610398 [DOI] [PMC free article] [PubMed] [Google Scholar]
Hügler M, Huber H, Stetter K.O, Fuchs G. Autotrophic CO2 fixation pathways in archaea (Crenarchaeota) Arch. Microbiol. 2003a;179:160–173. doi: 10.1007/s00203-002-0512-5. [DOI] [PubMed] [Google Scholar]
Hügler M, Krieger R.S, Jahn M, Fuchs G. Characterization of the acetyl-CoA/priopionyl-CoA carboxylase in Metallosphaera sedula. Eur. J. Biochem. 2003b;270:736–744. doi: 10.1046/j.1432-1033.2003.03434.x. doi:10.1046/j.1432-1033.2003.03434.x [DOI] [PubMed] [Google Scholar]
Ingalls A.E, Shah S.R, Hansman R.L, Aluwihare L.I, Santos G.M, Druffel E.R.M, Person A. Quantifying archaeal community autotrophy in the mesopelagic ocean using natural radiocarbon. Proc. Natl Acad. Sci. USA. 2006;103:6442–6447. doi: 10.1073/pnas.0510157103. doi:10.1073/pnas.0510157103 [DOI] [PMC free article] [PubMed] [Google Scholar]
Irvine W.M. Extraterrestrial organic matter: a review. Orig. Life Evol. Biosph. 1998;28:365–383. doi: 10.1023/a:1006574110907. doi:10.1023/A:1006574110907 [DOI] [PubMed] [Google Scholar]
Ishii M, Miyake T, Satoh T, Sugiyama H, Oshima Y, Kodama T, Igarashi Y. Autotrophic carbon dioxide fixation in Acidianus brierleyi. Arch. Microbiol. 1997;166:368–371. doi: 10.1007/BF01682981. doi:10.1007/s002030050397 [DOI] [PubMed] [Google Scholar]
Ivanovsky R.N, Krasilnikova E.N, Fal Y.I. A pathway of the autotrophic CO2 fixation in Chloroflexus aurantiacus. Arch. Microbiol. 1993;159:257–264. doi:10.1007/BF00248481 [Google Scholar]
Jacob F. Evolution and tinkering. Science. 1977;196:1161–1166. doi: 10.1126/science.860134. doi:10.1126/science.860134 [DOI] [PubMed] [Google Scholar]
Kanamori T, Kanou N, Atomi H, Imanaka T. Enzymatic characterization of prokaryotic urea carboxylase. J. Bacteriol. 2004;186:2532–2539. doi: 10.1128/JB.186.9.2532-2539.2004. doi:10.1128/JB.186.9.2532-2539.2004 [DOI] [PMC free article] [PubMed] [Google Scholar]
Kolber Z.S, et al. Contribution of aerobic photoheterotrophic bacteria to the carbon cycle in the ocean. Science. 2001;292:2492–2495. doi: 10.1126/science.1059707. doi:10.1126/science.1059707 [DOI] [PubMed] [Google Scholar]
Kopp R.E, Kirschvink J.L, hilburn I.A, Nash C.Z. The Paleoproterozoic Snowball Earth: a climate disaster triggered by the evolution of oxygenic photosynthesis. Proc. Natl Acad. Sci. USA. 2005;102:11 131–11 136. doi: 10.1073/pnas.0504878102. doi:10.1073/pnas.0504878102 [DOI] [PMC free article] [PubMed] [Google Scholar]
Kühl M, Chen M, Ralph P.J, Schreiber U, Larkum A.W.D. A niche for cyanobacteria containing chlorophyll d. Nature. 2005;433:820. doi: 10.1038/433820a. doi:10.1038/433820a [DOI] [PubMed] [Google Scholar]
Lazcano A, Miller S.L. On the origin of metabolic pathways. J. Mol. Evol. 1999;49:424–431. doi:10.1007/PL00006565 [PubMed] [Google Scholar]
Lin C, Shalitin D. Cryptochrome structure and signal transduction. Annu. Rev. Plant Biol. 2003;54:469–496. doi: 10.1146/annurev.arplant.54.110901.160901. doi:10.1146/annurev.arplant.54.110901.160901 [DOI] [PubMed] [Google Scholar]
Lovelock J.E. Thermodynamics and the recognition of alien biospheres. Proc. R. Soc. B. 1975;189:167–181. doi:10.1098/rspb.1975.0051 [Google Scholar]
McFadden B.A. Autotrophic CO2 fixation and the evolution of ribulose diphosphate carboxylase. Bacteriol. Rev. 1973;37:289–319. doi: 10.1128/br.37.3.289-319.1973. [DOI] [PMC free article] [PubMed] [Google Scholar]
Nielsen, R. 2005 Solar radiation. See
Nikolau B.J, Ohlrogge J.B, Wurtele E.S. Plant biotin-containing carboxylases. Arch. Biochem. Biophys. 2003;414:211–222. doi: 10.1016/s0003-9861(03)00156-5. doi:10.1016/S0003-9861(03)00156-5 [DOI] [PubMed] [Google Scholar]
Nisbet E.G, Carr J.R, Van Dover C.L. Origins of photosynthesis. Nature. 1995;373:479–480. doi:10.1038/373479a0 [Google Scholar]
Nisbet E.G, Grassineau N.V, Howe C.J, Abell P.I, Regelous M, Nisbet R.E.R. The age of rubisco: the evolution of oxygenic photosynthesis. Geobiology. 2007;5:311–335. doi:10.1111/j.1472-4669.2007.00127.x [Google Scholar]
Orgel L.E. Self-organizing biochemical cycles. Proc. Natl Acad. Sci. USA. 2000;97:12 503–12 507. doi: 10.1073/pnas.220406697. doi:10.1073/pnas.220406697 [DOI] [PMC free article] [PubMed] [Google Scholar]
Pain R.H. The evolution of enzyme activity. Nature. 1982;299:486–487. doi: 10.1038/299486a0. doi:10.1038/299486a0 [DOI] [PubMed] [Google Scholar]
Permentier H.P, Neerken S, Overmann J, Amesz J. A bacteriochlorophyll a antenna complex from purple bacteria absorbing at 963 nm. Biochemistry. 2001;40:5573–5578. doi: 10.1021/bi0024308. doi:10.1021/bi0024308 [DOI] [PubMed] [Google Scholar]
Raven J.A. Phagotrophy in phototrophs. Limnol. Oceanogr. 1997;42:198–205. [Google Scholar]
Raven J.A. Photosynthesis in watercolours. Nature. 2007;448:418. doi: 10.1038/448418a. doi:10.1038/448418a [DOI] [PubMed] [Google Scholar]
Raven J.A, Cockell C.S. Influence on photosynthesis of starlight, moonlight, planetlight, and light pollution (reflections on photosynthetically active radiation in the universe) Astrobiology. 2006;6:668–675. doi: 10.1089/ast.2006.6.668. doi:10.1089/ast.2006.6.668 [DOI] [PubMed] [Google Scholar]
Rosing M.T, Frei R. U-rich Archaean sea-floor sediments from Greenland indications of >3700 Ma oxygenic photosynthesis. Earth Planet. Sci. Lett. 2004;217:237–244. doi:10.1016/S0012-821X(03)00609-5 [Google Scholar]
Rothschild L.J. A model for diurnal patterns of carbon fixation in a Precambrian microbial mat based on a modern analog. Biosystems. 1991;25:13–23. doi: 10.1016/0303-2647(91)90009-a. doi:10.1016/0303-2647(91)90009-A [DOI] [PubMed] [Google Scholar]
Rothschild L.J, Giver L.J. Photosynthesis below the surface in a cryptic microbial mat. Int. J. Astrobiol. 2003;1:295–304. doi:10.1017/S1473550403001320 [Google Scholar]
Rothschild L.J, Mancinelli R.L. Model of carbon fixation in microbial mats from 3,500 Myr ago to the present. Nature. 1990;345:710–712. doi: 10.1038/345710a0. doi:10.1038/345710a0 [DOI] [PubMed] [Google Scholar]
Sancar A. Crytochrome: the second photoactive pigment in the eye and its role in circadian photoreception. Annu. Rev. Biochem. 2000;69:31–67. doi: 10.1146/annurev.biochem.69.1.31. doi:10.1146/annurev.biochem.69.1.31 [DOI] [PubMed] [Google Scholar]
Sato T, Atomi H, Imanaka T. Archaeal type III rubiscos function in a pathway for AMP metabolism. Science. 2007;315:1003–1006. doi: 10.1126/science.1135999. doi:10.1126/science.1135999 [DOI] [PubMed] [Google Scholar]
Schenk S, Decker K. Horizontal gene transfer involved in the convergent evolution of the plasmid-encoded enantioselective 6-hydroxynicotine oxidases. J. Mol. Evol. 1999;48:178–186. doi: 10.1007/pl00006456. doi:10.1007/PL00006456 [DOI] [PubMed] [Google Scholar]
Schidlowski M. A 3,800-million-year isotopic record of life from carbon in sedimentary rocks. Nature. 1988;333:313–318. doi:10.1038/333313a0 [Google Scholar]
Schopf J.W. Paleobiology of the Archean. In: Schopf J.W, Klein C, editors. The Proterozoic biosphere. Cambridge University Press; Cambridge, UK: 1992. pp. 25–39. [Google Scholar]
Shi T, Bibby T.S, Jiang L, Irwin A.J, Falkowski P.G. Protein interactions limit the rate of evolution of photosynthetic genes in cyanobacteria. Mol. Biol. Evol. 2005;22:2179–2189. doi: 10.1093/molbev/msi216. doi:10.1093/molbev/msi216 [DOI] [PubMed] [Google Scholar]
Shiba H, Kawasumi T, Igarashi Y, Kodama T, Minoda Y. The CO2 assimilation via the reductive tricarboxylic acid cycle in an autotrophic, aerobic hydrogen-oxidizing bacterium Hydrogenobacter thermophilus. Arch. Microbiol. 1985;141:198–203. doi:10.1007/BF00408058 [Google Scholar]
Shively J.M, Barton L.L. Variations in autotrophic life. Academic Press; London, UK: 1991. p. 346. [Google Scholar]
Shively J.M, van Keulen G, Meijer W.G. Something from almost nothing: carbon dioxide fixation in chemoautotrophs. Annu. Rev. Microbiol. 1998;52:191–230. doi: 10.1146/annurev.micro.52.1.191. doi:10.1146/annurev.micro.52.1.191 [DOI] [PubMed] [Google Scholar]
Smith E, Morowitz H.J. Universality in intermediary metabolism. Proc. Natl Acad. Sci. USA. 2004;101:13 168–13 173. doi: 10.1073/pnas.0404922101. doi:10.1073/pnas.0404922101 [DOI] [PMC free article] [PubMed] [Google Scholar]
Snow T.P, Witt A.N. The interstellar carbon budget and the role of carbon in dust and large molecules. Science. 1995;270:1455–1460. doi: 10.1126/science.270.5241.1455. doi:10.1126/science.270.5241.1455 [DOI] [PubMed] [Google Scholar]
Stomp M, Huisman J, de Jongh F, Veraart A.J, Gerla D, Rijkeboer M, Ibelings B.W, Wollenzien U.I.A, Stal L.J. Adaptive divergence in pigment composition promotes phytoplankton biodiversity. Nature. 2004;432:104–107. doi: 10.1038/nature03044. doi:10.1038/nature03044 [DOI] [PubMed] [Google Scholar]
Tarter J.C, et al. A re-appraisal of the habitability of planets around M dwarf stars. Astrobiology. 2007;7:30–65. doi: 10.1089/ast.2006.0124. doi:10.1089/ast.2006.0124 [DOI] [PubMed] [Google Scholar]
Tcherkez G.G.B, Farquhar G.D, Andrews T.J. Despite slow catalysis and confused substrate specificity, all ribulose bisphosphate carboxylases may be nearly perfectly optimized. Proc. Natl Acad. Sci. USA. 2006;203:7246–7251. doi: 10.1073/pnas.0600605103. doi:10.1073/pnas.0600605103 [DOI] [PMC free article] [PubMed] [Google Scholar]
Tice M.M, Lowe D.R. Hydrogen-based carbon fixation in the earliest photosynthetic organisms. Geology. 2006;34:37–40. doi:10.1130/G22012.1 [Google Scholar]
Vershinin A. Biological functions of carotenoids—diversity and evolution. BioFactors. 1999;10:99–104. doi: 10.1002/biof.5520100203. [DOI] [PubMed] [Google Scholar]
Visser C.M, Kellogg R.M. Biotin. Its place in evolution. J. Mol. Evol. 1978;11:171–187. doi: 10.1007/BF01733892. doi:10.1007/BF01733892 [DOI] [PubMed] [Google Scholar]
Vitkup D, Kharchenko P, Wagner A. Influence of metabolic network structure and function on enzyme evolution. Genome Biol. 2006;7:R39. doi: 10.1186/gb-2006-7-5-r39. doi:10.1186/gb-2006-7-5-r39 [DOI] [PMC free article] [PubMed] [Google Scholar]
Wachtershauser G. Evolution of the first metabolic cycles. Proc. Natl Acad. Sci. USA. 1990;87:200–204. doi: 10.1073/pnas.87.1.200. doi:10.1073/pnas.87.1.200 [DOI] [PMC free article] [PubMed] [Google Scholar]
Walsh M.M, Lowe D.R. Filamentous microfossils from the 3,500-Myr-old Onverwacht Group, Barberton Mountain Land, South Africa. Nature. 1985;314:530–532. doi: 10.1016/0301-9268(92)90074-x. doi:10.1038/314530a0 [DOI] [PubMed] [Google Scholar]
Walter M.R. Archean stromatolites: evidence of the Earth's earliest benthos. In: Schopf J.W, editor. Earth's earliest biosphere. Princeton University Press; Princeton, NJ: 1983. pp. 187–213. [Google Scholar]
Williams T.J, Zhang C.L, Scott J.H, Bazylinski D.A. Evidence for autotrophy via the reverse tricarboxylic acid cycle for the marine magnetotactic coccus strain MC-1. Appl. Environ. Microbiol. 2006;72:1322–1329. doi: 10.1128/AEM.72.2.1322-1329.2006. doi:10.1128/AEM.72.2.1322-1329.2006 [DOI] [PMC free article] [PubMed] [Google Scholar]
Wolstencroft R.D, Raven J.A. Photosynthesis: likelihood of occurrence and possibility of detection on Earth-like planets. Icarus. 2002;157:535–548. doi:10.1006/icar.2002.6854 [Google Scholar]
Wood H.G, Ljungdahl L.G. Autotrophic character of the acetogenic bacteria. In: Shively J.M, Barton L.L, editors. Variations in autotrophic life. Academic Press; London, UK: 1991. pp. 201–250. [Google Scholar]
Wu G, Fiser A, ter Kuile B, Sali A, Müller M. Convergent evolution of Trichomonas vaginalis lactate dehydrogenase from malate dehydrogenase. Proc. Natl Acad. Sci. USA. 1999;96:6285–6290. doi: 10.1073/pnas.96.11.6285. doi:10.1073/pnas.96.11.6285 [DOI] [PMC free article] [PubMed] [Google Scholar]
Xiong J, Bauer C.E. Complex evolution of photosynthesis. Annu. Rev. Plant Biol. 2002;53:503–521. doi: 10.1146/annurev.arplant.53.100301.135212. doi:10.1146/annurev.arplant.53.100301.135212 [DOI] [PubMed] [Google Scholar]
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870 | https://openoregon.pressbooks.pub/mhccmajorsbio/chapter/4-6-flagella-and-cilia/ | Skip to content
Flagella and Cilia
Flagella (singular = flagellum) are long, hair-like structures that extend from the plasma membrane and are used to move an entire cell, (for example, sperm, Euglena). When present, the cell has just one flagellum or a few flagella. Prokaryotes sometimes have flagella, but they are structurally very different from eukaryotic flagella. Prokaryotes can have more than one flagella. They serve the same function in both prokaryotes and eukaryotes (to move an entire cell).
When cilia (singular = cilium) are present, however, they are many in number and extend along the entire surface of the plasma membrane. They are short, hair-like structures that are used to move entire cells (such as paramecium) or move substances along the outer surface of the cell (for example, the cilia of cells lining the fallopian tubes that move the ovum toward the uterus, or cilia lining the cells of the respiratory tract that move particulate matter toward the throat that mucus has trapped). Cilia are not found on prokaryotes.
References
Unless otherwise noted, images on this page are licensed under CC-BY 4.0 by OpenStax.
Text adapted from: OpenStax, Concepts of Biology. OpenStax CNX. May 18, 2016 |
871 | https://www.albert.io/blog/semi-log-plots-a-key-tool-in-analyzing-exponential-data-for-ap-precalculus/ | Published Time: 2025-02-03T22:39:38+00:00
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Semi-Log Plots: A Key Tool in Analyzing Exponential Data for AP® Precalculus
The Albert Team
Last Updated On: May 5, 2025
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Introduction
What is a Semi-Log Plot?
Characteristics of a Semi-Log Plot
Advantages of Using Semi-Log Plots
Linearization of Exponential Data
Practical Applications of Semi-Log Plots
Quick Reference Chart: Vocabulary and Definitions
Conclusion
Need help preparing for your AP® Precalculus exam?
Introduction
Logarithms are fundamental in mathematics, offering a powerful way to manage exponential growth and decay. Sometimes, grappling with exponential data can be daunting. This is where the semi-log plot comes into play. Used extensively in various fields, these plots help us visualize data that grows or decays exponentially. This guide is designed to demystify section 2.15 Semi-Log Plots and arm students with the tools needed to tackle them confidently, especially when preparing for the AP® Precalculus exam.
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What is a Semi-Log Plot?
A semi-log plot is a graph where one axis—usually the y y y-axis—is on a logarithmic scale, while the other is linear. Unlike regular graphs, which have both axes on linear scales, semi-log plots represent exponential relationships in a simpler, more linear form.
The logarithmic axis is vital because it compresses the scale. Thus, differences in exponential data become more apparent. For example, in biological research, populations growing exponentially can be plotted more comfortably, revealing trends that a standard graph might obscure.
Characteristics of a Semi-Log Plot
In a semi-log plot:
One axis, usually the y y y-axis, has a logarithmic scale.
The x x x-axis remains linear.
Exponential relationships appear as straight lines.
This property is helpful because it makes trends and relationships easier to analyze. The real magic happens when exponential data transforms from curving steeply upwards into a tidy straight line once plotted on a semi-log plot.
Example: Converting Data Points to a Semi-Log Plot
A semi-log plot helps visualize exponential relationships by transforming one axis using logarithms. This technique is useful for identifying trends in exponential data, such as population growth, radioactive decay, or financial interest.
Scenario: Bacterial Growth
A bacterial culture triples every hour. Given the time (in hours) and population, we want to convert this dataset into a semi-log plot.
Step 1: List the Data
We collect the following data:
| Hours | Population |
--- |
| 1 | 3 |
| 2 | 9 |
| 3 | 27 |
| 4 | 81 |
Since the population follows exponential growth, we transform the y y y-values using logarithms.
Step 2: Apply Logarithm to y y y
We take the logarithm (base 10) of each y y y-value:
log(3)≈0.48\log(3) \approx 0.48 lo g(3)≈0.4 8
log(9)≈0.95\log(9) \approx 0.95 lo g(9)≈0.9 5
log(27)≈1.43\log(27) \approx 1.43 lo g(2 7)≈1.4 3
log(81)≈1.91\log(81) \approx 1.91 lo g(8 1)≈1.9 1
Step 3: Plot on a Semi-Log Scale
The x x x-axis (horizontal) remains unchanged, representing hours.
The y y y-axis (vertical) now represents log(y)\log(y)lo g(y) instead of y y y.
Image created using Desmos (CC BY-SA 4.0)
Advantages of Using Semi-Log Plots
Semi-log plots are superb for visualizing data with exponential trends because they simplify analysis. They streamline recognizing exponential trends by transforming the data into a straighter line.
There’s no need to adjust values manually. This easily readable format allows for quick modeling of exponential increases or decreases.
Example: Exponential Model with Semi-Log Plots
Consider an investment’s growth:
Initial amount: $100
Growth rate: 10% per year
Instead of seeing an upward curve over a 5-year period, the semi-log plot displays this as a clear line, aiding in easy prediction and extension.
Image created using Desmos (CC BY-SA 4.0)
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Linearization of Exponential Data
Linearization is when exponential data is converted into a linear form. Using semi-log plots, the exponential function y=a b x y = ab^x y=a b x becomes a straight line.
Here’s how to create a linear model:
Apply a Logarithmic Transformation: For y=a b x y = ab^x y=a b x, the equation in semi-log form becomes: log(y)=log(a)+x⋅log(b)\log(y) = \log(a) + x \cdot \log(b)lo g(y)=lo g(a)+x⋅lo g(b)
Plot Data on a Semi-Log Plot: After plotting, the linear pattern can be extrapolated.
Example: Linearization Using a Semi-Log Plot
For data y=5×2 x y = 5 \times 2^x y=5×2 x over x x x values (1, 2, 3), transform using steps similar to the bacterial growth example. This results in a linear graph on a semi-log plot.
Practical Applications of Semi-Log Plots
Leopoldo Martin R,CC BY-SA 4.0, via Wikimedia Commons
Semi-log plots appear in many real-world scenarios:
Biology:Analyzing cell growth
Finance:Understanding compound interest
Engineering:Signal processing
Each field relies on the power of simplifying exponential datasets into a manageable and insightful format.
Quick Reference Chart: Vocabulary and Definitions
TermDefinition
Semi-log Plot A graph with a logarithmic scale on one axis.
Logarithmic Axis An axis where equal distances reflect ratios and exponents.
Linearization Converting exponential data into a linear form.
Exponential Function A function in the form y=a b x y = ab^x y=a b x.
Explore these concepts thoroughly and gain confidence in visualizing and interpreting exponential data using semi-log plots.
Conclusion
Semi-log plots are essential for visualizing exponential data, making them invaluable for students in AP® Precalculus. Understanding their characteristics, advantages, and practical applications opens avenues to decipher complex data forms more straightforwardly. To master this, practice is key. Pursue exercises to strengthen these concepts and reveal underlying exponential relationships effortlessly.
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2.13 Exponential and Logarithmic Equations and Inequalities
2.14 Logarithmic Function Context and Data Modeling
3.1 Periodic Phenomena
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872 | https://www.giga-tires.com/blog/tread-patterns-on-tire-efficiency/?srsltid=AfmBOorjKMBgXkBhhp-XM09ux_F__Xf49tUpKFpsAn6Pt3RwyeudRTeu | Investigating the Influence of Tread Patterns on Tire Efficiency
Table of Contents
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Main Function of Tire Tread Patterns
Tires are fundamental components of any vehicle, critical for safety, efficiency, and overall performance. Among tires’ various features, tread patterns are essential elements that significantly influence their functionality. This article delves into the importance of tread patterns and provides an overview of tire performance metrics, offering insights into how these designs impact driving experiences.
The Importance of Tread Patterns
Tread patterns are not just aesthetic features; they are engineered designs that perform multiple critical functions:
Overview of Tire Performance Metrics
Understanding tire performance involves several key metrics:
The design and functionality of tread patterns are vital in determining the overall performance of a tire. They significantly affect not only the safety aspects but also the efficiency and comfort of the driving experience. Understanding these patterns and the critical performance metrics helps make informed decisions when choosing tires for any vehicle. This knowledge is essential for drivers, enthusiasts, and professionals, ensuring safety, reliability, and performance on the road.
Basics of Tread Patterns
Tire tread patterns are more than just the distinct designs you see on the outer surface of tires. They are sophisticated engineering solutions developed to ensure maximum vehicle safety, performance, and efficiency under various conditions. This chapter aims to unpack the basics of tread patterns, focusing on their design, the types available, and how they have evolved.
Understanding Tread Design
Tread design in tires is a complex science that balances multiple factors:
Types of Tread Patterns
There are several primary types of tread patterns, each designed for specific driving conditions and vehicle types:
The Evolution of Tread Designs
The evolution of tread patterns reflects the advancements in vehicle performance and road safety:
Understanding the basics of tread patterns is crucial for anyone interested in automotive technology or vehicle maintenance. The tire’s design and type of tread pattern significantly impact its performance, affecting everything from fuel efficiency and road noise to handling and safety in adverse conditions. The evolution of tread designs continues to push the boundaries of what tires can achieve, reflecting the ever-growing demands of modern vehicles and road conditions.
Tread Patterns and Road Grip
The interaction between a vehicle’s tires and the road surface is fundamental to driving safety and efficiency. This chapter explores how tread patterns influence this interaction, focusing on the science of traction and how tread performance varies on both dry and wet surfaces.
The Science of Traction
Traction is the friction between the tire and the road surface, allowing the vehicle to accelerate, decelerate, and maintain grip during cornering. Several factors in tread design influence traction:
Effect of Tread on Dry Surfaces
On dry surfaces, traction depends significantly on the tread pattern and compound:
Tread Performance on Wet Roads
The performance of tread patterns on wet roads is critical for safety, particularly in preventing hydroplaning:
The design and condition of tire treads are crucial in determining a vehicle’s road grip. Tread patterns must be carefully engineered for the best possible traction on dry and wet roads. This understanding is vital for drivers, vehicle manufacturers, and tire designers alike, emphasizing the importance of tire maintenance and selecting suitable tires for varying driving conditions. As technology advances, we can expect further enhancements in tread design, improving safety and efficiency in all driving conditions.
Tread Designs and Fuel Efficiency
The relationship between tire tread designs and fuel efficiency is critical to modern automotive engineering. This chapter explores how tread patterns influence rolling resistance, a key factor affecting fuel consumption, and discusses tread designs optimized for fuel efficiency.
Impact of Tread Patterns on Rolling Resistance
Rolling resistance is the force resisting the motion when a tire rolls on a surface. Various factors, including tread design, influence it:
Tread Designs for Optimized Fuel Consumption
To optimize fuel consumption, tire manufacturers have developed specific tread designs:
The design of tire treads plays a significant role in determining a vehicle’s fuel efficiency. The ongoing development of low-rolling resistance tires demonstrates the tire industry’s commitment to reducing fuel consumption and greenhouse gas emissions. Understanding the impact of tread design on fuel efficiency is crucial for consumers when selecting tires, as it can lead to cost savings and a reduced environmental footprint. As environmental concerns continue to grow, the importance of fuel-efficient tread designs is likely to become even more pronounced in the tire industry.
Tread Patterns and Tire Longevity
Tire longevity is crucial for vehicle owners, impacting safety and economic factors. The tread pattern significantly affects how a tire wears over time. This chapter explores the relationship between tread patterns and tire longevity, focusing on the natural wear and tear tire experience and the design features that can extend tire life.
Tread, Wear, and Tear Over Time
Several factors influence the way a tire’s tread wears over time:
Design Features for Extended Tire Life
Tire manufacturers incorporate various design features to enhance tire longevity:
Understanding how tread patterns and other design features impact tire longevity is essential for maintaining vehicle safety and optimizing cost-effectiveness. Regular maintenance and choosing the right tire for specific driving needs and conditions can significantly extend a tire’s life. As tire technology advances, we can expect further innovations that will enhance tire performance and longevity, benefiting consumers and the environment.
Innovations in Tread Technology
The tire industry is constantly evolving, driven by technological advancements and a growing emphasis on sustainability and efficiency. Innovations in tread technology play a significant role in shaping the future of automotive mobility. This chapter delves into the latest developments in tread patterns, focusing on the end of these designs and the emergence of eco-friendly and innovative tread technologies.
The Future of Tread Patterns
A combination of advanced engineering, materials science, and consumer needs is shaping the future of tread patterns:
Eco-Friendly and Smart Tread Designs
Eco-friendliness and intelligent technologies are at the forefront of tread innovation:
Innovations in tread technology are enhancing the performance and safety of tires and paving the way for a more sustainable and efficient future in transportation. As these technologies continue to develop, we can expect tires to be better suited to our driving needs and more in harmony with the environment. This evolution in tread technology will be a critical component in the broader shift towards more sustainable and intelligent transportation systems.
Choosing the Right Tread for Your Vehicle
Selecting the proper tire tread for your vehicle is a crucial decision that affects not only the performance and efficiency of your car but also your safety on the road. This chapter provides a comprehensive guide on the factors to consider when choosing tires and suitable tread patterns for different types of vehicles.
Factors to Consider When Selecting Tires
Choosing the correct tire involves considering several key factors:
Tread Patterns for Different Vehicle Types
Different vehicles and driving styles require specific tread patterns for optimal performance:
Passenger Vehicles typically use all-season tires with moderate tread depths and balanced patterns, which provide a smooth ride and good performance in most conditions.
Selecting the proper tread for your vehicle is a nuanced decision that requires careful consideration of various factors, including vehicle type, driving conditions, and personal preferences. Understanding the different types of tread patterns and their intended applications can help you choose tires that will offer the best performance, safety, and longevity for your specific needs. As technology advances, the range of available options grows, offering drivers more tailored choices for their unique driving requirements.
Conclusion & Recommendations
As we conclude our exploration into the world of tire treads, it’s evident that tread design is critical in determining a tire’s performance, safety, and efficiency. This final chapter recaps the impacts of tread design and looks forward to future perspectives in tire technology.
Recap of Tread Design Impacts
Tire treads have a profound impact on various aspects of driving:
Future Perspectives in Tire Technology
Looking ahead, tire technology is set to advance in several exciting directions:
Recommendations
Consider Giga Tires for their advanced grip technology and superior handling in wet and dry conditions, ensuring a safe and comfortable driving experience.
For consumers and professionals alike, staying informed about the latest developments in tire technology is crucial. It’s recommended to:
The significance of tire tread designs extends far beyond their immediate impact on a vehicle’s handling and safety. They are vital to the broader automotive technology landscape, evolving continuously to meet new challenges and opportunities. As we look forward to the future of tire technology, it’s clear that innovation in tread design will continue to play a pivotal role in shaping the future of transportation.
FAQs
What are the effects of tire tread design?
Tire tread design influences traction, handling, and safety. Different tread patterns affect grip on various road surfaces, water dispersal to prevent hydroplaning, and overall tire wear and noise, impacting performance and comfort.
Is it wrong to have tires with different tread patterns?
Yes, it’s generally not advisable. Different tread patterns can lead to uneven tire wear, reduced traction, and compromised handling, particularly in adverse weather conditions. Consistency in tread patterns is critical for balanced vehicle dynamics.
What effect do different tire treads have on performance?
Different tire treads can drastically alter a vehicle’s performance. For example, aggressive impressions are better for off-road traction but may reduce fuel efficiency, while finer tracks offer better fuel economy and smoother highway driving.
What is the primary function of the tread pattern on tires?
The primary function of tread patterns is to provide traction. They grip the road, channel water away to prevent hydroplaning and distribute the vehicle’s load, enhancing safety and stability.
What is the best tread for tires?
The best tread depends on your driving needs and conditions. All-season treads offer good overall performance in various shapes, while specialized impressions like winter or off-road patterns are better for specific challenges like snow or rough terrain.
What is the most common tread pattern?
The most common tread pattern is the all-season design, which provides a balanced performance in dry and wet conditions and offers moderate snow traction. It’s a versatile choice suitable for a wide range of driving conditions.
Vukasin Herbez
At Giga Tires, we’re all about making your tire shopping experience simple and convenient. Whether you want your tires delivered right to your door or to your local tire shop, we’ve got you covered. Plus, we’ve got a wide variety of tires to choose from, whether you prefer top-name brands or more budget-friendly options. So, when you need new tires, think Giga Tires, and hit the road with confidence!
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874 | https://www.cuemath.com/calculus/horizontal-scaling/ | Horizontal scaling
Horizontal scaling
There are various types of transformations of the graph of a function:
| Operation | Transformation of the Function f(x) |
--- |
| -f(x) | reflection over x-axis |
| f(-x) | reflection over y-axis |
| f(x) + k | Translates up by k units |
| f(x) - k | Translates down by k units |
| f(x + k) | Translates left by k units |
| f(x - k) | Translates right by k units |
| k f(x) | Vertically scales the graph in y-axis (k > 1 stretch, 0 < k < 1 shrink vertical) |
| f(kx) | Horizontally scales the graph in x-axis (k > 1 shrink, 0 < k < 1 stretch horizontal) |
In this section, we will learn about horizontal scaling in detail and practice solving questions around it.
You can check out the interactive simulations to know more about the lesson and try your hand at solving a few interesting interactive questions at the end of the page.
Lesson Plan
| | |
--- |
| 1. | What Do You Mean by Horizontal Scaling? |
| 2. | Important Notes on Horizontal Scaling |
| 3. | Solved Examples on Horizontal Scaling |
| 4. | Challenging Questions on Horizontal Scaling |
| 5. | Interactive Questions on Horizontal Scaling |
What Do You Mean by Horizontal Scaling?
Horizontal scaling means the stretching or shrinking the graph of the function along the x-axis.
Horizontal scaling can be done by multiplying the input with a constant.
Consider the following example:
Suppose, we have a function, y=f(x)
Horizontal scaling of the above function can be written as:y=f(Cx)
The graph stretches if the value of C < 1, and the graph will shink if the value of C > 1.
Coresponding points on the curves can be related from the table below:
| The point on f(x) | The point on y = f(Cx) |
--- |
| (x,y) | (Cx,y) |
Let us understand this with an example:
Example
We have a function y=f(x)
We want to shrink the graph of the function, hence we have to multiply the input or x-coordinate by a constant greater than 1, let us multiply the function with 2, i.e., y=f(2x)
We get,
We can see that the distance of the points on the curve gets closer to the y-axis.
How Does One Horizontally Scale a Graph?
Steps
Select the constant by which we want to scale the function.
Write the new function as g(x)=fC(x), where C is the constant.
Trace the new function graph by replacing each value of x with Cx.
X coordinate of each point in the graph is multiplied by ±C, and the curve shrinks/stretches accordingly.
Let us understand this by an example:
Example
Suppose we have a basic quadratic equation f(x)=x2 and the graphical representation of the graph is shown below.
We want to scale this function by a factor of +2.
So our constant becomes +2.
The equation of the new function will be:
g(x)=2f(x)=(2x)2
Now, we have to replace the value of x-coordinate by 2x.
Hence we have to plot the graph of a function: f(x)=(2x)2 and it is shrunk by a factor of +2 units in the x-direction.
Note: as we have scaled it with a factor of +2 units, it has made the graph steeper.
Horizontal Scaling in Graphs
Horizontal scaling of various graphs of the functions are shown below:
Horizontal scaling of function f(x) = x+2 by a factor of 2 units is shown in the graph below:
Horizontal scaling of function f(x)=(x2+3x+2) by a factor of 4 units is shown in the graph below:
Horizontal scaling of function f(x) = sin x by a factor of -3, is shown in the graph below:
Important Notes
Horizontal scaling refers to the shrinking or stretching of the curve along the x-axis by some specific units.
Horizontal scaling of function f(x) is given by g(x) = ± f(Cx).
Solved Examples
| |
| Example 1 |
Trevor wants to shrink the function y=sinx horizontally by a factor of three. Can you find out the new function by plotting them in the graph?
Solution
On horizontally shrinking the curve of y=sinx, the new function will be: y=sin3x
Graph of the function is:
| |
| ∴ The new function will be: y=sin3x |
| |
| Example 2 |
A task was given to Emma to plot the graph on horizontal scaling of a function y=f(x)=x3 to a function y=g(x)=f(\dfracx2). Can you help her plot this scaling on the graph.
Solution
The the graph on vertical scaling of a function y=f(x) to a function y=g(x)=f(\dfracx2x) is the graph of y=(x2)3:
| |
| ∴ y=(x2)3 |
Challenging Questions
What are the properties of vertical scaling?
If the function f(x) has coordinates as (x,y), can you tell what will be coordinates of the point on the curve after C units of vertical scaling?
Can you list out horizontal and vertical scaling differences?
Interactive Questions
Here are a few activities for you to practice.
Select/Type your answer and click the "Check Answer" button to see the result.
Let's Summarize
We hope you enjoyed learning about horizontal scaling with solved examples and interactive questions. Now, you will be able to easily solve problems on horizontal scalability, horizontal compression, horizontal stretch, horizontal scaling graph.
About Cuemath
At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!
Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.
Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.
Frequently Asked Questions (FAQs)
1. What is horizontal scaling in the graph?
In equation y = f ( k x ) If the magnitude of k is greater than 1, then the graph will compress horizontally, but if the magnitude of k is less than 1, then the graph will stretch out horizontally.
2. What do you mean by scaling?
In scaling, we measure and assign the objects to the numbers according to the specified rules.
3. What are the challenges of horizontal scaling?
In horizontal scaling, the graph gets stretched or shrink only on the x-axis, not on the y-axis.
4. When to use horizontal scaling?
A horizontal scaling uses when the function of x-coordinate changes its values.
5. How do you scale a function horizontally?
We can scale a function horizontally when the scale stretched or shrank on the x-axis.
6. What is on the horizontal scale of the graph?
The x-axis is on the vertical scale of the graph.
Download SOLVED Practice Questions of Horizontal scaling for FREE
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875 | https://www.pharmacy.umaryland.edu/media/SOP/wwwpharmacyumarylandedu/centers/mhp/pharmassist/pdf/Module%201%20-%20Opioids%20and%20Opioid%20Epidemic.pdf | Module 1: Introduction to Opioids and the Opioid Epidemic Learning Objectives 1.
Define opioid and identify examples of natural, semisynthetic, and synthetic opioids 2.
Describe the precipitating factors and scope of the opioid epidemic on society Opioid Definition What is an opioid?
Papaver somniferium: the opium poppy Opium is made by drying latex from the seed pods of the opium poppy, Papaver somniferium Burillo-Putze G, Miro O. Opioids. In: Tintinalli JE, Stapczynski J, Ma O, Yealy DM, Meckler GD, Cline DM. eds. Tintinalli’s Emergency Medicine: A Comprehensive Study Guide, 8e New York, NY: McGraw-Hill; 2016. Accessed January 05, 2019.
Substance Abuse and Mental Health Services Administration. Opioids. Last updated 2.23.2016. Accessed 1.10.2019.
4 Examples of opioids • An opioid is any compound that acts at one of the three opioid receptors: µ (mu), ĸ (kappa), and δ (delta) • Opioids can be natural or synthetic; prescription medications or illegal drugs • Opiates are the natural compounds found in opium poppies • The three main alkaloids in opium (morphine, codeine, and thebaine) can be used as is or modified to synthesize new compounds Semi-synthetic opioids – fentanyl – carfentanil - tramadol (Duragesic) (Ultram) – buprenorphrine (Suboxone) hydrocodone (Vicodin) What is an opioid?
Burillo-Putze G, Miro O. Opioids. In: Tintinalli JE, Stapczynski J, Ma O, Yealy DM, Meckler GD, Cline DM. eds. Tintinalli’s Emergency Medicine: A Comprehensive Study Guide, 8e New York, NY: McGraw-Hill; 2016. Accessed January 05, 2019.
5 Opioid: • Any compound that acts at one of the three opioid receptors: µ (mu), ĸ (kappa), and δ (delta) • When opioids bind to the various opioid receptors, they can cause various effects • Opioid overdose occurs when an opioid overwhelms a person’s system and impairs his/her ability to breathe • Have the potential for tolerance, dependence, and addiction Effects of opioids Analgesia Euphoria Sedation Decreased GI motility Miosis Hallucinations/dysphoria Respiratory depression Antidepressant effect Cough suppression Examples of prescription opioids • Centers for Disease Control and Prevention. Opioid Overdose. Opioid Basics. Prescription Opioids. Accessed 1.6.2019.
6 • Prescription medications indicated for pain management, cough suppression, and treatment of opioid use disorders Opioid Brand Name Opiates Morphine MS Contin ® Kadian ® Codeine Tylenol 3 ® Tylenol 4 ® Semisynthetic Buprenorphrine Suboxone ® Subutex ® Oxycodone OxyContin® Hydrocodone Vicodin® Synthetic Methadone Dolophine ® Fentanyl Actiq ® Duragesic ® Tramadol Ultram ® Illicit Opioids Source: National Institute on Drug Abuse; National Institutes of Health; U.S. Department of Health and Human Services.
United States Drug Enforcement Agency. Centers for Disease Control and Prevention. Opioid Overdose. Opioid Basics. Fentanyl. Accessed 1.6.2019.
Centers for Disease Control and Prevention. Opioid Overdose. Opioid Basics. Heroin. Accessed 1.6.2019 7 Non-pharmaceutical Fentanyl Heroin Black Tar Heroin Packets of fentanyl laced heroin • Made from morphine (semisynthetic) • Can be a black sticky substance, or a white or brown powder • Can be injected, sniffed, snorted, or smoked; sometimes mixed with crack cocaine, called speed balling • Slang terms: Big H, Horse, Hell Dust Asian Heroin Heroin Closeup • Illicitly produced, synthetic drug • Pill form package to look like prescription medications • Powder form looks similar to heroin • Fentanyl is often mixed with heroin or cocaine • Unclear what substances are truly in the illicitly manufactured drugs Heroin Powder Question 1 A 55-year old female with severe cancer pain comes to the pharmacy to fill fentanyl. The patient is concerned about taking fentanyl because she heard it is the same type of medication as heroin. What is a true statement about opioids?
A. Fentanyl is an opiate because it is a naturally occurring compound in the opium poppy B. Both fentanyl and heroin are opioids. Because both substances act on the opioid receptor and any compound that acts on opioid receptors is an opioid C. People only overdose from illicit fentanyl not prescription fentanyl D. Fentanyl is not an opioid because it is a prescription medication 8 Question 1 A 55-year old female with severe cancer pain comes to the pharmacy to fill fentanyl. The patient is concerned about taking fentanyl because she heard it is the same type of medication as heroin. What is a true statement about opioids?
A. Fentanyl is an opiate because it is a naturally occurring compound in the opium poppy.
B. Both fentanyl and heroin are opioids. Because both substances act on the opioid receptor and any compound that acts on opioid receptors is an opioid.
C. People only overdose from illicit fentanyl not prescription fentanyl D. Fentanyl is not an opioid because it is a prescription medication 9 Key Points • An opioid is any compound that acts on one of the three opioid receptors • Opioids can be categorized as (1) opiates - compounds that occur naturally in opium (morphine, codeine), (2) semisynthetics -chemical modified natural compounds (buprenorphine, oxycodone, hydrocodone), and (3) synthetics – completely artificial compounds (methadone, fentanyl, tramadol) • Opioids can be prescription medications or illicit substances • All opioids can lead to deadly overdoses • Fentanyl is 50 to 100x more potent than morphine and is not detectable by opiate toxicology screens. This increases the risk for overdose when produced on the street and mixed with heroin or other drugs. 10 Opioid Epidemic CDC Injury Center: Opioid Overdose Data Analysis • CDC identifies and classifies which types of drugs are involved in an overdose and how that changes overtime • Centers for Disease Control and Prevention. Opioid Overdose. Accessed 1.6.2019.
Three Waves in the Rise of Opioid Overdose Deaths • Centers for Disease Control and Prevention. Opioid Overdose. Accessed 1.6.2019 • Wide-ranging online data for epidemiologic research (WONDER). Atlanta, GA: CDC, National Center for Health Statistics; 2017. Available at
• Seth P, Scholl L, Rudd RA, Bacon S. Increases and Geographic Variations in Overdose Deaths Involving Opioids, Cocaine, and Psychostimulants with Abuse Potential – United States, 2015-2016. MMWR Morb Mortal Wkly Rep. ePub: 29 March 2018.
• Kolodny et al. 2015. The prescription opioid and heroin crisis: A public health approach to an epidemic of addiction. Annual Review of Public Health, 36, 559-74 • Rudd RA, Aleshire N, Zibbell JE, Gladden RM. Increases in Drug and Opioid Overdose Deaths – United States, 2000-2014. MMWR 2016, 64(50); 1378-82.
13 From 1999 to 2017, more than 399,000 people have died from a opioid overdose Wave 1: Began in the 1990s with an increase in the opioid overdose deaths from prescription opioids. Wave 2: Began in 2010 with increase in the number of overdose deaths involving heroin Wave 3: Began in 2013 with increases in overdose deaths involving synthetic opioids, particularly illegally manufactured fentanyl Opioid-Related Overdose Deaths in Maryland Maryland one of the top 5 states with the highest-rates of opioid overdose National Institute of Health. National Institute on Drug Abuse. Opioid-Related Overdose Deaths February 2018. Accessed 12.17.18.
14 Since 1999, the death-rate has consistently been above the average, ranging from roughly 1.5 to 3 times the rate In 2016, nearly 30 deaths per 100,000 persons were related to opioids in Maryland compared to a national average of 13.3 deaths per 100,000 persons Total Number of Drug- and Alcohol-Related Intoxication Deaths in Maryland, 2007 - 2017 • Maryland Department of Health. Unintentional Drug and Alcohol-Related Intoxication Deaths in Maryland Annual Report. 2017. Accessed 1.6.19.
15 Question 2 Which of the following is a true statement about the opioid epidemic.
A. Maryland is a state with one of the lowest rates of opioid overdose in the country B. The opioid epidemic is starting to cause less deaths than it has in previous years C. A majority of deaths from the opioid epidemic in the US and Maryland are driven by overdoses from synthetic opioids, including illicitly manufactured fentanyl D. Illicitly manufactured fentanyl is expensive and difficult to make; therefore, it is not contributing to the opioid epidemic 16 Question 2 Which of the following is a true statement about the opioid epidemic.
A. Maryland is a state with one of the lowest rates of opioid overdose in the country B. The opioid epidemic is starting to cause less deaths than it has in previous years C. A majority of deaths from the opioid epidemic in the US and Maryland are driven by overdoses from synthetic opioids, including illicitly manufactured fentanyl D. Illicitly manufactured fentanyl is expensive and difficult to make; therefore, it is not contributing to the opioid epidemic 17 Key Points • The opioid epidemic is a serious problem • Maryland is one of the top-five states with the highest rate of opioid overdose • Since 2013, there has been a rise in the increase of opioid overdoses due to fentanyl, often which is illicitly manufactured 18 You have completed Module 1: Opioids and the Opioid Epidemic Key Points • An opioid is any compound that acts on one of the three opioid receptors • Opioids can be categorized as (1) opiates - compounds that occur naturally in opium (morphine, codeine), (2) semisynthetics - chemical modified natural compounds (buprenorphine, oxycodone, hydrocodone), and (3) synthetics – completely artificial compounds (methadone, fentanyl, tramadol) • Opioids can be prescription medications or illicit substances • All opioids can lead to deadly overdoses • Fentanyl is 50 to 100x more potent than morphine and is not detectable by opiate toxicology screens. This increases the risk for overdose when produced on the street and mixed with heroin or other drugs. • The opioid epidemic is a serious problem • Maryland is one of the top-five states with the highest rate of opioid overdose • Since 2013, there has been a rise in the increase of opioid overdoses due to fentanyl, often which is illicitly manufactured 20 |
876 | https://www.combinatorics.org/ojs/index.php/eljc/article/download/v17i1r66/pdf/ | Published Time: 2010-04-23T03:08:41.000Z
A simple Havel–Hakimi type algorithm to realize graphical degree sequences of directed graphs ∗
P´ eter L. Erd˝ os and Istv´ an Mikl´ os
A. R´ enyi Institute of Mathematics, Hungarian Academy of Sciences, Budapest, PO Box 127, H-1364, Hungary
{elp } { miklosi }@renyi.hu
Zolt´ an Toroczkai
Interdisciplinary Center for Network Science and Applications and Department of Physics University of Notre Dame Notre Dame, IN, 46556, USA
toro@nd.edu
Submitted: May 29, 2009; Accepted: Apr 21, 2010; Published: Apr 30, 2010 Mathematics Subject Classification: 05C07, 05C20, 90B10, 90C35
Abstract
One of the simplest ways to decide whether a given finite sequence of positive integers can arise as the degree sequence of a simple graph is the greedy algorithm of Havel and Hakimi. This note extends their approach to directed graphs. It also studies cases of some simple forbidden edge-sets. Finally, it proves a result which is useful to design an MCMC algorithm to find random realizations of prescribed directed degree sequences.
Keywords . network modeling; directed graphs; degree sequences; greedy algorithm
1 Introduction
The systematic study of graphs (or more precisely the linear graphs , as it was called in that time) began sometimes in the late forties, through seminal works by P. Erd˝ os, P. Tur´ an, W.T. Tutte, and others. One problem which received considerable attention was
∗
PLE was partly supported by OTKA (Hungarian NSF), under contract Nos. AT048826 and K 68262. IM was supported by a Bolyai postdoctoral stipend and OTKA (Hungarian NSF) grant F61730. ZT was supported in part by the NSF BCS-0826958, HDTRA 201473-35045 and by Hungarian Bioinformatics MTKD-CT-2006-042794 Marie Curie Host Fellowships for Transfer of Knowledge.
the electronic journal of combinatorics 17 (2010), #R66
1the existence of certain subgraphs of a given graph. For example such a subgraph could be a perfect matching in a (not necessarily bipartite) graph, or a Hamiltonian cycle, etc. Generally these substructures are called factors . The first couple of important results of this kind are due to W.T. Tutte who gave necessary and sufficient conditions for the existence of 1-factors and f -factors. In the case of complete graphs, the existence problem of such factors is considerably easier. In particular, the existence problem of (sometimes simple) undirected graphs with given degree sequences even admits simple greedy algorithms for its solution. Subsequently, the theory was extended for factor problems of directed graphs as well, but the greedy type algorithm mentioned above, to the best knowledge of the authors, is missing even today. In this paper we fill this gap: after giving a short and comprehensive (but definitely not exhausting) history of the f -factor problem (Section 2), we describe a greedy algo-rithm to decide the existence of a directed simple graph possessing the prescribed degree sequence (Section 3). In Section 4 we prove a consequence of the previous existence the-orem, which is a necessary ingredient for the construction of edge-swap based Markov Chain Monte Carlo (MCMC) methods to sample directed graphs with prescribed degree sequence. Finally in Section 5 we discuss a slightly harder existence problem of directed graphs with prescribed degree sequences where some vertex-pairs are excluded from the constructions. This result can help to efficiently generate all possible directed graphs with a given degree sequence.
