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188000	https://artofproblemsolving.com/wiki/index.php/H%C3%B6lder%27s_Inequality?srsltid=AfmBOoof5Gh1D3AzI-txb5zUEA89ovMCt3u3k4kG5dDU7h5v774KLEvu	Art of Problem Solving Hölder's Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Hölder's Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Hölder's Inequality Contents [hide] 1 Elementary Form 2 Proof of Elementary Form 3 Statement 4 Proof 5 Examples Elementary Form If are nonnegativereal numbers and are nonnegative reals with sum of 1, then Note that with two sequences and , and , this is the elementary form of the Cauchy-Schwarz Inequality. We can state the inequality more concisely thus: Let be several sequences of nonnegative reals, and let be a sequence of nonnegative reals such that . Then Proof of Elementary Form We will use weighted AM-GM. We will disregard sequences for which one of the terms is zero, as the terms of these sequences do not contribute to the left-hand side of the desired inequality but may contribute to the right-hand side. For integers , let us define Evidently, . Then for all integers , by weighted AM-GM, Hence But from our choice of , for all integers , Therefore since the sum of the is one. Hence in summary, as desired. Equality holds when for all integers , i.e., when all the sequences are proportional. Statement If , , then and . Proof If then a.e. and there is nothing to prove. Case is similar. On the other hand, we may assume that for all . Let . Young's Inequality gives us These functions are measurable, so by integrating we get Examples Prove that, for positive reals , the following inequality holds: Retrieved from " Categories: Algebra Mathematics Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
188001	https://www.bbc.co.uk/bitesize/articles/z6xjdp3	KS3 Weight and mass Part of PhysicsForces and movement Save to My Bitesize Key points The terms weightA force that acts on mass, pulling it down, due to gravity. Weight is a force and forces are measured in newtons (N). and massA measure of the amount of matter an object is made out of. Mass is measured in kilograms (kg). are often wrongly used to describe the same thing. In science, weight and mass are very different. Back to top Weight versus mass The terms weight and mass are often used incorrectly. Phrases like ‘a bag of sugar weighs 1 kg’ are not scientifically correct. Mass is a measure of the amount of matterAny substance that has mass and takes up space by having volume. Matter can be solid, liquid or gas. an object is made out of. Mass is measured in kilograms (kg). Very small masses are sometimes measured in grams (g). There are 1000 g in one kg. Weight is the force that acts on mass due to gravity and is therefore measured in newtons (N). Back to top Calculating weight and mass The following equation can be used to calculate the weight of an object: (W = m \times g) (Weight~(N) = mass~(kg) \times gravitational~field~strength~(N/Kg)) The gravitational field strength The force that attracts one kilogram towards another massive object, like a planet. Gravitational field strength has the symbol g and it is measured in newtons per kilogram (N/kg). on Earth is 10 N/kg. This means that a 1 kg mass would be attracted to Earth by a force of 10 N. Back to top How to calculate weight An apple has a mass of 0.2 kg. Calculate the weight of the apple on Earth (g = 10 N/kg). W = ? m = 0.2 kg g = 10 N/kg Use the weight equation: (W = m \times g) and substitute in the values you know: (W = 0.2 \times 10) (W = 2~N) The weight of the apple on Earth is 2 N. Back to top How to calculate mass A big textbook weighs 90 N. Calculate the mass of the book (g = 10 N/kg). W = 90 N m = ? g = 10 N/kg Remember to use the equation: (W = m \times g) Substitute in the values you know: (90 = m \times10) Now divide both sides by 10: ( \frac{90}{10} = \frac{m \times 10}{10}) ( 9 = m ) ( m = 9~kg ) The mass of the book is 9 kg. Now have a go at answering this question. A student has a mass of 45 kg. The gravitational field strength on Earth is 10 N/kg. What is the weight of the student? The weight of the student is 450 N. This is calculated by multiplying their mass by the gravitational field strength. 