2 A brief history (of f -factors)
For a given function f : V (G) → N ∪ { 0}, an f -factor of a given simple graph G(V, E ) is a subgraph H such that dH (v) = f (v) for all v ∈ V. One of the very first key results of modern graph theory is due to W.T. Tutte: in 1947 he gave a complete characterization of simple graphs with an f -factor in case of f ≡ 1 (Tutte’s 1-factor theorem, ). Tutte later solved the problem of the existence of f -factors for general f ’s (Tutte’s f -factor theorem, ). In 1954 he also found a beautiful graph transformation to handle f -factor problems via perfect matchings in bipartite graphs . This also gave a clearly polynomial time algorithm for finding f -factors. In cases where G is a complete graph, the f -factor problem becomes easier: then we are simply interested in the existence of a graph with a given degree sequence (the exact definitions will come in Section 3). In 1955 P. Havel developed a simple greedy algorithm to solve the degree sequence problem for simple undirected graphs (). In 1960 P. Erd˝ os and T. Gallai studied the f -factor problem for the case of a complete graph G, and proved a simpler Tutte-type result for the degree sequence problem (see ). As they already pointed out, the result can be derived directly form the original f -factor theorem, taking into consideration the special properties of the complete graph G, but their proof was independent of Tutte’s proof and they referred to Havel’s theorem. In 1962 S.L. Hakimi studied the degree sequence problem in undirected graphs with multiple edges (). He developed an Erd˝ os-Gallai type result for this much simpler case,
the electronic journal of combinatorics 17 (2010), #R66 2
and for the case of simple graphs he rediscovered the greedy algorithm of Havel. Since then this algorithm is referred to as the Havel–Hakimi algorithm .For directed graphs the analogous question of recognizability of a bi-graphical-sequence
comes naturally. In this case we are given two n-element vectors d+, d− of non-negative integers. The problem is the existence of a directed graph on n vertices, such that the first vector represents the out-degrees and the second one the in-degrees of the vertices in this graph. In 1957 D. Gale and H. J. Ryser independently solved this problem for simple directed graphs (there are no parallel edges, but loops are allowed), see [5, 13]. In 1958 C. Berge generalized these results for p-graphs where at most p parallel edges are allowed (). (Berge calls the out-degree and in-degree together the demi-degrees .) Finally in 1973, the revised version of his book Graphs () gives a solution for the p-graph problem, loops excluded. To show some of the afterlife of these results: D. West in his renowned recent textbook (), discusses the case of simple directed graphs with loops allowed. The analog of f -factor problems for directed graphs has a sparser history. Øystein Ore started the systematic study of that question in 1956 (see [11, 12]). His method is rather algebraic, and the finite and infinite cases - more or less - are discussed together. The first part developed the tools and proved the directly analogous result of Tutte’s
f -factor problem for finite directed graphs (with loops), while the second part dealt with the infinite case. In 1962 L.R. Ford and D.R. Fulkerson studied, generalized and solved the “original”
f -factor problem for a directed graph ~G (). Here lower and upper bounds were given for both demi-degrees of the desired subgraph (no parallel edges, no loops) with the original question naturally corresponding to equal lower and upper bounds. The solutions (as well as in Berge’s cases) are based on network flow theory. Finally, in a later paper Hakimi also proves results for bi-graphical sequences, however, without presenting a directed version of his original greedy algorithm (see ).
3 Greedy algorithm to realize bi-graphical sequences
A sequence d = {d1, d 2, . . . , d n} of nonnegative integers is called a graphical sequence if a simple graph G(V, E ) exists on n nodes, V = {v1, v 2, . . . , v n}, whose degree sequence is d.In this case we say that G realizes the sequence d. For simplicity of the notation we will consider only sequences of strictly positive integers ( dn > 0) to avoid isolated points. The following, well-known result, was proved independently by V. Havel and S.L. Hakimi.
Theorem 1 (Havel , Hakimi ) There exists a simple graph with degree sequence
d1 > 0, d 2 > · · · > dn > 0 ( n > 3) if and only if there exists one with degree sequence
d2 − 1, . . . , d d1+1 − 1, d d1+2 , . . . , d n. (Note that there is no prescribed ordering relation between d1 and the other degrees. )This can be proved using a recursive procedure, which transforms any realization of the degree sequence into the form described in the Theorem 1, by a sequence of two-edge swaps.
the electronic journal of combinatorics 17 (2010), #R66 3
A bi-degree-sequence (or BDS for short) ( d+, d−) = ( {d+1 , d +2 , . . . , d +
n
}, {d−
1
, d −
2
, . . . , d −
n
})of nonnegative integers is called a bi-graphical sequence if there exists a simple (no parallel edges, no loops) directed graph ~G(V, ~E) on n nodes, V = {v1, v 2, . . . , v n}, such that the out-degree and in-degree sequences together form ( d+, d−). (That is the out-degree of vertex vj is d+
j
and its in-degree is d−
j
.) In this case we say that ~G realizes our BDS. For simplicity, we will consider only sequences of strictly positive integer BDS’s, that is each degree is > 0 and d+
j
d−
j
0, to avoid isolated points. Our goal is to prove a Havel–Hakimi type algorithm to realize bi-graphical sequences. To that end we introduce the notion of normal order: we say that the BDS is in normal order if the entries satisfy the following properties: for each i = 1 , . . . , n −2 we either have
d−
i
d −
i+1
or d−
i
= d−
i+1
and d+
i
d+
i+1
. Clearly, all BDS-s can be arranged into normal order. Note that we made no ordering assumption about node vn (the pair d+
n
, d −
n
).
Theorem 2 Assume that the BDS (d+, d−) ( with d+
j
d−
j
0, j ∈ [1 , n ]) is in normal order and d+
n
0 ( that is the out-degree of the last vertex is positive ). Then (d+, d−) is bi-graphical if and only if the BDS
∆+
k
=
{ d+
k
if k 6 = n
0 if k = n , (1) ∆−
k
=
{ d−
k
− 1 if k 6 d+
n
d−
k
if k > d +
n
, (2)
with zero elements removed (those j for which ∆+
j
= ∆ −
j
= 0) is bi-graphical.
Before starting the proof, we emphasize the similarity between this result and the original HH-algorithm. As in the undirected case, using Theorem 2, we can find in a greedy way a proper realization of graphical bi-degree sequences. Indeed: choose any vertex vn with non-zero out-degree from the sequence, arrange the rest in normal order, then make d−
n
connections from vn to nodes with largest in-degrees, thus constructing the out-neighborhood of vn in the (final) realization. Next, remove the vertices (if any) from the remaining sequence that have lost both their in- and out- degrees in the process, pick a node with non-zero out-degree, then arrange the rest in normal order. Applying Theorem 2 again, we find the final out-neighborhood of our second chosen vertex. Step by step we find this way the out-neighborhood of all vertices, while their in-neighborhoods get defined eventually (being exhausted by incoming edges). Note, that every vertex in this process is picked at most once, namely, when its out-neighborhood is determined by the Theorem, and never again after that. Our forthcoming proof is not the simplest, however, we use a more general setup to shorten the proofs of later results. First, we define the partial order among k-element vectors of increasing positive integers: we say a b iff for each j = 1 , . . . , k we have aj 6 bj .
A possible out-neighborhood (or PON for short) of vertex vn is a d+
n
-element subset of V \ { vn} which is a candidate for an out-neighborhood of vn in some graphical rep-resentation. (In essence, a PON can be any d+
n
-element subset of V \ { vn} but later on
the electronic journal of combinatorics 17 (2010), #R66 4
we may consider some restrictions on it.) Let A be a PON of vn. Then denote by i(A)the vector of the increasingly ordered subscripts of the elements of A. (For example, if
A = {v2, v 4, v 9}, then i(A) = (2 , 4, 9).) Let A and B be two PONs of vn. We write:
B A ⇔ iB iA . (3) In this case we also say that B is to the left of A. (For example, B = {v1, v 2, v 6, v 7} is to the left of A = {v2, v 4, v 6, v 9}.)
Definition 3 Consider a bi-graphical BDS sequence (d+, d−) and let A be a PON of vn.The A-reduced BDS (d+∣∣A, d−∣∣A
) is defined as:
d+
k
∣∣A =
{ d+
k
if k 6 = n
0 if k = n , (4)
d−
k
∣∣A =
{ d−
k
− 1 if k ∈ i(A)
d−
k
if k 6 ∈ i(A). (5) In other words, if A is a PON in a BDS, then the reduced degree sequence (d+∣∣A, d−∣∣A
)
is obtained by removing the out-edges of node vn (according to the set A). As usual, if for a particular subscript k in the A-reduced BDS we have d+
k
∣∣A = d−
k
∣∣A = 0 then the vertex with this index is to be removed from the bi-degree sequence.
Lemma 4 Let (d+, d−) be a BDS, and let A be a possible out-neighborhood of vn. Fur-thermore let B be another PON with B = A \ { vk} ∪ { vi} where d−
i
d−
k
and in case of d−
i
= d−
k
we have d+
i
d+
k
. Then if (D+, D−) := (d+∣∣A, d−∣∣A
) is bi-graphical, so is
(d+∣∣B , d−∣∣B
).Proof. Since our A-reduced BDS ( D+, D−) is bi-graphical, there exists a directed graph
~G which realizes the bi-degree sequence ( D+, D−). We are going to show that in this case there exists a directed graph ~G′ which realizes the BDS (d+∣∣B , d−∣∣B
). In the following,
vavb will always mean a directed edge from node va to node vb. Let us now construct the directed graph ~G1 by adding vnv directed edges for each v ∈ A. (Since, according to (4), in ( D+, D−) the out-degree of vn is equal to zero, no parallel edges are created.) The bi-degree-sequence of ~G1 is ( d+, d−). Our goal is to construct another realization ~G′
1
of ( d+, d−) such that the deletion of the out-edges of vn in the latter produces the BDS (d+∣∣B , d−∣∣B
).By definition we have vnvk ∈ ~E1, (the edge set of ~G1) but vnvi 6 ∈ ~E1. At first assume that there exists a vertex vℓ (ℓ 6 = i, k, n ), such that vℓvi ∈ ~E1 but vℓvk 6 ∈ ~E1. (When
d−
i
d −
k
then this happens automatically, however if d−
i
= d−
k
and vkvi ∈ ~E1 then it is possible that the in-neighborhood of vi and vk are the same - except of course
vk, vi themselves and vn.) This means that now we can swap the edges vnvk and vℓvi
into vnvi and vℓvk. (Formally we create the new graph ~G′
1
= ( V, ~E′
1
) such that ~E′
1
=
~E1 \ { vnvk, v ℓvi} ∪ { vnvi, v ℓvk}.) This achieves our wanted realization.
the electronic journal of combinatorics 17 (2010), #R66 5
Our second case is when d−
i
= d−
k
, v kvi ∈ ~E1, and furthermore for each ℓ 6 = i, k, n we have vℓvi ∈ ~E1 ⇔ vℓvk ∈ ~E1. (6) It is important to observe that in this case vivk 6 ∈ ~E1 : otherwise some vℓ would not satisfy (6) (in order to keep d−
i
= d−
k
). Now, if there exists a subscript m (different from k, i, n ) such that vivm ∈ ~E1 but
vkvm 6 ∈ ~E1, then we create the required new graph ~G′
1
by applying the following triple swap (or three-edge swap): we exchange the directed edges vnvk, v kvi and vivm into vnvi, v ivk
and vkvm.By our assumption we have d+
i
d+
k
. On one hand side if d+
i
d +
k
holds then due to the properties vkvi ∈ ~E and vivk 6 ∈ ~E, there exist at least two subscripts m1, m 2 6 = i, k
such that vivmj ∈ ~E but vkvmj 6 ∈ ~E and at least one of them differs from n. Thus, when
d+
i
d +
k
, we do find such an m for which the triple swap above can be performed. The final case is when d−
i
= d−
k
and d+
i
= d+
k
. If vertex vm does not exist, then we must have vivn ∈ ~E1 (to keep d+
i
= d+
k
), and in this case clearly, vkvn /∈ ~E1. Therefore, in this (final) case the graphical realization ~G1 has the properties vnvk, v kvi, v ivn ∈ ~E1 and
vnvi, v ivk, v kvn 6 ∈ ~E1. Then the triple swap
~E′
1
:= ~E1 \ { vnvk, v kvi, v ivn} ∪ { vnvi, v ivk, v kvn} (7) will produce the required new graphical realization ~G′
1
.
Observation 5 For later reference it is important to recognize that in all cases above, the transformations from one realization to the next one happened with the use of two-edge or three-edge swaps.
Lemma 6 Let (d+, d−) be a BDS and let A and C be two possible out-neighborhoods of vn. Furthermore assume that C A, that is C is to the left of A. Finally assume that vertices in A ∪ C are in normal order. Then if (d+∣∣A, d−∣∣A
) is bi-graphical, so is
(d+∣∣C , d−∣∣C
).Proof. Since C is to the left of A therefore, there is a (unique) bijection φ : C \ A →
A \ C such that ∀c ∈ C \ A : i({c}) < i({φ(c)}) (the subscript of vertex c is smaller than the subscript of vertex φ(c)). (For example, if A = {v4, v 5, v 6, v 7, v 8, v 9} and C =
{v1, v 2, v 3, v 5, v 7, v 8}, then C \ A = {v1, v 2, v 3}, A \ C = {v4, v 6, v 9}, and φ is the map
{v1 ↔ v4, v 2 ↔ v6, v 3 ↔ v9}). To prove Lemma 6 we apply Lemma 4 recursively for each c ∈ C \ A (in arbitrary order) to exchange φ(c) ∈ A with c ∈ C, preserving the graphical character at every step. After the last step we find that the sequence reduced by C is graphical.
Proof of Theorem 2 : We can easily achieve now the required graphical realization of (d+, d−) if we use Lemma 6 with the current A, and C = {v1, . . . , v d+
n
}. We can do that since ( d+, d−) is in normal order, therefore the assumptions of Lemma 6 always hold.
the electronic journal of combinatorics 17 (2010), #R66 6
4 A simple prerequisite for MCMC algorithms to sample directed graphs with given BDS
In practice it is often useful to choose uniformly a random element from a set of objects. A frequently used tool for that task is a well-chosen Markov-Chain Monte-Carlo method (MCMC for short). To that end, a graph is established on the objects and random walks are generated on it. The edges represent operations which can transfer one object to the other. If the Markov chain can step from an object x to object y with non-zero probability, then it must be able to jump to x from y with non-zero probability (reversibility). If the graph is connected, then applying the well-known Metropolis-Hastings algorithm, it will yield a random walk converging to the uniform distribution starting from an arbitrary (even fixed) object. To be able to apply this technique we have to define our graph (the Markov chain)
G(d+, d−) = ( V, E). The vertices are the different possible realizations of the bi-graphical sequence ( d+, d−). An edge represents an operation consisting of a two or three-edge swap which transforms the first realization into the second one. (For simplicity, sometimes we just say swap for any of them.) We will show:
Theorem 7 Let ~G1, ~G2 be two realizations of the same bi-graphical sequence (d+, d−).
Then there exists a sequence of swaps which transforms ~G1 into ~G2 through different realizations of the same bi-graphical sequence.
Remark : In the case of undirected graphs the (original) analogous observation (need-ing only two-edges swaps) was proved by H.J. Ryser ().
Proof. We prove the following stronger statement: (z) there exists a sequence of at most 2e swaps which transform ~G1 into ~G2, where e is the total number of out-edges in (d+, d−)by induction on e. Assume that ( z) holds for e′ < e . We can assume that our bi-graphical sequence is in normal order on the first n − 1 vertices and d+
n
By Theorem 2 there is a sequence T1 (T2) of d = d+
n
many swaps which transforms ~G1 ( ~G2) into a ~G′
1
( ~G′
2
) such that Γ +
~G′
1
(vn) = {v1, . . . , v d} (Γ +
~G′
2
(vn) = {v1, . . . , v d}). We consider now the directed graphs ~G′′
1
( ~G′′
2
) derived from directed graph ~G′
1
(directed graph ~G′
2
) by deleting all out-neighbors of vn. Then both directed graphs realize the bi-graphical sequence (∆ +, ∆−) which, in turn, satisfies relations (1) and (2). Therefore the total number of out-degrees is e − d in both directed graphs, and by the inductive assumption there is a sequence T of 2( e − d) many swaps which transforms ~G′′
1
into ~G′′
2
.Now observe that if a swap transforms ~H into ~H′, then the “inverse swap” (choosing the same edges and non-edges and swap them) transforms ~H′ into ~H. So the swap sequence T2
has an inverse T ′
2
which transforms ~G′
2
into ~G2. Hence the sequence T1T T ′
2
is the required swap sequence: it transforms ~G1 into ~G2 and its length is at most d + 2( e − d) + d = 2 e.
the electronic journal of combinatorics 17 (2010), #R66 7
5 Is a BDS bi-graphical when one of its vertex’s out-neighborhood is constrained?
In network modeling of complex systems (for a rather general reference see ) one usually defines a (di)graph with components of the system being represented by the nodes, and the interactions (usually directed) amongst the components being represented as the edges of this directed graph. Typical cases include biological networks, such as metabolic networks, signal transduction networks, gene transcription networks, etc. The graph is usually inferred from empirical observations of the system and it is uniquely determined if one can specify all the connections in the graph. Frequently, however, the data available from the system is incomplete, and one cannot uniquely determine this graph. In this case there will be a set D of (di)graphs satisfying the existing data, and one can be faced with: (i) finding a typical element of the class D,
(ii) or generating all elements of the class D.(A more complete analysis of this phenomenon can be found in .) In Section 4 we already touched upon problem (i) when D is the class of all directed graphs of a given BDS. The analogous Problem (ii) for undirected graphs was recently addressed in which provides an economical way of constructing all elements from D. In this Section we give a prescription based on the method from , to solve (ii) for the case of all directed graphs with prescribed BDS. This is particularly useful from the point of view of studying the abundance of motifs in real-world networks: one needs to know first all the (small) subgraphs, or motifs, before we study their statistics from the data. Before we give the details, it is perhaps worthy making the following remark: Clearly, one way to solve problem (i) would be to first solve problem (ii), then choose uniformly from D. However, in (those very small) cases when reasonable answers can be expected for problem (ii), problem (i) is rather uninteresting. In general, however, (i) cannot be solved efficiently by the use of (ii). We start the discussion of problem (ii) by pointing out that our new, directed Havel– Hakimi type algorithm is unable to generate all realization of a prescribed DBS (see Figure 1).
(0,3)
(2,1) (2,1) (2,1) (2,1) (2,1) (2,1) (0,3)
Figure 1: This graph cannot be obtained by the directed Havel–Hakimi procedure. The integers indicate node degrees.
The situation is very similar to the non-directed case, see . The directed HH-algorithm must start with a vertex with degree-pair (2 , 1), therefore the two vertices of degree-pair
the electronic journal of combinatorics 17 (2010), #R66 8
(0 , 3) must be out-neighbors of the same vertex - not as it is shown in the graph in the Figure. One possible way to overcome this shortfall is to discover systematically all possible out-connections from a given vertex v in all realizations of the prescribed graphical BDS. We do not know a greedy algorithm to achieve this. The next best thing we can do is to develop a greedy algorithm to decide whether a given (sub)set of prescribed out-neighbors of v would prevent to find a realization of the BDS containing those prescribed out-neighbors. In the following, we describe such a greedy algorithm. (It is perhaps interesting to note that this latter problem can be considered as a very special directed
f -factor problem.) To start, we consider a ( d+, d−) bi-degree sequence together with a forbidden vertex set F whose elements are not allowed to be out-neighbors of vertex vn. (Or, just oppositely, we can imagine that we already have decided that those vertices will become out-neighbors of vn and the BDS is already updated accordingly. The forbidden vertex set governs only the out-neighbors, since in the process the in-neighbors are born “automatically”.) It is clear that |F | + 1 + d−
n
6 n must hold for the existence of a graphical realization of this
F -restricted BDS. Assume that the vertices are enumerated in such a way that subset F consists of vertices vn−| F |, . . . , v n−1 and vertices V ′ = {v1, . . . , v n−| F |− 1} are in normal order. (We can also say that we apply a permutation on the subscripts accordingly.) Then we say that the BDS is in F -normal order .
Definition 8 Consider a bi-graphical BDS sequence (d+, d−) in F -normal order, and let
A be a PON. The A-reduced BDS (d+∣∣A, d−∣∣A
) is defined as in (4) and (5) , while keeping in mind the existence of an F set to the right of A.
In other words, if A is a PON in an F -restricted BDS, then the reduced degree sequence (d+∣∣A, d−∣∣A
) is still obtained by removing the out-edges of node vn (according to the possible out-neighborhood A). Finally, one more notation: let ( d+, d−) be a BDS, F a forbidden vertex subset of V
and denote by F [k] the set of the first k vertices in the F -normal order.
Theorem 9 Let A be any PON in the F -restricted (d+, d−) BDS, which is in F -normal order. Then if the A-reduced BDS (d+∣∣A, d−∣∣A
) is graphical, then the F [d+
n
]-reduced BDS
(
d+∣∣F [d+
n]
, d−∣∣F [d+
n]
)
is graphical as well. Proof. It is immediate: Lemma 6 applies.
This statement gives us indeed a greedy way to check whether there exists a graphical realization of the F -restricted bi-degree sequence ( d+, d−): all we have to do is to check only whether the F [d+
n
]-reduced BDS
(
d+∣∣F [d+
n]
, d−∣∣F [d+
n]
)
is graphical. Finally, we want to remark that, similarly to the undirected case, Theorem 9 is suitable to speed up the generation of all possible graphical realizations of a BDS. The details can be found in which is a joint work of these authors with Hyunju Kim and L´ aszl´ o A. Sz´ ekely.
the electronic journal of combinatorics 17 (2010), #R66 9
Acknowledgements
The authors acknowledge useful discussions with G´ abor Tusn´ ady, ´Eva Czabarka, L´ aszl´ o A. Sz´ ekely and Hyunju Kim. ZT would also like to thank for the kind hospitality extended to him at the Alfr´ ed R´ enyi Institute of Mathematics, where this work was completed. Finally we want to express our gratitude to Antal Iv´ anyi for his editorial help.
References
C. Berge: The Theory of Graphs , Methuen, London (1962), Chap. 9. C. Berge: Graphs and Hypergraphs , North Holland, Amsterdam (1973), Chap. 6. P. Erd˝ os - T. Gallai: Gr´ afok el˝ o´ ırt fok´ u pontokkal (Graphs with prescribed degree of vertices), Mat. Lapok , 11 (1960), 264–274. (in Hungarian) L.R. Ford - D.R. Fulkerson: Flows in Networks , RAND Corporation R-375-PR
(1962) Chapter 2 Section 11. D. Gale: A theorem on flows in networks, Pacific J. Math. 7 (2) (1957), 1073–1082. S.L. Hakimi: On the realizability of a set of integers as degrees of the vertices of a simple graph. J. SIAM Appl. Math. 10 (1962), 496–506. S.L. Hakimi: On the degrees of the vertices of a directed graph, J. Franklin Institute
279 (4) (1965), 290–308. V. Havel: A remark on the existence of finite graphs. (Czech), ˇCasopis Pˇ est. Mat.
80 (1955), 477–480. Hyunju Kim - Z. Toroczkai - P.L. Erd˝ os - I. Mikl´ os - L.A. Sz´ ekely: Degree-based graph construction, J. Phys. A: Math. Theor. 42 (2009) 392001 (10pp) M.E.J. Newman - A.L. Barab´ asi - D.J. Watts: The Structure and Dynamics of Net-works (Princeton Studies in Complexity, Princeton UP) (2006), 624 pp. Ø. Ore: Studies on directed graphs, I, Annals of Math. (Second Series) 63 (3) (1956), 383–406. Ø. Ore: Studies on directed graphs, II, Annals of Math. (Second Series) 64 (3) (1956), 142–153. H.J. Ryser: Combinatorial properties of matrices of zeros and ones, Canad. J. Math.
9 (1957), 371–377. W.T. Tutte: The factorization of linear graphs, J. London Math. Soc. 22 (1947), 107–111. W.T. Tutte: The factors of graphs, Canad. J. Math. 4 (1952), 314–328. W.T. Tutte: A short proof of the factors theorem for finite graphs, Canad. J. Math.
6 (1954), 347–352. D.B. West: Introduction to Graph Theory , Prentice Hall, Upper Saddle River, US, (2001), Section 1.4. |
877 | https://www.youtube.com/watch?v=RprPAlK3eEs | Embryology | Development of Pharyngeal Apparatus
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In this embryology lecture, Professor Zach Murphy discusses the development of the pharyngeal apparatus, a crucial structure that gives rise to many head and neck components. This lecture provides a step-by-step overview of the pharyngeal arches, pouches, grooves (clefts), and membranes, and their adult derivatives—linking embryologic origins to key anatomical and clinical concepts.
We begin by discussing the formation of the five pharyngeal arches (numbered I, II, III, IV, and VI), detailing the cartilage, muscle, nerve, and arterial components associated with each. Zach walks through the specific cranial nerves and skeletal structures derived from each arch, making it easier to associate embryologic layers with clinical anatomy.
The lecture also covers the pharyngeal pouches and their endodermal derivatives, including the middle ear, palatine tonsil, thymus, parathyroid glands, and ultimobranchial body, along with the pharyngeal grooves and their role in forming the external auditory canal.
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Intro all right ninja nerds in this video we're going to be talking about the development or the embryology of the pharyngeal apparatus before we get started make sure you hit that like button comment down in the comment section and please subscribe also down in the description box with links to our facebook instagram patreon account go check that out get in contact with us we appreciate you and love you and let's get into this all right ninja Pharyngeal Apparatus so when we talk about the pharyngeal apparatus the best thing that i want us to do is start off taking a look at an embryo in a sagittal section so first thing to recognize when you're taking a look here at this little baby embryo this is going to be your cranial end and this down here is going to be the caudal or the tail end right so that's the first thing that you need to know now up in the cranial end you're going to have this kind of like little interesting almost gill-like looking structure within the embryo right about here right and that's called your pharyngeal apparatus you know what's really cool is if we take a slice imagine i take and i make a slice down this pharyngeal apparatus like this and then open it up and take a look at it looky looky what that pharyngeal apparatus is going to look like this is the structure that you'll get so the first thing here is you see where we have this green layer which is your endoderm layer and then this blue layer which is your ectoderm layer you see where they meet right here where they kind of like connect that right there is called your bucco pharyngeal membrane alright so that's going to become the mouth so whenever that kind of busts open it's going to give way to the mouth now there's a couple different components of the pharyngeal apparatus the first one is this kind of blue layer here this like nice little beautiful layer here of blue on the outer parts this right here is called your pharyngeal particularly you see like these little divots in here these little divots in the pharyngeal apparatus right here that is going to be giving way to what's called your pharyngeal clefts or sometimes it's even referred to as your pharyngeal groove so it's called your pharyngeal clefts or the pharyngeal glute groove now the big thing to remember about these pharyngeal clefts is that they are covered with ectoderm and now we're not going to really focus too much on these pharyngeal clefts okay because they don't really give way to anything really special except for the first one i can give away to like some of the tissue that lines the external ear canal but other than that nothing else really specific about it the next one is this inner kind of like red layer this inner red layer is a mesoderm core and this is kind of what gives way to the big structure called the pharyngeal arches okay so these are your pharyngeal arch and this is kind of a core of mesoderm which is a beautiful little like mesenchymal tissue okay the last part of this pharyngeal apparatus is the most inner portion and you see how there's these like these little divots here right here another little divot here another little divot here these little divots here are very specific and very special little structures that give way to a lot of different things and this is called your pharyngeal pouches so these are called your pharyngeal pouches okay so that's really really really important and again what i want you to remember from this is that those good old pharyngeal pouches are made up of endoderm now the next thing i want us to do here is i want us to basically kind of set a a guideline of what we're going to be talking about throughout this video and what we're going to do is we're going to talk about the pharyngeal pouches right what they become pharyngeal arches what they become and the pharyngeal clefts and what they become all right so really quickly i want to give you guys just a basic understanding of what the pharyngeal pouches pharyngeal arches and pharyngeal clefts will become okay tissue wise let's just be very vague here so the pharyngeal clefts is the first one here the pharyngeal clefts remember we said are basically ectoderm right and again we can call them like grooves as well the really the only one that you need to know here is this first one this first little pharyngeal cleft but pretty much what this will become is epithelium so remember it is ectoderm but that ectoderm is going to give way to epithelium and really what it's going to do is it's going to cover a particular area and the first one covers the in the actual lining of the external ear canal so if you really want to remember it the first cleft actually gives way to the lining the external ear canal that's really kind of all i want you to know about the pharyngeal cleft other than that nothing really specific to it the big mama what's going to count for pretty much most of this actual lecture is going to be these good old pharyngeal arches their pain and the noggin and these things are a nice little mesoderm and the reason why it's a tough little thing is that this can give way to connective tissue and we talk about connective tissue that there's a bunch of different things that are related with that but one of the big ones is also going to be cartilage so it can give away a bunch of connective tissue it can give way to a lot of different cartilage especially like laryngeal cartilage it can also give way to muscle okay and to bones as well as nerves sometimes there's going to be some special cranial nerves that are going to be kind of developing from these actual arches as well the last one here is going to be the pharyngeal pouches and really again it kind of goes back to the epithelium from the pharyngeal clefts so the pharyngeal pouches they will give way to some epithelial structures so they will kind of it's endoderm and then endoderm will become epithelium that'll align particular areas particular cavities but also the other thing to remember is that it can also give way to glands some special glands parathyroid it also can give way to uh like the thymus and some other structures as well okay so very important to remember these pharyngeal pouches and again uh what they're going to become so we're going to cover all of each one of these particularly actually the pharyngeal arches and the pharyngeal pouches in more detail so let's move on to that all right so the next thing we have to do is kind of go through particularly what next we kind of really just discussed the basics of differential cleft and again if you really want to remember the first pharyngeal cleft or groove will become the epithelium that lines the external ear canal but other than that nothing else really specific the next one we have to talk about is these pharyngeal arches so again the primary focus here is going to be the pharyngeal arches and again this is going to be that mesoderm core here right so that's what we're really primarily going to focus on now remember what i told you the pharyngeal arch is the mesoderm or that mesenchymal tissue will differentiate into a bunch of different types of tissues what are some of those tissues and that's what we've got to focus on so the first thing that i want you to remember is it gives way to muscles so different types of muscles now we give particular names to the arches right and it's very simple thank the lord it's very simple this is your first arch this is the second this is the third this is the fourth the fifth usually either doesn't form or it digresses and then this last one here is the sixth pharyngeal arch so what we need to know is is what does the mesoderm within that first second third fourth and six pharyngeal arch differentiate into and the first thing is with respect to muscles so the first thing i like to remember is that the first pharyngeal arch is going to be responsible for your chewing muscles it's supplied by the trigeminal nerve so i just like to remember keep it simple right your muscles of mastication that's going to be the first one if you really want to know those that's going to be your pterygoid the medial lateral masseter as well as the temporalis the other muscles that it does also supply is going to be the myelohyoid and the digastric we're going to abbreviate this digastric anterior belly and then the last one is going to be that tensor vely palatini okay so these are the muscles that are supplied by the first pharyngeal arch these are for chewing these are basically going to be super hyoid muscles and then this is also going to play a role within movement of that soft palate during swallowing the second arch remember muscles of facial expression there's a ton of those we're not going to go through all of them but this is going to be your muscles a facial expression okay it also gives way to some other muscles okay such as the stylohyoid as well as the digastric posterior belly and then a tiny little muscle in the actual middle ear called the stapedius so called the stapedius muscle okay the next thing is the third arch the third arch is one muscle thank goodness all right and this muscle is called the stylopharyngeus it's called the stylopharynges and this one is supplied by the glossopharyngeal nerve again a little muscle that helps to be able to play within kind of the deglution or swallowing process and again second arch facial expression stylohyoid digastric posterior belly are super hyoid muscles and stapedius just kind of modulates the movement or bumping of stapedia stapes bone on the inner ear that controls kind of like the sensitivity of hearing the last one is your fourth and sixth arch and these supply your kind of a couple different muscles they supply what's called your pharyngeal constrictor muscles so we're just going to put your pharyngeal muscles right so that's your superior middle inferior fragile constrictors that control kind of the deglution process and it also acts on your laryngeal muscles and you've got a bunch of these little suckers but there's a bunch of these the cricothyroid the arytenoid muscles all those and then the last one is going to be what's called the levator vely palatini and again this is going to be again involved in that deglutition process so you kind of have an idea of what all these muscles that red portion there well all the mesoderm differentiates into muscles respective to their arches and those are these the next thing we've got to do Bones is talk about the next thing that the mesoderm can differentiate into which is bones the bones and connective tissue are cartilage of the first of the pharyngeal arches and again they're corresponding so what is the first second and thankfully nothing really from the third but from the fourth and sixth the first arch okay the first one is going to give way to a lot of the bones um pretty much of your entire skull okay so what are some of these bones it's going to give way to particularly your maxilla okay it's going to give way to the zygomatic bone well bones right and the maxillary bones it's going to give way to the mandible and it's also going to give way to you know this part over here what is this bone here this is that squamous part of the temporal bone so we'll just put temporal bone but again remember we'll put here we'll put squamous part of the temporal bone right so these are going to be some of them you can also have some other bones that come off of that temporal bone like your mastoid process and styloid process and stuff like that right the second arch you see this little bone here this highway bone kind of within the neck right there that's going to be one of them that are derived from the second arch right so the hyoid bone then look what the higher bone is actually connected to you see this ligament here this like bluish ligament this blue ligament here which we're going to kind of represent here there's a ligament that connects the holy bone to the styloid process that is called the stylohyoid ligament pretty original right so stylo hyoid ligament now there is also textbooks that say that the higher bone not only comes from the second arch but a little bit of it comes from the third arch for simplicity's sake we're going to say that most of it's going to be coming from the second arch but if you really want to remember some of it is also they can come from the third arch but the stylohyoid ligament is going to be coming from at least the second arch the next thing here is you have these little bones you see these little bones in your inner ear well actually in your middle ear these little bones here are called your ossicles this little one here which taps on the oval window is called stapes that is found in the that's actually derived from the second pharyngeal arch whereas if we're really being specific the malleus and incus which are the other two little bones in the inner ear those are technically coming from the first french arch so if you really wanted to add on here we can also say the malleus and incus whereas stapes is particularly from the second arch the last one here is going to be the actual cartilage from the larynx so you know the larynx you have a bunch of different types of laryngo cartilage right what are some of them we're not going to do all of them but we're going to mention them at least the first big one here is going to be the thyroid cartilage that's going to be one big one right another one is going to be the arytenoid the arytenoid cartilage and there is other ones there's like the cuny form and then the corniculate and these are pretty much going to give way to the cartilage of the larynx so to finish off with the actual bones that are derived from the mesoderm of the pharyngeal arches first will give way to these bones second to these bones and this ligament and the fourth and six will give way to the laryngeal cartilage thyroid aretenoid cuneiform and cornicule okay we've covered the muscles the bones the connective tissue and cartilage now we're going to do is we're going to move on to the nerves derived from the pharyngeal arches all right so the nerves of the Nerves pharyngeal arches so what nerves are going to be supplying the actual particular regions that are going to be derived from the pharyngeal arches so if we look at the nerves that are going to be derived here the first one that is going to be from the first arch is going to be cranial nerve 5. you know cranial nerve 5 is this is the trigeminal nerve so this is the trigeminal nerve the next one here which is going to be supplying the second pharyngeal arch that is going to be cranial nerve seven and cranial nerve seven is going to be the we're just gonna put facial nerve right the next one which is gonna be for the third pharyngeal arch the third pharyngeal arch is going to be cranial nerve nine and cranial nerve nine is going to be which one the glosso pharyngeal nerve and then last but not least the fourth and sixth pharyngeal arches are going to be supplied by cranial nerve ten and that is going to be the vagus nerve so again quick recap of these first pharyngeal arch will be the trigeminal nerve the second pharyngeal arch derivative is going to be the facial nerve third will be the glossopharyngeal and then fourth and sixth particularly if you really want to be specific to superior laryngeal nerve for the fourth and then the recurrent laryngeal nerve for the sixth will be for the vagus nerve so the last one here is going to be the arteries of the fringe arches if you guys watched our video on the development of the vascular system we talked about how from the aortic sac how through the process of angiogenesis blood vessels go from the aortic sac through the pharyngeal arches from the french arches you'll make some blood vessels and then that will sprout to the dorsal aorta and that made what's called the aortic arches so the aortic arches are sometimes even called the pharyngeal arteries right or the pharyngeal arch arteries and they have to become particular structures in the adult what are those structures do you guys remember from that video the first arch we said is what's called the maxillary artery so that'll become the maxillary remember the first is max so the max you can be is number one right so maxillary order for the first one the second one we said it's pretty rare that it actually remains but you said you have what's called the hyoidal artery and that hyoidal artery gives way to what's called the stapedial artery usually this does not actually remain in the adults pretty rare 10 people may keep it but you remember second and s stapedial all right so first is max so maxillary artery second is s stapedial third arch third letter in the alphabet is c so the third arch gives way to what's called the common carotid artery as well as the internal carotid artery but particularly the most proximal portion of the internal carotid artery the fourth arch you actually have the right and left of these arches right so you have technically a right and left first second third fourth and sixth the fifth one again either doesn't develop or digresses the fourth arch technically if we're being really specific here the right one will give way to what the right one gives way to what's called the the subclavian artery so it gives way technically to what's called the subclavian artery whereas the left one will give way to what the aortic arch all right and then the last one here is the sixth arch and again you have a right and you have a left one here the right one will specifically give way to the right pulmonary artery so the right pulmonary artery and then the left sixth arch will give way to what's called what the left pulmonary artery as well as what else the ductus arteriosus which is going to be a little connection between the pulmonary trunk and the aorta so these will be the derivatives of the mesoderm portion of the pharyngeal arches so whenever you kind of see in textbooks here they might have like a couple different like holes here of what's actually going to be coming from this right so you may have like for example there may be like a nerve kind of within that structure that's going to be coming from it there may be a vessel that's coming from that right and then again you're also going to have this portion here kind of transform into muscles into bones and cartilage and connective tissue and all that stuff like that so we're able to see that from these pharyngeal arches what they derive into with respect to muscles bones connective tissue nerves and arteries let's now move into the pharyngeal pouches and talk about what they become Pharyngeal pouches all right so we're almost done pharyngeal pouches right so this is the endoderm portion so we talked about the ectoderm like one of them right external acoustic meatus the uh particularly the epithelial lining then we talked about the french arches what all they become the inner part here is the endoderm which is going to be the pharyngeal pouches now there's again a couple different fragile pouches so if you look here the first one is right here this little like little divot in here that is going to be the first pharyngeal pouch this obviously will be the second pharyngeal pouch this will be the third pharyngeal pouch and this will be the fourth pharyngeal pouch now if you noticed the first and second they only have one little divot but then the third one it splits you see how it splits like this this portion here is going to be the third and what's particularly called the dorsal pouch so third dorsal pouch and this one here is called the third ventral pouch okay then over here you have the fourth dorsal pouch and then here you have the fourth ventral pouch what we need to know is is what does each one of these pouches first second third dorsal and ventral fourth dorsal and ventral what do they become all right so the first thing that we should talk about here is the first pharyngeal pouch the first pharyngeal pouch will actually have endoderm that will actually move into the inner i'm sorry the middle ear cavity so you know within the middle ear cavity you have this uh particular area here that we specifically don't call the middle ear cap but we actually call it the tympanic cavity right so it's the middle ear cavity or the tympanic cavity then you have some tissue that's going to be moving and kind of draining the middle ear or the tympanic cavity with actually the uh the nasal the nasal cavity right into the nasal cavity this structure here is called the eustachian tube so what is this one called eustacian tube so technically the first pharyngeal pouch will have epithelial tissue derived from the endoderm move and invade into these areas both the tympanic cavity and the eustachian tube the second pharyngeal pouch will actually become more specifically a type of lymphatic tissue if we're really being specific a type of lymphatic tissue and that kind of lines up in the areas of the actual nasal oral cavity as well and these are called your tonsils your tonsils they're kind of a lymphatic tissue all these green little structures here you're going to have the tonsils these are going to be derived from the second pharyngeal pouch the more particular one here is going to be the on the back of the nasopharynx and that's called your pharyngeal tonsils right but again you do have other tonsils here like your palatine tonsils your tubal tonsils and your lingual tonsils okay all right on to the good stuff which is your third and fourth pharyngeal patch but we actually have to talk about the ventral and we also have to talk about the dorsal components all right so when we talk about the third pharyngeal pouch again you got the ventral and the dorsal the dorsal part of the third pharyngeal pouch the dorsal part will give way to these tiny little very very special glands on the back of the thyroid but you know what's interesting is they're on the bottom part right of the thyroid well they'll eventually go on the bottom part of the thyroid and so these are called your inferior parathyroid glands okay your inferior parathyroid glands which seems odd right you would think that the the actual inferior parathyroid glands would come from like the fourth dorsal but what actually happens is during the development they actually move in a particular way where the uh there's movement of the fourth dorsal pouch upwards where the superior thyroid glands go above but anyway off my soap box the third dorsal pharyngeal pouch will become the inferior parathyroid glands the ventral portion right of that third pharyngeal pouch that is going to give way to a very special kind of lymphatic tissue that will actually start in the neck but then move down into the chest where t cells will actually mature this is called the thymus so this is called your thymus very important to actually understand this stuff the reason why is that sometimes there's a condition called digeorge syndrome they love to ask this stuff especially embryology where the third and fourth pharyngeal pouches don't actually develop and so because of that they lose their thymus and they also lose their parathyroid glands as well and so that becomes a very interesting kind of condition so very important to know this stuff let's go into the last one here the last one here is your fourth fringing pouch and again we got the dorsal and the ventral thankfully science has kept us nice and simple for us the dorsal the fourth dorsal friendship pouch becomes the superior parathyroid glands right so the superior parathyroid glands and then lastly the actual fourth ventral so then you have the fourth ventral pharyngeal pouch what does that become you see these little cells here they actually are particularly it starts off it's weird they start off as what's called these those these little structures here from the fourth ventral fringeal pouch start off what's called the ultimo uh pharyngeal body ultimopharyngeal body or brachial body right so ultimo pharyngeal body and then eventually they give way to what's called parafollicular cells or sometimes noted as c cells which secrete calcitonin so if you think about this let's say that somebody did have digeot syndrome and they didn't develop their parathyroid glands what would that affect their calcium metabolism if they didn't develop their thymus how would that what would that affect that would affect their t cell development so they may develop some immune system issues and also the parafollicular cells the calcitonin release that also is involved in the calcium regulation so you can see how something like this not developing properly could lead to some serious issues but that tells us about our pharyngeal pouches and our pharyngeal apparatus all right ninja nerds in this video we talked about the development of the pharyngeal apparatus i hope it made sense and i hope that you guys did like it and enjoy it as always ninja nerds until next time [Music] you |
878 | https://www.cuemath.com/trigonometry/secant-function/ | LearnPracticeDownload
Secant Function
The secant function that we are talking about is defined as one of the reciprocal of our basic three trigonometric functions. So, we have cosecantwhich is the reciprocal of sine, secantwhich is the reciprocal of cosine, and cotangent is the reciprocal of the tangent function. The value of the secant function can be determined by taking the ratio of the hypotenuse and base of a right-angled triangle.
In this article, we will explore the concept of the secant function and understand its formula using the unit circle, and angles, how to use the formula, and their various applications and properties. We will also plot the secant function graph and determine its value at various angles.
| | |
--- |
| 1. | What is Secant Function? |
| 2. | Secant Function Formula |
| 3. | Secant Function Values |
| 4. | Secant Function Graph |
| 5. | Domain and Range of Secant Function |
| 6. | Properties of Secant Function |
| 7. | FAQs on Secant Function |
What is Secant Function?
The secant function is a periodic function in trigonometry. The secant function or sec function can be defined as the ratio of the length of the hypotenuse to that of the length of the base in a right-angled triangle. It is the reciprocal of cosine function and hence, is also written as sec x = 1 / cos x. Let us try to understand the concept of secant function by analyzing a unit circle centered at the origin of the coordinate plane.
A variable point P is taken on the circumference of the circle and it continues to move on the circumference of this circle. From the figure, we observe that P is in the first quadrant, and OP makes an acute angle of x radians with the positive x-axis. PQ is the perpendicular dropped from P (a point on the circumference) to the x-axis. The triangle is thus formed by joining the points O, P, and Q as shown in the figure, where OQ is the base, and PQ is the height of the triangle. Hence, the sec function for the above case can be mathematically written as:
sec x = OP / OQ
Here, x is the acute angle formed between the hypotenuse and the base of a right-angled triangle.
Secant Function Formula
As discussed above, the formula for the secant function is given by the ratio of the hypotenuse and the base of a right-angled triangle. That is, we can write it mathematically as sec θ = Hypotenuse / Base. Also, we know that the secant function is the reciprocal of the cosine function. Therefore, we can also write its formula as sec θ = 1/ cos θ.
Secant Function Values
As we study the trigonometric table, we observe the values of the trigonometric functions at different angles. In this section, we will go through the value of the secant function for standard angles such as 0°, 30°, 45°, 60°, and 90° along with other angles like 180°, 270°, and 360° included. It is best to remember the values of the trigonometric ratios of these standard angles which help in various calculations.
| Secant Function | Value |
--- |
| sec 0° | 0 |
| sec 30° | 2/√3 |
| sec 45° | √2 |
| sec 60° | 2 |
| sec 90° | Not Defined |
| sec 120° | -2 |
| sec 150° | -2/√3 |
| sec 180° | -1 |
| sec 270° | Not Defined |
| sec 360° | 1 |
Secant Function Graph
As we observed the unit circle in the first section with center O at the origin, and a point P moving along the circumference of this circle. The angle that OP makes with the positive direction of the x-axis is x (radians). PQ is the perpendicular dropped from P to the horizontal axis. The secant function is the reciprocal of the cosine function, that is, sec x = 1 / cos x. It is also considered as the secant function formula. We note that: sec x = OP / OQ = 1 / OQ. As x varies, we note that the value of sec x varies with the variation of the length of OQ.
Graphing secant becomes very easy since we already know the cosine graph, so we can easily derive the graph for sec x by finding the reciprocal of each cosine value. When the value of cos x is very small, the value of sec x will become very large. That is, finding 1/y for each value of y on the curve y= cos x. The table below shows some angles in radians:
| x (radians) | cos x | sec x |
---
| 0 | 1 | 1 |
| π/6 | √3/2 | 2/√3 |
| π/4 | 1/√2 | √2 |
| π/3 | 1/2 | 2 |
| π/2 | 0 | Not Defined |
Also, we observe that whenever the value of the cosine function is zero, the secant function goes to infinity, which means when the value of cosine is 0 then, the secant is undefined. Thus, we obtain the following graph of sec x:
Domain and Range of Secant Function
Looking at the secant function on a domain centered at the y-axis helps us bring out its symmetry. Thus, as we can see in the secant function graph above, the secant function is symmetric about the origin. So the domain of secant is all real numbers except for points (2n + 1)π/2. The range of secant is the set of all real numbers with a magnitude greater than or equal to 1. Thus, we have:
Domain of secant function: R - (2n + 1)π/2
Range of secant function: (-∞,-1] U [1, ∞)
Properties of Secant Function
Properties of the secant function depend upon the quadrant in which the angle lies. From the above secant graph and circle relation, we can see that value of the secant function is positive in the first and fourth quadrant whereas, in the second and third quadrant, it bears a negative value. Let us list some of the basic properties of the secant function:
The secant function is a periodic function.
A periodic function is a function which, when meets a specific horizontal shift, P, results in a function equal to the original function, i.e. f(x + P) = f(x), for all values of x within the domain of f. The secant graph repeats itself after 2π, which suggests the function is periodic with a period of 2π.
So we can say that: sec(x+2nπ) = sec x, for every x
The secant function is an even function because sec(-x) = sec x, for all x.
Sec x has vertical asymptotes at all values of x = π/2 + nπ, n being an integer.
The secant function graph is symmetric with respect to the y-axis.
Important Notes on Secant Function
Sec function can be mathematically written as: Sec x = Hypotenuse / Base
It is a periodic function with a period of 2π.
The domain of the secant function is R - (2n + 1)π/2 and the range is (-∞,-1] U [1, ∞).
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Secant Function Examples
Example 1: Find the value of sec 780° using the secant function formula.
Solution: We know sec x = 1 / cos x. So,
sec (780°) = 1 / cos 780°
Now, cos 780° = cos (60° + 2 × 360°)
= cos 60° --- [Because cos (2×360° + x) = cos x]
= 1/2 --- [Because cos 60° = 1/2 using trigonometry table]
So, sec (780°) = 1 / cos 780°
= 1 / (1/2)
= 2
Answer: sec 780° = 2
2. Example 2: Find the length of the hypotenuse of a right-angled triangle when sec x = 0.6 and the length of the base is 5 units.
Solution: Using secant function formula, we know that sec x = Hypotenuse / Base. So, we have
0.6 = Hyp / 5
⇒ Hyp = 0.6 × 5
= 3
Answer: The length of the hypotenuse is equal to 3 units.
3. Example 3: What is the value of the secant function when the cos x is equal to 3/4 for the corresponding angle?
Solution: We know that sec x can be written as the reciprocal of cos x. So, we have
sec x = 1 / cos x
= 1 / (3/4)
= 4 / 3
Answer: sec x = 4/3
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FAQs on Secant Function
What is Secant Function in Trigonometry?
Secant function is one of the important trigonometric functions in trigonometry. It is equal to the ratio of the hypotenuse and the base of a right-angled triangle. Secant function is also known as the reciprocal of the cosine function.
How Do youWrite a Secant Function?
We can write the formula for the secant function as sec x = Hypotenuse / Base. It can also be written as sec x = 1 / cos x as it is the reciprocal of cos x.
What is the Definition of Secant?
The secant function or sec function can be defined as the ratio of the length of the hypotenuse to that of the length of the base in a right-angled triangle.
What is Secant in terms of Cos?
Secant function in terms of cos is written as sec x = 1 / cos x.
Where is the SecantFunction Not Defined?
The secant function is not defined at points where the value of the cosine function is equal to 0. The value of cos x is equal to zero at all real numbers except (2n + 1)π/2. So, the secant function is not defined at points (2n + 1)π/2, where n is an integer.
In What Quadrant is Secant Negative?
Secant function is negative in the second and third quadrant.
What is the Reciprocal of the Secant Function?
Using the reciprocal identities, we know that the reciprocal of the secant function is the cosine function.
Is Secant Function an Even Function?
Secant function is an even function because sec (-x) = sec x, for all x.
What is the Domain and Range of Secant Function?
The domain of the secant function is R - (2n + 1)π/2 and the range is (-∞,-1] U [1, ∞).
What is the Period of Secant Function?
The period of the secant function is equal to 2π.
Where are the Vertical Asymptotes on the Graph of the Secant Function?
The secant function has vertical asymptotes at points where the cosine function is equal to zero. Since sec x is the reciprocal of cos x, therefore sec x is not defined at points where cos x is 0. So, the secant function has vertical asymptotes at points (2n + 1)π/2, where n is an integer.
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Why are two definitions of ellipses equivalent?
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In classical geometry an ellipse is usually defined as the locus of points in the plane such that the distances from each point to the two foci have a given sum.
When we speak of an ellipse analytically, we usually describe it as a circle that has been squashed in one direction, i.e. something similar to the curve x 2+(y/b)2=1 x 2+(y/b)2=1.
"Everyone knows" that these two definitions yield the same family of shapes. But how can that be proved?
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asked Oct 29, 2013 at 10:28
hmakholm left over Monicahmakholm left over Monica
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I needed this fact when answering this question, but we don't seem to have a question specifically for this basic fact. My first attempt at proving it by mindless formula-crunching led me horribly astray, so it seems to be useful to have a way that works written down explicitly.hmakholm left over Monica –hmakholm left over Monica 2013-10-29 10:29:15 +00:00 Commented Oct 29, 2013 at 10:29
I'm interested in getting more answers to this, especially various intuitive explanations. Is it okay with you if I edit this question into a big-list question asking for more answers?Sophie Swett –Sophie Swett 2017-04-07 03:05:05 +00:00 Commented Apr 7, 2017 at 3:05
1 3Blue1Brown recently posted a video giving a beautiful geometric proof of the relationship between the 3 most common descriptions of ellipses.Paul Sinclair –Paul Sinclair 2019-01-04 00:38:03 +00:00 Commented Jan 4, 2019 at 0:38
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Suppose we have a classical ellipse with its two foci and major axis given. We want to prove that it's a squashed circle.
Select a coordinate system with its origin in the center if the ellipse, the x x-axis passing through the foci, and a scale such that the common sum-of-distances-to-the-foci is 2 2. The foci have coordinates (±c,0)(±c,0) for some c∈[0,1)c∈[0,1).
By considering the nodes at the end of the minor axis we find that the semiminor axis of the ellipse must be b=1−c 2−−−−−√b=1−c 2. We must then prove that the equations
x 2+(y 1−c 2−−−−−√)2=1(1)(1)x 2+(y 1−c 2)2=1
(x+c)2+y 2−−−−−−−−−−−√+(x−c)2+y 2−−−−−−−−−−−√=2(2)(2)(x+c)2+y 2+(x−c)2+y 2=2
are equivalent.
Rearranging (1) gives
y 2=(1−c 2)(1−x 2)(1')(1')y 2=(1−c 2)(1−x 2)
and therefore (x+c)2+y 2=(1+x c)2(x+c)2+y 2=(1+x c)2 by multiplying out each side and collecting terms. Similarly, (x−c)2+y 2=(1−x c)2(x−c)2+y 2=(1−x c)2. So for points where (1) holds, (2) reduces to (1+x c)+(1−x c)=2(1+x c)+(1−x c)=2, which is of course true.
Thus every solution of (1) is a solution of (2). On the other hand, for every fixed x∈(−1,1)x∈(−1,1), it is clear that the left-hand-side of (2) increases monotonically with |y||y| so it can have at most two solutions, which must then be exactly the two solutions for y y we get from (1'). (And neither equation has solutions with |x|>1|x|>1). So the equations are indeed equivalent.
Conversely, if we have a squashed circle with major and minor axes 2 a 2 a and 2 b 2 b, we can find the focal distance 2 c 2 c by c 2+b 2=a 2 c 2+b 2=a 2 and locate the foci on the major axis. Then the above argument shows that the classical ellipse with these foci coincides with our squashed circle.
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answered Oct 29, 2013 at 10:28
hmakholm left over Monicahmakholm left over Monica
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Awesome algebraic explanation! However, do you think there's a way to better see this connection geometrically? To somehow see that if we stretch a circle, there will be two points whose sum of the distances to any point on the ellipse will be a constant? I've been trying to come up with one and haven't been able to. Thank you!joshuaronis –joshuaronis 2019-01-05 18:51:19 +00:00 Commented Jan 5, 2019 at 18:51
@JoshuaRonis: If I had a good geometric proof, I wouldn't have needed to rely on mindless formula-crunching here.hmakholm left over Monica –hmakholm left over Monica 2019-01-06 14:38:44 +00:00 Commented Jan 6, 2019 at 14:38
Haha okay I'm asking in a bunch of forums and trying to find one myself, I'll post here if I do joshuaronis –joshuaronis 2019-01-06 15:30:18 +00:00 Commented Jan 6, 2019 at 15:30
Do you know why, given a squashed circle, do the foci lie on the major axis? It's seems so obvious, but I cannot explain it rigorously...Bman72 –Bman72 2019-06-26 13:38:50 +00:00 Commented Jun 26, 2019 at 13:38
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(Stretched circle ⇒⇒ Ellipse with focis)
It's easy to show that projection of a diagram at an angle is the same as stretching. Let E E be a stretched circle, then it is not hard to find the corresponding angle s.t. E E is the projection of a circle C C at that angle.
Let C 0 C 0 be the unqiue cyclinder s.t. it contains the circle and it is normal to the plane containing the circle. Clearly E E is the intersection of C 0 C 0 with the plane containing E E.
And there is a classical proof showing that the intersection of a cyclinder with a plane is an ellipse with focis, see @Element118 's answer.
(Ellipse with focis ⇒⇒ Stretched circle)
Let E E be an ellipse with focis F 1,F 2=(±c,0,0)F 1,F 2=(±c,0,0), center O=(0,0,0)O=(0,0,0), major axis 2 a 2 a and minor axis 2 b 2 b sitting on the plane z=0 z=0. Clearly a 2=b 2+c 2 a 2=b 2+c 2.
Consider two spheres of radius b b and centers respectively (c,0,b)(c,0,b) and (−c,0,−b)(−c,0,−b) as the following diagram. There exists an unique cyclinder C 0 C 0 inscribing both of the spheres. Clearly C 0∩{z=0}C 0∩{z=0} is an ellipse E 1 E 1 with focis F 1 F 1 and F 2 F 2 due to the classical proof mentioned above. To prove that E 1=E E 1=E, it reduces to showing that O M=a O M=a, which is equivalent to showing that ∠O A M=∠O M A∠O A M=∠O M A. Since ∠O A M=∠A M J∠O A M=∠A M J, it reduces to showing ∠O M A=∠A M J∠O M A=∠A M J which is clearly true. Thus E 1=E E 1=E. And we clearly have that E E is the projection of a circle, which is the same as a stretched circle. The result follows.
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answered Aug 26, 2022 at 23:05
Z WuZ Wu
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From 3Blue1Brown, at about 9:40, expanding the comment from @Paul Sinclair into an answer.
A squashed circle with major axis 2 a 2 a and minor axis 2 b 2 b can sit inside tangent to a cylinder with radius b b. By constructing spheres tangent to the cylinder and the squashed circle, they would be tangent to the squashed circle at the foci, let them be F 1 F 1 and F 2 F 2.
The spheres are tangent to the cylinder, touching the cylinder at a circle each. These circles are perpendicular to the axis of the cylinder.
Pick a point P P on the perimeter of the squashed circle. F 1 P F 1 P is tangent to a sphere, and another line l l from P P tangent to the sphere at Q 1 Q 1 runs along parallel to the axis of the cylinder. Likewise, F 2 P F 2 P is tangent to the other sphere, and there is another line m m from P P tangent to the sphere Q 2 Q 2 running along, parallel to the axis of the cylinder.
Since l l and m m are parallel with common point P P, they are the same line. It follows that F 1 P+F 2 P=Q 1 P+Q 2 P=Q 1 Q 2 F 1 P+F 2 P=Q 1 P+Q 2 P=Q 1 Q 2. It follows that no matter what point P P we picked, Q 1 Q 2 Q 1 Q 2 is of a constant length, which implies that F 1 F 1 and F 2 F 2 are indeed foci and the squashed circle is indeed an ellipse by the "sum of distances" definition.
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answered Apr 14, 2020 at 3:47
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I know this question is 10 years old, but here is a completely pre-algebraic solution.
I will start with the Foci- Formula
(x−e)2+y 2−−−−−−−−−−−√+(x+e)2+y 2−−−−−−−−−−−√=2.(x−e)2+y 2+(x+e)2+y 2=2.
To get rid of lefthand root, we shift the righthand root to the right and square. We obtain
(x−e)2+y 2=(2−(x+e)2+y 2−−−−−−−−−−−√)2=4−4(x+e)2+y 2−−−−−−−−−−−√+(x+e)2+y 2.(x−e)2+y 2=(2−(x+e)2+y 2)2=4−4(x+e)2+y 2+(x+e)2+y 2.
By eliminating y 2 y 2 and shifting (x+e)2(x+e)2 to the left we get
−4 x e=4(1−(x+e)2+y 2−−−−−−−−−−−√)−4 x e=4(1−(x+e)2+y 2)
Now dividing by 4, shifting 1 1 to the left and multiplying by −1−1 yields
x e+1=(x+e)2+y 2−−−−−−−−−−−√x e+1=(x+e)2+y 2
Now we are able to eliminate the second root by squaring and get
x 2 e 2+1=x 2+e 2+y 2.x 2 e 2+1=x 2+e 2+y 2.
Observe the identity b 2=1−e 2 b 2=1−e 2, eliminate all e 2 e 2 and rearrange:
x 2(1−b 2)+b 2=x 2+y 2.x 2(1−b 2)+b 2=x 2+y 2.
x 2 x 2 cancels and by dividing by b 2 b 2 we get
−x 2+1=y 2 b 2,−x 2+1=y 2 b 2,
which is obviously equivalent to the Squashed-Circle-Formula
x 2+(y b)2=1.x 2+(y b)2=1.
Of course all steps are indeed equivalences and so the equivalence between the two formulas is proven.
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answered Sep 7, 2023 at 15:21
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880 | https://web.ung.edu/media/chemistry/Chapter7/Chapter7-ChemicalBonding-MolecularGeometry.pdf | Chapter 7 Chemical Bonding and Molecular Geometry Figure 7.1 Nicknamed “buckyballs,” buckminsterfullerene molecules (C60) contain only carbon atoms. Here they are shown in a ball-and-stick model (left). These molecules have single and double carbon-carbon bonds arranged to form a geometric framework of hexagons and pentagons, similar to the pattern on a soccer ball (center). This unconventional molecular structure is named after architect R. Buckminster Fuller, whose innovative designs combined simple geometric shapes to create large, strong structures such as this weather radar dome near Tucson, Arizona (right). (credit middle: modification of work by “Petey21”/Wikimedia Commons; credit right: modification of work by Bill Morrow) Chapter Outline 7.1 Ionic Bonding 7.2 Covalent Bonding 7.3 Lewis Symbols and Structures 7.4 Formal Charges and Resonance 7.5 Strengths of Ionic and Covalent Bonds 7.6 Molecular Structure and Polarity Introduction It has long been known that pure carbon occurs in different forms (allotropes) including graphite and diamonds.
But it was not until 1985 that a new form of carbon was recognized: buckminsterfullerene, commonly known as a “buckyball.” This molecule was named after the architect and inventor R. Buckminster Fuller (1895–1983), whose signature architectural design was the geodesic dome, characterized by a lattice shell structure supporting a spherical surface. Experimental evidence revealed the formula, C60, and then scientists determined how 60 carbon atoms could form one symmetric, stable molecule. They were guided by bonding theory—the topic of this chapter—which explains how individual atoms connect to form more complex structures.
Chapter 7 Chemical Bonding and Molecular Geometry 345 7.1 Ionic Bonding By the end of this section, you will be able to: • Explain the formation of cations, anions, and ionic compounds • Predict the charge of common metallic and nonmetallic elements, and write their electron configurations As you have learned, ions are atoms or molecules bearing an electrical charge. A cation (a positive ion) forms when a neutral atom loses one or more electrons from its valence shell, and an anion (a negative ion) forms when a neutral atom gains one or more electrons in its valence shell.
Compounds composed of ions are called ionic compounds (or salts), and their constituent ions are held together by ionic bonds: electrostatic forces of attraction between oppositely charged cations and anions. The properties of ionic compounds shed some light on the nature of ionic bonds. Ionic solids exhibit a crystalline structure and tend to be rigid and brittle; they also tend to have high melting and boiling points, which suggests that ionic bonds are very strong. Ionic solids are also poor conductors of electricity for the same reason—the strength of ionic bonds prevents ions from moving freely in the solid state. Most ionic solids, however, dissolve readily in water. Once dissolved or melted, ionic compounds are excellent conductors of electricity and heat because the ions can move about freely.
Neutral atoms and their associated ions have very different physical and chemical properties. Sodium atoms form sodium metal, a soft, silvery-white metal that burns vigorously in air and reacts explosively with water. Chlorine atoms form chlorine gas, Cl2, a yellow-green gas that is extremely corrosive to most metals and very poisonous to animals and plants. The vigorous reaction between the elements sodium and chlorine forms the white, crystalline compound sodium chloride, common table salt, which contains sodium cations and chloride anions (Figure 7.2).
The compound composed of these ions exhibits properties entirely different from the properties of the elements sodium and chlorine. Chlorine is poisonous, but sodium chloride is essential to life; sodium atoms react vigorously with water, but sodium chloride simply dissolves in water.
Figure 7.2 (a) Sodium is a soft metal that must be stored in mineral oil to prevent reaction with air or water. (b) Chlorine is a pale yellow-green gas. (c) When combined, they form white crystals of sodium chloride (table salt).
(credit a: modification of work by “Jurii”/Wikimedia Commons) The Formation of Ionic Compounds Binary ionic compounds are composed of just two elements: a metal (which forms the cations) and a nonmetal (which forms the anions). For example, NaCl is a binary ionic compound. We can think about the formation of such compounds in terms of the periodic properties of the elements. Many metallic elements have relatively low ionization potentials and lose electrons easily. These elements lie to the left in a period or near the bottom of a group on the periodic table. Nonmetal atoms have relatively high electron affinities and thus readily gain electrons lost by metal 346 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at atoms, thereby filling their valence shells. Nonmetallic elements are found in the upper-right corner of the periodic table.
As all substances must be electrically neutral, the total number of positive charges on the cations of an ionic compound must equal the total number of negative charges on its anions. The formula of an ionic compound represents the simplest ratio of the numbers of ions necessary to give identical numbers of positive and negative charges. For example, the formula for aluminum oxide, Al2O3, indicates that this ionic compound contains two aluminum cations, Al3+, for every three oxide anions, O2−[thus, (2 × +3) + (3 × –2) = 0].
It is important to note, however, that the formula for an ionic compound does not represent the physical arrangement of its ions. It is incorrect to refer to a sodium chloride (NaCl) “molecule” because there is not a single ionic bond, per se, between any specific pair of sodium and chloride ions. The attractive forces between ions are isotropic—the same in all directions—meaning that any particular ion is equally attracted to all of the nearby ions of opposite charge. This results in the ions arranging themselves into a tightly bound, three-dimensional lattice structure. Sodium chloride, for example, consists of a regular arrangement of equal numbers of Na+ cations and Cl– anions (Figure 7.3).
Figure 7.3 The atoms in sodium chloride (common table salt) are arranged to (a) maximize opposite charges interacting. The smaller spheres represent sodium ions, the larger ones represent chloride ions. In the expanded view (b), the geometry can be seen more clearly. Note that each ion is “bonded” to all of the surrounding ions—six in this case.
The strong electrostatic attraction between Na+ and Cl– ions holds them tightly together in solid NaCl. It requires 769 kJ of energy to dissociate one mole of solid NaCl into separate gaseous Na+ and Cl– ions: NaCl(s) ⟶Na+(g) + Cl–(g) ΔH = 769 kJ Electronic Structures of Cations When forming a cation, an atom of a main group element tends to lose all of its valence electrons, thus assuming the electronic structure of the noble gas that precedes it in the periodic table. For groups 1 (the alkali metals) and 2 (the alkaline earth metals), the group numbers are equal to the numbers of valence shell electrons and, consequently, to the charges of the cations formed from atoms of these elements when all valence shell electrons are removed.
For example, calcium is a group 2 element whose neutral atoms have 20 electrons and a ground state electron configuration of 1s22s22p63s23p64s2. When a Ca atom loses both of its valence electrons, the result is a cation with 18 electrons, a 2+ charge, and an electron configuration of 1s22s22p63s23p6. The Ca2+ ion is therefore isoelectronic with the noble gas Ar.
For groups 12–17, the group numbers exceed the number of valence electrons by 10 (accounting for the possibility of full d subshells in atoms of elements in the fourth and greater periods). Thus, the charge of a cation formed by the loss of all valence electrons is equal to the group number minus 10. For example, aluminum (in group 13) forms 3+ ions (Al3+).
Chapter 7 Chemical Bonding and Molecular Geometry 347 Exceptions to the expected behavior involve elements toward the bottom of the groups. In addition to the expected ions Tl3+, Sn4+, Pb4+, and Bi5+, a partial loss of these atoms’ valence shell electrons can also lead to the formation of Tl+, Sn2+, Pb2+, and Bi3+ ions. The formation of these 1+, 2+, and 3+ cations is ascribed to the inert pair effect, which reflects the relatively low energy of the valence s-electron pair for atoms of the heavy elements of groups 13, 14, and 15. Mercury (group 12) also exhibits an unexpected behavior: it forms a diatomic ion, Hg2 2+ (an ion formed from two mercury atoms, with an Hg-Hg bond), in addition to the expected monatomic ion Hg2+ (formed from only one mercury atom).
Transition and inner transition metal elements behave differently than main group elements. Most transition metal cations have 2+ or 3+ charges that result from the loss of their outermost s electron(s) first, sometimes followed by the loss of one or two d electrons from the next-to-outermost shell. For example, iron (1s22s22p63s23p63d64s2) forms the ion Fe2+ (1s22s22p63s23p63d64s2) by the loss of the 4s electron and the ion Fe3+ (1s22s22p63s23p63d5) by the loss of the 4s electron and one of the 3d electrons. Although the d orbitals of the transition elements are—according to the Aufbau principle—the last to fill when building up electron configurations, the outermost s electrons are the first to be lost when these atoms ionize. When the inner transition metals form ions, they usually have a 3+ charge, resulting from the loss of their outermost s electrons and a d or f electron.
Example 7.1 Determining the Electronic Structures of Cations There are at least 14 elements categorized as “essential trace elements” for the human body. They are called “essential” because they are required for healthy bodily functions, “trace” because they are required only in small amounts, and “elements” in spite of the fact that they are really ions. Two of these essential trace elements, chromium and zinc, are required as Cr3+ and Zn2+. Write the electron configurations of these cations.
Solution First, write the electron configuration for the neutral atoms: Zn: [Ar]3d104s2 Cr: [Ar]3d34s1 Next, remove electrons from the highest energy orbital. For the transition metals, electrons are removed from the s orbital first and then from the d orbital. For the p-block elements, electrons are removed from the p orbitals and then from the s orbital. Zinc is a member of group 12, so it should have a charge of 2+, and thus loses only the two electrons in its s orbital. Chromium is a transition element and should lose its s electrons and then its d electrons when forming a cation. Thus, we find the following electron configurations of the ions: Zn2+: [Ar]3d10 Cr3+: [Ar]3d3 Check Your Learning Potassium and magnesium are required in our diet. Write the electron configurations of the ions expected from these elements.
Answer: K+: [Ar], Mg2+: [Ne] 348 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at Electronic Structures of Anions Most monatomic anions form when a neutral nonmetal atom gains enough electrons to completely fill its outer s and p orbitals, thereby reaching the electron configuration of the next noble gas. Thus, it is simple to determine the charge on such a negative ion: The charge is equal to the number of electrons that must be gained to fill the s and p orbitals of the parent atom. Oxygen, for example, has the electron configuration 1s22s22p4, whereas the oxygen anion has the electron configuration of the noble gas neon (Ne), 1s22s22p6. The two additional electrons required to fill the valence orbitals give the oxide ion the charge of 2– (O2–).
Example 7.2 Determining the Electronic Structure of Anions Selenium and iodine are two essential trace elements that form anions. Write the electron configurations of the anions.
Solution Se2–: [Ar]3d104s24p6 I–: [Kr]4d105s25p6 Check Your Learning Write the electron configurations of a phosphorus atom and its negative ion. Give the charge on the anion.
Answer: P: [Ne]3s23p3; P3–: [Ne]s23p6 7.2 Covalent Bonding By the end of this section, you will be able to: • Describe the formation of covalent bonds • Define electronegativity and assess the polarity of covalent bonds In ionic compounds, electrons are transferred between atoms of different elements to form ions. But this is not the only way that compounds can be formed. Atoms can also make chemical bonds by sharing electrons equally between each other. Such bonds are called covalent bonds. Covalent bonds are formed between two atoms when both have similar tendencies to attract electrons to themselves (i.e., when both atoms have identical or fairly similar ionization energies and electron affinities). For example, two hydrogen atoms bond covalently to form an H2 molecule; each hydrogen atom in the H2 molecule has two electrons stabilizing it, giving each atom the same number of valence electrons as the noble gas He.
Compounds that contain covalent bonds exhibit different physical properties than ionic compounds. Because the attraction between molecules, which are electrically neutral, is weaker than that between electrically charged ions, covalent compounds generally have much lower melting and boiling points than ionic compounds. In fact, many covalent compounds are liquids or gases at room temperature, and, in their solid states, they are typically much softer than ionic solids. Furthermore, whereas ionic compounds are good conductors of electricity when dissolved in water, most covalent compounds are insoluble in water; since they are electrically neutral, they are poor conductors of electricity in any state.
Chapter 7 Chemical Bonding and Molecular Geometry 349 Formation of Covalent Bonds Nonmetal atoms frequently form covalent bonds with other nonmetal atoms. For example, the hydrogen molecule, H2, contains a covalent bond between its two hydrogen atoms. Figure 7.4 illustrates why this bond is formed.
Starting on the far right, we have two separate hydrogen atoms with a particular potential energy, indicated by the red line. Along the x-axis is the distance between the two atoms. As the two atoms approach each other (moving left along the x-axis), their valence orbitals (1s) begin to overlap. The single electrons on each hydrogen atom then interact with both atomic nuclei, occupying the space around both atoms. The strong attraction of each shared electron to both nuclei stabilizes the system, and the potential energy decreases as the bond distance decreases. If the atoms continue to approach each other, the positive charges in the two nuclei begin to repel each other, and the potential energy increases. The bond length is determined by the distance at which the lowest potential energy is achieved.
Figure 7.4 The potential energy of two separate hydrogen atoms (right) decreases as they approach each other, and the single electrons on each atom are shared to form a covalent bond. The bond length is the internuclear distance at which the lowest potential energy is achieved.
350 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at It is essential to remember that energy must be added to break chemical bonds (an endothermic process), whereas forming chemical bonds releases energy (an exothermic process). In the case of H2, the covalent bond is very strong; a large amount of energy, 436 kJ, must be added to break the bonds in one mole of hydrogen molecules and cause the atoms to separate: H2(g) ⟶2H(g) ΔH = 436 kJ Conversely, the same amount of energy is released when one mole of H2 molecules forms from two moles of H atoms: 2H(g) ⟶H2(g) ΔH = −436 kJ Pure vs. Polar Covalent Bonds If the atoms that form a covalent bond are identical, as in H2, Cl2, and other diatomic molecules, then the electrons in the bond must be shared equally. We refer to this as a pure covalent bond. Electrons shared in pure covalent bonds have an equal probability of being near each nucleus.
In the case of Cl2, each atom starts off with seven valence electrons, and each Cl shares one electron with the other, forming one covalent bond: Cl + Cl ⟶Cl2 The total number of electrons around each individual atom consists of six nonbonding electrons and two shared (i.e., bonding) electrons for eight total electrons, matching the number of valence electrons in the noble gas argon. Since the bonding atoms are identical, Cl2 also features a pure covalent bond.
When the atoms linked by a covalent bond are different, the bonding electrons are shared, but no longer equally.
Instead, the bonding electrons are more attracted to one atom than the other, giving rise to a shift of electron density toward that atom. This unequal distribution of electrons is known as a polar covalent bond, characterized by a partial positive charge on one atom and a partial negative charge on the other. The atom that attracts the electrons more strongly acquires the partial negative charge and vice versa. For example, the electrons in the H–Cl bond of a hydrogen chloride molecule spend more time near the chlorine atom than near the hydrogen atom. Thus, in an HCl molecule, the chlorine atom carries a partial negative charge and the hydrogen atom has a partial positive charge.
Figure 7.5 shows the distribution of electrons in the H–Cl bond. Note that the shaded area around Cl is much larger than it is around H. Compare this to Figure 7.4, which shows the even distribution of electrons in the H2 nonpolar bond.
We sometimes designate the positive and negative atoms in a polar covalent bond using a lowercase Greek letter “delta,” δ, with a plus sign or minus sign to indicate whether the atom has a partial positive charge (δ+) or a partial negative charge (δ–). This symbolism is shown for the H–Cl molecule in Figure 7.5.
Figure 7.5 (a) The distribution of electron density in the HCl molecule is uneven. The electron density is greater around the chlorine nucleus. The small, black dots indicate the location of the hydrogen and chlorine nuclei in the molecule. (b) Symbols δ+ and δ– indicate the polarity of the H–Cl bond.
Chapter 7 Chemical Bonding and Molecular Geometry 351 Electronegativity Whether a bond is nonpolar or polar covalent is determined by a property of the bonding atoms called electronegativity. Electronegativity is a measure of the tendency of an atom to attract electrons (or electron density) towards itself. It determines how the shared electrons are distributed between the two atoms in a bond. The more strongly an atom attracts the electrons in its bonds, the larger its electronegativity. Electrons in a polar covalent bond are shifted toward the more electronegative atom; thus, the more electronegative atom is the one with the partial negative charge. The greater the difference in electronegativity, the more polarized the electron distribution and the larger the partial charges of the atoms.
Figure 7.6 shows the electronegativity values of the elements as proposed by one of the most famous chemists of the twentieth century: Linus Pauling (Figure 7.7). In general, electronegativity increases from left to right across a period in the periodic table and decreases down a group. Thus, the nonmetals, which lie in the upper right, tend to have the highest electronegativities, with fluorine the most electronegative element of all (EN = 4.0). Metals tend to be less electronegative elements, and the group 1 metals have the lowest electronegativities. Note that noble gases are excluded from this figure because these atoms usually do not share electrons with others atoms since they have a full valence shell. (While noble gas compounds such as XeO2 do exist, they can only be formed under extreme conditions, and thus they do not fit neatly into the general model of electronegativity.) Figure 7.6 The electronegativity values derived by Pauling follow predictable periodic trends with the higher electronegativities toward the upper right of the periodic table.
Electronegativity versus Electron Affinity We must be careful not to confuse electronegativity and electron affinity. The electron affinity of an element is a measurable physical quantity, namely, the energy released or absorbed when an isolated gas-phase atom acquires an electron, measured in kJ/mol. Electronegativity, on the other hand, describes how tightly an atom attracts electrons in a bond. It is a dimensionless quantity that is calculated, not measured. Pauling derived the first electronegativity values by comparing the amounts of energy required to break different types of bonds. He chose an arbitrary relative scale ranging from 0 to 4.
Portrait of a Chemist 352 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at Linus Pauling Linus Pauling, shown in Figure 7.7, is the only person to have received two unshared (individual) Nobel Prizes: one for chemistry in 1954 for his work on the nature of chemical bonds and one for peace in 1962 for his opposition to weapons of mass destruction. He developed many of the theories and concepts that are foundational to our current understanding of chemistry, including electronegativity and resonance structures.
Figure 7.7 Linus Pauling (1901–1994) made many important contributions to the field of chemistry. He was also a prominent activist, publicizing issues related to health and nuclear weapons.
Pauling also contributed to many other fields besides chemistry. His research on sickle cell anemia revealed the cause of the disease—the presence of a genetically inherited abnormal protein in the blood—and paved the way for the field of molecular genetics. His work was also pivotal in curbing the testing of nuclear weapons; he proved that radioactive fallout from nuclear testing posed a public health risk.
Electronegativity and Bond Type The absolute value of the difference in electronegativity (ΔEN) of two bonded atoms provides a rough measure of the polarity to be expected in the bond and, thus, the bond type. When the difference is very small or zero, the bond is covalent and nonpolar. When it is large, the bond is polar covalent or ionic. The absolute values of the electronegativity differences between the atoms in the bonds H–H, H–Cl, and Na–Cl are 0 (nonpolar), 0.9 (polar covalent), and 2.1 (ionic), respectively. The degree to which electrons are shared between atoms varies from completely equal (pure covalent bonding) to not at all (ionic bonding). Figure 7.8 shows the relationship between electronegativity difference and bond type.
Chapter 7 Chemical Bonding and Molecular Geometry 353 Figure 7.8 As the electronegativity difference increases between two atoms, the bond becomes more ionic.
A rough approximation of the electronegativity differences associated with covalent, polar covalent, and ionic bonds is shown in Figure 7.8. This table is just a general guide, however, with many exceptions. For example, the H and F atoms in HF have an electronegativity difference of 1.9, and the N and H atoms in NH3 a difference of 0.9, yet both of these compounds form bonds that are considered polar covalent. Likewise, the Na and Cl atoms in NaCl have an electronegativity difference of 2.1, and the Mn and I atoms in MnI2 have a difference of 1.0, yet both of these substances form ionic compounds.
The best guide to the covalent or ionic character of a bond is to consider the types of atoms involved and their relative positions in the periodic table. Bonds between two nonmetals are generally covalent; bonding between a metal and a nonmetal is often ionic.
Some compounds contain both covalent and ionic bonds. The atoms in polyatomic ions, such as OH–, NO3 −, and NH4 +, are held together by polar covalent bonds. However, these polyatomic ions form ionic compounds by combining with ions of opposite charge. For example, potassium nitrate, KNO3, contains the K+ cation and the polyatomic NO3 −anion. Thus, bonding in potassium nitrate is ionic, resulting from the electrostatic attraction between the ions K+ and NO3 −, as well as covalent between the nitrogen and oxygen atoms in NO3 −.
Example 7.3 Electronegativity and Bond Polarity Bond polarities play an important role in determining the structure of proteins. Using the electronegativity values in Figure 7.6, arrange the following covalent bonds—all commonly found in amino acids—in order of increasing polarity. Then designate the positive and negative atoms using the symbols δ+ and δ–: C–H, C–N, C–O, N–H, O–H, S–H Solution The polarity of these bonds increases as the absolute value of the electronegativity difference increases. The atom with the δ– designation is the more electronegative of the two. Table 7.1 shows these bonds in order of increasing polarity.
354 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at Bond Polarity and Electronegativity Difference Bond ΔEN Polarity C–H 0.4 C δ−−H δ+ S–H 0.4 S δ−−H δ+ C–N 0.5 C δ+ −N δ− N–H 0.9 N δ−−H δ+ C–O 1.0 C δ+ −O δ− O–H 1.4 O δ−−H δ+ Table 7.1 Check Your Learning Silicones are polymeric compounds containing, among others, the following types of covalent bonds: Si–O, Si–C, C–H, and C–C. Using the electronegativity values in Figure 7.6, arrange the bonds in order of increasing polarity and designate the positive and negative atoms using the symbols δ+ and δ–.
Answer: Bond Electronegativity Difference Polarity C–C 0.0 nonpolar C–H 0.4 C δ−−H δ+ Si–C 0.7 Si δ+ −C δ− Si–O 1.7 Si δ+ −O δ− 7.3 Lewis Symbols and Structures By the end of this section, you will be able to: • Write Lewis symbols for neutral atoms and ions • Draw Lewis structures depicting the bonding in simple molecules Thus far in this chapter, we have discussed the various types of bonds that form between atoms and/or ions. In all cases, these bonds involve the sharing or transfer of valence shell electrons between atoms. In this section, we will Chapter 7 Chemical Bonding and Molecular Geometry 355 explore the typical method for depicting valence shell electrons and chemical bonds, namely Lewis symbols and Lewis structures.
Lewis Symbols We use Lewis symbols to describe valence electron configurations of atoms and monatomic ions. A Lewis symbol consists of an elemental symbol surrounded by one dot for each of its valence electrons: Figure 7.9 shows the Lewis symbols for the elements of the third period of the periodic table.
Figure 7.9 Lewis symbols illustrating the number of valence electrons for each element in the third period of the periodic table.
Lewis symbols can also be used to illustrate the formation of cations from atoms, as shown here for sodium and calcium: Likewise, they can be used to show the formation of anions from atoms, as shown here for chlorine and sulfur: Figure 7.10 demonstrates the use of Lewis symbols to show the transfer of electrons during the formation of ionic compounds.
356 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at Figure 7.10 Cations are formed when atoms lose electrons, represented by fewer Lewis dots, whereas anions are formed by atoms gaining electrons. The total number of electrons does not change.
Lewis Structures We also use Lewis symbols to indicate the formation of covalent bonds, which are shown in Lewis structures, drawings that describe the bonding in molecules and polyatomic ions. For example, when two chlorine atoms form a chlorine molecule, they share one pair of electrons: The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs) and one shared pair of electrons (written between the atoms). A dash (or line) is sometimes used to indicate a shared pair of electrons: A single shared pair of electrons is called a single bond. Each Cl atom interacts with eight valence electrons: the six in the lone pairs and the two in the single bond.
The Octet Rule The other halogen molecules (F2, Br2, I2, and At2) form bonds like those in the chlorine molecule: one single bond between atoms and three lone pairs of electrons per atom. This allows each halogen atom to have a noble gas electron configuration. The tendency of main group atoms to form enough bonds to obtain eight valence electrons is known as the octet rule.
The number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons); this is especially true of the nonmetals of the second period of the periodic table (C, N, O, and F). For example, each atom of a group 14 element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds, as Chapter 7 Chemical Bonding and Molecular Geometry 357 illustrated here for carbon in CCl4 (carbon tetrachloride) and silicon in SiH4 (silane). Because hydrogen only needs two electrons to fill its valence shell, it is an exception to the octet rule. The transition elements and inner transition elements also do not follow the octet rule: Group 15 elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds, as in NH3 (ammonia). Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds: Double and Triple Bonds As previously mentioned, when a pair of atoms shares one pair of electrons, we call this a single bond. However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. A double bond forms when two pairs of electrons are shared between a pair of atoms, as between the carbon and oxygen atoms in CH2O (formaldehyde) and between the two carbon atoms in C2H4 (ethylene): A triple bond forms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion (CN–): Writing Lewis Structures with the Octet Rule For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. See these examples: For more complicated molecules and molecular ions, it is helpful to follow the step-by-step procedure outlined here: 358 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at 1.
Determine the total number of valence (outer shell) electrons. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge.
2.
Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. (Generally, the least electronegative element should be placed in the center.) Connect each atom to the central atom with a single bond (one electron pair).
3.
Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an octet around each atom.
4.
Place all remaining electrons on the central atom.
5.
Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.
Let us determine the Lewis structures of SiH4, CHO2 −, NO+, and OF2 as examples in following this procedure: 1.
Determine the total number of valence (outer shell) electrons in the molecule or ion.
• For a molecule, we add the number of valence electrons on each atom in the molecule: SiH4 Si: 4 valence electrons/atom × 1 atom = 4 + H: 1 valence electron/atom × 4 atoms = 4 = 8 valence electrons • For a negative ion, such as CHO2 −, we add the number of valence electrons on the atoms to the number of negative charges on the ion (one electron is gained for each single negative charge): CHO2 − C: 4 valence electrons/atom × 1 atom = 4 H: 1 valence electron/atom × 1 atom = 1 O: 6 valence electrons/atom × 2 atoms = 12 + 1 additional electron = 1 = 18 valence electrons • For a positive ion, such as NO+, we add the number of valence electrons on the atoms in the ion and then subtract the number of positive charges on the ion (one electron is lost for each single positive charge) from the total number of valence electrons: NO+ N: 5 valence electrons/atom × 1 atom = 5 O: 6 valence electron/atom × 1 atom = 6 + −1 electron (positive charge) = −1 = 10 valence electrons • Since OF2 is a neutral molecule, we simply add the number of valence electrons: OF2 O: 6 valence electrons/atom × 1 atom = 6 + F: 7 valence electrons/atom × 2 atoms = 14 = 20 valence electrons 2.
Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom and connecting each atom to the central atom with a single (one electron pair) bond. (Note that we denote ions with brackets around the structure, indicating the charge outside the brackets:) Chapter 7 Chemical Bonding and Molecular Geometry 359 When several arrangements of atoms are possible, as for CHO2 −, we must use experimental evidence to choose the correct one. In general, the less electronegative elements are more likely to be central atoms. In CHO2 −, the less electronegative carbon atom occupies the central position with the oxygen and hydrogen atoms surrounding it. Other examples include P in POCl3, S in SO2, and Cl in ClO4 −. An exception is that hydrogen is almost never a central atom. As the most electronegative element, fluorine also cannot be a central atom.
3.
Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen) to complete their valence shells with an octet of electrons.
• There are no remaining electrons on SiH4, so it is unchanged: 4.
Place all remaining electrons on the central atom.
• For SiH4, CHO2 −, and NO+, there are no remaining electrons; we already placed all of the electrons determined in Step 1.
• For OF2, we had 16 electrons remaining in Step 3, and we placed 12, leaving 4 to be placed on the central atom: 5.
Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.
• SiH4: Si already has an octet, so nothing needs to be done.
• CHO2 −: We have distributed the valence electrons as lone pairs on the oxygen atoms, but one oxygen atom and one carbon atom lack octets: • NO+: For this ion, we added eight valence electrons, but neither atom has an octet. We cannot add any more electrons since we have already used the total that we found in Step 1, so we must move electrons to form a multiple bond: This still does not produce an octet, so we must move another pair, forming a triple bond: • In OF2, each atom has an octet as drawn, so nothing changes.
360 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at Example 7.4 Writing Lewis Structures NASA’s Cassini-Huygens mission detected a large cloud of toxic hydrogen cyanide (HCN) on Titan, one of Saturn’s moons. Titan also contains ethane (H3CCH3), acetylene (HCCH), and ammonia (NH3). What are the Lewis structures of these molecules?
Solution Step 1.
Calculate the number of valence electrons.
HCN: (1 × 1) + (4 × 1) + (5 × 1) = 10 H3CCH3: (1 × 3) + (2 × 4) + (1 × 3) = 14 HCCH: (1 × 1) + (2 × 4) + (1 × 1) = 10 NH3: (5 × 1) + (3 × 1) = 8 Step 2.
Draw a skeleton and connect the atoms with single bonds. Remember that H is never a central atom: Step 3.
Where needed, distribute electrons to the terminal atoms: HCN: six electrons placed on N H3CCH3: no electrons remain HCCH: no terminal atoms capable of accepting electrons NH3: no terminal atoms capable of accepting electrons Step 4.
Where needed, place remaining electrons on the central atom: HCN: no electrons remain H3CCH3: no electrons remain HCCH: four electrons placed on carbon NH3: two electrons placed on nitrogen Step 5.
Where needed, rearrange electrons to form multiple bonds in order to obtain an octet on each atom: HCN: form two more C–N bonds H3CCH3: all atoms have the correct number of electrons HCCH: form a triple bond between the two carbon atoms Chapter 7 Chemical Bonding and Molecular Geometry 361 NH3: all atoms have the correct number of electrons Check Your Learning Both carbon monoxide, CO, and carbon dioxide, CO2, are products of the combustion of fossil fuels. Both of these gases also cause problems: CO is toxic and CO2 has been implicated in global climate change.
What are the Lewis structures of these two molecules?
Answer: Fullerene Chemistry Carbon soot has been known to man since prehistoric times, but it was not until fairly recently that the molecular structure of the main component of soot was discovered. In 1996, the Nobel Prize in Chemistry was awarded to Richard Smalley (Figure 7.11), Robert Curl, and Harold Kroto for their work in discovering a new form of carbon, the C60 buckminsterfullerene molecule (Figure 7.1). An entire class of compounds, including spheres and tubes of various shapes, were discovered based on C60. This type of molecule, called a fullerene, shows promise in a variety of applications. Because of their size and shape, fullerenes can encapsulate other molecules, so they have shown potential in various applications from hydrogen storage to targeted drug delivery systems. They also possess unique electronic and optical properties that have been put to good use in solar powered devices and chemical sensors.
Figure 7.11 Richard Smalley (1943–2005), a professor of physics, chemistry, and astronomy at Rice University, was one of the leading advocates for fullerene chemistry. Upon his death in 2005, the US Senate honored him as the “Father of Nanotechnology.” (credit: United States Department of Energy) How Sciences Interconnect 362 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at Exceptions to the Octet Rule Many covalent molecules have central atoms that do not have eight electrons in their Lewis structures. These molecules fall into three categories: • Odd-electron molecules have an odd number of valence electrons, and therefore have an unpaired electron.
• Electron-deficient molecules have a central atom that has fewer electrons than needed for a noble gas configuration.
• Hypervalent molecules have a central atom that has more electrons than needed for a noble gas configuration.
Odd-electron Molecules We call molecules that contain an odd number of electrons free radicals. Nitric oxide, NO, is an example of an odd-electron molecule; it is produced in internal combustion engines when oxygen and nitrogen react at high temperatures.
To draw the Lewis structure for an odd-electron molecule like NO, we follow the same six steps we would for other molecules, but with a few minor changes: 1.
Determine the total number of valence (outer shell) electrons. The sum of the valence electrons is 5 (from N) + 6 (from O) = 11. The odd number immediately tells us that we have a free radical, so we know that not every atom can have eight electrons in its valence shell.
2.
Draw a skeleton structure of the molecule. We can easily draw a skeleton with an N–O single bond: N–O 3.
Distribute the remaining electrons as lone pairs on the terminal atoms. In this case, there is no central atom, so we distribute the electrons around both atoms. We give eight electrons to the more electronegative atom in these situations; thus oxygen has the filled valence shell: 4.
Place all remaining electrons on the central atom. Since there are no remaining electrons, this step does not apply.
5.
Rearrange the electrons to make multiple bonds with the central atom in order to obtain octets wherever possible. We know that an odd-electron molecule cannot have an octet for every atom, but we want to get each atom as close to an octet as possible. In this case, nitrogen has only five electrons around it. To move closer to an octet for nitrogen, we take one of the lone pairs from oxygen and use it to form a NO double bond. (We cannot take another lone pair of electrons on oxygen and form a triple bond because nitrogen would then have nine electrons:) Electron-deficient Molecules We will also encounter a few molecules that contain central atoms that do not have a filled valence shell. Generally, these are molecules with central atoms from groups 2 and 12, outer atoms that are hydrogen, or other atoms that do not form multiple bonds. For example, in the Lewis structures of beryllium dihydride, BeH2, and boron trifluoride, BF3, the beryllium and boron atoms each have only four and six electrons, respectively. It is possible to draw a structure with a double bond between a boron atom and a fluorine atom in BF3, satisfying the octet rule, but experimental evidence indicates the bond lengths are closer to that expected for B–F single bonds. This suggests the best Lewis structure has three B–F single bonds and an electron deficient boron. The reactivity of the compound is also consistent with an electron deficient boron. However, the B–F bonds are slightly shorter than what is actually expected for B–F single bonds, indicating that some double bond character is found in the actual molecule.
Chapter 7 Chemical Bonding and Molecular Geometry 363 An atom like the boron atom in BF3, which does not have eight electrons, is very reactive. It readily combines with a molecule containing an atom with a lone pair of electrons. For example, NH3 reacts with BF3 because the lone pair on nitrogen can be shared with the boron atom: Hypervalent Molecules Elements in the second period of the periodic table (n = 2) can accommodate only eight electrons in their valence shell orbitals because they have only four valence orbitals (one 2s and three 2p orbitals). Elements in the third and higher periods (n ≥3) have more than four valence orbitals and can share more than four pairs of electrons with other atoms because they have empty d orbitals in the same shell. Molecules formed from these elements are sometimes called hypervalent molecules. Figure 7.12 shows the Lewis structures for two hypervalent molecules, PCl5 and SF6.
Figure 7.12 In PCl5, the central atom phosphorus shares five pairs of electrons. In SF6, sulfur shares six pairs of electrons.
In some hypervalent molecules, such as IF5 and XeF4, some of the electrons in the outer shell of the central atom are lone pairs: When we write the Lewis structures for these molecules, we find that we have electrons left over after filling the valence shells of the outer atoms with eight electrons. These additional electrons must be assigned to the central atom.
Example 7.5 Writing Lewis Structures: Octet Rule Violations Xenon is a noble gas, but it forms a number of stable compounds. We examined XeF4 earlier. What are the Lewis structures of XeF2 and XeF6?
Solution 364 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at We can draw the Lewis structure of any covalent molecule by following the six steps discussed earlier. In this case, we can condense the last few steps, since not all of them apply.
Step 1.
Calculate the number of valence electrons: XeF2: 8 + (2 × 7) = 22 XeF6: 8 + (6 × 7) = 50 Step 2.
Draw a skeleton joining the atoms by single bonds. Xenon will be the central atom because fluorine cannot be a central atom: Step 3.
Distribute the remaining electrons.
XeF2: We place three lone pairs of electrons around each F atom, accounting for 12 electrons and giving each F atom 8 electrons. Thus, six electrons (three lone pairs) remain. These lone pairs must be placed on the Xe atom. This is acceptable because Xe atoms have empty valence shell d orbitals and can accommodate more than eight electrons. The Lewis structure of XeF2 shows two bonding pairs and three lone pairs of electrons around the Xe atom: XeF6: We place three lone pairs of electrons around each F atom, accounting for 36 electrons. Two electrons remain, and this lone pair is placed on the Xe atom: Check Your Learning The halogens form a class of compounds called the interhalogens, in which halogen atoms covalently bond to each other. Write the Lewis structures for the interhalogens BrCl3 and ICl4 −.
Answer: 7.4 Formal Charges and Resonance By the end of this section, you will be able to: • Compute formal charges for atoms in any Lewis structure • Use formal charges to identify the most reasonable Lewis structure for a given molecule • Explain the concept of resonance and draw Lewis structures representing resonance forms for a given molecule Chapter 7 Chemical Bonding and Molecular Geometry 365 In the previous section, we discussed how to write Lewis structures for molecules and polyatomic ions. As we have seen, however, in some cases, there is seemingly more than one valid structure for a molecule. We can use the concept of formal charges to help us predict the most appropriate Lewis structure when more than one is reasonable.
Calculating Formal Charge The formal charge of an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms. Another way of saying this is that formal charge results when we take the number of valence electrons of a neutral atom, subtract the nonbonding electrons, and then subtract the number of bonds connected to that atom in the Lewis structure.
Thus, we calculate formal charge as follows: formal charge = # valence shell electrons (free atom) −# lone pair electrons −1 2 # bonding electrons We can double-check formal charge calculations by determining the sum of the formal charges for the whole structure.
The sum of the formal charges of all atoms in a molecule must be zero; the sum of the formal charges in an ion should equal the charge of the ion.
We must remember that the formal charge calculated for an atom is not the actual charge of the atom in the molecule.
Formal charge is only a useful bookkeeping procedure; it does not indicate the presence of actual charges.
Example 7.6 Calculating Formal Charge from Lewis Structures Assign formal charges to each atom in the interhalogen ion ICl4 −.
Solution Step 1.
We divide the bonding electron pairs equally for all I–Cl bonds: Step 2.
We assign lone pairs of electrons to their atoms. Each Cl atom now has seven electrons assigned to it, and the I atom has eight.
Step 3.
Subtract this number from the number of valence electrons for the neutral atom: I: 7 – 8 = –1 Cl: 7 – 7 = 0 The sum of the formal charges of all the atoms equals –1, which is identical to the charge of the ion (–1).