45 x 10 = 450. Back to top Weight and other planets weightA force that acts on mass, pulling it down, due to gravity. Weight is a force and forces are measured in newtons (N). is a result of gravity interacting with massA measure of the amount of matter an object is made out of. Mass is measured in kilograms (kg).. This means that an object’s weight will change depending on the strength of the gravitational field strength The force that attracts one kilogram towards another massive object, like a planet. Gravitational field strength has the symbol g and it is measured in newtons per kilogram (N/kg). acting on it. Different objects that are in our Universe have gravitational fields of varying strengths. Did you know? The planets in our solar system have different gravitational field strengths. For example, the gravitational field strength on Earth is 10 N/kg whereas on Jupiter it is 24.7 N/kg. This means that a mass of 50 kg will weigh just over double on Jupiter than it would on Earth. The table shows the different gravitational field strengths for each of the different planets. \| Planet \| Gravitational field strength (N/kg) \| --- \| \| Mercury \| 4 \| \| Venus \| 9 \| \| Earth \| 10 \| \| Mars \| 4 \| \| Jupiter \| 23 \| \| Saturn \| 9 \| \| Uranus \| 9 \| \| Neptune \| 11 \| We can also calculate how an object’s weight would be different on different planets. For example, because the gravitational field strength on Earth is 10 N/kg, if a person has a mass of 60 kg their weight on Earth would be calculated as follows: (W = m \times g) (W = 60 \times 10) (W = 600 N) However on Mars, which has a gravitational field strength of 4 N/kg, the same person’s weight would be: (W = m \times g) (W = 60 \times 4) (W = 240 N) Back to top More about weight and mass Watch the video of physicist Jon Chase explaining that the mass of an object is constant but its weight depends on the gravitational force. Back to top Measuring weight and mass As with most quantities in science you need to know how you can find the mass and weight of objects. To find an object’s mass you need to use a balanceA piece of scientific equipment used to measure mass. . It is important that you set the balance to 0 before placing any object on it so that your reading is accurateA measurement that is close to the true value.. Once you have used the balance to find the mass of an object you can then complete a calculation using the weight equation to calculate the weight of the object. (W = m \times g) You can also use a newton meter, sometimes called a force meter, to measure the weight of an object too. A newton meter has a spring in it, which extends due to the weight of the object you hang from the hook at the bottom. The scale on the newton meter then indicates the amount the object weighs. Back to top Atomic Labs game. game Atomic Labs game Try out practical experiments in this KS3 science game Back to top More on Forces and movement Find out more by working through a topic
188002	https://pmc.ncbi.nlm.nih.gov/articles/PMC3129714/	Sharp Quadratic Majorization in One Dimension - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Published in final edited form as: Comput Stat Data Anal. 2009 May 15;53(7):2471–2484. doi: 10.1016/j.csda.2009.01.002 Search in PMC Search in PubMed View in NLM Catalog Add to search Sharp Quadratic Majorization in One Dimension Jan de Leeuw Jan de Leeuw 1 Department of Statistics, University of California, Los Angeles, CA 90095 Find articles by Jan de Leeuw 1, Kenneth Lange Kenneth Lange 2 Department of Biomathematics, Human Genetics and Statistics, University of California, Los Angeles, CA 90095 Find articles by Kenneth Lange 2 Author information Copyright and License information 1 Department of Statistics, University of California, Los Angeles, CA 90095 2 Department of Biomathematics, Human Genetics and Statistics, University of California, Los Angeles, CA 90095 PMC Copyright notice PMCID: PMC3129714 NIHMSID: NIHMS90438 PMID: 21738282 The publisher's version of this article is available at Comput Stat Data Anal Abstract Majorization methods solve minimization problems by replacing a complicated problem by a sequence of simpler problems. Solving the sequence of simple optimization problems guarantees convergence to a solution of the complicated original problem. Convergence is guaranteed by requiring that the approximating functions majorize the original function at the current solution. The leading examples of majorization are the EM algorithm and the SMACOF algorithm used in Multidimensional Scaling. The simplest possible majorizing subproblems are quadratic, because minimizing a quadratic is easy to do. In this paper quadratic majorizations for real-valued functions of a real variable are analyzed, and the concept of sharp majorization is introduced and studied. Applications to logit, probit, and robust loss functions are discussed. Keywords: Successive approximation, iterative majorization, convexity 1 Introduction Majorization algorithms, including the EM algorithm, are used for more and more computational tasks in statistics [De Leeuw, 1994; Heiser, 1995; Hunter and Lange, 2004; Lange et al., 2000]. The basic idea is simple. A function g majorizes a function f at a point y if g ≥ f and g(y) = f(y). If we are minimizing a complicated objective function f iteratively, then we construct a majorizing function at the current best solution x(k). We then find a new solution x(k+1) by minimizing the majorization function. Then we construct a new majorizing function at x(k+1), and so on. Majorization