Check Your Learning Calculate the formal charge for each atom in the carbon monoxide molecule: Answer: C −1, O +1 Example 7.7 Calculating Formal Charge from Lewis Structures Assign formal charges to each atom in the interhalogen molecule BrCl3.
366 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at Solution Step 1.
Assign one of the electrons in each Br–Cl bond to the Br atom and one to the Cl atom in that bond: Step 2.
Assign the lone pairs to their atom. Now each Cl atom has seven electrons and the Br atom has seven electrons.
Step 3.
Subtract this number from the number of valence electrons for the neutral atom. This gives the formal charge: Br: 7 – 7 = 0 Cl: 7 – 7 = 0 All atoms in BrCl3 have a formal charge of zero, and the sum of the formal charges totals zero, as it must in a neutral molecule.
Check Your Learning Determine the formal charge for each atom in NCl3.
Answer: N: 0; all three Cl atoms: 0 Using Formal Charge to Predict Molecular Structure The arrangement of atoms in a molecule or ion is called its molecular structure. In many cases, following the steps for writing Lewis structures may lead to more than one possible molecular structure—different multiple bond and lone-pair electron placements or different arrangements of atoms, for instance. A few guidelines involving formal charge can be helpful in deciding which of the possible structures is most likely for a particular molecule or ion: 1.
A molecular structure in which all formal charges are zero is preferable to one in which some formal charges are not zero.
2.
If the Lewis structure must have nonzero formal charges, the arrangement with the smallest nonzero formal charges is preferable.
3.
Lewis structures are preferable when adjacent formal charges are zero or of the opposite sign.
4.
When we must choose among several Lewis structures with similar distributions of formal charges, the structure with the negative formal charges on the more electronegative atoms is preferable.
To see how these guidelines apply, let us consider some possible structures for carbon dioxide, CO2. We know from our previous discussion that the less electronegative atom typically occupies the central position, but formal charges allow us to understand why this occurs. We can draw three possibilities for the structure: carbon in the center and double bonds, carbon in the center with a single and triple bond, and oxygen in the center with double bonds: Comparing the three formal charges, we can definitively identify the structure on the left as preferable because it has only formal charges of zero (Guideline 1).
As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: CNS–, NCS–, or CSN–. The formal charges present in each of these Chapter 7 Chemical Bonding and Molecular Geometry 367 molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here: Note that the sum of the formal charges in each case is equal to the charge of the ion (–1). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).
Example 7.8 Using Formal Charge to Determine Molecular Structure Nitrous oxide, N2O, commonly known as laughing gas, is used as an anesthetic in minor surgeries, such as the routine extraction of wisdom teeth. Which is the likely structure for nitrous oxide?
Solution Determining formal charge yields the following: The structure with a terminal oxygen atom best satisfies the criteria for the most stable distribution of formal charge: The number of atoms with formal charges are minimized (Guideline 2), and there is no formal charge larger than one (Guideline 2). This is again consistent with the preference for having the less electronegative atom in the central position.
Check Your Learning Which is the most likely molecular structure for the nitrite ⎛ ⎝NO2 −⎞ ⎠ion?
Answer: ONO– Resonance You may have noticed that the nitrite anion in Example 7.8 can have two possible structures with the atoms in the same positions. The electrons involved in the N–O double bond, however, are in different positions: If nitrite ions do indeed contain a single and a double bond, we would expect for the two bond lengths to be different. A double bond between two atoms is shorter (and stronger) than a single bond between the same two atoms.
Experiments show, however, that both N–O bonds in NO2 −have the same strength and length, and are identical in all other properties.
368 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at It is not possible to write a single Lewis structure for NO2 −in which nitrogen has an octet and both bonds are equivalent. Instead, we use the concept of resonance: if two or more Lewis structures with the same arrangement of atoms can be written for a molecule or ion, the actual distribution of electrons is an average of that shown by the various Lewis structures. The actual distribution of electrons in each of the nitrogen-oxygen bonds in NO2 − is the average of a double bond and a single bond. We call the individual Lewis structures resonance forms. The actual electronic structure of the molecule (the average of the resonance forms) is called a resonance hybrid of the individual resonance forms. A double-headed arrow between Lewis structures indicates that they are resonance forms.
Thus, the electronic structure of the NO2 −ion is shown as: We should remember that a molecule described as a resonance hybrid never possesses an electronic structure described by either resonance form. It does not fluctuate between resonance forms; rather, the actual electronic structure is always the average of that shown by all resonance forms. George Wheland, one of the pioneers of resonance theory, used a historical analogy to describe the relationship between resonance forms and resonance hybrids. A medieval traveler, having never before seen a rhinoceros, described it as a hybrid of a dragon and a unicorn because it had many properties in common with both. Just as a rhinoceros is neither a dragon sometimes nor a unicorn at other times, a resonance hybrid is neither of its resonance forms at any given time. Like a rhinoceros, it is a real entity that experimental evidence has shown to exist. It has some characteristics in common with its resonance forms, but the resonance forms themselves are convenient, imaginary images (like the unicorn and the dragon).
The carbonate anion, CO3 2−, provides a second example of resonance: One oxygen atom must have a double bond to carbon to complete the octet on the central atom. All oxygen atoms, however, are equivalent, and the double bond could form from any one of the three atoms. This gives rise to three resonance forms of the carbonate ion. Because we can write three identical resonance structures, we know that the actual arrangement of electrons in the carbonate ion is the average of the three structures. Again, experiments show that all three C–O bonds are exactly the same.
The online Lewis Structure Make ( includes many examples to practice drawing resonance structures.
Link to Learning Chapter 7 Chemical Bonding and Molecular Geometry 369 7.5 Strengths of Ionic and Covalent Bonds By the end of this section, you will be able to: • Describe the energetics of covalent and ionic bond formation and breakage • Use the Born-Haber cycle to compute lattice energies for ionic compounds • Use average covalent bond energies to estimate enthalpies of reaction A bond’s strength describes how strongly each atom is joined to another atom, and therefore how much energy is required to break the bond between the two atoms. In this section, you will learn about the bond strength of covalent bonds, and then compare that to the strength of ionic bonds, which is related to the lattice energy of a compound.
Bond Strength: Covalent Bonds Stable molecules exist because covalent bonds hold the atoms together. We measure the strength of a covalent bond by the energy required to break it, that is, the energy necessary to separate the bonded atoms. Separating any pair of bonded atoms requires energy (see Figure 7.4). The stronger a bond, the greater the energy required to break it.
The energy required to break a specific covalent bond in one mole of gaseous molecules is called the bond energy or the bond dissociation energy. The bond energy for a diatomic molecule, DX–Y, is defined as the standard enthalpy change for the endothermic reaction: XY(g) ⟶X(g) + Y(g) DX−Y = ΔH° For example, the bond energy of the pure covalent H–H bond, DH–H, is 436 kJ per mole of H–H bonds broken: H2(g) ⟶2H(g) DH−H = ΔH° = 436 kJ Molecules with three or more atoms have two or more bonds. The sum of all bond energies in such a molecule is equal to the standard enthalpy change for the endothermic reaction that breaks all the bonds in the molecule. For example, the sum of the four C–H bond energies in CH4, 1660 kJ, is equal to the standard enthalpy change of the reaction: The average C–H bond energy, DC–H, is 1660/4 = 415 kJ/mol because there are four moles of C–H bonds broken per mole of the reaction. Although the four C–H bonds are equivalent in the original molecule, they do not each require the same energy to break; once the first bond is broken (which requires 439 kJ/mol), the remaining bonds are easier to break. The 415 kJ/mol value is the average, not the exact value required to break any one bond.
The strength of a bond between two atoms increases as the number of electron pairs in the bond increases. Generally, as the bond strength increases, the bond length decreases. Thus, we find that triple bonds are stronger and shorter than double bonds between the same two atoms; likewise, double bonds are stronger and shorter than single bonds between the same two atoms. Average bond energies for some common bonds appear in Table 7.2, and a comparison of bond lengths and bond strengths for some common bonds appears in Table 7.3. When one atom bonds to various atoms in a group, the bond strength typically decreases as we move down the group. For example, C–F is 439 kJ/mol, C–Cl is 330 kJ/mol, and C–Br is 275 kJ/mol.
370 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at Bond Energies (kJ/mol) Bond Bond Energy Bond Bond Energy Bond Bond Energy H–H 436 C–S 260 F–Cl 255 H–C 415 C–Cl 330 F–Br 235 H–N 390 C–Br 275 Si–Si 230 H–O 464 C–I 240 Si–P 215 H–F 569 N–N 160 Si–S 225 H–Si 395 N = N 418 Si–Cl 359 H–P 320 N ≡N 946 Si–Br 290 H–S 340 N–O 200 Si–I 215 H–Cl 432 N–F 270 P–P 215 H–Br 370 N–P 210 P–S 230 H–I 295 N–Cl 200 P–Cl 330 C–C 345 N–Br 245 P–Br 270 C = C 611 O–O 140 P–I 215 C ≡C 837 O = O 498 S–S 215 C–N 290 O–F 160 S–Cl 250 C = N 615 O–Si 370 S–Br 215 C ≡N 891 O–P 350 Cl–Cl 243 C–O 350 O–Cl 205 Cl–Br 220 C = O 741 O–I 200 Cl–I 210 C ≡O 1080 F–F 160 Br–Br 190 C–F 439 F–Si 540 Br–I 180 C–Si 360 F–P 489 I–I 150 C–P 265 F–S 285 Table 7.2 Chapter 7 Chemical Bonding and Molecular Geometry 371 Average Bond Lengths and Bond Energies for Some Common Bonds Bond Bond Length (Å) Bond Energy (kJ/mol) C–C 1.54 345 C = C 1.34 611 C ≡C 1.20 837 C–N 1.43 290 C = N 1.38 615 C ≡N 1.16 891 C–O 1.43 350 C = O 1.23 741 C ≡O 1.13 1080 Table 7.3 We can use bond energies to calculate approximate enthalpy changes for reactions where enthalpies of formation are not available. Calculations of this type will also tell us whether a reaction is exothermic or endothermic. An exothermic reaction (ΔH negative, heat produced) results when the bonds in the products are stronger than the bonds in the reactants. An endothermic reaction (ΔH positive, heat absorbed) results when the bonds in the products are weaker than those in the reactants.
The enthalpy change, ΔH, for a chemical reaction is approximately equal to the sum of the energy required to break all bonds in the reactants (energy “in”, positive sign) plus the energy released when all bonds are formed in the products (energy “out,” negative sign). This can be expressed mathematically in the following way: ΔH = ƩD bonds broken −ƩD bonds formed In this expression, the symbol Ʃ means “the sum of” and D represents the bond energy in kilojoules per mole, which is always a positive number. The bond energy is obtained from a table (like Table 7.3) and will depend on whether the particular bond is a single, double, or triple bond. Thus, in calculating enthalpies in this manner, it is important that we consider the bonding in all reactants and products. Because D values are typically averages for one type of bond in many different molecules, this calculation provides a rough estimate, not an exact value, for the enthalpy of reaction.
Consider the following reaction: H2(g) + Cl2(g) ⟶2HCl(g) or H–H(g) + Cl–Cl(g) ⟶2H–Cl(g) To form two moles of HCl, one mole of H–H bonds and one mole of Cl–Cl bonds must be broken. The energy required to break these bonds is the sum of the bond energy of the H–H bond (436 kJ/mol) and the Cl–Cl bond (243 kJ/mol). During the reaction, two moles of H–Cl bonds are formed (bond energy = 432 kJ/mol), releasing 2 × 432 kJ; or 864 kJ. Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it consumes: 372 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at ΔH = ƩD bonds broken −ƩD bonds formed ΔH = ⎡ ⎣DH−H + DCl−Cl ⎤ ⎦−2DH−Cl = [436 + 243] −2(432) = −185 kJ This excess energy is released as heat, so the reaction is exothermic. Appendix G gives a value for the standard molar enthalpy of formation of HCl(g), ΔHf ° , of –92.307 kJ/mol. Twice that value is –184.6 kJ, which agrees well with the answer obtained earlier for the formation of two moles of HCl.
Example 7.9 Using Bond Energies to Calculate Approximate Enthalpy Changes Methanol, CH3OH, may be an excellent alternative fuel. The high-temperature reaction of steam and carbon produces a mixture of the gases carbon monoxide, CO, and hydrogen, H2, from which methanol can be produced. Using the bond energies in Table 7.3, calculate the approximate enthalpy change, ΔH, for the reaction here: CO(g) + 2H2(g) ⟶CH3 OH(g) Solution First, we need to write the Lewis structures of the reactants and the products: From this, we see that ΔH for this reaction involves the energy required to break a C–O triple bond and two H–H single bonds, as well as the energy produced by the formation of three C–H single bonds, a C–O single bond, and an O–H single bond. We can express this as follows: ΔH = ƩD bonds broken −ƩD bonds formed ΔH = ⎡ ⎣DC ≡O + 2⎛ ⎝DH−H ⎞ ⎠ ⎤ ⎦−⎡ ⎣3⎛ ⎝DC−H ⎞ ⎠+ DC−O + DO−H ⎤ ⎦ Using the bond energy values in Table 7.3, we obtain: ΔH = ⎡ ⎣1080 + 2(436) ⎤ ⎦− ⎡ ⎣3(415) + 350 + 464 ⎤ ⎦ = −107 kJ We can compare this value to the value calculated based on ΔHf ° data from Appendix G: ΔH = ⎡ ⎣ΔHf ° CH3 OH(g)⎤ ⎦−⎡ ⎣ΔHf ° CO(g) + 2 × ΔHf ° H2 ⎤ ⎦ = [−201.0] −[−110.52 + 2 × 0] = −90.5 kJ Note that there is a fairly significant gap between the values calculated using the two different methods.
This occurs because D values are the average of different bond strengths; therefore, they often give only rough agreement with other data.
Check Your Learning Ethyl alcohol, CH3CH2OH, was one of the first organic chemicals deliberately synthesized by humans. It has many uses in industry, and it is the alcohol contained in alcoholic beverages. It can be obtained by the fermentation of sugar or synthesized by the hydration of ethylene in the following reaction: Chapter 7 Chemical Bonding and Molecular Geometry 373 Using the bond energies in Table 7.3, calculate an approximate enthalpy change, ΔH, for this reaction.
Answer: –35 kJ Ionic Bond Strength and Lattice Energy An ionic compound is stable because of the electrostatic attraction between its positive and negative ions. The lattice energy of a compound is a measure of the strength of this attraction. The lattice energy (ΔHlattice) of an ionic compound is defined as the energy required to separate one mole of the solid into its component gaseous ions. For the ionic solid MX, the lattice energy is the enthalpy change of the process: MX(s) ⟶Mn+(g) + Xn−(g) ΔHlattice Note that we are using the convention where the ionic solid is separated into ions, so our lattice energies will be endothermic (positive values). Some texts use the equivalent but opposite convention, defining lattice energy as the energy released when separate ions combine to form a lattice and giving negative (exothermic) values. Thus, if you are looking up lattice energies in another reference, be certain to check which definition is being used. In both cases, a larger magnitude for lattice energy indicates a more stable ionic compound. For sodium chloride, ΔHlattice = 769 kJ.
Thus, it requires 769 kJ to separate one mole of solid NaCl into gaseous Na+ and Cl– ions. When one mole each of gaseous Na+ and Cl– ions form solid NaCl, 769 kJ of heat is released.
The lattice energy ΔHlattice of an ionic crystal can be expressed by the following equation (derived from Coulomb’s law, governing the forces between electric charges): ΔHlattice = C(Z+)(Z−) Ro in which C is a constant that depends on the type of crystal structure; Z+ and Z– are the charges on the ions; and Ro is the interionic distance (the sum of the radii of the positive and negative ions). Thus, the lattice energy of an ionic crystal increases rapidly as the charges of the ions increase and the sizes of the ions decrease. When all other parameters are kept constant, doubling the charge of both the cation and anion quadruples the lattice energy. For example, the lattice energy of LiF (Z+ and Z– = 1) is 1023 kJ/mol, whereas that of MgO (Z+ and Z– = 2) is 3900 kJ/ mol (Ro is nearly the same—about 200 pm for both compounds).
Different interatomic distances produce different lattice energies. For example, we can compare the lattice energy of MgF2 (2957 kJ/mol) to that of MgI2 (2327 kJ/mol) to observe the effect on lattice energy of the smaller ionic size of F– as compared to I–.
Example 7.10 Lattice Energy Comparisons The precious gem ruby is aluminum oxide, Al2O3, containing traces of Cr3+. The compound Al2Se3 is used in the fabrication of some semiconductor devices. Which has the larger lattice energy, Al2O3 or Al2Se3?
Solution 374 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at In these two ionic compounds, the charges Z+ and Z– are the same, so the difference in lattice energy will depend upon Ro. The O2– ion is smaller than the Se2– ion. Thus, Al2O3 would have a shorter interionic distance than Al2Se3, and Al2O3 would have the larger lattice energy.
Check Your Learning Zinc oxide, ZnO, is a very effective sunscreen. How would the lattice energy of ZnO compare to that of NaCl?
Answer: ZnO would have the larger lattice energy because the Z values of both the cation and the anion in ZnO are greater, and the interionic distance of ZnO is smaller than that of NaCl.
The Born-Haber Cycle It is not possible to measure lattice energies directly. However, the lattice energy can be calculated using the equation given in the previous section or by using a thermochemical cycle. The Born-Haber cycle is an application of Hess’s law that breaks down the formation of an ionic solid into a series of individual steps: • ΔHf ° , the standard enthalpy of formation of the compound • IE, the ionization energy of the metal • EA, the electron affinity of the nonmetal • ΔHs ° , the enthalpy of sublimation of the metal • D, the bond dissociation energy of the nonmetal • ΔHlattice, the lattice energy of the compound Figure 7.13 diagrams the Born-Haber cycle for the formation of solid cesium fluoride.
Figure 7.13 The Born-Haber cycle shows the relative energies of each step involved in the formation of an ionic solid from the necessary elements in their reference states.
We begin with the elements in their most common states, Cs(s) and F2(g). The ΔHs ° represents the conversion of solid cesium into a gas, and then the ionization energy converts the gaseous cesium atoms into cations. In the next step, we account for the energy required to break the F–F bond to produce fluorine atoms. Converting one mole of Chapter 7 Chemical Bonding and Molecular Geometry 375 fluorine atoms into fluoride ions is an exothermic process, so this step gives off energy (the electron affinity) and is shown as decreasing along the y-axis. We now have one mole of Cs cations and one mole of F anions. These ions combine to produce solid cesium fluoride. The enthalpy change in this step is the negative of the lattice energy, so it is also an exothermic quantity. The total energy involved in this conversion is equal to the experimentally determined enthalpy of formation, ΔHf ° , of the compound from its elements. In this case, the overall change is exothermic.
Hess’s law can also be used to show the relationship between the enthalpies of the individual steps and the enthalpy of formation. Table 7.4 shows this for cesium chloride, CsCl2.
Enthalpy of sublimation of Cs(s) Cs(s) ⟶Cs(g) ΔH = ΔHs ° = 76.5kJ One-half of the bond energy of Cl2 1 2 Cl2(g) ⟶Cl(g) ΔH = 1 2 D = 122kJ Ionization energy of Na(g) Na(g) ⟶Na+(g) + e− ΔH = IE = 496kJ Negative of the electron affinity of Cl Cl(g) + e−⟶Cl−(g) ΔH = −EA = −368kJ Negative of the lattice energy of NaCl(s) Na+(g) + Cl−(g) ⟶NaCl(s) ΔH = −ΔHlattice = ?
Enthalpy of formation of NaCl(s), add steps 1–5 ΔH = ΔH f ° = ΔHs ° + 1 2 D + IE + (−EA) + ⎛ ⎝−ΔHlattice ⎞ ⎠ Na(s) + 1 2 Cl2(g) ⟶NaCl(s) = −411kJ Table 7.4 Thus, the lattice energy can be calculated from other values. For cesium chloride, using this data, the lattice energy is: ΔHlattice = (411 + 109 + 122 + 496 + 368) kJ = 770 kJ The Born-Haber cycle may also be used to calculate any one of the other quantities in the equation for lattice energy, provided that the remainder is known. For example, if the relevant enthalpy of sublimation ΔHs ° , ionization energy (IE), bond dissociation enthalpy (D), lattice energy ΔHlattice, and standard enthalpy of formation ΔHf ° are known, the Born-Haber cycle can be used to determine the electron affinity of an atom.
Lattice energies calculated for ionic compounds are typically much higher than bond dissociation energies measured for covalent bonds. Whereas lattice energies typically fall in the range of 600–4000 kJ/mol (some even higher), covalent bond dissociation energies are typically between 150–400 kJ/mol for single bonds. Keep in mind, however, that these are not directly comparable values. For ionic compounds, lattice energies are associated with many interactions, as cations and anions pack together in an extended lattice. For covalent bonds, the bond dissociation energy is associated with the interaction of just two atoms.
376 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at 7.6 Molecular Structure and Polarity By the end of this section, you will be able to: • Predict the structures of small molecules using valence shell electron pair repulsion (VSEPR) theory • Explain the concepts of polar covalent bonds and molecular polarity • Assess the polarity of a molecule based on its bonding and structure Thus far, we have used two-dimensional Lewis structures to represent molecules. However, molecular structure is actually three-dimensional, and it is important to be able to describe molecular bonds in terms of their distances, angles, and relative arrangements in space (Figure 7.14). A bond angle is the angle between any two bonds that include a common atom, usually measured in degrees. A bond distance (or bond length) is the distance between the nuclei of two bonded atoms along the straight line joining the nuclei. Bond distances are measured in Ångstroms (1 Å = 10–10 m) or picometers (1 pm = 10–12 m, 100 pm = 1 Å).
Figure 7.14 Bond distances (lengths) and angles are shown for the formaldehyde molecule, H2CO.
VSEPR Theory Valence shell electron-pair repulsion theory (VSEPR theory) enables us to predict the molecular structure, including approximate bond angles around a central atom, of a molecule from an examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement that minimizes repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom form either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far from each other as possible.
VSEPR theory predicts the arrangement of electron pairs around each central atom and, usually, the correct arrangement of atoms in a molecule. We should understand, however, that the theory only considers electron-pair repulsions. Other interactions, such as nuclear-nuclear repulsions and nuclear-electron attractions, are also involved in the final arrangement that atoms adopt in a particular molecular structure.
As a simple example of VSEPR theory, let us predict the structure of a gaseous BeF2 molecule. The Lewis structure of BeF2 (Figure 7.15) shows only two electron pairs around the central beryllium atom. With two bonds and no lone pairs of electrons on the central atom, the bonds are as far apart as possible, and the electrostatic repulsion between these regions of high electron density is reduced to a minimum when they are on opposite sides of the central atom.
The bond angle is 180° (Figure 7.15).
Chapter 7 Chemical Bonding and Molecular Geometry 377 Figure 7.15 The BeF2 molecule adopts a linear structure in which the two bonds are as far apart as possible, on opposite sides of the Be atom.
Figure 7.16 illustrates this and other electron-pair geometries that minimize the repulsions among regions of high electron density (bonds and/or lone pairs). Two regions of electron density around a central atom in a molecule form a linear geometry; three regions form a trigonal planar geometry; four regions form a tetrahedral geometry; five regions form a trigonal bipyramidal geometry; and six regions form an octahedral geometry.
Figure 7.16 The basic electron-pair geometries predicted by VSEPR theory maximize the space around any region of electron density (bonds or lone pairs).
378 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at Electron-pair Geometry versus Molecular Structure It is important to note that electron-pair geometry around a central atom is not the same thing as its molecular structure. The electron-pair geometries shown in Figure 7.16 describe all regions where electrons are located, bonds as well as lone pairs. Molecular structure describes the location of the atoms, not the electrons.
We differentiate between these two situations by naming the geometry that includes all electron pairs the electron-pair geometry. The structure that includes only the placement of the atoms in the molecule is called the molecular structure. The electron-pair geometries will be the same as the molecular structures when there are no lone electron pairs around the central atom, but they will be different when there are lone pairs present on the central atom.
For example, the methane molecule, CH4, which is the major component of natural gas, has four bonding pairs of electrons around the central carbon atom; the electron-pair geometry is tetrahedral, as is the molecular structure (Figure 7.17). On the other hand, the ammonia molecule, NH3, also has four electron pairs associated with the nitrogen atom, and thus has a tetrahedral electron-pair geometry. One of these regions, however, is a lone pair, which is not included in the molecular structure, and this lone pair influences the shape of the molecule (Figure 7.18).
Figure 7.17 The molecular structure of the methane molecule, CH4, is shown with a tetrahedral arrangement of the hydrogen atoms. VSEPR structures like this one are often drawn using the wedge and dash notation, in which solid lines represent bonds in the plane of the page, solid wedges represent bonds coming up out of the plane, and dashed lines represent bonds going down into the plane.
Figure 7.18 (a) The electron-pair geometry for the ammonia molecule is tetrahedral with one lone pair and three single bonds. (b) The trigonal pyramidal molecular structure is determined from the electron-pair geometry. (c) The actual bond angles deviate slightly from the idealized angles because the lone pair takes up a larger region of space than do the single bonds, causing the HNH angle to be slightly smaller than 109.5°.
As seen in Figure 7.18, small distortions from the ideal angles in Figure 7.16 can result from differences in repulsion between various regions of electron density. VSEPR theory predicts these distortions by establishing an order of repulsions and an order of the amount of space occupied by different kinds of electron pairs. The order of electron-pair repulsions from greatest to least repulsion is: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair This order of repulsions determines the amount of space occupied by different regions of electrons. A lone pair of electrons occupies a larger region of space than the electrons in a triple bond; in turn, electrons in a triple bond occupy more space than those in a double bond, and so on. The order of sizes from largest to smallest is: lone pair > triple bond > double bond > single bond Chapter 7 Chemical Bonding and Molecular Geometry 379 Consider formaldehyde, H2CO, which is used as a preservative for biological and anatomical specimens (Figure 7.14). This molecule has regions of high electron density that consist of two single bonds and one double bond. The basic geometry is trigonal planar with 120° bond angles, but we see that the double bond causes slightly larger angles (121°), and the angle between the single bonds is slightly smaller (118°).
In the ammonia molecule, the three hydrogen atoms attached to the central nitrogen are not arranged in a flat, trigonal planar molecular structure, but rather in a three-dimensional trigonal pyramid (Figure 7.18) with the nitrogen atom at the apex and the three hydrogen atoms forming the base. The ideal bond angles in a trigonal pyramid are based on the tetrahedral electron pair geometry. Again, there are slight deviations from the ideal because lone pairs occupy larger regions of space than do bonding electrons. The H–N–H bond angles in NH3 are slightly smaller than the 109.5° angle in a regular tetrahedron (Figure 7.16) because the lone pair-bonding pair repulsion is greater than the bonding pair-bonding pair repulsion (Figure 7.18). Figure 7.19 illustrates the ideal molecular structures, which are predicted based on the electron-pair geometries for various combinations of lone pairs and bonding pairs.
Figure 7.19 The molecular structures are identical to the electron-pair geometries when there are no lone pairs present (first column). For a particular number of electron pairs (row), the molecular structures for one or more lone pairs are determined based on modifications of the corresponding electron-pair geometry.
According to VSEPR theory, the terminal atom locations (Xs in Figure 7.19) are equivalent within the linear, trigonal planar, and tetrahedral electron-pair geometries (the first three rows of the table). It does not matter which 380 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at X is replaced with a lone pair because the molecules can be rotated to convert positions. For trigonal bipyramidal electron-pair geometries, however, there are two distinct X positions, as shown in Figure 7.20: an axial position (if we hold a model of a trigonal bipyramid by the two axial positions, we have an axis around which we can rotate the model) and an equatorial position (three positions form an equator around the middle of the molecule). As shown in Figure 7.19, the axial position is surrounded by bond angles of 90°, whereas the equatorial position has more space available because of the 120° bond angles. In a trigonal bipyramidal electron-pair geometry, lone pairs always occupy equatorial positions because these more spacious positions can more easily accommodate the larger lone pairs.
Theoretically, we can come up with three possible arrangements for the three bonds and two lone pairs for the ClF3 molecule (Figure 7.20). The stable structure is the one that puts the lone pairs in equatorial locations, giving a T-shaped molecular structure.
Figure 7.20 (a) In a trigonal bipyramid, the two axial positions are located directly across from one another, whereas the three equatorial positions are located in a triangular arrangement. (b–d) The two lone pairs (red lines) in ClF3 have several possible arrangements, but the T-shaped molecular structure (b) is the one actually observed, consistent with the larger lone pairs both occupying equatorial positions.
When a central atom has two lone electron pairs and four bonding regions, we have an octahedral electron-pair geometry. The two lone pairs are on opposite sides of the octahedron (180° apart), giving a square planar molecular structure that minimizes lone pair-lone pair repulsions (Figure 7.19).
Predicting Electron Pair Geometry and Molecular Structure The following procedure uses VSEPR theory to determine the electron pair geometries and the molecular structures: 1.
Write the Lewis structure of the molecule or polyatomic ion.
2.
Count the number of regions of electron density (lone pairs and bonds) around the central atom. A single, double, or triple bond counts as one region of electron density.
3.
Identify the electron-pair geometry based on the number of regions of electron density: linear, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral (Figure 7.19, first column).
4.
Use the number of lone pairs to determine the molecular structure (Figure 7.19). If more than one arrangement of lone pairs and chemical bonds is possible, choose the one that will minimize repulsions, remembering that lone pairs occupy more space than multiple bonds, which occupy more space than single bonds. In trigonal bipyramidal arrangements, repulsion is minimized when every lone pair is in an equatorial position. In an octahedral arrangement with two lone pairs, repulsion is minimized when the lone pairs are on opposite sides of the central atom.
The following examples illustrate the use of VSEPR theory to predict the molecular structure of molecules or ions that have no lone pairs of electrons. In this case, the molecular structure is identical to the electron pair geometry.
Example 7.11 Chapter 7 Chemical Bonding and Molecular Geometry 381 Predicting Electron-pair Geometry and Molecular Structure: CO2 and BCl3 Predict the electron-pair geometry and molecular structure for each of the following: (a) carbon dioxide, CO2, a molecule produced by the combustion of fossil fuels (b) boron trichloride, BCl3, an important industrial chemical Solution (a) We write the Lewis structure of CO2 as: This shows us two regions of high electron density around the carbon atom—each double bond counts as one region, and there are no lone pairs on the carbon atom. Using VSEPR theory, we predict that the two regions of electron density arrange themselves on opposite sides of the central atom with a bond angle of 180°. The electron-pair geometry and molecular structure are identical, and CO2 molecules are linear.
(b) We write the Lewis structure of BCl3 as: Thus we see that BCl3 contains three bonds, and there are no lone pairs of electrons on boron. The arrangement of three regions of high electron density gives a trigonal planar electron-pair geometry. The B–Cl bonds lie in a plane with 120° angles between them. BCl3 also has a trigonal planar molecular structure (Figure 7.21).
Figure 7.21 The electron-pair geometry and molecular structure of BCl3 are both trigonal planar. Note that the VSEPR geometry indicates the correct bond angles (120°), unlike the Lewis structure shown above.
Check Your Learning Carbonate, CO3 2−, is a common polyatomic ion found in various materials from eggshells to antacids.
What are the electron-pair geometry and molecular structure of this polyatomic ion?
Answer: The electron-pair geometry is trigonal planar and the molecular structure is trigonal planar. Due to resonance, all three C–O bonds are identical. Whether they are single, double, or an average of the two, each bond counts as one region of electron density.
Example 7.12 Predicting Electron-pair Geometry and Molecular Structure: Ammonium 382 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at Two of the top 50 chemicals produced in the United States, ammonium nitrate and ammonium sulfate, both used as fertilizers, contain the ammonium ion. Predict the electron-pair geometry and molecular structure of the NH4 + cation.
Solution We write the Lewis structure of NH4 + as: We can see that NH4 + contains four bonds from the nitrogen atom to hydrogen atoms and no lone pairs.
We expect the four regions of high electron density to arrange themselves so that they point to the corners of a tetrahedron with the central nitrogen atom in the middle (Figure 7.19). Therefore, the electron pair geometry of NH4 + is tetrahedral, and the molecular structure is also tetrahedral (Figure 7.22).
Figure 7.22 The ammonium ion displays a tetrahedral electron-pair geometry as well as a tetrahedral molecular structure.
Check Your Learning Identify a molecule with trigonal bipyramidal molecular structure.
Answer: Any molecule with five electron pairs around the central atoms including no lone pairs will be trigonal bipyramidal. PF5 is a common example.
The next several examples illustrate the effect of lone pairs of electrons on molecular structure.
Example 7.13 Predicting Electron-pair Geometry and Molecular Structure: Lone Pairs on the Central Atom Predict the electron-pair geometry and molecular structure of a water molecule.
Solution The Lewis structure of H2O indicates that there are four regions of high electron density around the oxygen atom: two lone pairs and two chemical bonds: Chapter 7 Chemical Bonding and Molecular Geometry 383 We predict that these four regions are arranged in a tetrahedral fashion (Figure 7.23), as indicated in Figure 7.19. Thus, the electron-pair geometry is tetrahedral and the molecular structure is bent with an angle slightly less than 109.5°. In fact, the bond angle is 104.5°.
Figure 7.23 (a) H2O has four regions of electron density around the central atom, so it has a tetrahedral electron-pair geometry. (b) Two of the electron regions are lone pairs, so the molecular structure is bent.
Check Your Learning The hydronium ion, H3O+, forms when acids are dissolved in water. Predict the electron-pair geometry and molecular structure of this cation.
Answer: electron pair geometry: tetrahedral; molecular structure: trigonal pyramidal Example 7.14 Predicting Electron-pair Geometry and Molecular Structure: SF4 Sulfur tetrafluoride, SF4, is extremely valuable for the preparation of fluorine-containing compounds used as herbicides (i.e., SF4 is used as a fluorinating agent). Predict the electron-pair geometry and molecular structure of a SF4 molecule.
Solution The Lewis structure of SF4 indicates five regions of electron density around the sulfur atom: one lone pair and four bonding pairs: We expect these five regions to adopt a trigonal bipyramidal electron-pair geometry. To minimize lone pair repulsions, the lone pair occupies one of the equatorial positions. The molecular structure (Figure 7.24) is that of a seesaw (Figure 7.19).
384 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at Figure 7.24 (a) SF4 has a trigonal bipyramidal arrangement of the five regions of electron density. (b) One of the regions is a lone pair, which results in a seesaw-shaped molecular structure.
Check Your Learning Predict the electron pair geometry and molecular structure for molecules of XeF2.
Answer: The electron-pair geometry is trigonal bipyramidal. The molecular structure is linear.
Example 7.15 Predicting Electron-pair Geometry and Molecular Structure: XeF4 Of all the noble gases, xenon is the most reactive, frequently reacting with elements such as oxygen and fluorine. Predict the electron-pair geometry and molecular structure of the XeF4 molecule.
Solution The Lewis structure of XeF4 indicates six regions of high electron density around the xenon atom: two lone pairs and four bonds: These six regions adopt an octahedral arrangement (Figure 7.19), which is the electron-pair geometry. To minimize repulsions, the lone pairs should be on opposite sides of the central atom (Figure 7.25). The five atoms are all in the same plane and have a square planar molecular structure.
Chapter 7 Chemical Bonding and Molecular Geometry 385 Figure 7.25 (a) XeF4 adopts an octahedral arrangement with two lone pairs (red lines) and four bonds in the electron-pair geometry. (b) The molecular structure is square planar with the lone pairs directly across from one another.
Check Your Learning In a certain molecule, the central atom has three lone pairs and two bonds. What will the electron pair geometry and molecular structure be?
Answer: electron pair geometry: trigonal bipyramidal; molecular structure: linear Molecular Structure for Multicenter Molecules When a molecule or polyatomic ion has only one central atom, the molecular structure completely describes the shape of the molecule. Larger molecules do not have a single central atom, but are connected by a chain of interior atoms that each possess a “local” geometry. The way these local structures are oriented with respect to each other also influences the molecular shape, but such considerations are largely beyond the scope of this introductory discussion.
For our purposes, we will only focus on determining the local structures.
Example 7.16 Predicting Structure in Multicenter Molecules The Lewis structure for the simplest amino acid, glycine, H2NCH2CO2H, is shown here. Predict the local geometry for the nitrogen atom, the two carbon atoms, and the oxygen atom with a hydrogen atom attached: Solution Consider each central atom independently. The electron-pair geometries: • nitrogen––four regions of electron density; tetrahedral • carbon (CH2)––four regions of electron density; tetrahedral 386 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at • carbon (CO2)—three regions of electron density; trigonal planar • oxygen (OH)—four regions of electron density; tetrahedral The local structures: • nitrogen––three bonds, one lone pair; trigonal pyramidal • carbon (CH2)—four bonds, no lone pairs; tetrahedral • carbon (CO2)—three bonds (double bond counts as one bond), no lone pairs; trigonal planar • oxygen (OH)—two bonds, two lone pairs; bent (109°) Check Your Learning Another amino acid is alanine, which has the Lewis structure shown here. Predict the electron-pair geometry and local structure of the nitrogen atom, the three carbon atoms, and the oxygen atom with hydrogen attached: Answer: electron-pair geometries: nitrogen––tetrahedral; carbon (CH)—tetrahedral; carbon (CH3)—tetrahedral; carbon (CO2)—trigonal planar; oxygen (OH)—tetrahedral; local structures: nitrogen—trigonal pyramidal; carbon (CH)—tetrahedral; carbon (CH3)—tetrahedral; carbon (CO2)—trigonal planar; oxygen (OH)—bent (109°) The molecular shape simulator ( lets you build various molecules and practice naming their electron-pair geometries and molecular structures.
Example 7.17 Molecular Simulation Using molecular shape simulator ( allows us to control whether bond angles and/or lone pairs are displayed by checking or unchecking the boxes under “Options” on the right. We can also use the “Name” checkboxes at bottom-left to display or hide the electron pair geometry (called “electron geometry” in the simulator) and/or molecular structure (called “molecular shape” in the simulator).
Build the molecule HCN in the simulator based on the following Lewis structure: H–C ≡N Click on each bond type or lone pair at right to add that group to the central atom. Once you have the complete molecule, rotate it to examine the predicted molecular structure. What molecular structure is this?
Solution Link to Learning Chapter 7 Chemical Bonding and Molecular Geometry 387 The molecular structure is linear.
Check Your Learning Build a more complex molecule in the simulator. Identify the electron-group geometry, molecular structure, and bond angles. Then try to find a chemical formula that would match the structure you have drawn.
Answer: Answers will vary. For example, an atom with four single bonds, a double bond, and a lone pair has an octahedral electron-group geometry and a square pyramidal molecular structure. XeOF4 is a molecule that adopts this structure.
Molecular Polarity and Dipole Moment As discussed previously, polar covalent bonds connect two atoms with differing electronegativities, leaving one atom with a partial positive charge (δ+) and the other atom with a partial negative charge (δ–), as the electrons are pulled toward the more electronegative atom. This separation of charge gives rise to a bond dipole moment. The magnitude of a bond dipole moment is represented by the Greek letter mu (µ) and is given by the formula shown here, where Q is the magnitude of the partial charges (determined by the electronegativity difference) and r is the distance between the charges: µ = Qr This bond moment can be represented as a vector, a quantity having both direction and magnitude (Figure 7.26).
Dipole vectors are shown as arrows pointing along the bond from the less electronegative atom toward the more electronegative atom. A small plus sign is drawn on the less electronegative end to indicate the partially positive end of the bond. The length of the arrow is proportional to the magnitude of the electronegativity difference between the two atoms.
Figure 7.26 (a) There is a small difference in electronegativity between C and H, represented as a short vector. (b) The electronegativity difference between B and F is much larger, so the vector representing the bond moment is much longer.
A whole molecule may also have a separation of charge, depending on its molecular structure and the polarity of each of its bonds. If such a charge separation exists, the molecule is said to be a polar molecule (or dipole); otherwise the molecule is said to be nonpolar. The dipole moment measures the extent of net charge separation in the molecule as a whole. We determine the dipole moment by adding the bond moments in three-dimensional space, taking into account the molecular structure.
For diatomic molecules, there is only one bond, so its bond dipole moment determines the molecular polarity.
Homonuclear diatomic molecules such as Br2 and N2 have no difference in electronegativity, so their dipole moment is zero. For heteronuclear molecules such as CO, there is a small dipole moment. For HF, there is a larger dipole moment because there is a larger difference in electronegativity.
When a molecule contains more than one bond, the geometry must be taken into account. If the bonds in a molecule are arranged such that their bond moments cancel (vector sum equals zero), then the molecule is nonpolar. This is the situation in CO2 (Figure 7.27). Each of the bonds is polar, but the molecule as a whole is nonpolar. From the Lewis structure, and using VSEPR theory, we determine that the CO2 molecule is linear with polar C=O bonds on opposite sides of the carbon atom. The bond moments cancel because they are pointed in opposite directions. In the case of the 388 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at water molecule (Figure 7.27), the Lewis structure again shows that there are two bonds to a central atom, and the electronegativity difference again shows that each of these bonds has a nonzero bond moment. In this case, however, the molecular structure is bent because of the lone pairs on O, and the two bond moments do not cancel. Therefore, water does have a net dipole moment and is a polar molecule (dipole).
Figure 7.27 The overall dipole moment of a molecule depends on the individual bond dipole moments and how they are arranged. (a) Each CO bond has a bond dipole moment, but they point in opposite directions so that the net CO2 molecule is nonpolar. (b) In contrast, water is polar because the OH bond moments do not cancel out.
The OCS molecule has a structure similar to CO2, but a sulfur atom has replaced one of the oxygen atoms. To determine if this molecule is polar, we draw the molecular structure. VSEPR theory predicts a linear molecule: Although the C–O bond is polar, C and S have the same electronegativity values as shown in Figure 7.6, so there is no C–S dipole. Thus, the two bonds do not have of the same bond dipole moment, and the bond moments do not cancel. Because oxygen is more electronegative than sulfur, the oxygen end of the molecule is the negative end.
Chloromethane, CH3Cl, is another example of a polar molecule. Although the polar C–Cl and C–H bonds are arranged in a tetrahedral geometry, the C–Cl bonds have a larger bond moment than the C–H bond, and the bond moments do not completely cancel each other. All of the dipoles have a downward component in the orientation shown, since carbon is more electronegative than hydrogen and less electronegative than chlorine: When we examine the highly symmetrical molecules BF3 (trigonal planar), CH4 (tetrahedral), PF5 (trigonal bipyramidal), and SF6 (octahedral), in which all the polar bonds are identical, the molecules are nonpolar. The bonds in these molecules are arranged such that their dipoles cancel. However, just because a molecule contains identical bonds does not mean that the dipoles will always cancel. Many molecules that have identical bonds and lone pairs on the central atoms have bond dipoles that do not cancel. Examples include H2S and NH3. A hydrogen atom is at the positive end and a nitrogen or sulfur atom is at the negative end of the polar bonds in these molecules: Chapter 7 Chemical Bonding and Molecular Geometry 389 To summarize, to be polar, a molecule must: 1.
Contain at least one polar covalent bond.
2.
Have a molecular structure such that the sum of the vectors of each bond dipole moment does not cancel.
Properties of Polar Molecules Polar molecules tend to align when placed in an electric field with the positive end of the molecule oriented toward the negative plate and the negative end toward the positive plate (Figure 7.28). We can use an electrically charged object to attract polar molecules, but nonpolar molecules are not attracted. Also, polar solvents are better at dissolving polar substances, and nonpolar solvents are better at dissolving nonpolar substances.
Figure 7.28 (a) Molecules are always randomly distributed in the liquid state in the absence of an electric field. (b) When an electric field is applied, polar molecules like HF will align to the dipoles with the field direction.
The molecule polarity simulation ( 16MolecPolarity) provides many ways to explore dipole moments of bonds and molecules.
Link to Learning 390 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at Example 7.18 Polarity Simulations Open the molecule polarity simulation ( and select the “Three Atoms” tab at the top. This should display a molecule ABC with three electronegativity adjustors. You can display or hide the bond moments, molecular dipoles, and partial charges at the right.
Turning on the Electric Field will show whether the molecule moves when exposed to a field, similar to Figure 7.28.
Use the electronegativity controls to determine how the molecular dipole will look for the starting bent molecule if: (a) A and C are very electronegative and B is in the middle of the range.
(b) A is very electronegative, and B and C are not.
Solution (a) Molecular dipole moment points immediately between A and C.
(b) Molecular dipole moment points along the A–B bond, toward A.
Check Your Learning Determine the partial charges that will give the largest possible bond dipoles.
Answer: The largest bond moments will occur with the largest partial charges. The two solutions above represent how unevenly the electrons are shared in the bond. The bond moments will be maximized when the electronegativity difference is greatest. The controls for A and C should be set to one extreme, and B should be set to the opposite extreme. Although the magnitude of the bond moment will not change based on whether B is the most electronegative or the least, the direction of the bond moment will.
Chapter 7 Chemical Bonding and Molecular Geometry 391 axial position bond angle bond dipole moment bond distance bond energy bond length Born-Haber cycle covalent bond dipole moment double bond electron-pair geometry electronegativity equatorial position formal charge free radical hypervalent molecule inert pair effect ionic bond lattice energy (ΔHlattice) Lewis structure Lewis symbol linear lone pair molecular structure Key Terms location in a trigonal bipyramidal geometry in which there is another atom at a 180° angle and the equatorial positions are at a 90° angle angle between any two covalent bonds that share a common atom separation of charge in a bond that depends on the difference in electronegativity and the bond distance represented by partial charges or a vector (also, bond length) distance between the nuclei of two bonded atoms (also, bond dissociation energy) energy required to break a covalent bond in a gaseous substance distance between the nuclei of two bonded atoms at which the lowest potential energy is achieved thermochemical cycle relating the various energetic steps involved in the formation of an ionic solid from the relevant elements bond formed when electrons are shared between atoms property of a molecule that describes the separation of charge determined by the sum of the individual bond moments based on the molecular structure covalent bond in which two pairs of electrons are shared between two atoms arrangement around a central atom of all regions of electron density (bonds, lone pairs, or unpaired electrons) tendency of an atom to attract electrons in a bond to itself one of the three positions in a trigonal bipyramidal geometry with 120° angles between them; the axial positions are located at a 90° angle charge that would result on an atom by taking the number of valence electrons on the neutral atom and subtracting the nonbonding electrons and the number of bonds (one-half of the bonding electrons) molecule that contains an odd number of electrons molecule containing at least one main group element that has more than eight electrons in its valence shell tendency of heavy atoms to form ions in which their valence s electrons are not lost strong electrostatic force of attraction between cations and anions in an ionic compound energy required to separate one mole of an ionic solid into its component gaseous ions diagram showing lone pairs and bonding pairs of electrons in a molecule or an ion symbol for an element or monatomic ion that uses a dot to represent each valence electron in the element or ion shape in which two outside groups are placed on opposite sides of a central atom two (a pair of) valence electrons that are not used to form a covalent bond arrangement of atoms in a molecule or ion 392 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at molecular structure octahedral octet rule polar covalent bond polar molecule pure covalent bond resonance resonance forms resonance hybrid single bond tetrahedral trigonal bipyramidal trigonal planar triple bond valence shell electron-pair repulsion theory (VSEPR) vector structure that includes only the placement of the atoms in the molecule shape in which six outside groups are placed around a central atom such that a three-dimensional shape is generated with four groups forming a square and the other two forming the apex of two pyramids, one above and one below the square plane guideline that states main group atoms will form structures in which eight valence electrons interact with each nucleus, counting bonding electrons as interacting with both atoms connected by the bond covalent bond between atoms of different electronegativities; a covalent bond with a positive end and a negative end (also, dipole) molecule with an overall dipole moment (also, nonpolar covalent bond) covalent bond between atoms of identical electronegativities situation in which one Lewis structure is insufficient to describe the bonding in a molecule and the average of multiple structures is observed two or more Lewis structures that have the same arrangement of atoms but different arrangements of electrons average of the resonance forms shown by the individual Lewis structures bond in which a single pair of electrons is shared between two atoms shape in which four outside groups are placed around a central atom such that a three-dimensional shape is generated with four corners and 109.5° angles between each pair and the central atom shape in which five outside groups are placed around a central atom such that three form a flat triangle with 120° angles between each pair and the central atom, and the other two form the apex of two pyramids, one above and one below the triangular plane shape in which three outside groups are placed in a flat triangle around a central atom with 120° angles between each pair and the central atom bond in which three pairs of electrons are shared between two atoms theory used to predict the bond angles in a molecule based on positioning regions of high electron density as far apart as possible to minimize electrostatic repulsion quantity having magnitude and direction Key Equations • formal charge = # valence shell electrons (free atom) −# one pair electrons −1 2 # bonding electrons • Bond energy for a diatomic molecule: XY(g) ⟶X(g) + Y(g) DX–Y = ΔH° • Enthalpy change: ΔH = ƩDbonds broken – ƩDbonds formed • Lattice energy for a solid MX: MX(s) ⟶Mn+(g) + Xn−(g) ΔHlattice • Lattice energy for an ionic crystal: ΔHlattice = C(Z+)(Z−) Ro Chapter 7 Chemical Bonding and Molecular Geometry 393 Summary 7.1 Ionic Bonding Atoms gain or lose electrons to form ions with particularly stable electron configurations. The charges of cations formed by the representative metals may be determined readily because, with few exceptions, the electronic structures of these ions have either a noble gas configuration or a completely filled electron shell. The charges of anions formed by the nonmetals may also be readily determined because these ions form when nonmetal atoms gain enough electrons to fill their valence shells.
7.2 Covalent Bonding Covalent bonds form when electrons are shared between atoms and are attracted by the nuclei of both atoms. In pure covalent bonds, the electrons are shared equally. In polar covalent bonds, the electrons are shared unequally, as one atom exerts a stronger force of attraction on the electrons than the other. The ability of an atom to attract a pair of electrons in a chemical bond is called its electronegativity. The difference in electronegativity between two atoms determines how polar a bond will be. In a diatomic molecule with two identical atoms, there is no difference in electronegativity, so the bond is nonpolar or pure covalent. When the electronegativity difference is very large, as is the case between metals and nonmetals, the bonding is characterized as ionic.
7.3 Lewis Symbols and Structures Valence electronic structures can be visualized by drawing Lewis symbols (for atoms and monatomic ions) and Lewis structures (for molecules and polyatomic ions). Lone pairs, unpaired electrons, and single, double, or triple bonds are used to indicate where the valence electrons are located around each atom in a Lewis structure. Most structures—especially those containing second row elements—obey the octet rule, in which every atom (except H) is surrounded by eight electrons. Exceptions to the octet rule occur for odd-electron molecules (free radicals), electron-deficient molecules, and hypervalent molecules.
7.4 Formal Charges and Resonance In a Lewis structure, formal charges can be assigned to each atom by treating each bond as if one-half of the electrons are assigned to each atom. These hypothetical formal charges are a guide to determining the most appropriate Lewis structure. A structure in which the formal charges are as close to zero as possible is preferred. Resonance occurs in cases where two or more Lewis structures with identical arrangements of atoms but different distributions of electrons can be written. The actual distribution of electrons (the resonance hybrid) is an average of the distribution indicated by the individual Lewis structures (the resonance forms).
7.5 Strengths of Ionic and Covalent Bonds The strength of a covalent bond is measured by its bond dissociation energy, that is, the amount of energy required to break that particular bond in a mole of molecules. Multiple bonds are stronger than single bonds between the same atoms. The enthalpy of a reaction can be estimated based on the energy input required to break bonds and the energy released when new bonds are formed. For ionic bonds, the lattice energy is the energy required to separate one mole of a compound into its gas phase ions. Lattice energy increases for ions with higher charges and shorter distances between ions. Lattice energies are often calculated using the Born-Haber cycle, a thermochemical cycle including all of the energetic steps involved in converting elements into an ionic compound.
7.6 Molecular Structure and Polarity VSEPR theory predicts the three-dimensional arrangement of atoms in a molecule. It states that valence electrons will assume an electron-pair geometry that minimizes repulsions between areas of high electron density (bonds and/ or lone pairs). Molecular structure, which refers only to the placement of atoms in a molecule and not the electrons, is equivalent to electron-pair geometry only when there are no lone electron pairs around the central atom. A dipole moment measures a separation of charge. For one bond, the bond dipole moment is determined by the difference in electronegativity between the two atoms. For a molecule, the overall dipole moment is determined by both the 394 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at individual bond moments and how these dipoles are arranged in the molecular structure. Polar molecules (those with an appreciable dipole moment) interact with electric fields, whereas nonpolar molecules do not.
Exercises 7.1 Ionic Bonding 1. Does a cation gain protons to form a positive charge or does it lose electrons?
2. Iron(III) sulfate [Fe2(SO4)3] is composed of Fe3+ and SO4 2−ions. Explain why a sample of iron(III) sulfate is uncharged.
3. Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: P, I, Mg, Cl, In, Cs, O, Pb, Co?
4. Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: Br, Ca, Na, N, F, Al, Sn, S, Cd?
5. Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds: (a) P (b) Mg (c) Al (d) O (e) Cl (f) Cs 6. Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds: (a) I (b) Sr (c) K (d) N (e) S (f) In 7. Write the electron configuration for each of the following ions: (a) As3– (b) I– (c) Be2+ (d) Cd2+ (e) O2– (f) Ga3+ (g) Li+ (h) N3– (i) Sn2+ Chapter 7 Chemical Bonding and Molecular Geometry 395 (j) Co2+ (k) Fe2+ (l) As3+ 8. Write the electron configuration for the monatomic ions formed from the following elements (which form the greatest concentration of monatomic ions in seawater): (a) Cl (b) Na (c) Mg (d) Ca (e) K (f) Br (g) Sr (h) F 9. Write out the full electron configuration for each of the following atoms and for the monatomic ion found in binary ionic compounds containing the element: (a) Al (b) Br (c) Sr (d) Li (e) As (f) S 10. From the labels of several commercial products, prepare a list of six ionic compounds in the products. For each compound, write the formula. (You may need to look up some formulas in a suitable reference.) 7.2 Covalent Bonding 11. Why is it incorrect to speak of a molecule of solid NaCl?
12. What information can you use to predict whether a bond between two atoms is covalent or ionic?
13. Predict which of the following compounds are ionic and which are covalent, based on the location of their constituent atoms in the periodic table: (a) Cl2CO (b) MnO (c) NCl3 (d) CoBr2 (e) K2S (f) CO (g) CaF2 (h) HI (i) CaO (j) IBr 396 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at (k) CO2 14. Explain the difference between a nonpolar covalent bond, a polar covalent bond, and an ionic bond.
15. From its position in the periodic table, determine which atom in each pair is more electronegative: (a) Br or Cl (b) N or O (c) S or O (d) P or S (e) Si or N (f) Ba or P (g) N or K 16. From its position in the periodic table, determine which atom in each pair is more electronegative: (a) N or P (b) N or Ge (c) S or F (d) Cl or S (e) H or C (f) Se or P (g) C or Si 17. From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity: (a) C, F, H, N, O (b) Br, Cl, F, H, I (c) F, H, O, P, S (d) Al, H, Na, O, P (e) Ba, H, N, O, As 18. From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity: (a) As, H, N, P, Sb (b) Cl, H, P, S, Si (c) Br, Cl, Ge, H, Sr (d) Ca, H, K, N, Si (e) Cl, Cs, Ge, H, Sr 19. Which atoms can bond to sulfur so as to produce a positive partial charge on the sulfur atom?
20. Which is the most polar bond?
(a) C–C (b) C–H (c) N–H (d) O–H Chapter 7 Chemical Bonding and Molecular Geometry 397 (e) Se–H 21. Identify the more polar bond in each of the following pairs of bonds: (a) HF or HCl (b) NO or CO (c) SH or OH (d) PCl or SCl (e) CH or NH (f) SO or PO (g) CN or NN 22. Which of the following molecules or ions contain polar bonds?
(a) O3 (b) S8 (c) O2 2− (d) NO3 − (e) CO2 (f) H2S (g) BH4 − 7.3 Lewis Symbols and Structures 23. Write the Lewis symbols for each of the following ions: (a) As3– (b) I– (c) Be2+ (d) O2– (e) Ga3+ (f) Li+ (g) N3– 24. Many monatomic ions are found in seawater, including the ions formed from the following list of elements.
Write the Lewis symbols for the monatomic ions formed from the following elements: (a) Cl (b) Na (c) Mg (d) Ca (e) K (f) Br (g) Sr 398 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at (h) F 25. Write the Lewis symbols of the ions in each of the following ionic compounds and the Lewis symbols of the atom from which they are formed: (a) MgS (b) Al2O3 (c) GaCl3 (d) K2O (e) Li3N (f) KF 26. In the Lewis structures listed here, M and X represent various elements in the third period of the periodic table.
Write the formula of each compound using the chemical symbols of each element: (a) (b) (c) (d) 27. Write the Lewis structure for the diatomic molecule P2, an unstable form of phosphorus found in high-temperature phosphorus vapor.
28. Write Lewis structures for the following: (a) H2 (b) HBr (c) PCl3 (d) SF2 (e) H2CCH2 (f) HNNH (g) H2CNH (h) NO– Chapter 7 Chemical Bonding and Molecular Geometry 399 (i) N2 (j) CO (k) CN– 29. Write Lewis structures for the following: (a) O2 (b) H2CO (c) AsF3 (d) ClNO (e) SiCl4 (f) H3O+ (g) NH4 + (h) BF4 − (i) HCCH (j) ClCN (k) C2 2+ 30. Write Lewis structures for the following: (a) ClF3 (b) PCl5 (c) BF3 (d) PF6 − 31. Write Lewis structures for the following: (a) SeF6 (b) XeF4 (c) SeCl3 + (d) Cl2BBCl2 (contains a B–B bond) 32. Write Lewis structures for: (a) PO4 3− (b) IC4 − (c) SO3 2− (d) HONO 33. Correct the following statement: “The bonds in solid PbCl2 are ionic; the bond in a HCl molecule is covalent.
Thus, all of the valence electrons in PbCl2 are located on the Cl– ions, and all of the valence electrons in a HCl molecule are shared between the H and Cl atoms.” 34. Write Lewis structures for the following molecules or ions: 400 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at (a) SbH3 (b) XeF2 (c) Se8 (a cyclic molecule with a ring of eight Se atoms) 35. Methanol, H3COH, is used as the fuel in some race cars. Ethanol, C2H5OH, is used extensively as motor fuel in Brazil. Both methanol and ethanol produce CO2 and H2O when they burn. Write the chemical equations for these combustion reactions using Lewis structures instead of chemical formulas.
36. Many planets in our solar system contain organic chemicals including methane (CH4) and traces of ethylene (C2H4), ethane (C2H6), propyne (H3CCCH), and diacetylene (HCCCCH). Write the Lewis structures for each of these molecules.
37. Carbon tetrachloride was formerly used in fire extinguishers for electrical fires. It is no longer used for this purpose because of the formation of the toxic gas phosgene, Cl2CO. Write the Lewis structures for carbon tetrachloride and phosgene.
38. Identify the atoms that correspond to each of the following electron configurations. Then, write the Lewis symbol for the common ion formed from each atom: (a) 1s22s22p5 (b) 1s22s22p63s2 (c) 1s22s22p63s23p64s23d10 (d) 1s22s22p63s23p64s23d104p4 (e) 1s22s22p63s23p64s23d104p1 39. The arrangement of atoms in several biologically important molecules is given here. Complete the Lewis structures of these molecules by adding multiple bonds and lone pairs. Do not add any more atoms.
(a) the amino acid serine: (b) urea: (c) pyruvic acid: (d) uracil: Chapter 7 Chemical Bonding and Molecular Geometry 401 (e) carbonic acid: 40. A compound with a molar mass of about 28 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound.
41. A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound.
42. Two arrangements of atoms are possible for a compound with a molar mass of about 45 g/mol that contains 52.2% C, 13.1% H, and 34.7% O by mass. Write the Lewis structures for the two molecules.
43. How are single, double, and triple bonds similar? How do they differ?
7.4 Formal Charges and Resonance 44. Write resonance forms that describe the distribution of electrons in each of these molecules or ions.
(a) selenium dioxide, OSeO (b) nitrate ion, NO3 − (c) nitric acid, HNO3 (N is bonded to an OH group and two O atoms) (d) benzene, C6H6: (e) the formate ion: 45. Write resonance forms that describe the distribution of electrons in each of these molecules or ions.
(a) sulfur dioxide, SO2 (b) carbonate ion, CO3 2− (c) hydrogen carbonate ion, HCO3 −(C is bonded to an OH group and two O atoms) (d) pyridine: 402 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at (e) the allyl ion: 46. Write the resonance forms of ozone, O3, the component of the upper atmosphere that protects the Earth from ultraviolet radiation.
47. Sodium nitrite, which has been used to preserve bacon and other meats, is an ionic compound. Write the resonance forms of the nitrite ion, NO2 – .
48. In terms of the bonds present, explain why acetic acid, CH3CO2H, contains two distinct types of carbon-oxygen bonds, whereas the acetate ion, formed by loss of a hydrogen ion from acetic acid, only contains one type of carbon-oxygen bond. The skeleton structures of these species are shown: 49. Write the Lewis structures for the following, and include resonance structures where appropriate. Indicate which has the strongest carbon-oxygen bond.
(a) CO2 (b) CO 50. Toothpastes containing sodium hydrogen carbonate (sodium bicarbonate) and hydrogen peroxide are widely used. Write Lewis structures for the hydrogen carbonate ion and hydrogen peroxide molecule, with resonance forms where appropriate.
51. Determine the formal charge of each element in the following: (a) HCl (b) CF4 (c) PCl3 (d) PF5 52. Determine the formal charge of each element in the following: (a) H3O+ (b) SO4 2− (c) NH3 (d) O2 2− (e) H2O2 53. Calculate the formal charge of chlorine in the molecules Cl2, BeCl2, and ClF5.
54. Calculate the formal charge of each element in the following compounds and ions: Chapter 7 Chemical Bonding and Molecular Geometry 403 (a) F2CO (b) NO– (c) BF4 − (d) SnCl3 − (e) H2CCH2 (f) ClF3 (g) SeF6 (h) PO4 3− 55. Draw all possible resonance structures for each of these compounds. Determine the formal charge on each atom in each of the resonance structures: (a) O3 (b) SO2 (c) NO2 − (d) NO3 − 56. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in nitrosyl chloride: ClNO or ClON?
57. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in hypochlorous acid: HOCl or OClH?
58. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in sulfur dioxide: OSO or SOO?
59. Draw the structure of hydroxylamine, H3NO, and assign formal charges; look up the structure. Is the actual structure consistent with the formal charges?
60. Iodine forms a series of fluorides (listed here). Write Lewis structures for each of the four compounds and determine the formal charge of the iodine atom in each molecule: (a) IF (b) IF3 (c) IF5 (d) IF7 61. Write the Lewis structure and chemical formula of the compound with a molar mass of about 70 g/mol that contains 19.7% nitrogen and 80.3% fluorine by mass, and determine the formal charge of the atoms in this compound.
62. Which of the following structures would we expect for nitrous acid? Determine the formal charges: 63. Sulfuric acid is the industrial chemical produced in greatest quantity worldwide. About 90 billion pounds are produced each year in the United States alone. Write the Lewis structure for sulfuric acid, H2SO4, which has two oxygen atoms and two OH groups bonded to the sulfur.
7.5 Strengths of Ionic and Covalent Bonds 64. Which bond in each of the following pairs of bonds is the strongest?
404 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at (a) C–C or C = C (b) C–N or C ≡N (c) C ≡O or C = O (d) H–F or H–Cl (e) C–H or O–H (f) C–N or C–O 65. Using the bond energies in Table 7.2, determine the approximate enthalpy change for each of the following reactions: (a) H2(g) + Br2(g) ⟶2HBr(g) (b) CH4(g) + I2(g) ⟶CH3 I(g) + HI(g) (c) C2 H4(g) + 3O2(g) ⟶2CO2(g) + 2H2 O(g) 66. Using the bond energies in Table 7.2, determine the approximate enthalpy change for each of the following reactions: (a) Cl2(g) + 3F2(g) ⟶2ClF3(g) (b) H2 C = CH2(g) + H2(g) ⟶H3 CCH3(g) (c) 2C2 H6(g) + 7O2(g) ⟶4CO2(g) + 6H2 O(g) 67. When a molecule can form two different structures, the structure with the stronger bonds is usually the more stable form. Use bond energies to predict the correct structure of the hydroxylamine molecule: 68. How does the bond energy of HCl(g) differ from the standard enthalpy of formation of HCl(g)?
69. Using the standard enthalpy of formation data in Appendix G, show how the standard enthalpy of formation of HCl(g) can be used to determine the bond energy.
70. Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbon-sulfur double bond in CS2.
71. Using the standard enthalpy of formation data in Appendix G, determine which bond is stronger: the S–F bond in SF4(g) or in SF6(g)?
72. Using the standard enthalpy of formation data in Appendix G, determine which bond is stronger: the P–Cl bond in PCl3(g) or in PCl5(g)?
73. Complete the following Lewis structure by adding bonds (not atoms), and then indicate the longest bond: 74. Use the bond energy to calculate an approximate value of ΔH for the following reaction. Which is the more stable form of FNO2?
Chapter 7 Chemical Bonding and Molecular Geometry 405 75. Use principles of atomic structure to answer each of the following: (a) The radius of the Ca atom is 197 pm; the radius of the Ca2+ ion is 99 pm. Account for the difference.
(b) The lattice energy of CaO(s) is –3460 kJ/mol; the lattice energy of K2O is –2240 kJ/mol. Account for the difference.
(c) Given these ionization values, explain the difference between Ca and K with regard to their first and second ionization energies.
Element First Ionization Energy (kJ/mol) Second Ionization Energy (kJ/mol) K 419 3050 Ca 590 1140 (d) The first ionization energy of Mg is 738 kJ/mol and that of Al is 578 kJ/mol. Account for this difference.
76. The lattice energy of LiF is 1023 kJ/mol, and the Li–F distance is 200.8 pm. NaF crystallizes in the same structure as LiF but with a Na–F distance of 231 pm. Which of the following values most closely approximates the lattice energy of NaF: 510, 890, 1023, 1175, or 4090 kJ/mol? Explain your choice.
77. For which of the following substances is the least energy required to convert one mole of the solid into separate ions?
(a) MgO (b) SrO (c) KF (d) CsF (e) MgF2 78. The reaction of a metal, M, with a halogen, X2, proceeds by an exothermic reaction as indicated by this equation: M(s) + X2(g) ⟶MX2(s). For each of the following, indicate which option will make the reaction more exothermic. Explain your answers.
(a) a large radius vs. a small radius for M+2 (b) a high ionization energy vs. a low ionization energy for M (c) an increasing bond energy for the halogen (d) a decreasing electron affinity for the halogen (e) an increasing size of the anion formed by the halogen 79. The lattice energy of LiF is 1023 kJ/mol, and the Li–F distance is 201 pm. MgO crystallizes in the same structure as LiF but with a Mg–O distance of 205 pm. Which of the following values most closely approximates the lattice energy of MgO: 256 kJ/mol, 512 kJ/mol, 1023 kJ/mol, 2046 kJ/mol, or 4008 kJ/mol? Explain your choice.
80. Which compound in each of the following pairs has the larger lattice energy? Note: Mg2+ and Li+ have similar radii; O2– and F– have similar radii. Explain your choices.
(a) MgO or MgSe 1. This question is taken from the Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.
406 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at (b) LiF or MgO (c) Li2O or LiCl (d) Li2Se or MgO 81. Which compound in each of the following pairs has the larger lattice energy? Note: Ba2+ and K+ have similar radii; S2– and Cl– have similar radii. Explain your choices.
(a) K2O or Na2O (b) K2S or BaS (c) KCl or BaS (d) BaS or BaCl2 82. Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?
(a) MgO (b) SrO (c) KF (d) CsF (e) MgF2 83. Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?
(a) K2S (b) K2O (c) CaS (d) Cs2S (e) CaO 84. The lattice energy of KF is 794 kJ/mol, and the interionic distance is 269 pm. The Na–F distance in NaF, which has the same structure as KF, is 231 pm. Which of the following values is the closest approximation of the lattice energy of NaF: 682 kJ/mol, 794 kJ/mol, 924 kJ/mol, 1588 kJ/mol, or 3175 kJ/mol?
Explain your answer.
7.6 Molecular Structure and Polarity 85. Explain why the HOH molecule is bent, whereas the HBeH molecule is linear.
86. What feature of a Lewis structure can be used to tell if a molecule’s (or ion’s) electron-pair geometry and molecular structure will be identical?
87. Explain the difference between electron-pair geometry and molecular structure.
88. Why is the H–N–H angle in NH3 smaller than the H–C–H bond angle in CH4? Why is the H–N–H angle in NH4 + identical to the H–C–H bond angle in CH4?
89. Explain how a molecule that contains polar bonds can be nonpolar.
90. As a general rule, MXn molecules (where M represents a central atom and X represents terminal atoms; n = 2 – 5) are polar if there is one or more lone pairs of electrons on M. NH3 (M = N, X = H, n = 3) is an example. There are two molecular structures with lone pairs that are exceptions to this rule. What are they?
91. Predict the electron pair geometry and the molecular structure of each of the following molecules or ions: Chapter 7 Chemical Bonding and Molecular Geometry 407 (a) SF6 (b) PCl5 (c) BeH2 (d) CH3 + 92. Identify the electron pair geometry and the molecular structure of each of the following molecules or ions: (a) IF6 + (b) CF4 (c) BF3 (d) SiF5 − (e) BeCl2 93. What are the electron-pair geometry and the molecular structure of each of the following molecules or ions?
(a) ClF5 (b) ClO2 − (c) TeCl4 2− (d) PCl3 (e) SeF4 (f) PH2 − 94. Predict the electron pair geometry and the molecular structure of each of the following ions: (a) H3O+ (b) PCl4 − (c) SnCl3 − (d) BrCl4 − (e) ICl3 (f) XeF4 (g) SF2 95. Identify the electron pair geometry and the molecular structure of each of the following molecules: (a) ClNO (N is the central atom) (b) CS2 (c) Cl2CO (C is the central atom) (d) Cl2SO (S is the central atom) (e) SO2F2 (S is the central atom) (f) XeO2F2 (Xe is the central atom) (g) ClOF2 + (Cl is the central atom) 408 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at 96. Predict the electron pair geometry and the molecular structure of each of the following: (a) IOF5 (I is the central atom) (b) POCl3 (P is the central atom) (c) Cl2SeO (Se is the central atom) (d) ClSO+ (S is the central atom) (e) F2SO (S is the central atom) (f) NO2 − (g) SiO4 4− 97. Which of the following molecules and ions contain polar bonds? Which of these molecules and ions have dipole moments?
(a) ClF5 (b) ClO2 − (c) TeCl4 2− (d) PCl3 (e) SeF4 (f) PH2 − (g) XeF2 98. Which of the molecules and ions in Exercise 7.93 contain polar bonds? Which of these molecules and ions have dipole moments?
(a) H3O+ (b) PCl4 − (c) SnCl3 − (d) BrCl4 − (e) ICl3 (f) XeF4 (g) SF2 99. Which of the following molecules have dipole moments?
(a) CS2 (b) SeS2 (c) CCl2F2 (d) PCl3 (P is the central atom) (e) ClNO (N is the central atom) 100. Identify the molecules with a dipole moment: (a) SF4 Chapter 7 Chemical Bonding and Molecular Geometry 409 (b) CF4 (c) Cl2CCBr2 (d) CH3Cl (e) H2CO 101. The molecule XF3 has a dipole moment. Is X boron or phosphorus?
102. The molecule XCl2 has a dipole moment. Is X beryllium or sulfur?
103. Is the Cl2BBCl2 molecule polar or nonpolar?
104. There are three possible structures for PCl2F3 with phosphorus as the central atom. Draw them and discuss how measurements of dipole moments could help distinguish among them.
105. Describe the molecular structure around the indicated atom or atoms: (a) the sulfur atom in sulfuric acid, H2SO4 [(HO)2SO2] (b) the chlorine atom in chloric acid, HClO3 [HOClO2] (c) the oxygen atom in hydrogen peroxide, HOOH (d) the nitrogen atom in nitric acid, HNO3 [HONO2] (e) the oxygen atom in the OH group in nitric acid, HNO3 [HONO2] (f) the central oxygen atom in the ozone molecule, O3 (g) each of the carbon atoms in propyne, CH3CCH (h) the carbon atom in Freon, CCl2F2 (i) each of the carbon atoms in allene, H2CCCH2 106. Draw the Lewis structures and predict the shape of each compound or ion: (a) CO2 (b) NO2 − (c) SO3 (d) SO3 2− 107. A molecule with the formula AB2, in which A and B represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion for each shape.
108. A molecule with the formula AB3, in which A and B represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion that has each shape.
109. Draw the Lewis electron dot structures for these molecules, including resonance structures where appropriate: (a) CS3 2− (b) CS2 (c) CS (d) predict the molecular shapes for CS3 2−and CS2 and explain how you arrived at your predictions 110. What is the molecular structure of the stable form of FNO2? (N is the central atom.) 410 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at 111. A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen. What is its molecular structure?
112. Use the simulation ( to perform the following exercises for a two-atom molecule: (a) Adjust the electronegativity value so the bond dipole is pointing toward B. Then determine what the electronegativity values must be to switch the dipole so that it points toward A.
(b) With a partial positive charge on A, turn on the electric field and describe what happens.
(c) With a small partial negative charge on A, turn on the electric field and describe what happens.
(d) Reset all, and then with a large partial negative charge on A, turn on the electric field and describe what happens.
113. Use the simulation ( to perform the following exercises for a real molecule. You may need to rotate the molecules in three dimensions to see certain dipoles.
(a) Sketch the bond dipoles and molecular dipole (if any) for O3. Explain your observations.
(b) Look at the bond dipoles for NH3. Use these dipoles to predict whether N or H is more electronegative.
(c) Predict whether there should be a molecular dipole for NH3 and, if so, in which direction it will point. Check the molecular dipole box to test your hypothesis.
114. Use the Molecule Shape simulator ( to build a molecule. Starting with the central atom, click on the double bond to add one double bond. Then add one single bond and one lone pair. Rotate the molecule to observe the complete geometry. Name the electron group geometry and molecular structure and predict the bond angle. Then click the check boxes at the bottom and right of the simulator to check your answers.
115. Use the Molecule Shape simulator ( to explore real molecules. On the Real Molecules tab, select H2O. Switch between the “real” and “model” modes. Explain the difference observed.
116. Use the Molecule Shape simulator ( to explore real molecules. On the Real Molecules tab, select “model” mode and S2O. What is the model bond angle? Explain whether the “real” bond angle should be larger or smaller than the ideal model angle.
Chapter 7 Chemical Bonding and Molecular Geometry 411 412 Chapter 7 Chemical Bonding and Molecular Geometry This content is available for free at |
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| | | | | --- | | | | Students / Subjects | | | | | | | | | | | Handbook >> Market Equilibrium >> Impact of Shifts in Demand and Supply Whenever there is a change in one of the factors of either supply or demand, market equilibrium will be affected. Shift in Demand When there is a change of one of the factors of demand- like the price of the product and related goods, consumer preferences, or income- there is a corresponding change in the demand curve. For instance, if someone's income grows, then his demand for goods will increase, shifting his demand curve to the right. This will lead to a higher quantity being consumed at a higher price, ceteris paribus. Conversely, there can be a negative effect that shifts the supply curve to the left where a lower quantity is consumed at a lower price, ceteris paribus. This can occur when the price of substitutes falls or consumers begin to lose their taste for the product. Shift in Supply When there is a change of one of the factors of supply- like changes in the prices of production inputs like labor or capital; a change in production technology and its associated productivity change; or the amount of competition in a specific product market- there is a corresponding change in the supply curve. For example, if worker productivity improves due to some human capital or technology investment, then the costs of production decrease. This exerts a positive effect on the supply curve shifting it right, where the new market equilibrium is at a higher quantity and a lower price, holding everything else constant. There can also be a negative shift that moves the supply curve to the left, with the resulting market clearing price being higher and quantity lower, ceteris paribus. This type of change can occur when the price of an input like labor or raw material jumps. Shifts in Demand and Supply Realistically speaking, ceteris paribus doesn't hold in the real world marketplace as many things are happening at once that either have complimentary or contrary influences upon the market equilibrium. You can't gauge what the new market equilibrium might be as you are not holding everything constant, but two things are being changed simultaneously. To find out what the new market equilibrium is you need detailed information on the magnitude of the supply and demand factor changes and the corresponding shifts in the graph, along with knowledge of the shapes of the curves. Take the market for apples for instance. If both supply and demand increase (on the graph this would be represented by the supply and demand curves both shifting to the right)- if orchard productivity rises while a new medical reports touts the discovery of the newly added health benefits of apples- then the quantity will definitely go up but the new price is indefinite. It could go up if the increase in demand is significant enough, or it could go down if it's not. Similarly, a certain quantity reduction but an uncertain price will pertain when the both demand and supply curves shift to the left. This could happen if the price of apple substitutes like plums drops dramatically, while farm labor becomes much more expensive. There will be certainty about the price, but not the quantity when the supply and demand curves move in opposite directions. For instance, another medical report could come out detailing the unsanitary apple harvesting conditions, shifting demand curve to the left. Simultaneously, genetic engineers have produced an apple that doesn't require as much costly care as before, shifting the supply curve to the right. Price will certainly go down, but the quantity consumed will demand on how relatively large the shifts of each curve are. Similarly, it could be certain that price would go up, but whether quantity consumed would go up or down is uncertain. This could happen if the demand curve shifts to the right while the supply curve shifts to the left- say if everyone's income increases, thereby increasing their consumption of apples and new government regulations curtail farmer's dependence on cheap illegal alien labor. The price of apples would rise, but it would depend on magnitude of the changes in both the supply and demand curves. Back to Market Equilibrium |
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Angles and parallel lines
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Opposite angles I
Opposite angles II
Opposite angles III
When two lines intersect they form two pairs of opposite angles, A + C and B + D. Another word for opposite angles are vertical angles.
Vertical angles are always congruent, which means that they are equal.
Adjacent angles are angles that come out of the same vertex. Adjacent angles share a common ray and do not overlap.
The size of the angle xzy in the picture above is the sum of the angles A and B.
Two angles are said to be complementary when the sum of the two angles is 90°.
Two angles are said to be supplementary when the sum of the two angles is 180°.
If we have two parallel lines and have a third line that crosses them as in the ficture below - the crossing line is called a transversal
When a transversal intersects with two parallel lines eight angles are produced.
The eight angles will together form four pairs of corresponding angles. Angles 1 and 5 constitutes one of the pairs. Corresponding angles are congruent. All angles that have the same position with regards to the parallel lines and the transversal are corresponding pairs e.g. 3 + 7, 4 + 8 and 2 + 6.
Angles that are in the area between the parallel lines like angle 2 and 8 above are called interior angles whereas the angles that are on the outside of the two parallel lines like 1 and 6 are called exterior angles.
Angles that are on the opposite sides of the transversal are called alternate angles e.g. 1 + 8.
All angles that are either exterior angles, interior angles, alternate angles or corresponding angles are all congruent.
Example
The picture above shows two parallel lines with a transversal. The angle 6 is 65°. Is there any other angle that also measures 65°?
6 and 8 are vertical angles and are thus congruent which means angle 8 is also 65°.
6 and 2 are corresponding angles and are thus congruent which means angle 2 is 65°.
6 and 4 are alternate exterior angles and thus congruent which means angle 4 is 65°.
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883 | https://www.freemathhelp.com/forum/threads/finding-a-term-in-this-series.79474/ | Finding a term in this series | Free Math Help Forum
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Finding a term in this series
Thread starterSimonSpacz
Start dateFeb 20, 2013
S
SimonSpacz
New member
Joined Feb 10, 2013 Messages 3
Feb 20, 2013
#1
The problem: Given this sequence Tn-1 = (Tn)^2 – 3n , given that T1=1, find T5
Tn referring to the nth term.
My work so far; Manipulated the formula of Tn=a+(n-1)d for T4 and T5 and filled them in to try and work out d (figuring that a=1, from t1=1)
I got decimal place answers for d that didnt work for the formula given in the question so I'm kinda lost and realize my lack of knowledge in this sector.
Any help?
L
lookagain
Elite Member
Joined Aug 22, 2010 Messages 3,266
Feb 20, 2013
#2
SimonSpacz said:
The problem: Given this sequence >> Tn-1 = (Tn)^2 – 3n <<, given that T1=1, find T5
Tn referring to the nth term.
My work so far; Manipulated the formula of
>> Tn=a+(n-1)d <<
for T4 and T5 and filled them in to try and work out d
(figuring that a=1, from t1=1) I got decimal place answers for d that didn't work for the formula given in the
question so I'm kinda lost and realize my lack of knowledge in this sector.
Any help?
Click to expand...
If your first equation is
T n−1=(T n)2−3 n,\displaystyle T_{n - 1} \ = \ (T_n)^2 \ - \ 3n, T n−1=(T n)2−3 n,
then the first few terms show that the formula in the second
equation can't be used. The first few terms in the recursive equation do not have a common difference.
The common differences show up in arithmetic sequences. The second equation is used for
arithmetic sequences.
D
daon2
Senior Member
Joined Aug 17, 2011 Messages 1,003
Feb 20, 2013
#3
Is it possible you have the indices confused?
For example, T 2=7\displaystyle T_2 = \sqrt{7}T 2=7 and T 2=−7\displaystyle T_2 = -\sqrt{7}T 2=−7 both are acceptable answers.
D
DrPhil
Senior Member
Joined Nov 29, 2012 Messages 1,383
Feb 20, 2013
#4
SimonSpacz said:
The problem: Given this sequence Tn-1 = (Tn)^2 – 3n , given that T1=1, find T5
Tn referring to the nth term.
My work so far; Manipulated the formula of Tn=a+(n-1)d for T4 and T5 and filled them in to try and work out d (figuring that a=1, from t1=1)
I got decimal place answers for d that didnt work for the formula given in the question so I'm kinda lost and realize my lack of knowledge in this sector.
Any help?
Click to expand...
Because of the square (or square root if you look the other direction) in the recursion formula, this is NOT an arithmetic sequence with equal spacing "d" between terms, That is, the terms don't have the form Tn=a+(n-1)d (as you discovered). Please do check the problem to be sure that what you type really looks like what was given. If you have it right, solving for T_n would give
T n=T n−1+3 n\displaystyle T_n = \sqrt{T_{n-1} + 3n} T n=T n−1+3 n
As daon2 pointed out, this is ambiguous unless there is something that tells you always to use the positive square root. If you made that assumption, you could in principle write down the sequence up through T 5\displaystyle T_5 T 5. But this doesn't look "right" - very messy with the square roots.
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5
How do you differentiate (t-1/t^2-1)?
George Ivey
Former Math Professor at Gallaudet University
·
Author has 23.7K answers and 2.6M answer views
·
1y
What you wrote, t-1/t^2- 1= t- t^-2– 1, has derivative 1+ t^-3= 1+ 1/t^3/
-If you MEAN (t- 1)/(t^2- 1) that is NOT differentiable at t= 1 or t= -1.
For all other x, (t- 1)/(t^2–1)= (t-1)/((t-1)(t+ 1)= 1/(t+1)= (t+ 1)^-1 has derivative -(t+ 1)^-2= -1/(t+ 1)^2= -1/(t^2+ 2t+ 1).
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Is [x^2] differentiable in (-1, 1)?
Darren Lorent
Author has 5.1K answers and 594.6K answer views
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Be careful.
f(t) = t – 1/t² – 1 = t – t⁻² – 1
f’(t) = 1 – (–2)t⁻³ = 1 + 2/t³
g(t) = (t – 1)/(t² – 1) = 1/(t + 1), t ≠ 1
g’(t) = –1/(t + 1)², t ≠ 1
Gordon M. Brown
Math Tutor at San Diego City College (2018-Present)
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Related
How do you differentiate Y= - sqrt (1-y^2/1-x^2)?
Let’s begin by pointing out that you could not have done much worse to make clear to your audience the expression you want to differentiate! Firstly, Y is not the same variable as y. This error alone would be embarrassing in a seventh-grader, let alone a student of calculus! Second, your radicand does not contain sufficient grouping symbols to make clear what you mean by it. You are literally making us all guess at what you want. Lastly, you never tell us whether you intend to differentiate with respect to x, or with respect to y.
In the image that follows, I am assuming that you want to differ
Let’s begin by pointing out that you could not have done much worse to make clear to your audience the expression you want to differentiate! Firstly, Y is not the same variable as y. This error alone would be embarrassing in a seventh-grader, let alone a student of calculus! Second, your radicand does not contain sufficient grouping symbols to make clear what you mean by it. You are literally making us all guess at what you want. Lastly, you never tell us whether you intend to differentiate with respect to x, or with respect to y.
In the image that follows, I am assuming that you want to differentiate implicitly the equation
y = -√[(1 - y^2) / (1 - x^2)]
with respect to x. If that is not what you intended, well, whose fault is that?
(Click on the image below to expand it as necessary.)
Mann Vora
B Tech from A D Patel Institute of Technology, Vallabh Vidhyanagar
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Author has 58 answers and 131.1K answer views
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Related
What is the differentiation and double differentiation of x=(t+5) ^(-1) w.r.t t?
Hope this helps..
Hope this helps..
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What is the differentiation and double differentiation of x=(t+5) ^(-1) w.r.t t?
Why is
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Rajeev Raj
Former Study at Resonance 〈kota〉 (2012–2015)
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Author has 8.4K answers and 2M answer views
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Related
How do you differentiate Y= - sqrt (1-y^2/1-x^2)?
solⁿ Given y=-√((1-y²)/(1-x²))
squaring
y² =(1-y²)/(1-x²)
(1-x²) y² =1- y²
1 - x² =y⁻² - 1
Differentiate both sides w.r.t. x
∴ 0 - 2x =-2y⁻²⁻¹ × dy/dx - 0
or - 2x =-2 y⁻³ × dy/dx
or x =y⁻³ × dy/dx
or x/y⁻³ =dy/dx
or xy³ =dy/dx
dy/dx =xy³
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Franck Salindres
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How do I integrate ∫_0 ^∞ t (t^2 +1) ^ (− (t+(1/t))) dt?
Quick convergence check of the integral:
1 - When t → 0+ , we have t+1/t → +oo ,
so, (t^2+1) power - (t+1/t) ~ exp(- t) => f(t) ~ t → 0 and for all p < 1, we have f(t)/t^p ~ t^(t-1) → 0 fast (because 1- p > 0),
so f is negligible compared to t^p for all p < 1 when t → 0+.
Thus, the convergence of I is guaranteed at 0.
2 - When t → oo , t+1/t ~ t,
so (t^2+1) power (- t) ~ t^(- 2t) → 0 fast, because it is negligible compared to t^p with p > 1 (compared to t^2 for example).
The factor t is not enough to compensate, thus, the function decreases rapidly.
Remark : both are sufficient conditions (but not
Quick convergence check of the integral:
1 - When t → 0+ , we have t+1/t → +oo ,
so, (t^2+1) power - (t+1/t) ~ exp(- t) => f(t) ~ t → 0 and for all p < 1, we have f(t)/t^p ~ t^(t-1) → 0 fast (because 1- p > 0),
so f is negligible compared to t^p for all p < 1 when t → 0+.
Thus, the convergence of I is guaranteed at 0.
2 - When t → oo , t+1/t ~ t,
so (t^2+1) power (- t) ~ t^(- 2t) → 0 fast, because it is negligible compared to t^p with p > 1 (compared to t^2 for example).
The factor t is not enough to compensate, thus, the function decreases rapidly.
Remark : both are sufficient conditions (but not necessarily)
Therefore, 1 and 2 => the integral is convergent.
3 - Exact result ?
This integral has no known antiderivative in terms of elementary functions, even with special functions (like polylogarithms, the gamma function, etc.), no closed-form expression is known in the current mathematical literature. It can be numerically evaluated if needed, but not analytically using simple expressions. We can apply (for example) the trapézoidal method to have an approximated value of I (see the picture to have a view of this method)
You can read: n = 100 (subdivisions) => I ~ 0,366
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How do you integrate (x^2+1) / (x ^2-1)?
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How do you integrate (x^2+1) / (x ^2-1)?
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Niall Thompson
MA (Camb) Maths, MSc Applied Maths, EdD Maths Education
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How would you differentiate fractionally, like d^(1/2) /dx^(1/2)?
This idea is called fractional differentiation.
The half derivative of a function is simply the operator that, when applied twice, is equivalent to the first derivative. More generally, the “a/b th derivative,” where a and b are coprime integers, is the operator that, when applied b times, is equivalent to the ath derivative.
In general, these are non-trivial to calculate. Sometimes this is possible for functions using analytic continuation. For example, if you express the nth derivative of a monomial using the gamma function instead of factorials, you will find that you can allow n to take non-
This idea is called fractional differentiation.
The half derivative of a function is simply the operator that, when applied twice, is equivalent to the first derivative. More generally, the “a/b th derivative,” where a and b are coprime integers, is the operator that, when applied b times, is equivalent to the ath derivative.
In general, these are non-trivial to calculate. Sometimes this is possible for functions using analytic continuation. For example, if you express the nth derivative of a monomial using the gamma function instead of factorials, you will find that you can allow n to take non-integer values quite easily and that the resulting expressions are consistent with the above definition.
Deniz
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What's the integral of 1/ t - √ (1 - t^2)?
Since there are no parentheses, I’m assuming the question is asking to evaluate [math]\int \frac{1}{x} - \sqrt{1-x^{2}} \, \mathrm{dx}[/math].
Let’s split it into [math]\int \frac{1}{x} \, \mathrm{dx} - \int \sqrt{1-x^{2}} \, \mathrm{dx}[/math]. The first part is equal to [math]\ln x[/math], so let’s focus on the second part.
[math]x=\sin u \implies \mathrm{dx}=\cos u \, \mathrm{du} \[/math]
[math]\begin{align} \int \sqrt{1-x^{2}} \, \mathrm{dx} &= \int \cos u \sqrt{1-\sin^{2}u} \, \mathrm{du} \ &= \int \cos^{2}u \, \mathrm{du} \end{align}[/math]
Let’s use a little trigonometry to evaluate this.
[math]\cos^{2}x=\frac{\cos(2x)+1}{2} \implies \int \cos^{2}u \, \math[/math]
Since there are no parentheses, I’m assuming the question is asking to evaluate [math]\int \frac{1}{x} - \sqrt{1-x^{2}} \, \mathrm{dx}[/math].
Let’s split it into [math]\int \frac{1}{x} \, \mathrm{dx} - \int \sqrt{1-x^{2}} \, \mathrm{dx}[/math]. The first part is equal to [math]\ln x[/math], so let’s focus on the second part.
[math]x=\sin u \implies \mathrm{dx}=\cos u \, \mathrm{du} \[/math]
[math]\begin{align} \int \sqrt{1-x^{2}} \, \mathrm{dx} &= \int \cos u \sqrt{1-\sin^{2}u} \, \mathrm{du} \ &= \int \cos^{2}u \, \mathrm{du} \end{align}[/math]
Let’s use a little trigonometry to evaluate this.
[math]\cos^{2}x=\frac{\cos(2x)+1}{2} \implies \int \cos^{2}u \, \mathrm{du} = \frac{1}{2} \int \cos(2u)+1 \, \mathrm{du} \[/math]
[math]2u=v \implies 2\mathrm{du}=\mathrm{dv} \[/math]
[math]\begin{align} \frac{1}{2} \int \cos(2u)+1 \, \mathrm{du} &= \frac{1}{4} \int \cos v +1 \, \mathrm{dv} \ &= \frac{\sin v +v}{4} \end{align}[/math]
Undoing all substitutions, we get:
[math]\ln x - \frac{\arcsin x}{2} - \frac{x\sqrt{1-x^{2}}}{2} + C[/math].
Mohammad Afzaal Butt
B.Sc in Mathematics & Physics, Islamia College Gujranwala (Graduated 1977)
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How do I differentiate -3e^ (-t/2)?
[math]\dfrac{d}{dt} \left(-3 e^{\left(\dfrac{-t}{2}\right)}\right)[/math]
[math]= -3 e^{\left(\dfrac{-t}{2}\right)}\dfrac{d}{dt} \left(\dfrac{-t}{2}\right)[/math]
Ragu Rajagopalan
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What is the step-by-step solution for t^4-t^2-2=0?
[math]t^4-t^2-2 = 0 [/math]
[math]\text{Completing the squares : } \left(t^2 - \dfrac{1}{2} \right )^2 -2 - \dfrac{1}{4} = 0[/math]
[math]\implies \left(t^2 - \dfrac{1}{2} \right )^2 = \left(\dfrac{3}{2} \right )^2 \implies t^2 - \dfrac{1}{2} = \pm \dfrac{3}{2}[/math]
[math]\text{Case - 1 : Considering +ve value : }[/math]
[math]t^2 - \dfrac{1}{2} = \dfrac{3}{2} \implies t = \pm \sqrt{2}[/math]
[math]\text{Case - 2 : Considering +ve value : }[/math]
[math]t^2 - \dfrac{1}{2} = \dfrac{-3}{2} \implies t = \pm i[/math]
[math]\mathbf{\text{Ans : }} \boxed{t = \pm \sqrt{2} \text{ OR } t = \pm i }[/math]
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885 | https://math.stackexchange.com/questions/1025729/expanding-1-x-2y3-in-powers-of-x-1-and-y-2-with-a-taylor-series | multivariable calculus - Expanding $(1 - x + 2y)^3$ in powers of $x-1$ and $y-2$ with a Taylor series - Mathematics Stack Exchange
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Expanding (1−x+2 y)3(1−x+2 y)3 in powers of x−1 x−1 and y−2 y−2 with a Taylor series
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I would like to do this.
I observe that I can write
f(x,y)=(1−x+2 y)3=(2(y−2)−(x−1)+4)3.f(x,y)=(1−x+2 y)3=(2(y−2)−(x−1)+4)3.
It's easy to do this via algebra directly.
However, I'm asked to do it by computing the Taylor series. Is it correct to say that I could expand the function
f(a,b)=(2 a−b+4)3 f(a,b)=(2 a−b+4)3
(where a=y−2 a=y−2 and b=x−1 b=x−1) around the point (2,1)(2,1) to the third-order term like so:
f(a,b)=f(2,1)+f a(2,1)a+f b(2,1)b+1 2 f a a(2,1)a 2+1 2 f b b(2,1)b 2+f a b(2,1)a b+⋯f(a,b)=f(2,1)+f a(2,1)a+f b(2,1)b+1 2 f a a(2,1)a 2+1 2 f b b(2,1)b 2+f a b(2,1)a b+⋯
(where I've neglected to write the third order term for simplicity's sake)?
Or am I missing the point of the question / is my approach incorrect?
multivariable-calculus
taylor-expansion
partial-derivative
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edited Nov 17, 2014 at 12:09
Git Gud
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asked Nov 17, 2014 at 9:47
Marcus EmilssonMarcus Emilsson
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You are not expanding around (a,b)=(2,1)(a,b)=(2,1), but (a,b)=(0,0)(a,b)=(0,0). The rest is fine.user65203 –user65203 2014-11-17 12:42:58 +00:00 Commented Nov 17, 2014 at 12:42
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I don't understand what you're doing, you seem to be using f f with two different meanings.
The function is a third degree polynomial, therefore the fourth order partials are constantly zero and the Taylor series of f f of order 3 3 (or above) coincides with f f (i.e. the remainder is 0 0).
Thus, for all (x,y)∈R 2(x,y)∈R 2, f(x,y)=f((1,2)+(x−1,y−2))=(−(x−1)+2(y−2)+4)3 f(x,y)=f((1,2)+(x−1,y−2))=(−(x−1)+2(y−2)+4)3 and it's easy to get the taylor expansion with powers of (x−1)(x−1) and (y−2)(y−2).
If you must use the formula for Taylor series, in the notation at the bottom of the linked section set x=(a+(x−1,y−2))x=(a+(x−1,y−2)) where a=(1,2)a=(1,2) to get
f(x,y)=f(1,2)+∂f∂x(1,2)(x−1)+∂f∂y(1,2)(y−2)+1 2![∂2 f∂x 2(1,2)(x−1)2+2∂2 f∂y∂x(1,2)(x−1)(y−2)+∂2 f∂y 2(1,2)(y−2)2]+R 2(x−1,y−2),f(x,y)=f(1,2)+∂f∂x(1,2)(x−1)+∂f∂y(1,2)(y−2)+1 2![∂2 f∂x 2(1,2)(x−1)2+2∂2 f∂y∂x(1,2)(x−1)(y−2)+∂2 f∂y 2(1,2)(y−2)2]+R 2(x−1,y−2),
where R 2 R 2 is a function such that lim(x,y)→(0,0)[R 2(x,y)∥(x,y)∥2]=0 lim(x,y)→(0,0)[R 2(x,y)‖(x,y)‖2]=0.
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edited Nov 17, 2014 at 12:15
answered Nov 17, 2014 at 12:08
Git GudGit Gud
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Come back to me for confirmation of the final result.Git Gud –Git Gud 2014-11-17 12:09:07 +00:00 Commented Nov 17, 2014 at 12:09
Should there not be factors of (x−1)2(x−1)2, (y−2)2(y−2)2, and (x−1)(y−2)(x−1)(y−2) in the second-order term of the Taylor expansion?Marcus Emilsson –Marcus Emilsson 2014-11-17 12:14:54 +00:00 Commented Nov 17, 2014 at 12:14
1 @ElizabethLin Of course xD Thanks.Git Gud –Git Gud 2014-11-17 12:15:16 +00:00 Commented Nov 17, 2014 at 12:15
Right -- thanks to you too. I think I got a bit confused on the notation and overcomplicated it for some reason.Marcus Emilsson –Marcus Emilsson 2014-11-17 12:15:58 +00:00 Commented Nov 17, 2014 at 12:15
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Since the given function f f is a polynomial of degree 3 3 in x x and y y its Taylor series of order 3 3 set up at any point (a,b)(a,b) gives the exact value of f f for all increments (X,Y)(X,Y). Now the third order Taylor expansion in general looks like
f(a+X,b+Y)=f(a,b)+(f x X+f y Y)+1 2(f x x X 2+2 f x y X Y+f y y Y 2)+1 6(f x x x X 3+3 f x x y X 2 Y+3 f x y y X Y 2+f y y y Y 3)+R,(1)(1)f(a+X,b+Y)=f(a,b)+(f x X+f y Y)+1 2(f x x X 2+2 f x y X Y+f y y Y 2)+1 6(f x x x X 3+3 f x x y X 2 Y+3 f x y y X Y 2+f y y y Y 3)+R,
where the partial derivatives of f f have to be taken at (a,b)(a,b), whence are constants in (1)(1), and R=0 R=0 in the case at hand.