algorithms are worth considering if the majorizing functions can be chosen to be much easier to minimize than the original objective function, for instance linear or quadratic. In this paper we will look in more detail at majorization with quadratic functions. We restrict ourselves to functions of a single real variable. This is not as restrictive as it seems, because many functions F(x 1, ⋯ , x n) in optimization and statistics are separable in the sense that and majorization of the univariate functions f i automatically gives a majorization of F. Many of our results generalize without much trouble to real-valued functions on R n and to constrained minimization over subsets of R n. The univariate context suffices to explain most of the basic ideas. 2 Majorization 2.1 Definitions We formalize the definition of majorization at a point. Definition 2.1 Suppose f and g are real-valued functions on R n. We say that g majorizes f at y if g(x) ≥ f(x) for all x, g(y) = f(y). If the first condition can be replaced by g(x) >f(x) for all x ≠ y, we say that majorization is strict. Thus g majorizes f at y if d = g − f has a minimum, equal to zero, at y. And majorization is strict if this minimizer is unique. If g majorizes f at y, then f minorizes g at y. Alternatively we also say that f supports g at y. It is also useful to have a global definition, which says that f can be majorized at all y. Definition 2.2 Suppose f is a real-valued function on R n and g is a real-valued function on R n⨂R n. We say that g majorizes f if g(x,y) ≥ f(x) for all x and all y, g(x,x) = f(x) for all x. Majorization is strict if the first condition is g(x,y) >f(x) for all x ≠ y. 2.2 Majorization Algorithms The basic idea of majorization algorithms is simple. Suppose our current best approximation to the minimum of f is x(k), and we have a g that majorizes f in x(k). If x(k) already minimizes g we stop, otherwise we update x(k) to x(k+1) by minimizing g. If we do not stop, we have the sandwich inequality and in the case of strict majorization Repeating these steps produces a decreasing sequence of function values, and appropriate additional compactness and continuity conditions guarantee convergence of both sequences x(k) and f(x(k)) [De Leeuw, 1994]. In fact, it is not necessary to actually minimize the majorization function; it is sufficient to have a continuous update function h such that g[h(y)] <g(y) for all y. In that case the sandwich inequality still applies with x(k+1) = h(x(k)). 2.3 Majorizing Differentiable Functions We first show that majorization functions must have certain properties at the point where they touch the target. Theorem 2.1 Suppose f and g are differentiable at y. If g majorizes f at y, then g(y) = f(y), g′(y) = f′(y). If f and g are twice differentiable at y, then in addition g″(y) ≥ f″(y). Proof If g majorizes f at y then d = g − f has a minimum at y. Now use the familiar necessary conditions for the minimum of a differentiable function, which say the derivative at the minimum is zero and the second derivative is non-negative. Theorem 2.1 can be generalized in many directions if differentiability fails. If f has a left and right derivatives in y, for instance, and g is differentiable, then If f is convex, then f′L(y) ≤ f′R(y), and f′(y) must exist in order for a differentiable g to majorize f at y. In this case g′(y) = f′(y). For nonconvex f more general differential inclusions are possible using the four Dini derivatives of f at y [see, for example, McShane, 1944, Chapter V]. 3 Quadratic Majorizers As we said, it is desirable that the subproblems in which we minimize the majorization function are simple. One way to guarantee this is to try to find a convex quadratic majorizer. We limit ourselves mostly to convex quadratic majorizers because concave ones have no minima and are of limited use for algorithmic purposes. The first result, which has been widely applied, applies to functions with a continuous and uniformly bounded second derivative [Böhning and Lind-say, 1988]. Theorem 3.1 If f is twice differentiable and there is an B> 0 such that f″(x) ≤ B for all x, then for each y the convex quadratic function Majorizes f at y. Proof Use Taylor’s theorem in the form with ξ on the line connecting x and y. Because f″(ξ) ≤ B, this implies f(x) ≤ g(x), where g is defined above. This result is very useful, but it has some limitations. In the first place we would like a similar result for functions that are not everywhere twice differentiable, or even those that are not everywhere differentiable. Second, the bound does take into account that we only need to bound the second derivative on the interval between x and y, and not on the whole line. This may result in a bound which is not sharp. In particular we shall see below that substantial improvements can result from a non-uniform bound B(y) that depends on the support point y. Why