To solve your problem (the hard way!) you have to put
(a,b):=(1,2),X:=x−1,Y:=y−2(a,b):=(1,2),X:=x−1,Y:=y−2
in the above formula. You will then obtain
f(x,y)=f(1,2)+f x(1,2)(x−1)+f y(1,2)(y−2)+1 2(f x x(1,2)(x−1)2+2 f x y(1,2)(x−1)(y−2)+f y y(1,2)(y−2)2)+…,f(x,y)=f(1,2)+f x(1,2)(x−1)+f y(1,2)(y−2)+1 2(f x x(1,2)(x−1)2+2 f x y(1,2)(x−1)(y−2)+f y y(1,2)(y−2)2)+…,
where I have omitted the third degree terms.
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answered Nov 17, 2014 at 12:31
Christian BlatterChristian Blatter
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Denoting s=2 a−b+4 s=2 a−b+4 for conciseness, the successive partial derivatives are:
s 3 s 3
3.2.s 2,3.1¯.s 2 3.2.s 2,3.1¯.s 2
3.2.2.2.s,3.2.2.1¯.s,3.2.1¯.1¯.s 3.2.2.2.s,3.2.2.1¯.s,3.2.1¯.1¯.s
3.2.1.2.2.2,3.2.1.2.2.1¯,3.2.1.2.1¯.1¯,3.2.1.1¯.1¯.1¯3.2.1.2.2.2,3.2.1.2.2.1¯,3.2.1.2.1¯.1¯,3.2.1.1¯.1¯.1¯
Evaluated at (0,0)(0,0), s=4 s=4:
64 64
96,−48 96,−48
96,−48,24 96,−48,24
48,−24,12,−6 48,−24,12,−6
Hence the Taylor development
64+96 a−48 b+1 2(96 a 2−2.48 a b+24 b 2)+1 6(48 a 3−3.24 a 2 b+3.12 a b 2−6 b 3).64+96 a−48 b+1 2(96 a 2−2.48 a b+24 b 2)+1 6(48 a 3−3.24 a 2 b+3.12 a b 2−6 b 3).
=64+96 a−48 b+48 a 2−48 a b+12 b 2+8 a 3−12 a 2 b+6 a b 2−b 3,=64+96 a−48 b+48 a 2−48 a b+12 b 2+8 a 3−12 a 2 b+6 a b 2−b 3,
as you can obtain by direct expansion of s 3 s 3.
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edited Nov 17, 2014 at 13:43
answered Nov 17, 2014 at 12:56
user65203 user65203
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887 | https://stats.libretexts.org/Bookshelves/Introductory_Statistics/Support_Course_for_Elementary_Statistics/The_Number_Line/Plotting_Points_and_Intervals_on_the_Number_Line | Skip to main content
Plotting Points and Intervals on the Number Line
Last updated
: Apr 9, 2022
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Distance between Two Points on a Number Line
Represent an Inequality as an Interval on a Number Line
Page ID
: 4729
Larry Green
Lake Tahoe Community College
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Outcomes
Plot a point on the number line
Plot an interval on the number line
The number line is of fundamental importance and is used repeatedly in statistics. It is a tool to visualize all of the possible outcomes of a study and to organize the results of the study. Often a diagram is placed above the number line to provide us with a picture of the results. By the end of this section, you will be able to plot points and intervals on a number line and use these plots to understand the possible outcomes and actual outcomes of studies.
Drawing Points on a Number Line
A number line is just a horizontal line that is used to display all the possible outcomes. It is similar to a ruler in that it helps us describe and compare numbers. Similar to a ruler that can be marked with many different scales such as inches or centimeters, we get to choose the scale of the number line and where the center is.
Example
The standard normal distribution is plotted above a number line. The most important values are the integers between -3 and 3. The number 0 is both the mean (average) and median (center).
Plot the number line that best displays this information.
Plot the value -1.45 on this number line.
Solution
We sketch a line, mark 0 as the center, and label the numbers -3, -2, -1, 0, 1, 2, 3 from left to right.
To plot the point -1.45, we first have to understand that this number is between -1 and -2. It is close to half way between -1 and -2. We put a circle on the number line that is close to halfway between these values as shown below.
Example
When working with box plots, we need to first set up a number line that labels what is called the five point summary: Minimum, First Quartile, Median, Third Quartile, and Maximum. Suppose the five point summary for height in inches for a basketball team is: 72,74,78,83,89. Plot these points on a number line
Solution
When plotting points on a number line, we first have to decide what range of the line we want to show in order to best display the points that appear. Technically all numbers are on every number line, but that does not mean we show all numbers. In this example, the numbers are all between 70 and 90, so we certainly don't need to display the number 0. A good idea is to let 70 be on the far left and 90 be on the far right and then plot the points between them. We also have to decide on the spacing of the tick marks. Since the range from 70 to 90 is 20, this may be too many numbers to display. Instead we might want to count by 5's. Below is the number line that shows the numbers 70 to 90 and counts by 5's. The five point summary is plotted on this line.
Exercise
A histogram will be drawn to display the annual income that experienced registered nurses make. The boundaries of the bars of the histogram are: 108,000, 162,000, and $189,000. Plot these points on a number line.
Plotting an Interval on a Number Line
Often in statistics, instead of just having to plot a few points on a number line, we need to instead plot a whole interval on the number line. This is especially useful when we want to exhibit a range of values between two numbers, to the left of a number or to the right of a number.
Example
A 95% confidence interval for the proportion of Americans who work on weekends is found to be 0.24 to 0.32, with the center at 0.28. Use a number line to display this information.
Solution
We just draw a number line, include the three key numbers: 0.24, 0.32, and 0.28 and highlight the part of the interval between 0.23 and 0.31.
Example : rejection region
In Hypothesis testing, we sketch something called the rejection region which is an interval that goes off to infinity or to negative infinity. Suppose that the mean number of hours to work on the week's homework is 4.2. The rejection region for the hypothesis test is all numbers larger than 7.3 hours. Plot the mean and sketch the rejection region on a number line.
Solution
We plot the point 4.2 on the number line and shade everything to the right of 7.3 on the number line.
Plot Integers on the Number Line
Intervals: Given an Inequality, Graph the Interval and State Using Interval Notation
Plotting Points on a Number Line Application
Distance between Two Points on a Number Line
Represent an Inequality as an Interval on a Number Line |
888 | https://www.jaadcasereports.org/article/S2352-5126(22)00525-2/fulltext | A case of Lyme disease complicated by the Jarisch-Herxheimer reaction and coinfection with Babesia - JAAD Case Reports
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Case reportVolume 32p68-70 February 2023 Open access
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A case of Lyme disease complicated by the Jarisch-Herxheimer reaction and coinfection with Babesia
Maria Karim, BA
Maria Karim, BA
Correspondence
Correspondence to: Maria Karim, BA, Hackensack Meridian School of Medicine, 340 Kingsland St Nutley, NJ 07110.
maria.karim@hmhn.org
Affiliations
Hackensack Meridian School of Medicine, Nutley, New Jersey
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amaria.karim@hmhn.org ∙ Allen N.Sapadin, MD
Allen N.Sapadin, MD
Affiliations
Department of Dermatology, Icahn School of Medicine at Mount Sinai, New York, New York
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b
Affiliations & Notes Article Info
a Hackensack Meridian School of Medicine, Nutley, New Jersey
b Department of Dermatology, Icahn School of Medicine at Mount Sinai, New York, New York
Publication History:
Published online November 30, 2022
Footnotes:
Funding sources: None.
IRB approval status: Not applicable.
Patient Consent: Consent for the publication of all patient photographs and medical information was provided by the authors at the time of article submission to the journal stating that all patients gave consent for their photographs and medical information to be published in print and online and with the understanding that this information may be publicly available.
DOI: 10.1016/j.jdcr.2022.11.023 External LinkAlso available on ScienceDirect External Link
Copyright: © 2022 by the American Academy of Dermatology, Inc. Published by Elsevier, Inc.
User License: Creative Commons Attribution (CC BY 4.0) | Elsevier's open access license policy
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Key words
babesiosis coinfection
Jarisch-Herxheimer reaction
Lyme disease
Abbreviation used
JHR (Jarisch-Herxheimer reaction)
Introduction
Lyme disease is a prevalent zoonosis transmitted by the spirochete Borrelia burgdorferi. Acute Lyme disease is characterized by the hallmark rash of erythema migrans at the site of the tick bite, which is often the diagnostic clue. The Jarisch-Herxheimer reaction (JHR) is a transient clinical phenomenon that develops within 24 hours after the initiation of antibiotic therapy for spirochetal infections, rarely reported in association with Lyme disease.1
1.
Dhakal, A. ∙ Sbar, E.
Jarisch Herxheimer Reaction
StatPearls Publishing, 2022
Google Scholar
Lyme disease is the most frequently transmitted infection by the Ixodes tick, although other pathogens may be simultaneously transmitted.2
2.
Swanson, S.J. ∙ Neitzel, D. ∙ Reed, K.D. ...
Coinfections acquired from ixodes ticks
Clin Microbiol Rev. 2006; 19:708-727
Crossref
Scopus (0)
PubMed
Google Scholar
Patients coinfected with multiple tick-borne illnesses demonstrate nonspecific symptoms and follow an unpredictable disease course, presenting a diagnostic challenge. Herein, the authors report a case of Lyme disease complicated by the JHR and coinfection with Babesia occurring in an endemic area.
Case Report
A healthy 68-year-old man presented for evaluation of an acute non-pruritic rash on his abdomen. The rash was initially noticed 3 days before, after returning from a trip to Connecticut. Over the last 3 days, the lesion rapidly enlarged and deepened in color. The patient denied having any fever, chills, nausea, vomiting, headache, myalgias, or joint pain. He denied a history of a recent tick bite, although activities on his trip included gardening in a wooded area.
Physical examination revealed an 11-cm red, annular plaque with defined borders and central clearing on the left side of the lateral abdomen and a central tick bite. This was warmer to the touch than the surrounding skin (Fig 1). Differential diagnoses included tinea corporis and erythema annulare centrifugum. Laboratory evaluation revealed negative Lyme IgG and immunoglobulin M antibodies. A 6-mm punch biopsy sample revealed changes consistent with spongiotic dermatitis, acanthosis and intercellular edema in the epidermis, and a perivascular mononuclear infiltrate in the dermis. Because of the patient’s clinical history, classic appearance of the lesion, and biopsy findings, suspicion of Lyme disease was high and the patient was empirically treated with doxycycline 100 mg twice daily for 21 days.
Figure viewer
Fig 1 Erythema migrans in a patient with Lyme disease at initial presentation. An 11-cm red, annular plaque with defined borders and central clearing on the left side of the lateral abdomen.
However, 18 hours after starting doxycycline, the patient’s condition deteriorated and he became acutely ill. He complained of overwhelming fatigue, fever, chills, nausea, vomiting, and pain in his knees and ankles. He was encouraged to go to the emergency department, and on examination, significant scleral icterus was noted. Laboratory results revealed a decreased hemoglobin of 8.8 g/dL (reference normal, 13.0-17.7 g/dL), decreased platelet count of 74 × 10E3/uL (reference normal, 150-450 × 10E3/uL), elevated total bilirubin of 3.4 mg/dL (reference normal, 0.0-1.2 mg/dL), elevated direct bilirubin of 2.2 mg/dL (reference normal, 0.0-0.3 mg/dL), and elevated lactate dehydrogenase of 764 U/L (reference normal, <240 U/L). Lyme antibodies were now positive. As the laboratory findings were consistent with hemolytic anemia, acute Lyme disease, and constitutional symptoms, a blood smear was performed. His blood smear showed 8% parasitemia, with Babesia microti identified by rapid Giemsa stain. The patient was subsequently treated with doxycycline 100 mg twice daily, atovaquone 750 mg daily, and azithromycin 500 mg daily for 21 days, with a consequent resolution of his presenting symptoms and laboratory abnormalities.
Discussion
Lyme disease is a prevalent zoonosis transmitted by the spirochete Borrelia burgdorferi, with various clinical manifestations at distinct stages of infection.3
3.
Bhate, C. ∙ Schwartz, R.A.
Lyme disease: Part I. Advances and perspectives
J Am Acad Dermatol. 2011; 64:619-638
Full Text
Full Text (PDF)
Scopus (69)
PubMed
Google Scholar
The first stage of acute Lyme disease is characterized by erythema migrans at the site of the tick bite, although a history of a recent tick bite may be absent.3
3.
Bhate, C. ∙ Schwartz, R.A.
Lyme disease: Part I. Advances and perspectives
J Am Acad Dermatol. 2011; 64:619-638
Full Text
Full Text (PDF)
Scopus (69)
PubMed
Google Scholar
This characteristic rash is present in up to 70% to 80% of infected patients, and often serves as the diagnostic clue, as positive serology can be delayed for up to several weeks after initial exposure to become detectable and histopathology is nonspecific, evidenced by this patient.4
4.
Steere, A.C. ∙ Sikand, V.K.
The presenting manifestations of Lyme disease and the outcomes of treatment
N Engl J Med. 2003; 348:2472-2474
Crossref
Scopus (153)
PubMed
Google Scholar
,5
5.
Reed, K.D.
Laboratory testing for Lyme disease: possibilities and practicalities
J Clin Microbiol. 2002; 40:319-324
Crossref
Scopus (64)
PubMed
Google Scholar
Treatment should be empirically initiated in cases with high suspicion of Lyme after phlebotomy.
The JHR is a transient though potentially life-threatening phenomenon that develops within 24 hours after initiating antibiotic therapy for spirochetal infections.1
1.
Dhakal, A. ∙ Sbar, E.
Jarisch Herxheimer Reaction
StatPearls Publishing, 2022
Google Scholar
Although the reaction is most commonly associated with syphilis, it can occur on rare occasion after treating Lyme disease, as Borrelia burgdorferi is also a spirochete.6
6.
Butler, T.
The Jarisch-Herxheimer reaction after antibiotic treatment of spirochetal infections: a review of recent cases and our understanding of pathogenesis
Am J Trop Med Hyg. 2017; 96:46-52
Crossref
Scopus (117)
PubMed
Google Scholar
The JHR has been reported to occur in 7% to 30% of treated Lyme disease cases.6
6.
Butler, T.
The Jarisch-Herxheimer reaction after antibiotic treatment of spirochetal infections: a review of recent cases and our understanding of pathogenesis
Am J Trop Med Hyg. 2017; 96:46-52
Crossref
Scopus (117)
PubMed
Google Scholar
The exact pathomechanism has not been elucidated, although the release of endotoxin-like materials and elevated levels of inflammatory cytokines from the lysis of spirochetal organisms are thought to play a role.7
7.
Belum, G.R. ∙ Belum, V.R. ∙ Chaitanya Arudra, S.K. ...
The Jarisch-Herxheimer reaction: revisited
Travel Med Infect Dis. 2013; 11:231-237
Crossref
Scopus (75)
PubMed
Google Scholar
Clinically, JHR presents as an acute exacerbation of cutaneous symptoms and the abrupt onset of systemic symptoms, including fever, tachycardia, headache, nausea, and myalgia.1
1.
Dhakal, A. ∙ Sbar, E.
Jarisch Herxheimer Reaction
StatPearls Publishing, 2022
Google Scholar
,7
7.
Belum, G.R. ∙ Belum, V.R. ∙ Chaitanya Arudra, S.K. ...
The Jarisch-Herxheimer reaction: revisited
Travel Med Infect Dis. 2013; 11:231-237
Crossref
Scopus (75)
PubMed
Google Scholar
This typically resolves without intervention and antibiotic therapy can be continued, although severe cases may require hospitalization.
Lyme disease is the most frequently transmitted pathogen by the Ixodes tick, although other pathogens, including anaplasmosis and babesiosis, are carried by the same tick.2
2.
Swanson, S.J. ∙ Neitzel, D. ∙ Reed, K.D. ...
Coinfections acquired from ixodes ticks
Clin Microbiol Rev. 2006; 19:708-727
Crossref
Scopus (0)
PubMed
Google Scholar
Rarely, coinfection of babesiosis with Lyme disease can occur.8
8.
Bhesania, S. ∙ Arora, K.S. ∙ Tokarski, M. ...
A case of tick bite induced babesiosis with lyme disease
Cureus. 2021; 13, e17401
Crossref
Google Scholar
In a study of 52 patients with erythema migrans, evidence of coinfection with Babesia was detected in 4 (7.7%) of patients.9
9.
Wormser, G.P. ∙ McKenna, D. ∙ Scavarda, C. ...
Co-infections in persons with early lyme disease, New York, USA
Emerg Infect Dis. 2019; 25:748-752
Crossref
Scopus (33)
PubMed
Google Scholar
Babesiosis presents with nonspecific constitutional symptoms, including fever, malaise, myalgia, or headache, occurring after parasite-mediated hemolysis of erythrocytes. Laboratory findings reveal hemolytic anemia, leukopenia, and thrombocytopenia.10
10.
Sanchez, E. ∙ Vannier, E. ∙ Wormser, G.P. ...
Diagnosis, treatment, and prevention of lyme disease, human granulocytic anaplasmosis, and babesiosis: a review
JAMA. 2016; 315:1767-1777
Crossref
Scopus (243)
PubMed
Google Scholar
Early identification and treatment of babesiosis is critical to prevent potential complications of acute respiratory, renal, and congestive heart failures and disseminated intravascular coagulation.8
8.
Bhesania, S. ∙ Arora, K.S. ∙ Tokarski, M. ...
A case of tick bite induced babesiosis with lyme disease
Cureus. 2021; 13, e17401
Crossref
Google Scholar
This case was remarkable in that not only did the JHR occur in association with Lyme disease, but also coinfection with babesiosis was subsequently identified. Although the symptoms of acute hemolytic anemia can occur with either the JHR or babesiosis, the timeframe of initiation of systemic symptoms after initiation of antibiotic treatment and exacerbation of a rash is characteristic of the JHR. To our knowledge, this is the first reported case of both the JHR and coinfection with Babesia presenting with Lyme disease. Eighteen hours after the first dose of doxycycline, the patient developed fatigue, fever, chills, nausea, and vomiting symptoms consistent with the JHR. Laboratory findings of hemolytic anemia and thrombocytopenia prompted evaluation for coinfection with Babesia. This case highlights the importance of maintaining high clinical suspicion for coinfection with additional tick-borne diseases when patients with Lyme disease acutely deteriorate or fail to respond to conventional therapy. Dermatologists should be cognizant of the potential of the JHR to occur in patients treated for Lyme disease in addition to its classic association with syphilis.
Conflicts of interest
None disclosed.
References
1.
Dhakal, A. ∙ Sbar, E.
Jarisch Herxheimer Reaction
StatPearls Publishing, 2022
Google Scholar
2.
Swanson, S.J. ∙ Neitzel, D. ∙ Reed, K.D. ...
Coinfections acquired from ixodes ticks
Clin Microbiol Rev. 2006; 19:708-727
Crossref
Scopus (0)
PubMed
Google Scholar
3.
Bhate, C. ∙ Schwartz, R.A.
Lyme disease: Part I. Advances and perspectives
J Am Acad Dermatol. 2011; 64:619-638
Full Text
Full Text (PDF)
Scopus (69)
PubMed
Google Scholar
4.
Steere, A.C. ∙ Sikand, V.K.
The presenting manifestations of Lyme disease and the outcomes of treatment
N Engl J Med. 2003; 348:2472-2474
Crossref
Scopus (153)
PubMed
Google Scholar
5.
Reed, K.D.
Laboratory testing for Lyme disease: possibilities and practicalities
J Clin Microbiol. 2002; 40:319-324
Crossref
Scopus (64)
PubMed
Google Scholar
6.
Butler, T.
The Jarisch-Herxheimer reaction after antibiotic treatment of spirochetal infections: a review of recent cases and our understanding of pathogenesis
Am J Trop Med Hyg. 2017; 96:46-52
Crossref
Scopus (117)
PubMed
Google Scholar
7.
Belum, G.R. ∙ Belum, V.R. ∙ Chaitanya Arudra, S.K. ...
The Jarisch-Herxheimer reaction: revisited
Travel Med Infect Dis. 2013; 11:231-237
Crossref
Scopus (75)
PubMed
Google Scholar
8.
Bhesania, S. ∙ Arora, K.S. ∙ Tokarski, M. ...
A case of tick bite induced babesiosis with lyme disease
Cureus. 2021; 13, e17401
Crossref
Google Scholar
9.
Wormser, G.P. ∙ McKenna, D. ∙ Scavarda, C. ...
Co-infections in persons with early lyme disease, New York, USA
Emerg Infect Dis. 2019; 25:748-752
Crossref
Scopus (33)
PubMed
Google Scholar
10.
Sanchez, E. ∙ Vannier, E. ∙ Wormser, G.P. ...
Diagnosis, treatment, and prevention of lyme disease, human granulocytic anaplasmosis, and babesiosis: a review
JAMA. 2016; 315:1767-1777
Crossref
Scopus (243)
PubMed
Google Scholar
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889 | https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/14%3A_Chemical_Kinetics | Skip to main content
14: Chemical Kinetics
Last updated
: Jul 12, 2023
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Index
14.1: The Rate of a Chemical Reaction
Page ID
: 11739
( \newcommand{\kernel}{\mathrm{null}\,})
Template:HideTOC
14.1: The Rate of a Chemical Reaction
: In this Module, the quantitative determination of a reaction rate is demonstrated. Reaction rates can be determined over particular time intervals or at a given point in time. A rate law describes the relationship between reactant rates and reactant concentrations.
14.2: Measuring Reaction Rates
: The method for determining a reaction rate is relatively straightforward. Since a reaction rate is based on change over time, it must be determined from tabulated values or found experimentally. With the obtained data, it is possible to calculate the reaction rate either algebraically or graphically.
14.3: Effect of Concentration on Reaction Rates: The Rate Law
: Typically, reaction rates decrease with time because reactant concentrations decrease as reactants are converted to products. Reaction rates generally increase when reactant concentrations are increased. This section examines mathematical expressions called rate laws, which describe the relationships between reactant rates and reactant concentrations. Rate laws are mathematical descriptions of experimentally verifiable data.
14.4: Zero-Order Reactions
: The rates of zero-order reactions is apparently independent of reactant concentrations. This means that the rate of the reaction is equal to the rate constant, k, of that reaction, but clearly a zero-order process cannot continue after a reactant has been exhausted. Just before this point is reached, the reaction will revert to another rate law instead of falling directly to zero Situations which are apparently zero order occur when a reaction is catalyzed by attachment to a solid surface (hete
14.5: First-Order Reactions
: In a first-order reaction, the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products.
14.6: Second-Order Reactions
: The simplest kind of second-order reaction is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form 2A → products. A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products.
14.7: Reaction Kinetics: A Summary
: For a zeroth-order reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of −k. For a first-order reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of −k. For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.
14.8: Theoretical Models for Chemical Kinetics
14.9: The Effect of Temperature on Reaction Rates
14.10: Reaction Mechanisms
14.11: Catalysis
14.E: Exercises
14.1: The Rate of a Chemical Reaction |
890 | https://courses.lumenlearning.com/calculus1/chapter/hyperbolic-functions/ | Hyperbolic Functions
Learning Outcomes
Identify the hyperbolic functions, their graphs, and basic identities
The hyperbolic functions are defined in terms of certain combinations of ex and e−x. These functions arise naturally in various engineering and physics applications, including the study of water waves and vibrations of elastic membranes. Another common use for a hyperbolic function is the representation of a hanging chain or cable, also known as a catenary. If we introduce a coordinate system so that the low point of the chain lies along the y-axis, we can describe the height of the chain in terms of a hyperbolic function. First, we define the hyperbolic functions.
Figure 6. The shape of a strand of silk in a spider’s web can be described in terms of a hyperbolic function. The same shape applies to a chain or cable hanging from two supports with only its own weight. (credit: “Mtpaley”, Wikimedia Commons)
Definition
Hyperbolic cosine
coshx=ex+e−x2
Hyperbolic sine
sinhx=ex−e−x2
Hyperbolic tangent
tanhx=sinhxcoshx=ex−e−xex+e−x
Hyperbolic cosecant
cschx=1sinhx=2ex−e−x
Hyperbolic secant
sechx=1coshx=2ex+e−x
Hyperbolic cotangent
cothx=coshxsinhx=ex+e−xex−e−x
The name cosh rhymes with “gosh,” whereas the name sinh is pronounced “cinch.” Tanh, sech, csch, and coth are pronounced “tanch,” “seech,” “coseech,” and “cotanch,” respectively.
Using the definition of cosh(x) and principles of physics, it can be shown that the height of a hanging chain, such as the one in Figure 6, can be described by the function h(x)=acosh(x/a)+c for certain constants a and c.
But why are these functions called hyperbolic functions? To answer this question, consider the quantity cosh2t−sinh2t. Using the definition of cosh and sinh, we see that
cosh2t−sinh2t=e2t+2+e−2t4−e2t−2+e−2t4=1
This identity is the analog of the trigonometric identity cos2t+sin2t=1. Here, given a value t, the point (x,y)=(cosht,sinht) lies on the unit hyperbola x2−y2=1 (Figure 7).
Figure 7. The unit hyperbola cosh2t−sinh2t=1.
Graphs of Hyperbolic Functions
To graph coshx and sinhx, we make use of the fact that both functions approach (12)ex as x→∞, since e−x→0 as x→∞. As x→−∞,coshx approaches 12e−x, whereas sinhx approaches −12e−x. Therefore, using the graphs of 12ex,12e−x, and −12e−x as guides, we graph coshx and sinhx. To graph tanhx, we use the fact that tanh(0)=0,−1<tanh(x)<1 for all x,tanhx→1 as x→∞, and tanhx→−1 as x→−∞. The graphs of the other three hyperbolic functions can be sketched using the graphs of coshx,sinhx, and tanhx (Figure 8).
Figure 8. The hyperbolic functions involve combinations of ex and e−x.
Identities Involving Hyperbolic Functions
The identity cosh2t−sinh2t, shown in Figure 7, is one of several identities involving the hyperbolic functions, some of which are listed next. The first four properties follow easily from the definitions of hyperbolic sine and hyperbolic cosine. Except for some differences in signs, most of these properties are analogous to identities for trigonometric functions.
Identities Involving Hyperbolic Functions
cosh(−x)=coshx
sinh(−x)=−sinhx
coshx+sinhx=ex
coshx−sinhx=e−x
cosh2x−sinh2x=1
1−tanh2x=sech2x
coth2x−1=csch2x
sinh(x±y)=sinhxcoshy±coshxsinhy
cosh(x±y)=coshxcoshy±sinhxsinhy
Example: Evaluating Hyperbolic Functions
Simplify sinh(5lnx).
If sinhx=34, find the values of the remaining five hyperbolic functions.
Show Solution
Using the definition of the sinh function, we write
sinh(5lnx)=e5lnx−e−5lnx2=eln(x5)−eln(x−5)2=x5−x−52.
2. Using the identity cosh2x−sinh2x=1, we see that
cosh2x=1+(34)2=2516.
Since coshx≥1 for all x, we must have coshx=5/4. Then, using the definitions for the other hyperbolic functions, we conclude that tanhx=3/5,cschx=4/3,sechx=4/5, and cothx=5/3.
Watch the following video to see the worked solution to Example: Evaluating Hyperbolic Functions
Closed Captioning and Transcript Information for Video
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “1.5 Exponential and Logarithmic Functions” here (opens in new window).
Try It
Simplify cosh(2lnx).
Hint
Use the definition of the cosh function and the power property of logarithm functions.
Show Solution
(x2+x−2)2
Inverse Hyperbolic Functions
From the graphs of the hyperbolic functions, we see that all of them are one-to-one except coshx and sechx. If we restrict the domains of these two functions to the interval [0,∞), then all the hyperbolic functions are one-to-one, and we can define the inverse hyperbolic functions. Since the hyperbolic functions themselves involve exponential functions, the inverse hyperbolic functions involve logarithmic functions.
Definition
Inverse Hyperbolic Functions:
sinh−1x=arcsinh x=ln(x+√x2+1)cosh−1x=arccosh x=ln(x+√x2−1)tanh−1x=arctanh x=12ln(1+x1−x)coth−1x=arccot x=12ln(x+1x−1)sech−1x=arcsech x=ln(1+√1−x2x)csch−1x=arccsch x=ln(1x+√1+x2|x|)
Let’s look at how to derive the first equation. The others follow similarly. Suppose y=sinh−1x. Then, x=sinhy and, by the definition of the hyperbolic sine function, x=ey−e−y2. Therefore,
ey−2x−e−y=0
Multiplying this equation by ey, we obtain
e2y−2xey−1=0
This can be solved like a quadratic equation, with the solution
ey=2x±√4x2+42=x±√x2+1
Since ey>0, the only solution is the one with the positive sign. Applying the natural logarithm to both sides of the equation, we conclude that
y=ln(x+√x2+1)
Example: Evaluating Inverse Hyperbolic Functions
Evaluate each of the following expressions
sinh−1(2)
tanh−1(14)
Show Solution
sinh−1(2)=ln(2+√22+1)=ln(2+√5)≈1.4436
tanh−1(14)=12ln(1+1/41−1/4)=12ln(5/43/4)=12ln(53)≈0.2554
Try It
Evaluate tanh−1(12).
Hint
Use the definition of tanh−1x and simplify.
Show Solution
12ln(3)≈0.5493.
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891 | https://www.quora.com/How-can-we-tell-if-a-quadratic-equation-has-two-distinct-real-solutions | How can we tell if a quadratic equation has two distinct real solutions? - Quora
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How can we tell if a quadratic equation has two distinct real solutions?
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11mo
To determine if a quadratic equation has two distinct real solutions, you can use the discriminant of the quadratic equation. A quadratic equation is generally expressed in the form:
a x 2+b x+c=0 a x 2+b x+c=0
where a a, b b, and c c are constants, and a≠0 a≠0.
The discriminant (D D) is calculated using the formula:
D=b 2−4 a c D=b 2−4 a c
The nature of the roots of the quadratic equation is determined by the value of the discriminant:
Two distinct real solutions: If D>0 D>0
One real solution (repeated): If D=0 D=0
No real solutions (complex solutions): If D<0 D<0
Summary:
To check if a quadratic equation has two distinct real solutio
Continue Reading
To determine if a quadratic equation has two distinct real solutions, you can use the discriminant of the quadratic equation. A quadratic equation is generally expressed in the form:
a x 2+b x+c=0 a x 2+b x+c=0
where a a, b b, and c c are constants, and a≠0 a≠0.
The discriminant (D D) is calculated using the formula:
D=b 2−4 a c D=b 2−4 a c
The nature of the roots of the quadratic equation is determined by the value of the discriminant:
Two distinct real solutions: If D>0 D>0
One real solution (repeated): If D=0 D=0
No real solutions (complex solutions): If D<0 D<0
Summary:
To check if a quadratic equation has two distinct real solutions, calculate the discriminant D=b 2−4 a c D=b 2−4 a c. If D>0 D>0, the equation has two distinct real solutions.
Upvote ·
John Fryer
Author has 890 answers and 872.4K answer views
·3y
ax² + bx + c = 0
Is a quadratic equation
We could plot the graph of y = ax² + bx + c
Here with a = 1 , b = 6 and c = 5 we can see the function equals zero when x is -5 and x = -1
Hence it is clear there are two solutions.
We get the same result using the quadratic formula to solve for x
(-b + or - √(b²-4ac))/2a
(-6 + or - √(36 - 4x1x5))/2
= 3 + or - √16
= 3 + or - 4
Ie x = -1 or x = -5
So if the term under the square root is zero we get one solution and if it is negative we get no real solutions.
Continue Reading
ax² + bx + c = 0
Is a quadratic equation
We could plot the graph of y = ax² + bx + c
Here with a = 1 , b = 6 and c = 5 we can see the function equals zero when x is -5 and x = -1
Hence it is clear there are two solutions.
We get the same result using the quadratic formula to solve for x
(-b + or - √(b²-4ac))/2a
(-6 + or - √(36 - 4x1x5))/2
= 3 + or - √16
= 3 + or - 4
Ie x = -1 or x = -5
So if the term under the square root is zero we get one solution and if it is negative we get no real solutions.
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Devash Roy
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There are 2 ways we could do this:
Looking at the discriminant “b^2 - 4ac”
If the discriminant is greater than 0, then it has 2 real solutions
Using Descartes Rule of Signs: (Keep in mind that the total number of solutions is the highest degree, which is 2 in ax^2)
In f(x) = ax^2 + bx + c, look for any positive to negative changes or vise versa
a to b is +a to +b, which is + to +, meaning no sign change
b to c is +b to +c, which is + to +, meaning no sign change
Make a table:
Now, write an equation for f(-x):
f(-x) = ax^2 - bx + c
a to -b is + to -, meaning 1 sign change
-b to c is - to +, meaning anothe
Continue Reading
There are 2 ways we could do this:
Looking at the discriminant “b^2 - 4ac”
If the discriminant is greater than 0, then it has 2 real solutions
Using Descartes Rule of Signs: (Keep in mind that the total number of solutions is the highest degree, which is 2 in ax^2)
In f(x) = ax^2 + bx + c, look for any positive to negative changes or vise versa
a to b is +a to +b, which is + to +, meaning no sign change
b to c is +b to +c, which is + to +, meaning no sign change
Make a table:
Now, write an equation for f(-x):
f(-x) = ax^2 - bx + c
a to -b is + to -, meaning 1 sign change
-b to c is - to +, meaning another sign change
Now put this new info into the Negative part of the table (Since we already have 2 solutions, there are no imaginary solutions possible):
And there you have it!
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How can you tell when a quadratic equation has no real solutions?
What is an example of a quadratic equation with two real solutions?
How do you determine if a quadratic equation has two solutions or no solutions?
What are some examples of quadratic equations that have no real solutions? How can we prove that a quadratic equation has no real solutions mathematically?
How can we determine if a quadratic equation has two real solutions, one real solution, or no real solutions (not complex numbers)?
Aaron Briseno
B.S in Mathematics&Teaching, University of California, Los Angeles (Graduated 2010) · Author has 1.3K answers and 3M answer views
·2y
Related
How can we determine if a quadratic equation has two real solutions, one real solution, or no real solutions (not complex numbers)?
If we have a quadratic equation then we can quickly get it into the form
a x 2+b x+c=0 a x 2+b x+c=0
And now we can use our knowledge of the quadratic formula, particularly the part that looks like
√b 2–4 a c b 2–4 a c
To determine precisely what we are being asked.
We know that when the square root’s value is zero we have one real root, when the square root’s value is real, but not zero we have 2 real roots, and when the square root’s value is complex/imaginary, we have 2 complex solutions.
The square root returns one real value when the portion under the radical (discriminant) is zero. So if and only if b 2–4 a c=0 b 2–4 a c=0
The sq
Continue Reading
If we have a quadratic equation then we can quickly get it into the form
a x 2+b x+c=0 a x 2+b x+c=0
And now we can use our knowledge of the quadratic formula, particularly the part that looks like
√b 2–4 a c b 2–4 a c
To determine precisely what we are being asked.
We know that when the square root’s value is zero we have one real root, when the square root’s value is real, but not zero we have 2 real roots, and when the square root’s value is complex/imaginary, we have 2 complex solutions.
The square root returns one real value when the portion under the radical (discriminant) is zero. So if and only if b 2–4 a c=0 b 2–4 a c=0
The square root will return 2-real values if and only if the discriminant is positive namely: b 2–4 a c>0 b 2–4 a c>0
The square root will return 2-imaginary values if and only if the discriminant is negative, namely: b 2–4 a c<0 b 2–4 a c<0
So you can quickly check quadratic equations of the form: a x 2+b x+c=0 a x 2+b x+c=0
By checking the relationship between b 2 b 2 and 4 a c 4 a c
b 2<4 a c b 2<4 a c means discriminant is less than zero so two complex roots
b 2=4 a c b 2=4 a c means discriminant will be zero, so one real root
b 2>4 a c b 2>4 a c means discriminant will be positive, so two real roots.
EXAMPLES:
2 x 2–3 x+2=0 2 x 2–3 x+2=0
We see b 2 b 2 is nine, and 4 a c=16 4 a c=16 so b 2<4 a c b 2<4 a c so two complex roots, graph shown has no x interecepts.
Next one: 3 x 2+7 x+1=0 3 x 2+7 x+1=0
We see b 2 b 2 is 49, and 4 a c=12 4 a c=12 so b 2>4 a c b 2>4 a c so 2 real roots, graph shown has two x-intercepts:
Last one: −x 2+6 x–9=0−x 2+6 x–9=0
We see b 2 b 2 is 36, and 4 a c=36 4 a c=36 so b 2=4 a c b 2=4 a c so 1 real (repeated) root, graph shown has one x-intercept where the x-intercept is just touching (a hint at evenly repeated roots):
If you scrolled to the bottom just to get to the answer, then scroll back up a bit the answer portion is above, you just left the example section, and entered the ‘warning’ about skipping around section.
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9 2
Thomas Schürger
Senior software architect at CGI Group (company) · Author has 2.3K answers and 2.5M answer views
·3y
A quadratic equation x 2+p x+q=0 x 2+p x+q=0 has two distinct real solutions if the discriminant D=p 2−4 q>0 D=p 2−4 q>0. If D=0 D=0, there is exactly one real solution, if D<0 D<0, there are two distinct complex non-real solutions.
For the more generic case a x 2+b x+c=0 a x 2+b x+c=0 (a≠0 a≠0), the discriminant is D=b 2−4 a c D=b 2−4 a c, and the same rules as above apply.
Upvote ·
9 1
Andro
Knows Slovenian · Author has 472 answers and 128.9K answer views
·3y
The quadratic equation a x 2+b x+c=0 a x 2+b x+c=0 will have two distinct real solutions if D=b 2−4 a c D=b 2−4 a c will be >0>0.
If D=0 D=0, the quadratic equation has 1 1 double solution.
And if D<0 D<0, the quadratic equation has 2 2 distinct complex solution.
This is a theory behind the quadratic equation.
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More answers below
How can you tell if a quadratic equation has real solutions without actually solving it?
What are real solutions in quadratic equations?
What are some real life examples which require the quadratic equation?
Why are there always two equal solutions for any quadratic equation in real life?
Why are there usually two solutions to a quadratic equation?
Steve Amato
Author has 30.2K answers and 5.6M answer views
·3y
ax^2+bx+c=0
x=(-b +/-sqrt(b^2–4ac)/2a)
So it has two distinct real solutions if b^2–4ac > 0
That is b^2 > 4ac
Otherwise you’d be taking a square root of a negative number.
Upvote ·
Michael Ejercito
Studied at California State University, Long Beach · Author has 1.3K answers and 265.9K answer views
·1y
if a,b a,b, and c c are real numbers, not necessarily distinct,
a x 2+b x+c=0 a x 2+b x+c=0 have two distnict real solutions if and only if b 2–4 a c>0 b 2–4 a c>0
Upvote ·
9 1
Harvey Becker
AA in Mathematics&Russian (language), College of San Mateo (CSM) (Graduated 1969) · Author has 3.4K answers and 1.9M answer views
·3y
Check the discriminant which is D = b^2 - 4ac
If D > 0 we will have 2 distinct real roots
If D =0 we will have one real root (double root)
If D < 0 we will have two complex roots
Upvote ·
The Chosen One
Service Desk Analyst at Samsung SDS (2007–present) · Author has 14.7K answers and 5.5M answer views
·1y
A quadratic has the form:
ax^2 + bx + c = 0
If b^2 - 4ac > 0, the equation has two distinct real solutions
If b^2 - 4ac = 0, there is only one solution
If b^2 - 4ac < 0, the equation has imaginary numbers as solutions.
Upvote ·
Hugh Bothwell
Studied Electrical Engineering · Author has 2.6K answers and 1.5M answer views
·3y
A quadratic equation will have 2 distinct real solutions IFF the discriminant b 2−4 a c>0 b 2−4 a c>0.
Upvote ·
Paul Holloway
Author has 2.1K answers and 3.3M answer views
·3y
Related
What are real solutions in quadratic equations?
Real solutions to a quadratic equation are solutions which are real numbers.
Real numbers:
Real numbers are numbers that can represent real things. We learn about them in school, numbers like 1, 2, 3… all the way up to infinity. Also 0 and -1, -2, -3… all the way down to negative infinity are real numbers. Fractions are also real numbers as are irrational numbers. So 1/2, 1/5, 1/10 along with pi or e or the square root of 2 are all real numbers.
Here is a representation of the real numbers on a number line:
Because a negative number multiplied by itself is a positive number there is no real number
Continue Reading
Real solutions to a quadratic equation are solutions which are real numbers.
Real numbers:
Real numbers are numbers that can represent real things. We learn about them in school, numbers like 1, 2, 3… all the way up to infinity. Also 0 and -1, -2, -3… all the way down to negative infinity are real numbers. Fractions are also real numbers as are irrational numbers. So 1/2, 1/5, 1/10 along with pi or e or the square root of 2 are all real numbers.
Here is a representation of the real numbers on a number line:
Because a negative number multiplied by itself is a positive number there is no real number that represents the square root of a negative number. That is to say there is no real value for x such that x^2 = -1. You can’t point to a position on the number line and say here is x.
Imaginary numbers:
But what if there was a number that when multiplied by itself came to a negative number? Let’s imagine such numbers exist and let’s say the square root of -1 is i where i stands for imaginary. Does mathematics allow for this or will mathematics break down?
i + i = 2i well so far we can add imaginary numbers.
2i - i = i, we can subtract them well enough.
2i/2 = i, we can divide them well enough.
i i = -1, because of our definition of i.
So far, so good. It turns out we can do the same sort of operations on imaginary numbers as we can on real numbers.
We can draw a similar number line for imaginary numbers but let’s put this number line at right angles to the real number line.
Complex Numbers:
We know we can add or subtract real numbers and we just learned we can add or subtract imaginary numbers but can we add real numbers to imaginary numbers or subtract them? It turns out we can and the maths works just as well. We call these combinations of real and imaginary numbers complex numbers.
2 + 3i is a complex number. 2 - 3i is also a complex number. -2 + 3i is likewise a complex number.
What happens if we try to square a complex number though?
(2+3i)^2 = (2+3i)(2+3i) = 2^2 + (3i)^2 + 2(2)(3i) = 4 + 9i^2 + 12i but we know i^2 = -1 so 9i^2 becomes 9(-1) which is -9. So (2+3i)^2 becomes 4 - 9 + 12i which is -5 + 12i which is also a complex number. If we think of the real number line and the imaginary number line as the axes on a cartesian graph we can plot the position of a complex number on the complex plane.
Thus 1+i would be represented thus:
Now it turns out that we can plot any real or imaginary number on this graph, that is to say, the complex plane. This means all real and imaginary numbers can be considered a subset of complex numbers. If it helps, we can think of i as a complex number (0 + 1i) or we can think of 1 as a complex number (1 + 0i).
Now that we have an idea of what real, imaginary and complex numbers are let’s look again at your question.
The real solutions to a quadratic equation are solutions to that equation that are real.
The solution to x^2 = 1 are 1 and -1. Both these numbers can be plotted on the real number line and so are real numbers and thus are real solutions to the quadratic equation x^2 = 1.
The solution to x^2 = -1 are i and -i. Both these numbers can be plotted on the imaginary number line but not on the real number line so they are not real solutions to the quadratic equation x^2 = -1. Remember a square root of a negative number can’t be plotted anywhere on the real number line.
Now let’s look at the quadratic formula (-b +/- sqrt(b^2–4ac))/2a. For any quadratic equation in the form ax^2 + bx + c = 0.
Note the square root bit. If what is inside the square root is negative we can’t plot the solution on the real number line.
This means that if b^2 < 4ac then b^2 - 4ac will be less than zero and therefore negative and its square root will be imaginary. If b^2 = 4ac then b^2 - 4ac = 0 and the square root of zero is zero which means the only real solution is -b/2a.
If b^2 > 4ac then b^2 - 4ac will be a positive number and so its square root will be real and there will be two solutions to the quadratic equation.
Thus by looking at the bit under the square root of the quadratic equation (which we call the discriminant) we can figure out whether there are no real solutions, one real solution or two real solutions to any quadratic equation in the form ax^2 + bx + c = 0.
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Nikola Pavkovic
Studied at A Regular High School.
·4y
Related
Why are there usually two solutions to a quadratic equation?
Well, let’s look at a quadratic very quickly in a graphical representation:
This is y=x ²+2 x−3 y=x ²+2 x−3
Let’s consider x ²+2 x−3=0 x ²+2 x−3=0 very quickly.
Factorize and we get: (x+3)(x−1)=0(x+3)(x−1)=0 Two solutions: x=−3,x=1 x=−3,x=1
But look at yonder parabola and ye will see:
Our solutions correspond to the intercepts on the x-axis. Because, consider what we just did, we solved for “which x coordinates of this equation have a y position at zero?”. And for parabolas, that’s often two different x coordinates. However, parabolas have a vertex:
This is where you will find only one solution, if you solved for:
x ²+2 x−3=−4 x ²+2 x−3=−4
You would f
Continue Reading
Well, let’s look at a quadratic very quickly in a graphical representation:
This is y=x ²+2 x−3 y=x ²+2 x−3
Let’s consider x ²+2 x−3=0 x ²+2 x−3=0 very quickly.
Factorize and we get: (x+3)(x−1)=0(x+3)(x−1)=0 Two solutions: x=−3,x=1 x=−3,x=1
But look at yonder parabola and ye will see:
Our solutions correspond to the intercepts on the x-axis. Because, consider what we just did, we solved for “which x coordinates of this equation have a y position at zero?”. And for parabolas, that’s often two different x coordinates. However, parabolas have a vertex:
This is where you will find only one solution, if you solved for:
x ²+2 x−3=−4 x ²+2 x−3=−4
You would find one answer: -1.
As a sidenote:
If you tried to solve for x ²+2 x−3=−5 x ²+2 x−3=−5
You would delve into the mysterious and confusing world of “imaginary” and complex numbers, the result of square root of a negative number, and a topic I’m not familiar with.
Hope this helps!
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Related questions
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3.S: Interference (Summary)
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3.E: Interference (Exercises)
4: Diffraction
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OpenStax
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Key Terms
| | |
--- |
| coherent waves | waves are in phase or have a definite phase relationship |
| fringes | bright and dark patterns of interference |
| incoherent | waves have random phase relationships |
| interferometer | instrument that uses interference of waves to make measurements |
| monochromatic | light composed of one wavelength only |
| Newton’s rings | circular interference pattern created by interference between the light reflected off two surfaces as a result of a slight gap between them |
| order | integer m used in the equations for constructive and destructive interference for a double slit |
| principal maximum | brightest interference fringes seen with multiple slits |
| secondary maximum | bright interference fringes of intensity lower than the principal maxima |
| thin-film interference | interference between light reflected from different surfaces of a thin film |
Key Equations
| | |
--- |
| Constructive interference | , for m = 0, ±1, ±2, ±3… |
| Destructive interference | , for m = 0, ±1, ±2, ±3… |
| Path length difference for waves from two slits to a common point on a screen | |
| Constructive interference | , for m = 0, ±1, ±2, ±3… |
| Destructive interference | , for m = 0, ±1, ±2, ±3… |
| Distance from central maximum to the m-th bright fringe | |
| Displacement measured by a Michelson interferometer | |
Summary
3.1: Young's Double-Slit Interference
Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).
3.E: Interference (Exercises)
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Primality testing for 64 bit numbers
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For very small numbers, say 32 bits unsigned, testing all divisors up to the square root is a very decent approach. Some optimizations can be made to avoid trying all divisors, but these yield marginal improvements. The complexity remains O(n−−√).
On the other hand, much faster primality tests are available, but they are pretty sophisticated and deploy their efficiency for much longer numbers.
Is there an intermediate solution, i.e. a relatively simple algorithm, that is of practical use for, say, 64 bits unsigned, with a target running time under 1 ms ?
I am not after micro-optimization of the exhaustive division method. I am after a better working principle, of a reasonable complexity (and of the deterministic type).
Update:
Using a Python version of the Miller-Rabin test from Rosetta code, the time for the prime 264−59=18446744073709551557 is 0.7 ms. (Though this is not a sufficient test because nothing says we are in a worst case.)
And I guess that this code can be improved for speed.
primality-test
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edited Jan 30, 2018 at 7:27
asked Oct 20, 2017 at 8:22
user65203user65203
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You can build a sieve filtering out multiples of any primes you have already tried as you go. But it will of course require more memory to keep track of.
– mathreadler
Commented
Oct 20, 2017 at 8:37
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@Zubzub: for a prime close to 2^64, the function will try close to 2^32 divisions, so no. (By the way, for 18446744073709551557 it takes six minutes).
– user65203
Commented
Oct 20, 2017 at 8:38
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@mathreadler: I doubt this is usable for 64 bit integers. And if I am right, the gain will not exceed a factor log(n), not counting overhead.
– user65203
Commented
Oct 20, 2017 at 8:40
1
1 second is absolutely no proeblem. Even 100-digit numbers can be proven to be prime with PARI/GP using the Adleman-Pomerance-Rumely-test within about 200 milliseconds. But 1ms ? Perhaps the BPSW-test is a good idea, which is correct upto at least 264. Trial division makes only sense, if we want to test many numbers, to reduce the number of candidates, but verifying 64-bit numbers via trial division actually is not efficient. But checking the first few primes (lets say upto 100) could slightly improve the test.
– Peter
Commented
Oct 20, 2017 at 10:51
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A deterministic implementation with a maximum of 7 base tests, here.
– Brett Hale
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Sep 5, 2018 at 9:33
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I think some version of the (deterministic) Miller-Rabin Test should do the trick. It is usually used as a probabilistic test for whopping big numbers, but it can be repurposed as a deterministic test for smaller fixed ranges.
The core of the Miller-Rabin test is the notion of a "probable prime" for some base a. Specifically, let n be an odd prime, and write n−1=d×2s for some odd d. Then, it follows that ad≡1(modn) or ad2r≡−1(modn) for some 0≤r<s. (The wikipedia page has reasoning for this).