do we want the bounds on the second derivative to be sharp? The majorization algorithm corresponding to this result is Which converges linearly, say to x∞, by ostrowski’s Theorem [De Leeuw, 1994]. More precisely Thus the smaller we choose B, the faster our convergence. We mention some simple properties of quadratic majorizers. Property 1 If a quadratic g majorizes a twice-differentiable convex function f at y, then g is convex. This follows from g″(y) ≥ f″(y) ≥ 0. Property 2 Quadratic majorizers are not necessarily convex. In fact, they can even be concave. Take f(x) = −x 2 and . Property 3 If a concave quadratic g majorizes a twice-differentiable function f at y, then f is concave at y. This follows from 0 ≥ g″(y) ≥ f″(y). Property 4 For some functions quadratic majorizers may not exist. Suppose, for example, that f is a cubic. If g is quadratic and majorizes f, then we must have d = g − f ≥ 0. But d = g − f is a cubic, and thus d< 0 for at least one value of x. Property 5 Quadratic majorizers may exist almost everywhere, but not everywhere. Suppose, for example, that f(x) = \|x\|. Then f has a quadratic majorizer at each y except y = 0. If y ≠ 0 we can use, following Heiser , the arithmetic mean-geometric mean inequality in the form and find That a quadratic majorizer does not exist at y = 0 follows from the discussion at the end of Section 2: f is convex and f′L(0) = −1 <f′R(0) = +1. Example 1 For a nice regular example we use the celebrated functions Then To obtain quadratic majorizers we must bound the second derivatives. We can bound Φ″(x) by setting its derivative equal to zero. We have Φ‴(x) = 0 for x = ±1. Moreover Φ″″(1) < 0 and thus Φ″(x) ≤ ϕ(1). In the same way ϕ‴(x) = 0 for x = 0 and . At x = 0 the function ϕ″(x) has a minimum, at it has two maxima. Thus . More precisely, it follows that Thus we have the quadratic majorizers and The majorizers is illustrated for both Φ and ϕ at the points y = 0 and y = −3 in Figures 1 and 2. The inequalities in this section may be useful in majorizing multivariate functions involving ϕ and Φ. They are mainly intended, however, to illustrate construction of quadratic majorizers in the smooth case. Figure 1. Open in a new tab Quadratic majorization of cumulative normal Figure 2. Open in a new tab Quadratic majorization of normal density 4 Sharp Quadratic Majorization We now drop the assumption that the objective function is twice differentiable, even locally, and we try to improve our bound estimates at the same time. 4.1 Differentiable Case Let us first deal with the case in which f is differentiable in y. Consider all a> 0 for which for a fixed y and for all x. Equivalently, we must have, for all x, (1) Define the function for all x ≠ y. The system of inequalities (1) has a solution if and only if If this is the case, then any a ≥ A(y) will satisfy (1). Because we want a to be as small as possible, we will usually prefer to choose a = A(y). This is what we mean by the sharp quadratic majorization. If the second derivative is uniformly bounded by B, we have A(y) ≤ B, and thus our bound improves on the uniform bound considered before. The function δ has some interesting properties. For differentiable convex f we have f(x) ≥ f(y) + f′(y)(x − y) and thus δ(x,y) ≥ 0. In the same way for concave f we have δ(x,y) ≤ 0. For strictly convex and concave f these inequalities are strict. If δ(x,y) ≤ 0 for all x and y, then f must be concave. Consequently A(y) ≤ 0 only if f is concave, and without loss of generality we can exclude this case from consideration. The function δ(x,y) is closely related to the second derivative at or near y. If f is twice differentiable at y, then, by the definition of the second derivative, (2) If f is three times differentiable, we can use the Taylor Expansion to sharpen this to Moreover, in the twice differentiable case, the Mean Value Theorem implies there is a ξ in the interval extending from x to y with δ(x,y) = f″(ξ). We can also derive an integral representation of δ(x,y) and its first derivative with respect to x [Tom Ferguson, Personal Communication, 03/12/04]. Lemma 4.1 δ(x,y) can written as the expectation where the random variable V follows a β(2,1) distribution. Likewise where the random variable W follows a β(2,2) distribution. Thus δ(x,y) and δ′(x,y) can be interpreted as smoothed versions of f″ and f‴. Proof The first representation follows from the second-order Taylor’s expansion with integral remainder [Lange, 2004]. This can be rewritten as (3) Since the density of β(2,1) at v is 2 v this gives the first result in the lemma. Differentiation under the integral sign of 3 yields the second representation. In view of Lemma 4.1, δ(x,y) is jointly continuous in x and y when f″(x) is continuous. Furthermore, if f″(x) tends to ∞ as x tends to −∞ or +∞, then δ(x,y) is unbounded in x for each fixed y. Thus, quadratic majorizations do not exist for any y if the second derivative grows unboundedly. It also follows from Lemma 4.1 that the best quadratic majorization does not exist if the third derivative f‴ is always positive (or always negative). This happens, for instance, if the first derivative f′ is strictly convex or strictly concave. Thus as mentioned earlier, cubics do not have quadratic majorizations. Property 6 Majorization may be possible at all points y without the function A(y) being bounded. Suppose the graph of f″(x) is 0 except for an isosceles triangle centered at each integer n ≥ 2. If we let the base of the triangle be 2 n−3 and the height of the triangle be n, then the area under the triangle is n−2. The formulas define a nonnegative convex function f(x) satisfying To prove the A(y) is finite for every y, recall the limit (2) and observe that for some w between x and y. It follows that δ(x,y) tends to 0 as \|x\| tends to ∞. Because A(n) ≥ f″(n) = n, it is clear that A(y) is unbounded. 4.2 Computing the Sharp Quadratic Majorization Let us study the case in which the supremum of δ(x,y) over x ≠ y is attained at, say, z ≠ y. In our earlier notation A(y) = δ(z,y). Differentiating δ(x,y) with respect to x gives and (4) is a necessary and sufficient condition for δ′(z,y) to vanish. At the optimal z we have (5) It is interesting that the fundamental theorem of calculus allows us to recast equations (4) and (5) as When f is convex, A(y) ≥ 0. For the second derivative at z, we have At a maximum we must have δ″(z,y) ≤ 0, which is equivalent to (6) We can achieve more clarity by viewing these questions from a different angle. If the quadratic g majorizes f at y, then it satisfies for some a. If z is a second support point, then g not only intersects f at z, but it also majorizes f at z. The condition g′(z) = f′(z) yields If we match this value with the requirement δ(z,y) = a, then we recover the second equality in (5). Conversely, if a point z satisfies the second equality in (5), then it is a second support point. In this case, one can easily check condition (4) guaranteeing that z is a stationary point of δ(x,y). 4.3 Optimality with Two Support Points Building on earlier work by Groenen et al. , Van Ruitenburg proves that a quadratic function g majorizing a differentiable function f at two points must be a sharp majorizer. The idea of looking for quadratic majorizers with two support points has been used earlier by Heiser and others. Van Ruitenburg, however, is the first to present the result in full generality. Our approach is more analytical and computational, and designed to be applied eventually to multivariate quadratic majorizations. For completeness, we now summarize in our language Van Ruitenburg’s lovely proof of the two-point property. Lemma 4.2 Suppose two quadratic functions g 1 ≠ g 2 both majorize the differentiable function f at y. Then either g 1 strictly majorizes g 2 at y or g 1 strictly majorizes g 2 at y. Proof We have (7) (8) with a 1 ≠ a 2. Subtracting (7) and(8) proves the lemma. Lemma 4.3 Suppose the quadratic function g 1 majorizes a differentiable function f at y and z 1 ≠ y and that the quadratic function g 2 majorizes f at y and z 2 ≠ y. Then g 1 = g 2. Proof Suppose g 1 ≠ g 2. Since both g 1 and g 2 majorize f at y, Lemma 4.2 applies. If g 2 strictly majorizes g 1 at y, then g 1(z 2) <g 2(z 2) = f(z 2), and g 1 does not majorize f. If g 1 strictly majorizes g 2 at y, then similarly g 2(z 1) <g 1(z 1) = f(z 1), and g 2 does not majorize f. Unless g 1 = g 2, we reach a contradiction. We now come to Van Ruitenburg’s main result. Theorem 4.4 Suppose a quadratic function g 1 majorizes a differentiable function f at y and at z ≠ y, and suppose g 2 ≠ g 1 is a quadratic majorizing f at y. Then g 2 strictly majorizes g 1 at y. Proof Suppose g 1 strictly majorizes g 2. Then g 2(z) <g 1(z) = f(z) and thus g 2 does not majorize f. The result now follows from Lemma 4.2. Property 7 It is not true, by the way, that a quadratic majorizer can have at most two support points. There can even be an infinite number of them. Consider the function h(x) = c sin 2(x) for some c> 0. Clearly h(x) ≥ 0 and h(x) = 0 for all integer multiples of π. Now define f(x) = x 2 − h(x) and g(x) = x 2. Then g is a quadratic majorizer of f at all integer multiples of π. This is plotted in Figure 3 for c = 10. Figure 3. Open in a new tab Many support points. Property 8 There is no guarantee that a second support point z ≠ y exists. Consider the continuously differentiable convex function and fix y> 1. For x> 1 For x ≤ 1 It follows that lim x→−∞δ(x,y) = 2. On the other hand, one can easily demonstrate that δ(x,y) < 2 whenever x ≤ 1. Hence, A(y) = 2, but δ(x,y) < 2 for all x ≠ y. 4.4 Even Functions Assuming that f(x) is even, i.e. f(x) = f(−x) for all x, simplifies the construction of quadratic majorizers. If an even quadratic g satisfies g(y) = f(y) and g′(y) = f′(y), then it