If n is any number, and a<n, we could run the same test, and if that test passed we would call n a strong pseudoprime for base a. The usual Miller-Rabin test is based on doing this for a lot of different (randomly chosen) a, and some number-theoretic argument saying that if n is composite, the probability of not finding an a demonstrating this is vanishingly small (after trying a lot of them).
However, if Wikipedia is to be believed (I haven't followed up the reference), then for n<264 it is sufficient to test a∈{2,3,5,7,11,13,17,19,23,29,31,37}. Somehow there are no composite numbers below 264 which are a strong pseudoprime for all these a.
The above test is very fast, and altogether prime testing would require at most 12×64 modular exponentiations (this will be well below 1ms). For smaller n, you could even use a smaller list of possible a's.
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answered Oct 20, 2017 at 10:40
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I guess that this is exactly what I was looking for: nice tradeoff between complexity and efficiency.
– user65203
Commented
Oct 20, 2017 at 11:01
@YvesDaoust For so small numbers, the adleman-pomerance-rumely-test is not much slower (even with the slow program PARI/GP on my also slow computer 64-bit digit-.numbers require less than 1 millisecond for a decision.
– Peter
Commented
Oct 20, 2017 at 11:12
@Peter: The only description of that algorithm I can find goes for pages and pages though. It doesn't look "relatively simple". The algorithm I described I can implement in about 20 lines of Python.
– Joppy
Commented
Oct 20, 2017 at 11:24
3
@JiK: Well, I meant reasonably. In another language like C, you might have to write your own modular exponentiation, but that's about it.
– Joppy
Commented
Oct 20, 2017 at 11:58
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According to miller-rabin.appspot.com, it is sufficient to run the Miller–Rabin test for a∈{2,325,9375,28178,450775,9780504,1795265022}; that’s 7 trials instead of 12.
– Emil Jeřábek
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oeis/A014233 says:
The primality of numbers <264 can be determined by asserting strong
pseudoprimality to all prime bases ≤37.
The reference is the recent paper Strong pseudoprimes to twelve prime bases by Sorenson and Webster.
For code, see Prime64 and also the primes programs in FreeBSD, especially spsp.c.
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edited Oct 20, 2017 at 12:28
answered Oct 20, 2017 at 10:29
lhflhf
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This is true, but depends on previous work (i.e. somebody has already checked this works). Does that make it any different to downloading a list of primes?
– Henry
Commented
Oct 20, 2017 at 10:36
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@Henry, yes, because the list is too long. The OP only wants to test, not to list.
– lhf
Commented
Oct 20, 2017 at 10:37
@Henry: lhf is right. I am looking for an algorithm that avoids usage of big tables.
– user65203
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Oct 20, 2017 at 10:59
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Testing for 64-bit integers can be done deterministically in very short time using modifications of the Baille-PSW test or the Miller-Rabin test.
The fastest variant of the Miller-Rabin test involves constructing a list of bases that removes all pseudoprimes within the interval, this can take a large list of bases, 262,144 in the case of a test that uses two bases. A description of such a test is provided by Forisek and Jancina. Alternately one can use a slightly slower modification of the Baille-PSW test that does not require such a large set of bases. The worst case complexity of this test is approximately 2.5 fermat tests.
There is an highly optimised implementation of both of these algorithms in the machine-prime library that is in the U.S public domain so it can be easily copied or translated. Bindings are also provided to multiple languages, including Python, as copying is somewhat tricky and not necessarily even feasible (as many of the optimisations exploit finite ring arithmetic that CPU's use,but may not be available in interpreted languages).
Machine-prime easily satisfies the required speed even in the hardest case, on nearly any machine. On a moderate-end cpu at present day (2024), it is approximately 1 million times faster in the hardest case (2^64-59) than the algorithm in the post. (About 700K times faster for the Python binding)
(Disclaimer: I am the author of machine-prime).
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894 | https://www.youtube.com/watch?v=1zaoRUTkv7A | How to use YouGlish to improve your English Pronunciation - for FREE!
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hello everyone welcome back to Billy English where I help you to take your English skills to the next level today we're going to explore an amazing tool called yish if you're trying to improve your pronunciation or learn how native speakers use certain words this website will be your new best friend [Music] so what is yish and how can it help you let's take a look yish is a website that lets you search for any word or phrase and instantly find videos of native speakers using that word or phrase in real life situations it's a bit like having a personal tutor who can show you how words are actually used in everyday conversations and how they are pronounced so whether you're intermediate or Advanced it can help you improve your pronunciation and help you refine your vocabulary use now let's go through how to use uish step by step first open your browser and go to ug.com the website is simple and easy to navigate you will see a search bar right at the top of the page step one is to type a Word or phrase you want to learn more about for example let's say you want to learn how to pronounce the word entrepreneur just type it in and hit enter uish will now show you a list of videos when native speakers are using the word entrepreneur you can even see a transcript of what the speaker is saying underneath the search word is always highlighted in yellow step two is to choose a video and listen carefully what's really cool is that uish doesn't just give you one example it gives you many you can click through multiple videos to hear different people using the same word in various accents and contexts like a closing thought do you consider yourself an entrepreneur we're the young entrepreneur in Peru who teaching rural women digital skills and now I'm an entrepreneur I'm an investor but technology entrepreneur who gave an amazing presentation and within that statement is really what makes a great entrepreneur I'm interested in how the new entrepreneur either in that place and unique I think it was the massively successful entrepreneur Nal listen closely to how the speaker pronounces the word notice the intonation and stress and Rhythm they use in the sentence this is a great way to improve your pronunciation and sound more fluent and natural there's an additional tool I really like if the person speaks too fast you can slow down the speed of the video Let's listen to an example at normal speed and some studies suggest that conscientious speakers you and some studies suggest that conscientious speakers you and now again at a slowest and some studies suggest that conscientious speakers and some studies suggest that conscientious speakers I think this is a really helpful tool as well especially if you have a lower level of English and find it difficult to understand fast speech there's just one thing that you have to remember when you click play the video won't stop after you've heard the word it will continue playing so you can understand in which context the word was used to stop the video either click pause or click next to move to the next video step three is to practice alongside the video after you've listened to a few examples pause the video and try to repeat the word or phrase yourself uish also has a handy replay button so you can listen to the same part as many times as you need add it you will notice that Thomas Jefferson had taken a razor and scissors at it you will notice that Thomas Jefferson had taken a razor and scissors repeating after the speaker will help you get used to the Natural pronunciation and rhythm of English don't worry if it's difficult at first practice makes perfect you can search for words but also for short phrases or a combination of words for example native speakers often use does that make sense after they have said something to look for confirmation although most of the time the speaker is not really waiting for an answer from you and the sentence is also often said very fast and some sounds in the words get shortened let's look it up on yish and listen to some examples does that make sense we kind of just ask a lot of questions does that make sense kind of choice does that make sense yeah you know um does that make sense no it it does because so does that make sense to you at first so does that make sense to you at first does that make sense okay does that make sense okay does that make sense they separated does that make sense to everybody and one half went this dire they separated does that make sense to everybody did you notice how does that gets reduced quite often so it sounds more like one word rather than two separate words try repeating the phrase after each speaker you will notice that you will gain influency so you can see yish can also help you mimic English with features of connected speech but uish isn't just for pronunciation it's also an excellent tool for learning vocabulary in context let's say you come across an adjective like laidback and you want to see how it's used in real conversations type laidback into uish and watch how native speakers use it in different sentences it feels quite laidback and relaxed with plenty of sand for everyone to un on and at number one is flamco the city has wide open spaces and a laid-back atmosphere and there is an endless number of activities to keep visitors entertained to a laidback lifestyle and you lived we on Mo we lived in sprecklesville in a um old um plantation house that we were just renting it worked I mean um it's a pretty laid-back organization you know you wear you wear jeans and a t-shirt and and um you know people are pretty pretty geeky and nerdy at the NSA it's not like that in a lot of other agencies I worked for these examples help you understand the meaning of the phrase more deeply and how it fits into natural conversation for example after watching a few videos of laidback we can see that the adjective laidback is often used to describe an atmosphere in a place or the attitude of a person and in quite a Rel and informal setting finally step five is to customize your learning experience UGL lets you filter results by accent American British Australian and many more so you can focus on the accent that you're most interested in and there you have it uish is a powerful and free tool that can help you sound more natural in English and expand your vocabulary at the same time so give it a try and see how it can boost your learning thanks for watching and I'll see you in the next video |
895 | https://pmc.ncbi.nlm.nih.gov/articles/PMC7474787/ | Total Vitamin C, Ascorbic Acid, Dehydroascorbic Acid, Antioxidant Properties, and Iron Content of Underutilized and Commonly Consumed Fruits in Sri Lanka - PMC
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Total Vitamin C, Ascorbic Acid, Dehydroascorbic Acid, Antioxidant Properties, and Iron Content of Underutilized and Commonly Consumed Fruits in Sri Lanka
Hashini I Abeysuriya
Hashini I Abeysuriya
1 Department of Chemistry, Faculty of Science, University of Ruhuna, Matara 81000, Sri Lanka
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1, Vajira P Bulugahapitiya
Vajira P Bulugahapitiya
1 Department of Chemistry, Faculty of Science, University of Ruhuna, Matara 81000, Sri Lanka
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1,✉, Jayatissa Loku Pulukkuttige
Jayatissa Loku Pulukkuttige
2 Department of Botany, Faculty of Science, University of Ruhuna, Matara 81000, Sri Lanka
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1 Department of Chemistry, Faculty of Science, University of Ruhuna, Matara 81000, Sri Lanka
2 Department of Botany, Faculty of Science, University of Ruhuna, Matara 81000, Sri Lanka
Academic Editor: Mitsuru Yoshida
✉
Corresponding author.
Received 2020 Mar 7; Revised 2020 Jul 25; Accepted 2020 Jul 31; Collection date 2020.
Copyright © 2020 Hashini I. Abeysuriya et al.
This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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PMCID: PMC7474787 PMID: 32908860
Abstract
Sri Lanka is rich in a wide diversity of fruits, but many are underutilized by the people in Sri Lanka despite their nutritional value. This is mainly due to little awareness of the palatability of many fruits and hence low popularity in the market. The present study aimed at providing comparative data on the main biochemical and nutritional parameters of thirty-seven (37) species of fruits grown in Sri Lanka, including 22 underutilized fruits and 15 commonly consumed fruits. The main parameters of the comparison were the contents of ascorbic acid (AA), total vitamin C (TVC), total phenolic content (TPC), total flavonoid content (TFC), total iron (Fe), and antioxidant capacities (ACs). The mean AA, TVC, TPC, TFC, and Fe contents in 100 g of fresh edible portions of fruits ranged from 2.0 to 185.0 mg, 8.1 to 529.6 mg, 12.9 to 2701.7 mg gallic acid equivalent, 0.2 to 117.5 mg quercetin equivalents, and 0.1 to 1.1 mg, respectively. The IC 50 values in a DPPH assay varied between 0.8 to 1856.7 mg/mL and FRAP values in a FRAP assay ranged from 4.2 to 2070 μ mol FeSO 4/g in the studied fruits. Fruits were ranked based on the levels of the abovementioned biochemical properties. Using this ranking, 12 of the top 15 fruits were underutilized. Phyllanthus emblica (Indian gooseberry) is at the top of these underutilized fruits, and Psidium guajava (guava) is the best among commonly consumed fruits. These results indicate that underutilized fruits in Sri Lanka can be recommended as high quality and low-cost alternatives for securing nutritional requirements. Hence, underutilized fruits can be promoted as healthy additional fruits in Sri Lanka.
1. Introduction
Fruits and plant-based products have been used for food and medicinal purposes since the first human civilizations and indeed since the evolutionary origin of humans [1–3]. There is an emerging trend to consume more fruit on a regular basis, in response to the fact that fruits help to promote human health through supplying essential nutrients, improving immunity functions, and reducing the risk of many noncommunicable diseases (NCDs) such as cancers, diabetes mellitus, arthritis, Alzheimer's disease, Parkinson's disease in both developed, and developing countries .
Sri Lanka is a tropical country and a biodiversity hotspot with a wide array of fruits including underutilized and commonly consumed species. Though Sri Lanka has a long history of using fruits as food supplements, nutrient supplements, and for managing health, the fruits termed “underutilized fruits” are those that remain generally unrecognized with little market penetration or popularity within Sri Lanka. This is mainly due to less awareness of their nutritional value and their safety, which is linked to these fruits being less widely grown or commercially cultivated which restricts access to them. Underutilized fruits of Sri Lanka are poorly studied and appreciated. The few studies available, published by various local researchers, have highlighted, for example, the antioxidant capacity of selected fruits grown in Sri Lanka [3–13]. In addition, one recent study has been reported on vitamin C contents and in-vitro antioxidant activities of selected Sri Lankan fruits including some underutilized fruits by Silva and Sirasa .
The total vitamin C (TVC) is considered as the sum of ascorbic acid (AA) and dehydroascorbic acid (DHA). Dehydroascorbic acid (DHA) is the oxidized form of the ascorbate, and it has been found that DHA can be reduced reversibly into ascorbate in an enzymatic function and can be stored to increase ascorbate stores in the tissues of guinea pigs; a similar process can be anticipated in humans as well . Since fruits are rich with metabolizing enzymes, recycling of DHA can happen within the fruits, behaving similar to guinea pigs. Scientific studies have yet to be directed to that end. Vitamin C deficiency is a globally significant health issue, particularly in developing countries, with severe deficiency resulting in scurvy. Due to the many pharmacological activities associated with fruits researchers in natural product, chemistry have paid much attention to fruits as preventive measures for highly prevalent NCDs [16, 17]. It is well documented that the phytochemicals present in plant parts including fruits exhibit synergetic pharmacological effects to improve the human immune function which is directly correlated with a reduced risk of NCDs . A suggested cause for many NCDs is excessive oxidative stress in the cells resulting from an imbalance between the generation and the quenching of free radicals, namely, reactive oxygen (ROS) and reactive nitrogen (RNS) species in cells [1, 18, 19]. In that sense, the natural antioxidants such as polyphenols itself identified as nutraceuticals, and ascorbic acids present in the fruits may therefore act as nonenzymatic pathways to quench the harmful radicals and consequently to reduce the excess oxidative stress in cells . Iron deficiency is a very common health issue in the low and middle-income countries including Sri Lanka, and there is a known correlation between dietary iron absorption and ascorbic acid content [21, 22]. Therefore, knowing the data on iron content is equally important along with vitamin C and antioxidant properties of fruits.
.
Although Sri Lanka is a habitat for vast diversity of underutilized fruits, those are less popular among people and hence less market value despite their healthcare and nutritional value. Less availability of data on their nutritional value comparing to the commonly consumed fruits must be the main reason associated with this situation. This should be of great concern for making popular the healthier fruits at low cost. Therefore, this study was carried out with the prime objective of comparing important biochemical parameters which are known to be contributed to human health such as ascorbic acid (AA) content, dehydroascorbic acid (DHA) content, total vitamin C content (TVC), total phenolic content (TPC), total flavonoid content (TFC), total iron (Fe) content, and antioxidant capacities (ACs) of underutilized and commonly consumed fruits grown in Sri Lanka. The purpose is to explore the potential of promoting underutilized fruits among the Sri Lankan community as healthy additional and alternative sources of nutrition and to develop them as economical crops and products.
2. Materials and Methods
2.1. Chemicals
All the chemicals and reagents including methanol, L-ascorbic acid, 2,6-dichlorophenolindophenol (DCPIP), 2,4-dinitrophenylhydrazine (2,4-DNP), Folin-Ciocalteu's reagent, gallic acid, quercetin (≥95%), 2,2-diphenyl-1-picrylhydrazyl (DPPH), 2,4,6-tris(2-pyridyl)-s-triazine (≥98%) (TPTZ), and iodine used in this study were of analytical grade and purchased from Sigma-Aldrich.
2.2. Fruit Samples
Mature fruits from 37 locally grown fruit species (Table 1) were freshly harvested from home-gardens and commercial fruit growers in Sri Lanka. Harvesting or collecting of fruits was done by personal judgment based on experience of the farmer or fruit collector. Subjective criteria for assessing fruit maturity based on physical features of fruits were done. Characteristics of fruits such as skin colour, flesh colour, firmness of the skin, flesh, size, and shape of the fruit and sound when tapped were used. Fruits harvested from different mother plants were kept separately. Just after harvesting, samples were transported in ice boxes to the laboratory. The analyses of those fruit species which are generally harvested at edible maturity (e.g., guava) were started within 24 hours of harvesting, and the other fruit species (e.g., banana) were carried out after keeping them for natural ripening at ambient room temperature.
Table 1.
Harvested time, harvested location, and edible part(s) of fruit species used for this study.
| Scientific name | Common name(s) | Edible part(s)1 | Harvested location | Harvested month 4 |
:---: :---:
| Common/utilized fruits |
| |
| 1 . Ananas comosus (L.) Murr. | Pineapple (“Murusi”) | WPS | MT 2, SP 3 | March, 2018 |
| 2 . Carica papaya L. | Papaya (“red lady”) | WPS | MT, SP | March, 2018 |
| 3 . Citrullus lantanus (Thumb.) Matsum & Nakai | Watermelon | WPS | HB, SP | January, 2018 |
| 4 . Citrus aurantifolia (Christm. & Panzer) Swingle | Lime | WPS | HB, SP | March, 2017 |
| 5 . Citrus sinensis (L.) Osbeck | Sweet orange (“Bibila sweet”) | WPS | MN, UP | January, 2018 |
| 6 . Garcinia mangostana L. | Mangosteen | WPS | MT, SP | August, 2018 |
| 7 . Limonia acidissima L. | Wood apple | WP | MT, SP | October, 2017 |
| 8 . Mangifera indica L. | Mango | WPS | MT, SP | May, 2017 |
| 9. Musa paradisiaca L. AAB “Mysore” | Banana (“Embul”) | WP | MT, SP | November, 2017 |
| 10. Musa paradisiaca L.AAB, “Silk” | Banana (“Kolikuttu”) | WP | MT, SP | January, 2018 |
| 11. Nephelium lappaceum L. | Rambutan | WPS | CL, WP | June, 2018 |
| 12. Passiflora edulis Sims | Passion fruit | WP | HB, SP | November, 2017 |
| 13. Persea americana Miller. | Avocado | WPS | MT, SP | August, 2018 |
| 14. Psidium guajava L. | Guava (white flesh) | WF | HB, SP | Jun, 2017 |
| 15. Psidium guajava L. | Guava (pink flesh) | WF | GL, SP | May, 2018 |
| |
| Underutilized fruits |
| |
| 16. Aegle marmelos (L.) Correa | Bael fruit | WPS | HB, SP | March, 2017 |
| 17. Anacardium occidentale L. | Cashew apple (red) | WS | MT, SP | April, 2018 |
| 18. Anacardium occidentale L. | Cashew apple (yellow) | WS | MT, SP | April, 2018 |
| 19. Annona muricata L. | Sour sop | WPS | HB, SP | August, 2018 |
| 20. Artocarpus heterophyllus Lam. | Ripened jack fruit (“Wela”, i.e., texture soft or loose when ripe) | WPS | MT, SP | June, 2018 |
| 21. Artocarpus heterophyllus Lam. | Ripened jack fruit (“Waraka”, i.e., texture somewhat hard when ripe) | WPS | MT, SP | July, 2018 |
| 22. Averrhoa bilimbi L. | Bilimbi | WF | MT, SP | March, 2017 |
| 23. Averrhoa carambola L. | Star fruit | WF | MT, SP | February, 2017 |
| 24. Baccaurea motleyana Mull.-Arg. | Lansone | WPS | GL, SP | July, 2018 |
| 25. Carissa carandas L. | Jamson | WS | MT, SP | April, 2018 |
| 26. Citrus aurantium L. | Sour orange | WPS | MN, UP | January, 2018 |
| 27. Cynometra cauliflora L. | Nam nam | WS | MT, SP | November, 2017 |
| 28. Dialium ovoideum Thw. | Velvet tamarind | WPS | HB, SP | Oct, 2017 |
| 29. Elaeocarpus serratus L. | Ceylon olive | WS | MT, SP | March, 2017 |
| 30. Flacourtia inermis Roxb. | Lovi-lovi, Sapida | WS | MT, SP | July, 2018 |
| 31. Phyllanthus emblica | Indian gooseberry, Amla | WS | HB, SP | October, 2017 |
| 32. Pouteria campechiana (Kunth) Baehni | Yellow sapote | WPS | MT, SP | July, 2018 |
| 33. Punica granatum L. | Pomegranate (local variety) | WPS | HB, SP | September, 2018 |
| 34. Sandoricum koetjape (Burm. f.) Merr. | Santol/Cotton fruit | WPS | GL, SP | July, 2018 |
| 35. Sonneratia caseolaris L. | Mangrove apple | WF | MT, SP | June, 2018 |
| 36. Spondias dulcis Sol. ex Parkinson | Jamaica plum | WPS | HB, SP | February, 2017 |
| 37. Syzygium jambos (L.) Alston | Rose apple (“Malaysian”) | WS | MT, SP | March, 2017 |
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1 Edible part(s) of the fruit: WPS: without peel and seed(s); WP: without peel; WS: without seed(s); WF: whole fruit; 2 Fruits harvested district in Sri Lanka: MT: Matara; HB: Hambantota; CL: Colombo; MN: Monaragala; GL: Galle; 3 Fruits harvested province in Sri Lanka: SP: Southern province; WP: Western province; UP: Uva province.
2.3. Preparation of Fruit Samples
After sorting, the fruits were first washed with tap water and secondly with distilled water and then wiped with tissues to dryness. The edible mass of three fruit samples harvested from three different mother plants of the same species were pooled together and a known weight from that edible mass was used to prepare an extract in triplicate. In cases where the fruit is generally consumed as a fruit drink (e.g., C. sinensis, S. caseolaris), a known weight of juice/pulp was extracted and used for analysis.
2.4. Extraction of Vitamin C
Extraction of Vitamin C from plant materials was done according to the reported method (AOAC method 967.21, 45.1.14) (2, 9) as explained by Nielsen . Vitamin C was extracted into a freshly prepared solution containing 3% (w/v) meta-phosphoric acid and 8% (v/v) glacial acetic acid. In this method, the groundfruit sample was passed through a muslin cloth and filtrate was collected. The process was done three times and finally collected filtrate was made up to 100 mL with the meta-phosphoric acid-acetic acid solution. The extracts were stored at –10°C until its use for analysis of vitamin C. The extracts were protected from light by covering them with aluminum foil, and all appropriate measures were taken to prevent the loss of ascorbic acid during the extraction process and storage.
2.5. Preparation of Methanolic Extract of Fruits to Be Used in the Analysis
Methanolic extracts of fruits were prepared according to the method described by Ikram et al. with slight modifications . The definite amount of fruits was weighed from the homogenized or ground sample of the edible mass of the each fruit species and mixed with 10 mL of 80% (v/v) of methanol and stirred at 1500 rpm using “AREC Heating Magnetic Stirrer (VELP SCIENTIFICA®)” for an hour. Then, the mixture was centrifuged for 10 min at 5000 rpm (PLC-012E Universal Centrifuge), and supernatant was collected. The extraction process was repeated with the same sample of fruits to make replicates, and the volume was made up to 50 mL. The extracts were stored at –10°C until the analysis was done. These extracts were used for the determination of TPC, TFC, 2,2-diphenyl-1-picrylhydrazyl (DPPH) scavenging activity, and ferric reducing antioxidant power (FRAP) value of fruits.
2.6. Sample Analysis
2.6.1. Determination of Total Vitamin C Content (TVC)
Although most past researchers have used the DCPIP titrimetric method to determine the vitamin C content, it is successful only for the ascorbic acid form of TVC and it is also limited to the coloured fruit extract. Due to such limitations in the DCPIP titrimetric method, five other different methods were tested to select the most suitable method for the determination of vitamin C content of fruits in this study. Out of them, we used a slightly modified method to that elaborated by Ranganna adopting from Roe and Oesterling . In this method, 0.5 mL of 3% bromine water (v/v) was added into 8 mL of sample extract (which is further diluted if necessary) in order to oxidize ascorbic acid into dehydroascorbic acid, and 0.25 mL of 10% thiourea solution (w/v) was added to remove excess bromine. Then, 2 mL of 2,4-dinitrophenylhydrazine (2,4-DNP) solution (2 g of 2,4-DNP and 4 g of thiourea in 100 mL 4.5 M H 2 SO 4) was added, and all samples, standards, and blank were kept at 37°C for 3 hours in a thermostatic bath. After cooling in an ice bath for 30 minutes, samples were treated with 10 mL chilled 85% sulfuric acid (v/v) with constant stirring. Absorbance was measured at 520 nm using a spectrophotometer (HITACHI UH5300 Spectrophotometer). TVC content of each sample was determined as mg per 100 g of fresh weight of the fruit, using a standard curve prepared with L-ascorbic acid, the standard (0.005–0.025 mg/mL).
2.6.2. Determination of Ascorbic Acid (AA) Content
Two methods were used for the determination of AA. In the first method, a titrimetric method described by Suntornsuk et al. was used with slight modifications. Here, an aliquot of 10 mL of sample extract was titrated against iodine (0.005 mol L–1) solution, containing 25 mL of 2 N H 2 SO 4, using a 1% starch (w/v) as the indicator. Iodine solution was previously standardized using 5 mL of L-ascorbic acid solution (1 mg mL–1). In the second method, the AOAC's official titrimetric method (AOAC method 967.21, 45.1.14) (2, 9) was used, as explained by Nielsen . Accordingly, a 5 mL aliquot of extract was titrated with DCPIP reagent until a light but distinct rose pink colour appears and persists for more than 5 seconds. Each analysis was performed in triplicate, and ascorbic acid was expressed as mg of L-ascorbic acid equivalents (AAE) per 100 g of fresh weight of the fruit. In both methods, the iodine and DCPIP solutions were standardized daily with a standard L-AA solution (1 mg mL–1). The iodine titration method was applied to all fruit extracts, whereas the DCPIP method was not applied to fruit extracts with a colour (C. carandas, F. inermis, and P. granatum).
2.6.3. Determination of Dehydroascorbic Acid (DHA) Content
DHA content was calculated by subtracting mean AA content by TVC content, and percentage DHA content was determined compared to TVC content.
2.6.4. Determination of Total Phenolic Content (TPC)
The TPC contents of the extracts were determined using the Folin-Ciocalteu's (FC) reagent . Properly diluted fruit extracts (0.5 mL, in triplicate) were stand for 5 min after adding 2.5 mL of FC reagent (10%), followed by addition of 2 mL of Na 2 CO 3 (7.5% w/v). The samples were kept in the dark for 30 min, and absorbance was measured at 765 nm. Total phenol content was estimated from a standard curve of gallic acid (0.02–0.1 mg/mL), and TPC contents of fruits were expressed in gallic acid equivalents (GAE) (mg per 100 g of fresh fruit).
2.6.5. Determination of Total Flavonoid Content (TFC)
Aluminum chloride colorimetric method was used to determine the TFC of fruit extracts . Briefly, 1 mL of the fruit extract was mixed with 3 mL of methanol, 0.2 mL of 10% AlCl 3 (w/v), 0.2 mL of 1 M potassium acetate, and 5.6 mL of distilled water and stand in the dark for 30 min. The absorbance was measured at 420 nm. TFC of each fruit extract was determined using a standard curve prepared for quercetin (0.01–0.1 mg/mL) and expressed as mg quercetin equivalents (QE) per 100 g of fresh fruit.
2.6.6. DPPH· (2,2′-Diphenyl-1-Picrylhydrazyl Radical) Radical Scavenging Assay
The free radical scavenging activity of fruit extracts was determined by the modified DPPH method . The DPPH solution in methanol (0.06 mM, 3.9 mL) was mixed with 100 μ L of fruit extract at different concentrations. The samples were kept in the dark for 30 min, and absorbance was measured at 517 nm. The free radical scavenging activity was expressed as IC 50 value, calculated using % disappearance vs. concentration plot (concentration means the mg of fruit extract into 1 mL of solution). The % disappearance was calculated from [(A control–A sample)/A control] × 100. A control is the absorbance of DPPH without extract, and A sample is the absorbance of the DPPH solution containing plant extract.
2.6.7. Ferric Reducing Antioxidant Power (FRAP) Assay
The ferric reducing power of fruit extracts was determined using a modified version of the FRAP assay as originally reported by Benzie and Strain . Three milliliters of freshly prepared FRAP reagent (300 mM acetate buffer (pH 3.6): 10 mM TPTZ (in 40 mM HCl): 20 mM FeCl 3.6H 2 O in 10 : 1 : 1 ratio) was mixed with 100 μ L of diluted sample and absorbance at 593 nm was recorded after 30 min incubation at 37°C. An aqueous solution of FeSO 4.7H 2 O (100–1200 mM) was used for calibration.
2.6.8. Determination of Total Iron Content (Fe)
(1) Sample Preparation. Total iron content was determined for dry digested samples of fruits. Accurately weighed 10–20 g of the edible part of the fruit was first dried in an oven at 100°C and then kept at 450°C in a muffle furnace (Yamato FM-36) for 8 hours or overnight until get ash with white/gray colour. The residue in the crucible was treated with 5 mL of 6 M HCl, and then evaporated on a hot plate. The remaining content was dissolved in 15 mL of 0.1 M HNO 3 and covered with a watch glass and let stand for 1–2 h. The solution was stirred with a glass rod and then the content was transferred to a volumetric flask and volume was made up to 25 mL with 0.1 M HNO 3 . These extracts were stored at 4°C and used for the total iron determination.
(2) Total Iron Determination. The total iron contents of fruit samples were determined by converting iron to the ferric form using an oxidizing agent, potassium persulphate, and treating with potassium thiocyanate to form red coloured ferric thiocyanate, as elaborated by Ranganna . Five milliliters from the fruit extraction was mixed with 0.5 mL of Conc. H 2 SO 4, 1.0 mL of K 2 S 2 O 8 (saturated), and 2 mL of 3N KSCN, and then the volume was made up to 15 mL with deionized water. The absorbance of this solution was measured at 480 nm soon after mixing. Iron standards, ranging from 5 to 25 mg/L, were used for the calibration curve.
2.7. Statistical Analysis
One-way analysis of variance (ANOVA) and Tukey's post-hoc tests were used to evaluate the significant differences (p< 0.05) of the means between different fruits. The dependent variables considered are TVC, mean AA, TPC, TFC, antiradical power (ARP), FRAP value, and Fe content, and the independent variable was fruit type. Principal component analysis (PCA) was carried out using the IBM SPSS 25.0 statistical software package for Windows (SPSS Inc., Chicago, USA). PCA was performed to classify and discriminate between fruits. In PCA, DPPH radical scavenging activity data were fed as antiradical power (ARP). ARP is reciprocal of IC 50 (ARP = 1/IC 50).
3. Results and Discussion
3.1. Total Vitamin C, Ascorbic Acid, and Dehydroascorbic Acid Contents
3.1.1. Total Vitamin C (TVC) Content
Vitamin C can be defined as the generic term for all compounds exhibiting equivalent biological activity of L-ascorbic acid (AA) and Dehydroascorbic acid (DHA). TVC contents for the studied fruits are given in Table 2. According to that, TVC contents of the fruits tested varied from 8.1 to 529.6 mg/100 g fresh weight (FW). Among the 37 species of fruits studied, the highest TVC content was observed in P. emblica followed by A. marmelos and apple of A. occidentale. The lowest TVC was reported in S. jambos. Among the commonly consumed fruits, P. guajava (76.2 mg/100 g) had the highest TVC content followed by C. papaya (73.2 mg/100 g) and C. sinensis (53.6 mg/100 g). Among the underutilized fruits, the TVC content in P. emblica was more than 10 times higher than that in P. guajava (white). Of the common fruits studied, G. mangostana had the lowest TVC content (10.3 mg/100 g). Though total vitamin C content is highly significant when discussing vitamin C content, many previous researchers have not reported TVC, focusing instead on AA in most of studies [14, 33–35].
Table 2.
Contents of total Vitamin C (TVC), ascorbic acid (AA), dehydroascorbic acid (DHA), and total iron in common and underutilized fruits of Sri Lanka.
| Fruits | TVC (mg/100 g FW) | AA-I 2 (mg AAE/100 g FW) | AA-DCPIP (mg AAE/100 g FW) | DHA (mg/100 g FW) | Classification by TVC contents | Fe (mg/100 g FW) |
:---: :---: :---:
| Common fruits | | |
| |
| G. mangostana | 10.3 ± 1.1 b | 7.8 ± 0.2 defg | 7.5 ± 0.5 ef | 2.6 | Low (<50 mg/100 g FW) | 0.2 ± 0.1 |
| P. edulis | 12.0 ± 0.5 bc | 10.1 ± 0.2 ghi | 10.9 ± 0.1 hi | 1.5 | 0.3 ± 0.2 |
| L. acidissima | 12.3 ± 0.2 cd | 9.4 ± 1.0 fgh | 9.0 ± 0.2 fg | 3.1 | 0.4 ± 0.1 |
| C. lantanus | 12.4 ± 0.8 cde | 2.8 ± 0.5 a | 3.1 ± 0.4 b | 9.5 | 0.2 ± 0.1 |
| P. americana | 14.3 ± 0.6 def | 5.0 ± 0.8 bc | 4.8 ± 0.8 d | 9.4 | 0.2 ± 0.1 |
| M. paradisiaca AAB, “Silk” | 15.1 ± 2.5 fg | 2.3 ± 0.5 a | 1.7 ± 0.2 a | 13.1 | 0.2 ± 0.1 |
| M. paradisiaca AAB, “Mysore” | 17.0 ± 0.4 gh | 2.2 ± 0.5 a | 1.8 ± 0.1 a | 15.0 | 0.4 ± 0.1 |
| P. guajava (pink flesh) | 23.3 ± 0.7 i | 11.0 ± 0.2 hi | 10.0 ± 0.3 gh | 12.8 | 0.3 ± 0.1 |
| A. comosus | 31.2 ± 2.4 k | 15.1 ± 0.3 jkl | 12.2 ± 0.4 i | 17.6 | 0.3 ± 0.1 |
| C. aurantifolia | 33.3 ± 0.4 kl | 21.1 ± 0.9 mn | 31.1 ± 1.1 mn | 7.2 | 0.3 ± 0.2 |
| M. indica | 36.8 ± 0.4 l | 30.8 ± 0.4 op | 28.8 ± 0.2 m | 7.1 | 0.2 ± 0.1 |
| N. lappaceum | 49.4 ± 0.6 mn | 18.5 ± 0.2 klm | 11.3 ± 0.3 hi | 34.5 | 0.2 ± 0.1 |
| |
| C. sinensis | 53.6 ± 1.2 n | 48.6 ± 1.4 q | 43.9 ± 0.2 o | 7.4 | Medium (50–100 mg/100 g FW) | 0.1 ± 0.1 |
| C. papaya | 73.2 ± 1.6 o | 69.5 ± 1.7 r | 64.9 ± 1.8 p | 6.0 | 0.3 ± 0.0 |
| P. guajava (white flesh) | 76.2 ± 0.7 o | 68.8 ± 1.0 r | 70.3 ± 0.3 p | 6.7 | 1.1 ± 0.1 |
| |
| Underutilized fruits |
| |
| S. jambos | 8.1 ± 0.4 a | 6.1 ± 0.3 cd | 5.0 ± 0.2 d | 2.6 | Low (<50 mg/100 g FW) | 0.1 ± 0.1 |
| S. caseolaris | 8.4 ± 0.7 a | 7.1 ± 0.4 de | 7.2 ± 0.3 e | 1.2 | 0.5 ± 0.1 |
| A. bilimbi | 11.8 ± 0.2 bc | 8.6 ± 0.3 efgh | 7.7 ± 0.4 ef | 3.7 | 0.2 ± 0.1 |
| B. motleyana | 14.5 ± 0.5 efg | 4.3 ± 0.6 b | 3.9 ± 0.4 c | 10.4 | 0.2 ± 0.1 |
| F. inermis | 18.3 ± 0.5 h | 13.1 ± 0.4 ij | NA | 5.3 | 0.1 ± 0.0 |
| A. heterophyllus (Wela) | 18.7 ± 1.1 h | 4.8 ± 0.3 bc | 4.5 ± 0.3 cd | 14.1 | 0.3 ± 0.1 |
| P. granatum | 23.6 ± 0.5 i | 14.5 ± 0.5 jk | NA | 9.1 | 0.9 ± 0.1 |
| C. carandas | 25.3 ± 0.9 i | 24.4 ± 0.3 no | NA | 0.9 | 0.2 ± 0.1 |
| A. carambola | 25.5 ± 0.5 ij | 18.8 ± 0.3 klm | 16.0 ± 0.7 jk | 8.1 | 0.2 ± 0.1 |
| S. koetjape | 25.8 ± 0.3 ij | 7.7 ± 0.2 def | 5.3 ± 0.4 d | 19.3 | 0.4 ± 0.1 |
| A. muricata | 30.0 ± 1.2 jk | 19.6 ± 0.8 lmn | 18.2 ± 0.5 kl | 11.2 | 0.2 ± 0.0 |
| D. ovoideum | 32.9 ± 1.8 kl | 20.3 ± 2.1 mn | 17.5 ± 2.2 jkl | 14.0 | 0.9 ± 0.1 |
| C. aurantium | 34.2 ± 0.3 kl | 20.2 ± 0.3 mn | 19.9 ± 0.2 l | 14.2 | 0.5 ± 0.1 |
| A. heterophyllus (Waraka) | 34.4 ± 1.4 kl | 16.3 ± 1.3 jklm | 15.0 ± 0.3 j | 18.8 | 0.3 ± 0.1 |
| C. cauliflora | 37.9 ± 1.8 l | 31.3 ± 0.2 op | 29.2 ± 1.1 m | 7.7 | 0.4 ± 0.1 |
| P. campechiana | 44.8 ± 2.2 m | 32.1 ± 2.0 p | 29.6 ± 0.5 m | 14.0 | 0.6 ± 0.2 |
| E. serratus | 45.3 ± 0.3 m | 38.0 ± 0.4 pq | 37.0 ± 0.3 no | 7.9 | 0.5 ± 0.1 |
| |
| S. dulcis | 51.2 ± 0.1 mn | 32.0 ± 0.5 p | 29.6 ± 1.5 m | 20.5 | Medium (50–100 mg/100 g FW) | 0.2 ± 0.1 |
| |
| A. occidentale (yellow) | 202.3 ± 2.9 p | 189.2 ± 0.9 t | 180.8 ± 0.2 r | 17.3 | High (>100 mg/100 g FW) | 0.9 ± 0.3 |
| A. occidentale (red) | 203.3 ± 7.2 p | 188.3 ± 11.9 t | 181.3 ± 5.6 r | 18.5 | 0.2 ± 0.1 |
| A. marmelos | 516.6 ± 0.5 q | 31.1 ± 0.7 op | 29.7 ± 0.3 m | 484.8 | 0.3 ± 0.1 |
| P. emblica | 529.6 ± 57.5 q | 96.5 ± 3.8 s | 89.3 ± 2.1 q | 436.7 | 0.7 ± 0.3 |
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Means with different superscript letters in individual column are significantly (p< 0.05) different from each other. Data are expressed as Mean ± Standard deviation (n = 3). TVC: total vitamin C content; AA-I 2: L-ascorbic acid content determined by Iodine titration method; AA-DCPIP: L-ascorbic acid content determined by DCPIP titration; DHA: dehydroascorbic acid content; AAE: L-ascorbic acid equivalents; FW: fresh weight; NA: not applicable.
3.1.2. Ascorbic Acid (AA) Content
In the current study, two methods were applied for the determination of AA to get more accurate results as individual methods have their own limitations such as masking of colour change at the endpoint in fruit extracts with intense fruit colours and interferences of some naturally available substances during the titration, etc. It is difficult to choose a unique method to determine the vitamin C content in all fruits . Although many of past researchers have used only the DCPIP titrimetric method to express the vitamin C content [14, 33, 35], it has some limitations like only the AA form of TVC is measured and difficulties to perform with coloured fruit extracts . Ascorbic acid contents of fruits in this study are given in Table 2, using both I 2 titration method and the DCPIP titration method. The values varied between 2.2-189.2 and 1.7-181.3 mg AAE/100 g FW in I 2 method and DCPIP method, respectively. The highest AA content was given for A. occidentale, and it is the only fruit that had more than 100 mg of AA per 100 g out of all studied fruits. The most commonly consumed banana had the lowest AA (nearly 2.0 mg/100 g) content. Most of the fruits had lower AA contents, i.e., less than 50 mg/100 g, and only 4 species had AA contents higher than 50 mg/100 g. Out of commonly consumed fruits, P. guajava (white) and C. papaya had the highest and second-highest AA contents. Although citrus fruits are well known as a rich source of vitamin C, P. emblica, A. occidentale, C. papaya, and P. guajava (white) showed significantly higher AA and TVC contents compared to C. sinensis.
3.1.3. Dehydroascorbic Acid (DHA) Content
Ascorbic acid (AA) is the main biologically active vitamin C among different forms of vitamin C, and it is reversibly oxidized into DHA, which also involves in biological activities responsible to maintain human health. DHA content of the fruits studied is given in Table 2, and the values are less than 35 mg/100 g except for A. marmelos and P. emblica which showed 484.8 and 436.7 mg/100 g DHA contents, respectively. The lowest DHA content was observed in C. carandas (0.9 mg/100 g). Hernandez, Lobo and Gonzalez has reported lower DHA amounts, than the values obtained in this study for the fruits of C. sinensis (2.32 mg/100 g), M. acuminata (0.61 mg/100 g), and A. comosus (0.36 mg/100 g), but similar values received for C. papaya (5.32 mg/100 g) and M. indica (5.73 mg/100 g). Gil, Aguayo and Kader also reported similar DHA content for M. indica (about 6 mg/100 g) but contrasting results for A. comosus and C. lantanus. As noted previously, DHA can be converted into reduced form, ascorbate under enzymatic reactions, therefore, presence of higher amount DHA is important to serve as reservoir of ascorbic acid through reversible transformation for the continuous supply and DHA itself plays role in biochemical functions . However, many of previous researchers have not considered both AA and DHA contents when reporting vitamin C contents [34, 37, 38], and our study get highlighted for first reporting of it for Sri Lankan fruits.
3.2. Total Phenolic Content, Total Flavonoid Content, and Antioxidant Capacities (ACs)
3.2.1. Total Phenolic Content (TPC)
TPC of studied fruits (Table 3) varied in a broad range as 12.9-2701.7 mg GAE/100 g of FW, and P. emblica and C. lantanus show the highest and the lowest TPC, respectively. Of common fruits, P. guajava (white) (180.6 mg/100 g) showed the highest TPC. Only P. emblica, A. marmelos, and D. ovoideum gave TPC higher than 500 mg/100 g. All the fruits that have very high TPC are underutilized fruits, and the common fruits only have moderate to low TPCs. Polyphenolic compounds in plant including fruits are best known to act as powerful antioxidants and responsible with many pharmacological activities exerted in plants/fruits such as anti-inflammatory and antiglycemic properties .
Table 3.
Total phenolic content (TPC), total flavonoid content (TFC), of common and underutilized fruits in Sri Lanka.
| Fruits | TPC (mg GAE/100 g FW) | TFC (mg QE/100 g FW) | Classification by TPC's |
:---: :---: |
| Common fruits | | | |
| |
| C. Lantanus | 12.9 ± 0.2 a | 0.2 ± 0.1 a | Low (<100 mg GAE/100 g FW) |
| M. paradisiaca AAB, “Silk” | 24.6 ± 0.5 b | ND |
| G. mangostana | 26.4 ± 0.5 bc | 15.0 ± 0.3 fg |
| N. lappaceum | 29.9 ± 0.3 cd | 20.7 ± 0.3 hij |
| A. comosus | 31.3 ± 0.6 d | 14.5 ± 0.5 fg |
| C. aurantifolia | 49.9 ± 0.9 efg | 10.6 ± 0.5 de |
| C. papaya | 57.4 ± 1.1 hi | 17.3 ± 0.4 gh |
| C. sinensis | 66.8 ± 1.4 jk | 13.2 ± 0.2 ef |
| P. americana | 81.2 ± 1.1 lm | 20.1 ± 0.5 hij |
| M. paradisiaca AAB “Mysore” | 92.0 ± 1.9 no | 28.4 ± 0.5 klm |
| P. edulis | 93.5 ± 1.8 no | 50.1 ± 0.2 pqr |
| |
| L. acidissima | 103.1 ± 6.3 o | 2.4 ± 0.6 b | Medium (100–500 mg GAE/100 g FW) |
| M. indica | 103.8 ± 15.4 o | 62.2 ± 2.8 rs |
| P. guajava (pink flesh) | 120.6 ± 1.3 p | 43.8 ± 3.5 opq |
| P. guajava (white flesh) | 180.6 ± 4.3 s | 92.0 ± 0.3 tu |
| |
| Underutilized fruits | | | |
| |
| S. koetjape | 32.7 ± 2.0 d | 14.7 ± 0.4 fg | Low (<100 mg GAE/100 g FW) |
| S. dulcis | 44.7 ± 2.8 e | 23.8 ± 0.8 ijk |
| B. motleyana | 46.2 ± 0.2 e | 17.4 ± 0.5 gh |
| A. carambola | 48.1 ± 1.2 ef | 39.2 ± 1.0 nop |
| A. bilimbi | 53.9 ± 1.4 fgh | 29.4 ± 0.6 klm |
| S. jambos | 55.3 ± 1.9 gh | 8.4 ± 0.5 cd |
| C. aurantium | 58.0 ± 1.3 hi | 7.1 ± 0.2 c |
| A. occidentale (yellow) | 63.3 ± 0.8 ij | 56.0 ± 1.6 qrs |
| A. heterophyllus (Waraka) | 72.3 ± 0.6 kl | 36.8 ± 0.4 mno |
| A. muricata | 86.1 ± 0.9 mn | 28.4 ± 0.5 klm |
| |
| F. inermis | 117.1 ± 2.1 p | 32.6 ± 0.6 lmn | Medium (100 – 500 mg GAE/100 g FW) |
| P. campechiana | 144.6 ± 4.1 q | 113.3 ± 0.4 u |
| P. granatum | 150.3 ± 1.4 qr | 64.5 ± 2.2 rs |
| A. occidentale (red) | 153.8 ± 1.6 r | 117.5 ± 0.6 u |
| S. caseolaris | 164.6 ± 2.1 rs | 6.4 ± 0.4 c |
| C. carandas | 207.7 ± 2.6 t | 10.6 ± 0.5 de |
| E. serratus | 212.3 ± 1.3 t | 7.7 ± 0.8 c |
| A. heterophyllus (Wela) | 221.0 ± 3.9 t | 11.6 ± 0.7 ef |
| C. cauliflora | 428.5 ± 1.3 u | 26.1 ± 1.0 jkl |
| |
| D. ovoideum | 804.3 ± 61.1 v | 18.8 ± 1.3 ghi | High (>500 mg GAE/100 g FW) |
| A. marmelos | 1549.2 ± 16.1 w | 56.6 ± 0.4 qrs |
| P. emblica | 2701.7 ± 2.9 x | 73.9 ± 0.3 st |
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Means with different superscript letters in individual columns are significantly (p< 0.05) different from each other. Data are expressed as Mean ± Standard deviation (n = 3). TPC: total phenolic content; TFC: total flavonoid content; FW: fresh weight; GAE: gallic acid equivalents; QE: Quercetin equivalents; FW: fresh weight; ND: not detected.
TPCs of fruits grown in Sri Lanka reported by Silva and Sirasa showed similar findings to this study except for gooseberry, star fruit, and pomegranate. However, they have pointed out that the TPC values reported can be higher than the actual values, as they have not used any correction factor for the interfering substances in TPC determination. The TPC values reported by Ellong et al. are comparable with the results in this study except that they have obtained considerably higher TPCs for cashew apple, star fruit, lime, and guava compared.
3.2.2. Total Flavonoid Content (TFC)
The values of TFC are given in Table 3, and they varied between 0.2 to 117.5 mg QE/100 g of FW. The fruits C. lantanus and A. occidentale (red) showed the lowest and the highest TFCs, respectively. The second highest TFC showed P. campechiana, followed by P. guajava (white), P. emblica, P. granatum, and M. indica. The TFC values higher than 100 mg/100 g could be observed only in P. campechiana and A. occidentale (red) which are underutilized fruits, while most of the common fruits had low TFCs (<50 mg/100 g). Flavonoids are very diverse compounds with vast structural diversity as well as great diversity in pharmacological activities such as antioxidant effect and inhibition of cell proliferation .
3.2.3. DPPH (2,2′-Diphenyl-1-Picrylhydrazyl Radical) Radical Scavenging Assay
DPPH assay is a widely used in-vitro antioxidant assay and based on the ability of DPPH, a stable free radical, to change its colour in the presence of antioxidants. This is a direct and reliable method for determining radical scavenging action of plant extracts. Original DPPH solution is purple colour, and it changed to yellow when plant secondary metabolites reduced it by donating electrons as hydrogen radical. As given in Table 4, IC 50 values of DPPH assay varied greatly in between 0.8 to 1856.7 mg/mL. According the results, the highest radical scavenging activity (as characterized by the lowest IC 50) was observed in E. serratus followed by P. emblica, A. occidentale (red), D. ovoideum, A. occidentale (yellow), C. cauliflora, and P. guajava (white), and interestingly except P. guajava (white) the others are underutilized fruits. Among the common fruits, only P. guajava (white) and M. indica, and L. acidissima and M. paradisiaca (embul) showed high radical scavenging activities. The lowest radical scavenging activity (as characterized by the highest IC 50) was observed in C. lantanus followed by P. americana and C. aurantifolia, and those are commonly consumed fruits. Therefore, these results evidence the greater free radical scavenging activity of locally grown underutilized fruits and comparatively lower activity for commonly consumed fruits.