also satisfies g(−y) = f(−y) and g′(−y) = f′(−y). If in addition g majorizes f at either y or −y, then it majorizes f at both y and −y, and Theorem 4.4 implies that it is the best possible quadratic majorization at both points. This means we only need an extra condition to guarantee that g majorizes f. The next theorem, essentially proved in the references [Groenen et al., 2003; Jaakkola and Jordan, 2000; Hunter and Li, 2005] by other techniques, highlights an important sufficient condition. Theorem 4.5 Suppose f(x) is an even, differentiable function on R such that the ratio f′(x)/x is decreasing on (0,∞). Then the even quadratic is the best quadratic fimajorizer of f (x) at the point y. Proof It is obvious that g(x) is even and satisfies the tangency conditions g(y) = f(y) and g′(y) = f′(y). For the case 0 ≤ x ≤ y, we have where the inequality comes from the assumption that f′(x)/x is decreasing. It follows that g(x) ≥ f(x). The case 0 ≤ y ≤ x is proved in similar fashion, and all other cases reduce to these two cases given that f(x) and g(x) are even. There is an condition equivalent to the sufficient condition of Theorem 4.5 that is sometimes easier to check. Theorem 4.6 The ratio f′(x)/x is decreasing on (0,∞) if and only is concave. The set of functions satisfying this condition is a closed under the formation of (a) positive multiples, (b) convex combinations, (c) limits, and (d) composition with a concave increasing function g(x). Proof Suppose is concave in x and x>y. Then the two inequalities are valid. Adding these, subtracting the common sum from both sides, and rearranging give Dividing by (x − y)/2 yields the desired result Conversely, suppose the ratio is decreasing and x>y. Then the mean value expansion for z ∈ (y,x) leads to the concavity inequality. The asserted closure properties are all easy to check. As examples of property (d) of Theorem 4.6, note that the functions g(x) = ln x and are concave and increasing. Hence, if is concave, then and are concave as well. The above discussion suggests that we look at more general transformations of the argument of f. If we define f͂(x) = f(α + βx) for an arbitrary function f(x), then a brief calculation shows that using the identity δ͂(x,y) = β 2 δ(α+βx,α+βy). An even function f(x) satisfies f͂(x) = f(x) for α = 0 and β = −1. 4.5 Non-Differentiable Functions If f is not differentiable at y, then we must find a and b such that for all x. This is an infinite system of linear inequalities in a and b, which means that the solution set is a closed convex subset of the plane. Analogous to the differentiable case we define as well as If A(y,b) < +∞, we have the sharpest quadratic majorization for given y and b. The sharpest quadratic majorization at y is given by 5 Examples As we explained in the introduction, majorizing univariate functions is usually not useful in itself. The results become relevant for statistics if they are used in the context of separable multivariate problems. In this section we first illustrate how to compute sharp quadratic majorizers for some common univariate function ocurring in maximum likelihood problems, and then we apply these majorizers to the likelihood problems themselves. 5.1 Logistic Our first example is the negative logarithm of the logistic cdf Thus Clearly and Thus f″(x) > 0 and f(x) is strictly convex. Since f″(x) ≤ 1/4, a uniform bound is readily available. The symmetry relations demonstrate that z = −y satisfies equation (4) and hence maximizes δ(x,y). The optimum value is determined by (5) as The same result was derived, using quite different methods, by Jordan and Groenen et al. . We plot the function δ(x,y) for y = 1 and y = 8 in Figure 4. Observe that the uniform bound 1/4 is not improved much for y close to 0, but for large values of y the improvement is huge. This is because A(y) ≈ (2\|y\|)−1 for large \|y\|. Thus for large values of y we will see close to superlinear convergence if we use A(y). Figure 4. Open in a new tab δ for logistic at y = 1 (left) and y = 8 (right). Alternatively, we can majorize f(x) = log(1+e−x) by writing and majorizing the even function h(x) = log(e x/2 + e−x/2). Straightforward but tedious differentiation shows that Hence, h′(x)/x is decreasing on (0,∞), and Theorem 4.5 applies. 5.2 The Absolute Value Function Because \|x\| is even, Theorem 4.5 yields the majorization which is just the result given by the arithmetic/geometric mean inequality in Property 5. When y = 0, recall that no quadratic majorization exists. If we approach majorization of \|x\| directly, we need to find a> 0 and b such that for all x. Let us compute A(y,b). If y< 0 then b = −1, and thus If y> 0 then b = +1, and again In both cases, the best quadratic majorizer can be expressed as 5.3 The Huber Function Majorization for the Huber function, specifically quadratic majorization, has been studied earlier by Heiser and Verboon and Heiser . In those papers quadratic majorization functions appear more or less out of the blue, and it is then verified that they are indeed majorization functions. This is not completely satisfactory. Here we attack the problem by applying Theorem 4.5. This leads to the sharpest quadratic majorization. The Huber function is defined by Thus we really deal with a family of even functions, one for each c> 0. The Huber functions are differentiable with derivative Since it is obvious that f′(x)/x is decreasing (0,∞), Theorem 4.5 immediately gives the sharpest majorizer 5.4 General Logit and Probit Problems Suppose we observe independent counts from n binomials counts n i, with parameters N i and π i(x), where for given functions h i(x) of p unknown parameters x. This covers both linear and nonlinear logistic regression problems. The deviance, i.e. twice the negative log-likelihood, is where as before, f(x) = log(1+exp(−x)). Using f′(−h i(x)) = −Ψ(h i(x)) = −π i(x) we see that a quadratic majorization of f leads to the quadratic majorization of the deviance by a weighted least squares function of the form where This means we can solve the logistic problem by solving a sequence of weighted least squares problems. If the h i are linear, then these are just linear regression problems. If the h i are bilinear, the subproblems are weighted singular value decompositions, and if the h i are distances the subproblems are least squares multidimensional scaling problems. If we have algorithms to solve the weighted least squares problems, then we automatically have an algorithm to solve the corresponding logistic maximum likelihood problem. In fact, it is shown by De Leeuw that essentially the same results apply if we replace the logit Ψ by the probit function Φ. The difference is that for the probit we have A(y) ≡ 1 and thus uniform quadratic majorization is sharp. As one of the reviewers correctly points out, sharp univariate quadratic majorization of f does not imply sharp multivariate quadratic majorization of D. Although, it is of course true that sharp univariate majorization gives better results that unsharp univariate majorization. The problem of sharp multivariate majorization is basically unexplored, although we do have some tentative results. 5.5 Application to Discriminant Analysis Discriminant analysis is another attractive application. In discriminant analysis with two categories, each case i is characterized by a feature vector z i and a category membership indicator y i taking the values −1 or 1. In the machine learning approach to discriminant analysis (Vapnik, 1995), the hinge loss function [1 −y i(α+z i t β)]+ plays a prominent role. Here (u)+ is short-hand for the convex function max{u,0}. Just as in ordinary regression, we can penalize the overall separable loss where θ = (α, β), by imposing a lasso or ridge penalty +λθ′θ. Most strategies for estimating θ pass to the dual of the original minimization problem. A simpler strategy, proposed by Groenen et al. (2007) is to majorize each contribution to the loss by a quadratic and minimize the surrogate loss plus penalty. In Groenen et al. (2008) this approach is extended to quadratic and Huber hinges, still maintaining the idea of using quadratric majorizers. A little calculus shows that the absolute value hinge (u)+ is majorized at u n ≠ 0 by the quadratic (9) In fact, by the same reasoning as for the absolute value function, this is the best quadratic majorizer. To avoid the singularity at 0, we recommend replacing q(u \| u n) by In double precision, a good choice of ε is 10−5. Of course, the dummy variable u is identified in case i with 1 −y i(α + z i t β). If we impose a ridge penalty, then the majorization (9) leads to a majorization algorithm exploiting weighted least squares. 6 Iterative Computation of A(y) In general, one must find A(y) numerically. We do not suggest here that the combination of finding A(y) by an iterative algorithm, and then using this A(y) in the iterative quadratic majorization algorithm, is necessarily an efficient way to construct overall algorithms. It requires an infinite number of iterations within each of an infinite number of iterations. The results in this section can be used, however, for computing A(y) for specific functions to find out how it behaves as a function of y. This has been helpful to us in finding A(y) for the logit and probit, where the computations suggested the final analytic result. For a convex function f, two similar iterative algorithms are available. They both depend on minorizing f by the linear function f(x(k))+f′(x(k))(x−x(k)) at the current point x(k) in the search for the maximum z of δ(x,y). This minorization propels the further minorization Maximizing the displayed minorizer drives δ(x,y) uphill. Fortunately, the minorizer is a function of the form with w = x − y. The stationary point w = −2 d/c furnishes