Table 4.
DPPH assay of common and underutilized fruits in Sri Lanka.
| Fruits | DPPH-IC 50 (mg/mL) | Classification of AC measured by DPPH |
:---:
| Common fruits | | |
| |
| C. Lantanus | 1856.7 ± 51.3 a | Very low (IC 50> 500 mg/mL) |
| P. americana | 1244.3 ± 11.2 b |
| |
| C. aurantifolia | 467.0 ± 14.7 c | Low (IC 50 100–500 mg/mL) |
| G. mangostana | 465.0 ± 1.2 c |
| A. comosus | 453.3 ± 15.3 cd |
| P. guajava (pink flesh) | 385.7 ± 31.1 de |
| N. lappaceum | 347.7 ± 4.9 e |
| P. edulis | 230.7 ± 8.6 gh |
| M. paradisiaca AAB, “Silk” | 196.0 ± 15.1 hi |
| C. sinensis | 167.0 ± 11.1 i |
| C. papaya | 120.0 ± 10.0 j |
| |
| M. paradisiaca AAB “Mysore” | 94.8 ± 3.0 k | High (IC 50 20–100 mg/mL) |
| L. acidissima | 40.1 ± 0.5 no |
| |
| M. indica | 12.9 ± 2.1 r | Very high (IC 50< 20 mg/mL) |
| P. guajava (white flesh) | 9.8 ± 0.1 s |
| |
| Underutilized fruits |
| |
| S. koetjape | 526.5 ± 5.8 c | Very low (IC 50>500 mg/mL) |
| |
| B. motleyana | 276.3 ± 16.5 f | Low (IC 50 100–500 mg/mL) |
| C. aurantium | 236.6 ± 7.7 fg |
| S. jambos | 120.5 ± 16.6 gh |
| A. heterophyllus (Waraka) | 130.2 ± 10.0 j |
| |
| P. granatum | 94.4 ± 0.6 k | High (IC 50 20–100 mg/mL) |
| A. bilimbi | 93.1 ± 0.2 k |
| P. campechiana | 70.7 ± 5.5 l |
| A. muricata | 67.4 ± 0.3 l |
| A. heterophyllus (Wela) | 51.8 ± 3.5 m |
| F. inermis | 43.9 ± 2.0 mn |
| A. carambola | 43.9 ± 0.1 mn |
| A. marmelos | 39.1 ± 0.2 nop |
| S. dulcis | 35.1 ± 0.3 op |
| S. caseolaris | 33.0 ± 1.7 p |
| C. carandas | 26.4 ± 0.7 q |
| |
| C. cauliflora | 8.7 ± 0.3 s | Very high (IC 50< 20 mg/mL) |
| A. occidentale (yellow) | 8.4 ± 0.3 s |
| D. ovoideum | 6.6 ± 0.6 t |
| A. occidentale (red) | 4.4 ± 0.1 u |
| P. emblica | 1.0 ± 0.1 v |
| E. serratus | 0.8 ± 0.1 v |
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Means with different superscript letters in individual column are significantly (p< 0.05) different from each other. Data are expressed as Mean ± Standard deviation (n = 3). DPPH: 2,2-diphenyl-1-picrylhydrazyl; FW: fresh weight; AC: antioxidant capacity.
3.2.4. Ferric Reducing Antioxidant Power (FRAP) Assay
The FRAP assay treats the antioxidants contained in the samples as reductants in a redox-linked colorimetric reaction, and the value reflects the reducing power of the antioxidants, in which antioxidant reacts with Fe 3+-TPTZ to produce a coloured Fe 2+ TPTZ complex, which is measured at 593 nm . FRAP values of this study are given in Table 5, and the values range from 4.2 to 2070 μ mol FeSO 4/g FW. P. emblica and P. americana showed the highest FRAP values, indicating the highest AC. Second highest FRAP value was obtained for A. occidentale (red) followed by A. marmelos and A. occidentale (yellow). Except P. americana, all the other fruits are locally grown underutilized fruits.
Table 5.
FRAP values of common and underutilized fruits in Sri Lanka.
| Fruits | FRAP (μ mol FeSO 4/g FW) | Classification of AC measured by FRAP |
:---:
| Common fruits |
| |
| P. americana | 4.2 ± 0.7 a | Very low (<50 μ mol FeSO 4/g FW) |
| C. aurantifolia | 4.8 ± 0.3 a |
| A. comosus | 6.2 ± 0.2 b |
| P. edulis | 6.3 ± 0.6 b |
| G. mangostana | 7.3 ± 2.5 bc |
| C. sinensis | 11.1 ± 0.7 d |
| M. paradisiaca AAB, “silk” | 14.5 ± 0.2 ef |
| M. paradisiaca AAB “Mysore” | 23.9 ± 0.7 h |
| N. lappaceum | 24.4 ± 0.5 h |
| L. acidissima | 47.6 ± 0.5 j |
| |
| C. Lantanus | 79.2 ± 0.9 klm | Low (50–100 μ mol FeSO 4/g FW) |
| |
| C. papaya | 108.3 ± 7.6 no | High (100–1000 μ mol FeSO 4/g FW) |
| P. guajava (white flesh) | 131.5 ± 0.5 o |
| P. guajava (pink flesh) | 746.6 ± 3.0 r |
| M. indica | 950.0 ± 78.1 s |
| |
| Underutilized fruits |
| |
| C. aurantium | 8.0 ± 0.4 c | Very low (<50 μ mol FeSO 4/g FW) |
| A. heterophyllus (Wela) | 11.7 ± 0.3 de |
| S. koetjape | 11.9 ± 0.3 de |
| A. bilimbi | 14.2 ± 0.3 e |
| B. motleyana | 18.3 ± 0.7 fg |
| A. muricata | 20.8 ± 0.4 gh |
| S. dulcis | 33.2 ± 0.2 i |
| P. campechiana | 40.3 ± 1.0 ij |
| A. carambola | 40.4 ± 0.4 ij |
| |
| C. cauliflora | 63.2 ± 2.9 k | Low (50–100 μ mol FeSO 4/g FW) |
| S. caseolaris | 66.2 ± 2.6 kl |
| P. granatum | 81.0 ± 0.9 lm |
| A. heterophyllus (Waraka) | 88.3 ± 1.6 mn |
| F. inermis | 90.8 ± 0.6 mn |
| E. serratus | 92.8 ± 2.5 mn |
| |
| D. ovoideum | 130.0 ± 10.0 o | High (100–1000 μ mol FeSO 4/g FW) |
| S. jambos | 264.2 ± 5.2 p |
| C. carandas | 527.6 ± 2.5 q |
| |
| A. occidentale (yellow) | 1388.0 ± 12.5 t | Very high (>1000 μ mol FeSO 4/g FW) |
| A. marmelos | 1634.5 ± 18.6 tu |
| A. occidentale (red) | 1770.7 ± 26.1 u |
| P. emblica | 2070.0 ± 61.4 u |
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Means with different superscript letters in individual column are significantly (p< 0.05) different from each other. Data are expressed as Mean ± Standard deviation (n = 3). FRAP: Ferrous reducing antioxidant power; FW: fresh weight; AC: antioxidant capacity.
The only detailed study on antioxidant properties of fruits grown in Sri Lanka has been reported by Silva and Sirasa . The FRAP values reported by Silva and Sirasa are far below the values reported in this study for the most fruits. One possible reason for this deviation could be the loss of phytochemicals at elevated temperatures of 60°C as the authors have used the preparation of fruit extracts at 60°C. Pantelidis et al. have shown that, although cornelian cherry contains high amounts of AA, anthocyanin, and phenolic compounds, their AC measured by the FRAP assay was low. The authors claimed that the reason for this drop is loss of significant part of the AA and anthocyanin during air drying of the sample at 55°C for FRAP assay. As evidence, Piga et al. reported 55% loss in AA and 90% loss in anthocyanins in plum fruits, during the drying at 60°C. Miean and Mohamed had observed that increasing the temperature above 60°C decreased the phenolic amount considerably. At high temperatures, certain phenolics may decompose or combine with other plant components. As this study conducted controlling all the limiting factors, the results obtained in this study are compatible with correct situation of the fruits.
3.3. Total Iron (Fe) Content
Total Fe contents of common and underutilized fruits studied varied between 0.1 and 1.1 mg/100 g FW (Table 2). The highest Fe was reported from P. guajava (white variety), and the second highest Fe content was recorded for three of the fruit species, P. granatum, A. occidentale, and D. ovoideum. As AA has the ability to enhance the nonheme iron absorption, fruit sources rich in both AA and iron may help in alleviating iron deficiency among people . P. guajava, A. occidentale, and P. emblica, which have high amounts of both AA and total iron, would be potential sources in this regard.
3.4. Principal Component Analysis (PCA)
Principal component analysis (PCA) is a statistical dimensional reduction method which employs reducing the larger number of original dependent variables, to a smaller number of orthogonal new set of variables called principal components (PCs) . According to the Kaiser's rule, two principal components were extracted having eigenvalues > 1.0 from the original data set. The Kaiser-Meyer-Olkin measure of sampling adequacy is 0.645. Loading values, eigenvalues, and percent cumulative variance obtained for PCs are as in Table 6. The percent cumulative variance of the first two principal components was almost 70% of the total variance, which meets the general requirement of 70–85% for PCA. Loading values higher than 0.7 are marked in boldface type in “Table 6”. The PC1 correlates strongly with the original variables in descending order as FRAP value, TVC, TPC, and mean AA. These 4 variables of fruits, positively loaded heavily on the PC1, as determined based on the guideline provided by Stevens (factor loading >0.72). However, TFC, ARP, and Fe did not meet Steven's guideline.
Table 6.
Loading values, eigenvalues, and percent cumulative variance obtained for the two principal components.
| Variable | PC1 | PC2 |
:---:
| TVC | 0.897 | –0.006 |
| Mean AA | 0.722 | –0.260 |
| TPC | 0.793 | –0.503 |
| TFC | 0.656 | 0.473 |
| ARP | 0.593 | 0.578 |
| FRAP | 0.900 | –0.395 |
| Fe | 0.451 | 0.337 |
| Eigenvalue | 3.749 | 1.147 |
| % cumulative | 53.558 | 69.948 |
Open in a new tab
TVC: total vitamin C; AA: ascorbic acid; TPC: total phenolic content; TFC: total flavonoid content; ARP: antiradical power; FRAP: ferric reducing antioxidant power.
The score plot resulting from PCA is shown in “Figure 1”, and according to that plot, PC2 has separated all the fruits into two clusters. Most of the fruits are placed on the zero of the PC2 and in the negative side of PC1. The fruits which are in the positive side of PC1 are underutilized fruits except P. guajava (White flesh) and M. indica. The fruit P. emblica has the highest PC1 value, and it can be considered as the best fruit, in terms of studied variables, followed by A. marmelos, A. occidentale (Red), and A. occidentale (Yellow). Interestingly, all these fruits are underutilized fruits. Among commonly consumed fruits, P. guajava (White flesh) is found as the best fruit as shown in the positive side of the PC1. The results of the current study emphasizes that the underutilized fruits are with high antioxidant properties comparative to common fruits. These results are quite justifiable with previously reported studies [14, 18, 33, 35].
Figure 1.
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Score plot obtained from principal component analysis.
4. Conclusion
This study confirmed that the locally available underutilized fruits P. emblica, A. occidentale, A. marmelos, and E. serratus are rich sources of ascorbic acid contents, total vitamin C content, phenolic, flavonoid, iron, and comparably high antioxidant capacity. Among them, P. emblica is the highest. Hence, these results suggest that underutilized fruits could be used as a good alternative or addition to common fruits in promoting and securing better health in human populations and reducing the risk of many NCDs and iron deficiency. The production, marketing, and consumption of food depend on many factors other than nutritional quality, such as ease of cultivation and palatability. This work shows the nutritional value of these fruits and suggests the value of efforts in these other areas to encourage greater consumption of these valuable foodstuffs.
Acknowledgments
The authors would like to acknowledge the UGC Block Grant (RU/PG-R/16/13) of University of Ruhuna, Sri Lanka for the financial support and Prof. Mark Huxham, Edinburgh Napier University, Edinburgh, UK, for his suggestion to improve the manuscript.
Data Availability
The data used to support the findings of this study are available from the corresponding author upon request.
Disclosure
We declare that the funding resource did not involve in study design; in the collection, analysis and interpretation of data; in the writing of the report; and in the decision to submit the article for publication. The authors had full access to all of the data in this study and we take complete responsibility for the integrity of the data and the accuracy of the data analysis.
Conflicts of Interest
The authors declare that they have no conflicts of interest regarding the publication of this paper.
Funding
Funding resources provided funds for the research work and monitored the progress frequently.
References
1.Yahia E. M. The contribution of fruit and vegetable consumption to human health. In: Rosa L. A., Alvarez-Parrilla E., González-Aguilar G. A., editors. Fruit and Vegetable Phytochemicals: Chemistry, Nutritional Value, and Stability. Boboken: Blackwell Publishing; 2010. pp. 1–49. [DOI] [Google Scholar]
2.Khoo H. E., Azlan A., Kong K. W., Ismail A. Phytochemicals and medicinal properties of indigenous tropical fruits with potential for commercial development. Evidence-Based Complementary and Alternative Medicine. 2016;2016:20. doi: 10.1155/2016/7591951.7591951 [DOI] [PMC free article] [PubMed] [Google Scholar]
3.Piyathunga A. L. I., Mallawaarachchi M. A. L. N., Madhujith W. M. T. Phenolic content and antioxidant capacity of selected underutilized fruits grown in Sri Lanka. Tropical Agricultural Research. 2016;27(3):277–286. [Google Scholar]
4.Gunawardena H. P., Silva K. D. R. R. Int. Symposium IFS. Kandy, Sri Lanka: Natural Products; 2011. Determination of total antioxidant capacity and vitamin C content of selected local under-utilized and commonly consumed fresh fruits; pp. 3–8. [Google Scholar]
5.Bopitiya D., Madhujith T. Antioxidant potential of pomegranate (Punica granatum L.) cultivars grown in Sri Lanka. Tropical Agricultural Research. 2012;24(1):71–81. [Google Scholar]
6.Hewage S., Premakumara G. A. S., Madhujith T. Study on the antioxidant activity of mangosteen (Garcinia mangostana) Sri Lanka Association for the Advancement of Science Proceedings of the 64th Annual Sessions – 2008 (Pp. 105/B) 2008.
7.Mallawaarachchi M., Dissanayake K., Madhujith T. Anti-oxidant potential of selected underutilized fruit species grown in Sri Lanka. Proceedings of 8th International Research Conference, KDU; 2015; Colombo: KDU. pp. 103–107. [Google Scholar]
8.Mallawaarachchi M. A., Pushpakumara D. K. Antioxidant potential of selected underutilized fruit crop species grown in Sri Lanka. Tropical Agricultural Research. 2019;30(3):1–12. doi: 10.4038/tar.v30i3.8315. [DOI] [Google Scholar]
9.Nadeesha M. K. F., Bamunuarachchi A., Edirisinghe E. M. R. K. B. M., Weerasinghe W. M. S. K. Studies on antioxidant activity of indian gooseberry fruit and seed. Journal of Science of University of Kelaniya Sri Lanka. 2007;3:83–92. [Google Scholar]
10.Padmini S. M. P. C., Samarasekera R., Pushpakumara D. K. N. G. Antioxidant capacity and total phenol content of Sri Lankan Annona muricata L. Tropical Agricultural Research. 2015;25(2):252–260. doi: 10.4038/tar.v25i2.8146. [DOI] [Google Scholar]
11.Sandamali W. M. N., Abeysinghe D. C. Vitamin C, phenolics, antioxidant capacity and some physical and chemical properties of cashew apples of eight varieties. Proceedings of 10th Agricultural Research Symposium. 2010:193–197. [Google Scholar]
12.Sarananda K. H., Thillakawardane T. U., Alexander B. Production of health-friendly, ready-to-serve fruit drinks from under-utilized local fruits from Sri Lanka. Sri Lanka Journal of Food and Agriculture. 2017;3(2):37–48. doi: 10.4038/sljfa.v3i2.50. [DOI] [Google Scholar]
13.Sirasa M. S. F., Rukiya M. J. F., Silva K. D. R. R. Antioxidant properties of selected commonly consumed underutilized fruits in Sri Lanka. Proceedings of the Peradeniya Unv. International Research Sessions. 2014;18 [Google Scholar]
14.Silva K. D. R. R., Sirasa M. S. F. Antioxidant properties of selected fruit cultivars grown in Sri Lanka. Food Chemistry. 2018;238:203–208. doi: 10.1016/j.foodchem.2016.08.102. [DOI] [PubMed] [Google Scholar]
15.Frikke-Schmidt H., Tveden-Nyborg P., Lykkesfeldt J. L-dehydroascorbicacid can substitute L-ascorbic acid as dietary vitamin C source in Guinea pigs. Redox Biology. 2016;7:8–13. doi: 10.1016/j.redox.2015.11.003. [DOI] [PMC free article] [PubMed] [Google Scholar]
16.Cha J., Niedzwiecki A., Rath M. Hypoascorbemia induces atherosclerosis and vascular deposition of lipoprotein (a) in transgenic mice. American Journal of Cardiovascular Disease. 2015;5(1):53–62. [PMC free article] [PubMed] [Google Scholar]
17.Chambial S., Dwivedi S., Kant Shukla K., John P. J., Sharm P. Vitamin C in disease prevention and cure: an overview. Indian Journal of Clinical Biochemistry. 2013;28(4):314–328. doi: 10.1007/s12291-013-0375-3. [DOI] [PMC free article] [PubMed] [Google Scholar]
18.Luximon-Ramma A., Bahorun T., Crozier A. Antioxidant actions and phenolic and vitamin C contents of common Mauritian exotic fruits. Journal of the Science of Food and Agriculture. 2003;83(5):496–502. doi: 10.1002/jsfa.1365. [DOI] [Google Scholar]
19.Wang S., Melnyk J. P., Tsao R., Marcone M. F. How natural dietary antioxidants in fruits, vegetables and legumes promote vascular health. Food Research International. 2011;44(1):14–22. doi: 10.1016/j.foodres.2010.09.028. [DOI] [Google Scholar]
20.Venkatachalam K., Rangasamy R., Krishnan V. Total antioxidant activity and radical scavenging capacity of selected fruits and vegetables from South India. International Food Research Journal. 2014;21(3):1039–1043. [Google Scholar]
21.Amarasinghe G. S., Naottunna N. P. G. C. R., Agampodi T. C., Agampodi S. B. Factors associated with anemia among Sri Lankan primary school children in rural North Central Province. BMC Pediatrics. 2017;17:1–10. doi: 10.1186/s12887-017-0841-9. [DOI] [PMC free article] [PubMed] [Google Scholar]
22.Zimmermann M. B., Hurrell R. F. Nutritional iron deficiency. Lancet. 2007;370(9586):511–520. doi: 10.1016/S0140-6736(07)61235-5. [DOI] [PubMed] [Google Scholar]
23.Nielsen S. S. Food Analysis. 4th. New York: Springer; 2010. [DOI] [Google Scholar]
24.Ikram E. H. K., Eng K. H., Jalil A. M. M., et al. Antioxidant capacity and total phenolic content of Malaysian underutilized fruits. Journal of Food Composition and Analysis. 2009;22(5):388–393. doi: 10.1016/j.jfca.2009.04.001. [DOI] [Google Scholar]
25.Ranganna S. Hand Book of Analysis and Quality Control for Fruit and Vegetable Products. 2nd. New Delhi: Tata McGraw-Hill Publishing Company Limited; 1999. [Google Scholar]
26.Roe J. H., Oesterling M. J. The determination of dehydroascorbic acid and ascorbic acid in plant tissues by the 2,4-dinitrohydrazine method. Journal of Biological Chemistry. 1944;152:511–517. [Google Scholar]
27.Suntornsuk L., Gritsanapun W., Nilkamhank S., Paochom A. Quantitation of vitamin C content in herbal juice using direct titration. Journal of Pharmaceutical and Biomedical Analysis. 2002;28(5):849–855. doi: 10.1016/S0731-7085(01)00661-6. [DOI] [PubMed] [Google Scholar]
28.Lim Y. Y., Lim T. T., Tee J. J. Anti-oxidant properties of several tropical fruits: a comparative study. Food Chemistry. 2007;103(3):1003–1008. doi: 10.1016/j.foodchem.2006.08.038. [DOI] [Google Scholar]
29.Yadav R. N. S., Agarwala M. Phytochemical analysis of some medicinal plants. Journal of Phytology. 2011;3(12):567–576. [Google Scholar]
30.Brand-Williams W., Cuvelier M., Berset C. Use of a free radical method to evaluate antioxidant activity. LWT-Food Science and Technology. 1995;28(1):25–30. doi: 10.1016/S0023-6438(95)80008-5. [DOI] [Google Scholar]
31.Benzie I. F. F., Strain J. J. Ferric reducing/anti-oxidant power assay: direct measure of total anti-oxidant activity of biological fluids and modified version for simultaneous measurement of total anti-oxidant power and ascorbic acid concentration. Methods in Enzymology. 1998;299:15–27. doi: 10.1016/s0076-6879(99)99005-5. [DOI] [PubMed] [Google Scholar]
32.Jorhem L. Determination of metals in foods by atomic absorption spectrometry after dry ashing: NMKL1 collaborative study. Journal of AOAC International. 2000;83(5):1204–1211. doi: 10.1093/jaoac/83.5.1204. [DOI] [PubMed] [Google Scholar]
33.Contreras-Calderón J., Calderón-Jaimes L., Guerra-Hernández E., García-Villanova B. Antioxidant capacity, phenolic content and vitamin C in pulp, peel and seed from 24 exotic fruits from Colombia. Food Research International. 2011;44(7):2047–2053. doi: 10.1016/j.foodres.2010.11.003. [DOI] [Google Scholar]
34.Lee S. K., Kader A. A. Preharvest and postharvest factors influencing vitamin C content of horticultural crops. Postharvest Biology and Technology. 2000;20(3):207–220. doi: 10.1016/S0925-5214(00)00133-2. [DOI] [Google Scholar]
35.Rufino M. D. S. M., Alves R. E., Brito E. S. D., Pérez-Jiménez J., Saura-Calixto F., Mancini-Filho J. Bioactive compounds and antioxidant capacities of 18 non-traditional tropical fruits from Brazil. Food Chemistry. 2010;121(4):996–1002. doi: 10.1016/j.foodchem.2010.01.037. [DOI] [Google Scholar]
36.Arya S., Mahajan M., Jain P. Non-spectrophotometric methods for the determination of vitamin C. Analytica Chimica Acta. 2000;417(1):1–14. doi: 10.1016/S0003-2670(00)00909-0. [DOI] [Google Scholar]
37.Hernandez Y., Lobo M. G., Gonzalez M. Determination of vitamin C in tropical fruits : a comparative evaluation of methods. Food Chemistry. 2006;96(4):654–664. doi: 10.1016/j.foodchem.2005.04.012. [DOI] [Google Scholar]
38.Gil M. I., Aguayo E., Kader A. A. Quality changes and nutrient retention in fresh-cut versus whole fruits during storage. Journal of Agricultural and Food Chemistry. 2006;54(12):4284–4296. doi: 10.1021/jf060303y. [DOI] [PubMed] [Google Scholar]
39.Ellong E. N., Billard C., Adenet S., Rochefort K. Polyphenols, carotenoids, vitamin C content in tropical fruits and vegetables and impact of processing methods. Food and Nutrition Sciences. 2015;6(3, article 54466):299–313. doi: 10.4236/fns.2015.63030. [DOI] [Google Scholar]
40.Bravo L. Polyphenols: chemistry, dietary sources, metabolism, and nutritional significance. Nutrition Reviews. 1998;56(11):317–333. doi: 10.1111/j.1753-4887.1998.tb01670.x. [DOI] [PubMed] [Google Scholar]
41.Huang D., Ou B., Prior R. L. The chemistry behind antioxidant capacity assays. Journal of Agricultural and Food Chemistry. 2005;53(6):1841–1856. doi: 10.1021/jf030723c. [DOI] [PubMed] [Google Scholar]
42.Pantelidis G. E., Vasilakakis M., Manganaris G. A., Diamantidis G. Antioxidant capacity, phenol, anthocyanin and ascorbic acid contents in raspberries, blackberries, red currants, gooseberries and cornelian cherries. Food Chemistry. 2007;102(3):777–783. doi: 10.1016/j.foodchem.2006.06.021. [DOI] [Google Scholar]
43.Piga A., Caro A. D., Corda G. From plums to prunes: influence of drying parameters on polyphenols and antioxidant activity. Journal of Agricultural and Food Chemistry. 2003;51(12):3675–3681. doi: 10.1021/jf021207+. [DOI] [PubMed] [Google Scholar]
44.Miean K. H., Mohamed S. Flavonoid (Myricetin, Quercetin, Kaempferol, Luteolin, and Apigenin) content of edible tropical plants. Journal of Agricultural and Food Chemistry. 2001;49(6):3106–3112. doi: 10.1021/jf000892m. [DOI] [PubMed] [Google Scholar]
45.Abdi H., Williams L. J. Principal Component Analysis. Wiley Interdisciplinary Reviews: Computational Statistics. 2010;2(4):433–459. [Google Scholar]
46.Stevens J. Applied Multivariate Statistics for the Social Sciences. 2nd. Hillsdale, NJ: Lawrence Erlbaum Associates, Inc; 1992. [Google Scholar]
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The data used to support the findings of this study are available from the corresponding author upon request.
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5
Where (a+b) ² = a²+b² +2ab is used in real life?
·
The equation (a+b)2=a2+b2+2ab is a fundamental algebraic identity that can be found in various real-life applications across different fields. Here are some examples:
Architecture and Engineering
Area Calculations: When designing structures, architects often need to calculate areas. For instance, if you have a square room with side length a and a square extension with side length b, the area of the combined shape can be calculated using this identity.
Finance
Investment Growth: In finance, the formula can be used to model the growth of investments. If a represents the initial inve
The equation (a+b)2=a2+b2+2ab is a fundamental algebraic identity that can be found in various real-life applications across different fields. Here are some examples:
Architecture and Engineering
Area Calculations: When designing structures, architects often need to calculate areas. For instance, if you have a square room with side length a and a square extension with side length b, the area of the combined shape can be calculated using this identity.
Finance
Investment Growth: In finance, the formula can be used to model the growth of investments. If a represents the initial investment and b represents the additional investment over time, the total value after a certain period can be understood using this formula.
Physics
Kinematics: In physics, particularly in kinematics, this identity can be used to derive equations of motion. For example, when calculating distances traveled under uniform acceleration, the identity helps simplify expressions involving sums of velocities.
Computer Science
Algorithm Efficiency: In computer science, particularly in algorithms that involve nested loops, the identity can help analyze time complexity. Understanding how the square of a sum behaves can be crucial in optimizing algorithms.
Statistics
Variance Calculation: In statistics, the identity is often used in the calculation of variance. The variance of a sum of random variables can be expressed using this identity, which aids in data analysis.
Everyday Problem Solving
Budgeting and Planning: When budgeting, if you have fixed costs a and variable costs b, understanding the total cost can be simplified using this identity, especially when considering savings or expenditures that compound.
The identity (a+b)2=a2+b2+2ab is versatile and widely applicable in various domains, making it a crucial tool for problem-solving in both theoretical and practical contexts.
Related questions
Where is (a+b)² used in real life?
How do you prove that (a+b) ² = a² + b² + 2ab?
How is (a+b) ² =a²+2ab+b²? Can you explain?
What is the solution to this problem: (a+ b) 2 = a2 + 2ab + b2?
How do I prove that (a-b) ²= a²- 2ab+b²?
Max Gretinski
Studied Mathematics
·
Author has 6.5K answers and 2.5M answer views
·
1y
Any time we are dealing with the space that something occupies, we are usually looking at the area that it sits on (rather than its volume). For example, the living space of a home is expressed in square feet (or square meters). Likewise, the size of a warehouse is expressed in square units.
Most rooms are rectangular, or are composed of the sums of rectangles. Whenever we want to change the amount of space that we have, we are looking at the FOIL property. That formula (for perfect square trinomials) is a special case of FOIL.
Suppose we have a 10′ x 10′ room, but we find that it is a bit too s
Any time we are dealing with the space that something occupies, we are usually looking at the area that it sits on (rather than its volume). For example, the living space of a home is expressed in square feet (or square meters). Likewise, the size of a warehouse is expressed in square units.
Most rooms are rectangular, or are composed of the sums of rectangles. Whenever we want to change the amount of space that we have, we are looking at the FOIL property. That formula (for perfect square trinomials) is a special case of FOIL.
Suppose we have a 10′ x 10′ room, but we find that it is a bit too small for our king-sized bed and all of our furniture. How much larger is a 12′ x 12′ room?
Well, the original room measures A = 1010 = 100 square feet. The larger room measures A = 1212 = 144 square feet, which is quite a bit bigger!
Does enlarging a room by two feet on each side always add 44 square feet to the area? NO!
If you had a 6′ x 6′ walk-in closet, its area is 36 sq. ft. Enlarging it to 8′ x 8′ makes the area 64 sq. ft., and you have only added 64 - 36 = 28 sq. ft. to the original closet. Why is it different?
Because the larger area depends on the original size, too. If the original size is “a” by “a”, and if we enlarge that space by “b” units on each side, then the new area is
(a+b)2 = a2 + 2ab + b2
That is…
New Area = Old Area + 2ab + b2
Change in Area = 2ab + b2
The appearance of “a” in the formula for the change in area means that the change in area depends on the size of the original room. When we are dealing with rectangles instead of just squares, the issue becomes slightly more complicated, but the same thing happens.
The FOIL method is simply the distributive property twice (in two dimensions). Whenever there is multiplication, you might find it lurking.
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Mitchell Schoenbrun
MS in Mathematics, San Francisco State University (SFSU) (Graduated 2005)
·
Author has 4.5K answers and 6M answer views
·
4y
In high schools all over the world, students in Algebra, Trigonometry and Calculus classes are using (a+b)2=a2+2ab+b2. Oh, maybe you don’t think learning mathematics is “real life”.
OK, physicists, chemists, and biologists the world over are using mathematics in their jobs. Their understanding of mathematics is based on many things. One of these things is (a+b)2=a2+2ab+b2. Oh, maybe you are not planning to be a scientist, so this doesn’t count?
Well if you want an example that has to do with grocery shopping, plumbing or carpentry, you will be disappointed. Do you think that means it is no
In high schools all over the world, students in Algebra, Trigonometry and Calculus classes are using (a+b)2=a2+2ab+b2. Oh, maybe you don’t think learning mathematics is “real life”.
OK, physicists, chemists, and biologists the world over are using mathematics in their jobs. Their understanding of mathematics is based on many things. One of these things is (a+b)2=a2+2ab+b2. Oh, maybe you are not planning to be a scientist, so this doesn’t count?
Well if you want an example that has to do with grocery shopping, plumbing or carpentry, you will be disappointed. Do you think that means it is not imporant? Well if you don’t mind living in the stone age, it definitely isn’t important.
Anurag M
Bibliophile | Blogger | political analyst |
·
7y
Originally Answered: How is (a+b) ² =a²+2ab+b²? Can you explain?
·
By using Geometry we can easily prove the equation .
Explanation
Consider a square PQRS of side length (a+b) . Now devide the square in to 4 rectangles as shown in figure . The area of the Bigger square will be the some of the areas of these rectangles.
That is (a+b)^2 = a^2 + 2ab + b^2
By using Geometry we can easily prove the equation .
Explanation
Consider a square PQRS of side length (a+b) . Now devide the square in to 4 rectangles as shown in figure . The area of the Bigger square will be the some of the areas of these rectangles.
That is (a+b)^2 = a^2 + 2ab + b^2
Related questions
What is (a+b) ²+(a-b) ²?
What is the reason or logic behind (a+b) ²=a²+b²+2ab?
What is (a-b) ² + 2ab?
What is the use of (a+b) ^2 in practical life?
How do I prove (a+b)² ≠ a²+b² ?
Dhivagar Dharmarajan
Former RF Application Engineer at DT Techsolutions (2018–2021)
·
Updated 2y
Related
Where is (a+b)² used in real life?
Before going to the formula let's start from the basics.For example 5^2 is 5 square.Which means all the four sides of the square is equal to 5(always imagine in this way for better understanding this formula in real time applications).
Please refer the below figure.
To apply this formula, consider you have a farm in your village .In older days people use to say the size of farm in squares(still in use) eg. 5 square land ,10square land.
Dividing a farm into squares will be useful in doing calculations simple.
Consider you have a 5 square farm.That is, all the side of the farm equals to 5.
Now to cal
Before going to the formula let's start from the basics.For example 5^2 is 5 square.Which means all the four sides of the square is equal to 5(always imagine in this way for better understanding this formula in real time applications).
Please refer the below figure.
To apply this formula, consider you have a farm in your village .In older days people use to say the size of farm in squares(still in use) eg. 5 square land ,10square land.
Dividing a farm into squares will be useful in doing calculations simple.
Consider you have a 5 square farm.That is, all the side of the farm equals to 5.
Now to calculate how much fertilizer you need for your entire farm,you just divide your entire farm into 5 rows and 5 columns(now your farm totally has 25 small squares which is nothing but 5^2).
If one square need 1 pack of fertilizer then absolutely you need 25pack of fertilizers for your entire farm.
If you have a 10square farm then you need 10^2=100pack of fertilizers.
Now we go for (a+b)^2.Consider a line like the one below.
The pink side is a and the blue is b. So the total length of the line will be ( a+b)
To understand (a+b)^2 please refer the below picture.
Then the area of the inner square will be a^2.
Likewise all other areas will be like the below picture.
So the total area of the quare will be the addition of all these areas a^2+b^2+ab+ab
Which is a^2+2ab+b^2.
Think now you are cleared.
Content credit:
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Updated Sep 16
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Moiz Kharodawala
Btech EEE from Vellore Institute of Technology, Vellore (Graduated 2023)
·
9y
Originally Answered: How the formula (a+b) ^2 =a^2+b^2+2ab is used in our daily life?
·
Well
Example:
You are at a famous place. You have 19 friends with you.The ticket cost 19$.
Well, 19×19. Pretty hard isn't it?
Then,
Do: (10+9)^2
=100+81+2×9×10
=181+180
=361
Isn't it easy than calculating 19^2 ?
Try the same with 28^2 , 29^2 , 39^2 , etc.
Vaughan Pratt
Professor (Emeritus) at Stanford University (1981–present)
·
Upvoted by
Michael Jørgensen
, PhD in mathematics · Author has 9.4K answers and 4.7M answer views
·
4y
Millennia before John Napier invented logarithms as a way of multiplying two numbers by table lookup, in ancient times (a + b)² = a² + 2ab + b² was used to achieve the same goal via what has been called the Quarter-Squares method.
In place of Napier’s two tables of respectively logarithms and antilogarithms, this method can be described as using a single table of squares. To multiply a and b, look up the squares of their mean μ = (a+b)/2 and standard deviation σ = (a-b)/2, giving respectively (a² + 2ab + b²)/4 and (a² - 2ab + b²)/4. The difference of those two squares, μ² - σ², that is, mean sq
Millennia before John Napier invented logarithms as a way of multiplying two numbers by table lookup, in ancient times (a + b)² = a² + 2ab + b² was used to achieve the same goal via what has been called the Quarter-Squares method.
In place of Napier’s two tables of respectively logarithms and antilogarithms, this method can be described as using a single table of squares. To multiply a and b, look up the squares of their mean μ = (a+b)/2 and standard deviation σ = (a-b)/2, giving respectively (a² + 2ab + b²)/4 and (a² - 2ab + b²)/4. The difference of those two squares, μ² - σ², that is, mean squared minus variance, is then the desired product ab.
To avoid the initial two divisions by 2, the table can instead give quarter-squares, thereby precomputing those divisions. That is, one looks up the quarter-squares of a+b and a-b, leading to the name
This method has two advantages over Napier’s: whereas Napier requires two tables (or one if you do a reverse lookup in the log table but two is what is usually used) and three lookups, the quarter-squares method requires only one table and two lookups. But whereas Napier requires only one addition (to sum the two logs), the quarter-squares method requires an addition and two subtractions.
The article I linked to in the first paragraph gives more of the history of this method.
Another application is to give a short proof of the equivalence of Pythagorean Theorem and Heron’s formula for the area of a triangle, which I published under the title Factoring Heron in The College Mathematics Journal for January 2009. (I learned the quarter-squares method from David Smith’s history of mathematics, who attributed it to the Arab mathematician Al-Karkhi; I only later learned of its earlier history.)
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Sebastian Stan
5y
Originally Answered: How is (a+b) ² =a²+2ab+b²? Can you explain?
·
(a+b)²
= (a+b)×(a+b)
= (a+b)(a+b)
= [a×(a+b)]+[b×(a+b)]
= [a(a+b)]+[b(a+b)]
= [{(a×a)+(a×b)}] + [{(b×a)+(b×b)}]
= [(a²)+(ab)] + [(ba)+(b²)]
= (a²)+(ab)+(ba)+(b²)
Since a×b = b×a (commutative property), ba = ab.
= (a²)+(ab)+(ab)+(b²)
= (a²)+(2×ab)+(b²)
= (a²)+(2ab)+(b²)
Without Algebra, proof that (a+b)² is not equal to a²+b²
Let a = 5, and b = 4.
In (a+b)²,
= (5+4)²
= (9)²
= 9×9
= 81,
Whereas in a² + b²
= 5² + 4²
= 25 + 16
= 41.
Order of operations makes all the difference, if you add 5 and 4 before squaring the term, the answer is 81, whereas if you square 5 and then square 4 and add the two results, the final answ
(a+b)²
= (a+b)×(a+b)
= (a+b)(a+b)
= [a×(a+b)]+[b×(a+b)]
= [a(a+b)]+[b(a+b)]
= [{(a×a)+(a×b)}] + [{(b×a)+(b×b)}]
= [(a²)+(ab)] + [(ba)+(b²)]
= (a²)+(ab)+(ba)+(b²)
Since a×b = b×a (commutative property), ba = ab.
= (a²)+(ab)+(ab)+(b²)
= (a²)+(2×ab)+(b²)
= (a²)+(2ab)+(b²)
Without Algebra, proof that (a+b)² is not equal to a²+b²
Let a = 5, and b = 4.
In (a+b)²,
= (5+4)²
= (9)²
= 9×9
= 81,
Whereas in a² + b²
= 5² + 4²
= 25 + 16
= 41.
Order of operations makes all the difference, if you add 5 and 4 before squaring the term, the answer is 81, whereas if you square 5 and then square 4 and add the two results, the final answer is different. And the Parentheses are placed in (a+b)² is just so as to signify that the two terms, “a” and “b” are to be added before squaring them. (a+b)² will always be greater than a² + b².
George Tchutchulashvili
Playing in metal bands for 20 years.
·
Author has 1.2K answers and 368.8K answer views
·
3y
I have used it for example in construction.
When you are building something from diffrent wooden planks or laying ceramic plates of different color/size on the floor, or covering building which has weird shape. You may need to quickly estimate in your mind how much planks you need to cover the roof or the floor. I imagine, you would not want to start running around building with measuring tape or take out calculator and start sweating looking at the numbers while other workers laugh at you and owner who placed an order gets angry seeing your incompetence. Saves a lot of work better spent drinki
I have used it for example in construction.
When you are building something from diffrent wooden planks or laying ceramic plates of different color/size on the floor, or covering building which has weird shape. You may need to quickly estimate in your mind how much planks you need to cover the roof or the floor. I imagine, you would not want to start running around building with measuring tape or take out calculator and start sweating looking at the numbers while other workers laugh at you and owner who placed an order gets angry seeing your incompetence. Saves a lot of work better spent drinking beer after job is done.
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Nabarun Mondal
Spent 70% of my life alone in a room
·
Author has 9.1K answers and 32.2M answer views
·
8y
Related
Where is (a+b)² used in real life?
This is known as the carpet area problem. Last time we heard, this is a gazillion dollar problem because this defines the price of any house/flat, at the base.
Try calculating the carpet area, without using the formula. Yes. be my guest. Hmm.
Reasoning is overrated. The real world should be.. PUSH.
This is known as the carpet area problem. Last time we heard, this is a gazillion dollar problem because this defines the price of any house/flat, at the base.
Try calculating the carpet area, without using the formula. Yes. be my guest. Hmm.
Reasoning is overrated. The real world should be.. PUSH.
Dethan N
Retired Mechanical Engineer at Public Sector
·
Author has 193 answers and 584.7K answer views
·
9y
Originally Answered: How the formula (a+b) ^2 =a^2+b^2+2ab is used in our daily life?
·
A and B were friends once rash derived by squaring the speed. The burden of the two families squared and two cremations were done for both. (A+B)^2 = A^2 + B^2 +2AB
Two couples squared their love. Nature given them a twin of combined gene (A+B)^2 = A^2 + B^2 +2AB
Two neighbors squared the intensity of their fight. Which squared their mental tension along with 2 police cases each by them. (A+B)^2 = A^2 + B^2 +2AB
There are enormous examples for (A+B)^2 = A^2 + B^2 +2AB in our daily life.
Shawn Young
Read books related to mathematics
·
2y
Originally Answered: How is (a+b) ² =a²+2ab+b²? Can you explain?
·
(a+b)² = a²+2ab+b² is derived using the distributive property of multiplication. We expand (a+b)² as (a+b)(a+b), then apply the distributive property twice to get a²+2ab+b².
Ashwin Seetharaman
Final Year Mechanical Engineering Student
·
10y
this link.....(a+b)^2 doesn't mean directly...it is a unit of Pascal's triangle...we can find probability using this...and this triangle has a lot of hidden patterns in it like Fibonacci sequence..It is also useful in combinatorics..lots more to say about this...for your reference I have given the above link...
Related questions
Where is (a+b)² used in real life?
How do you prove that (a+b) ² = a² + b² + 2ab?
How is (a+b) ² =a²+2ab+b²? Can you explain?
What is the solution to this problem: (a+ b) 2 = a2 + 2ab + b2?
How do I prove that (a-b) ²= a²- 2ab+b²?
What is (a+b) ²+(a-b) ²?
What is the reason or logic behind (a+b) ²=a²+b²+2ab?
What is (a-b) ² + 2ab?
What is the use of (a+b) ^2 in practical life?
How do I prove (a+b)² ≠ a²+b² ?
How do l prove that
a
2
+
b
2
=
(
a
+
b
)
2
−
2
a
b
?
What is the formula for a²+b²?
What is the practical use of a²+b²+2ab?
Who invented the perfect square formula (a+b) ² = (a²+2ab+b²)?
What is the relevance of the equation (a+b) ² =a²+2ab+b² in real life?
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897 | https://www.youtube.com/watch?v=JjAw6wLUU8U | Google Sheets SLN Function | Calculate Depreciation | Straight Line Method
Software Spring
12400 subscribers
12 likes
Description
1690 views
Posted: 23 Sep 2020
The Google Sheets SLN function returns the depreciation of an asset for a single period. This function uses the straight-line method to calculate the depreciation. The three data you should input to the SLN function are the cost of the asset, its salvage value, and the life of the asset.
If you are interested in calculating the depreciation using the DB function,
which uses the arithmetic declining balance method, here is the step-by-step video tutorial:
In the straight-line method, a fixed depreciation amount is subtracted from the value of the asset at the end of each year, till the last year, to get the value of the asset at the end of the last year of its life.
The formula to calculate the depreciation amount, using the straight-line method is:
Depreciation = cost of asset-salvage value/life of the asset
Here is the format of the Google Sheets SLN function:
=SLN(cost, salvage, life)
Start the formula with an equal-to symbol.
SLN is the name of the function and stands for straight line.
Cost is the cost of the asset.
Salvage is the value of the asset at the end of the last year of its useful life.
Life is the number of periods over which the asset depreciates.
Let's look at an example of the SLN function:
=SLN(1000, 300, 3)
The SLN function returns $233.33 as the fixed depreciation amount for each year of the useful life of the asset.
At the end of the first year the value of the asset is 1000-$233.33=$766.67. The value of the asset at the end of the second and third year of its life are $766.67-$233.33=$533.33 and $533.33-$233.33=$300.00 respectively.
Please take a look at this video tutorial, which gives the steps to use the Google Sheets SLN function with examples.
Transcript:
[Music] software spring presence how to use sln in google sheets hi folks greetings welcome to the tutorial on the sln function here is an introduction to sln sln stands for straight line sln uses the straight-line method to calculate the depreciation of an asset sln returns the depreciation amount of an asset for a single period the depreciation amount is the same for each year of the life of the asset okay it's time to look at the format of the sln function here's an example cost is the cost of the asset salvage is the value of the asset at the end of the assets life life is the periods over which the asset depreciates moving forward here is an objective for using the sln function to calculate the depreciation amount of a printer whose cost is rupees 10 000 salvage value rupees 2000 and useful life 7 years all right let's create the data table [Music] let's create another table to calculate the depreciation and value of the printer [Music] okay we are all set to type the sln function formula in this cell first type equal to s l select sln from the menu [Music] for cost click this cell type comma for salvage click this cell type comma for life click this cell press enter there you go the sln function has returned the depreciation amount it is a fixed amount over the lifespan of the printer all right the next step is to copy this formula down these cells however we have to first lock the addresses in this formula so they will not change as we copy the formula click this cell select the addresses [Music] press f4 press enter okay let's copy the formula down these cells [Music] we are now all set to find the value of the printer at the end of each year of its life period let's input the value of the printer in this cell by typing the address of this cell type equal to click this cell and press [Music] enter in this cell subtract this amount from this amount and press enter [Music] copy the formula down these cells [Music] this is the salvage value of the printer at the end of the year of its useful life to remove the dollar symbol highlight this range click this down arrow click number that's it folks thank you for watching this video please subscribe to this channel |
898 | https://vmlc.tamu.edu/course-selection/calculus-i-for-biological-sciences/section-2-1-%E2%80%93-exponential-growth-and-decay/exponential-growth-in-discrete-time-example | Exponential Growth in Discrete Time Example - Virtual Math Learning Center
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Exponential Growth in Discrete Time Example
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899 | https://www.youtube.com/watch?v=hyaHxTzR8B4 | Set Theory Chapter: Definition of Set Equality and How to Determine if Sets are Equal
Ms. Hearn
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In this video we define set equality and practice identifying equal sets. Two sets are equal when they contain exactly the same elements. Often the challenge is to determine whether two sets are equal when one is written in set builder notation while the other is written using the listing method.
6 comments
Transcript:
hi i'm miss hearn let's get started in this video we're going to talk about set equality so if we have two sets a and b set a is equal to set b if you have two conditions that are met first every element of a has to also be an element of b and second every element of b has to be an element of a let's look at an example state whether the sets in each pair are equal so in part a we have the set a b c d and the set a c d b to determine if they're equal i'm going to check if every element of the first set is also an element of the second set and vice versa so in the first set we have the element a which is in the second set we have the element b which is also in the second set we have the element c which is in the second set we have the element d which is in the second set so all the elements of the first set are in the second now we go the other way around in the second set we have an a which is in the first set we check if c is in the first set and it is we check d it's there we check b it's there so we can conclude that the first set so we can conclude that the first set is equal to the second set so the answer to part a is yes so we can see here that the order of the elements doesn't matter only whether all of the elements are the same now let's look at part b in part b the first set is 2 4 6 and the second set is the set of all x such that x is an even number so we're going to determine if they're equal by first looking at the set 2 4 6 and checking each of its elements the number 2 is in the first set that is an even number so it's in the second set the number four is in the first set is it in the second set yes because it's also an even number the element six is in the first set and it's even two so it's also in the second set so all of the elements of the first set 2 4 and 6 are in the set of all x such that x is an even number now let's check the other way around so the set of all x such that x is an even number would include the number 2 the number 4 the number six the number eight the number ten and so on all multiples of two are even numbers so now this set is larger than the set two four six and if we can find even one element in this set that's not in the set 2 4 6 then they're not equal so for example the number 8 is in this set but it's not in 2 4 6. so no 2 4 6 does not represent all the even numbers for example eight is not an element of two four six so these two sets are not equal one of the times that you're going to find the concept of equality of sets useful is in some of your my labs homework for example if you're asked to express the set using the listing method and you're given a multiple choice exercise then really what you're being asked to do is to determine which of these sets on the right a b c or d is equal to the set you were given so let's find out so the set we're given is the set of all x such that x is a natural number remember that bold end means natural number and x is between 7 and 11 inclusive i say inclusive because we have or equal to here which means that x could be at the smallest 7 and at the largest 11. all right so going through our options here is this set equivalent to the set in option a well no a is incorrect because 7 is not included but the less than or equal to means that it should be so since 7 is not in set a these two sets are not equal now let's check option b notice that in the set b there are numbers that are not between 7 and 11. so b is incorrect because for example 4 is not between 7 and 11 that would be false now let's look at c c is indicating the empty set which means there are no elements but the set we were given has at least one element in it in fact it has several so there's no way that c could be the answer it's not equivalent to the set we were given so it's got to be d right so let's check d d includes all of the natural numbers between 7 and 11 inclusive 7 8 9 10 and 11 are in d and they're in the original set so those two are equal and the answer is d i hope you found this video helpful if you did please give it a thumbs up that helps other students to find the video |
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