the maximum of h(w) provided is negative. If f(x) is strictly convex, then is negative, and the test for a maximum succeeds. The update can be phrased as A brief calculation based on equations (4) and (5) shows that the iteration map x(k+1) = g(x(k)) has derivative at the optimal point z. On the other hand, the Dinkelbach maneuver for increasing h(w) considers the function e(w) = cw+d−h(w(k))w 2 with value e(w(k)) = 0. If we choose to maximize e(w), then it is obvious that h(w(k+1) ≥ h(w(k)). This gives the iteration map with derivative at z equal to f″(z)/A(y) by virtue of equations (4) and (5). Hence, the two algorithms have the same local rate of convergence. We recommend starting both algorithms near y. In the case of the Dinkelbach algorithm, this entails for f(x) strictly convex. Positivity of h(w(0)) is required for proper functioning of the algorithm. In view of the convexity of f(x), it is clear that f″(z)/A(y) ≥ 0. The inequality f″(z) ≤ A(y) follows from the condition A(y) = A(z) determined by Theorem 4.4 and inequality (6). Ordinarily, strict inequality f″(z) <A(y) prevails, and the two iteration maps just defined are locally contractive. Globally, the standard convergence theory for iterative majorization (MM algorithms) suggests that lim n→∞\|x(k+1)−x(k)\| = 0 and that the limit of every convergent subsequence must be a stationary point of δ(x,y) [Lange, 2004]. 7 Discussion In separable problems in which quadratic majorizers exist we have shown that it is often possible to use univariate sharp quadratic majorizers to increase convergence speed of iterative majorization (or MM) algorithms. This is true even for multivariate problems with a potentially very large number of parameters, such as the logit and probit problems in Section 5.4. There is, however, still plenty of room for improvement. We do not have, at the moment, a satisfactory theory of multivariate sharp quadratic majorization, and such a theory would obviously help to boost convergence rates even more. Such a theory should be based on the fact that a multivariate quadratic majorizer must satisfy for all x. This is an infinite system of linear inqualities in A, and thus the solution set is either empty or convex. Sharp quadratic majorization will concentrate on the extreme points of this convex set, although it is unrealistic to expect that we can find and A(y) which is sharp in all directions. Clearly both the theoretical and the implementation aspects of sharp multivariate majorization are a useful topic for further research. Acknowledgments This research was supported in part by NIH grants GM53275 and MH59490 to KL. Comments by two anonymous reviewers have been very helpful in preparing this final version. 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[Google Scholar] ACTIONS View on publisher site PDF (1.1 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract 1 Introduction 2 Majorization 3 Quadratic Majorizers 4 Sharp Quadratic Majorization 5 Examples 6 Iterative Computation of A(y) 7 Discussion Acknowledgments Footnotes References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
188003	https://bytes.usc.edu/files/ee109/documents/EE109AppendixB.pdf	1 2. Practice Problem Solutions The following are solutions to the set of practice problems presented separately. Use these only after attempting the problems on your own. 2 HW2 - Number Systems and Conversions 1. a. 1110110.0101112 = 126 + 125 +124 + 122 + 121 + 12-2 + 12-4 + 12-5 + 12-6 = 118.359375 10 = 001 \| 110 \| 110.010 \| 1112 = 166.278 = 0111 \| 0110.0101 \| 11002 = 76.5C16 b. 15B.3516 = 1162 + 5161 +11160 + 316-1 + 516-2 = 347.2070312510 = 0001 \| 0101 \| 1011.0011 \| 01012 = 101 \| 011 \| 011.001 \| 101 \| 0102 = 533.1528 2. Use the "Making Change" method a. Start by listing the powers of 2 and then find the coefficients of the number by starting with the largest powers and working toward lower powers determining which ones sum to the desired value. 0 1 1 1 0 0 1 1 0 1 1 . 1 0 1 1024 512 256 128 64 32 16 8 4 2 1 .5 .25 .125 (923.625)10 = (1110011011.101)2 3. a. 4530.1528 = 100 101 011 000.001 101 012 = 1001 \| 0101 \| 1000.0011 \| 01012 = 958.3516 4. a. DABB.AD0016 = 1101 1010 1011 1011.1010 11012 = 001 \| 101 \| 101 \| 010 \| 111 \| 011.101 \| 011 \| 0102 = 155273.5328 b. BAD.A16 = 11162 + 10161 +13160 + 1016-1 = 2989.62510 5. Use the "Making Change" method: Start by listing the powers of 5 and then find the coefficients of the number by starting with the largest powers and working toward lower powers determining which ones sum to the desired value a. 0 3 2 0 3 3 . 3 1 3125 625 125 25 5 1 .2 .04 (2143.64)10 = (32033.31)5 3 HW3 - Boolean Algebra, Logic Functions, and Canonical Representation, 2-Level Implementations 1. Probably the easiest method is perfect induction (i.e. a truth table) F = X + X’ = 1 X X’ F 0 1 1 1 0 1 2. a. F = WXYZ • (WXYZ' + WX'YZ + W'XYZ +WXY'Z) = WWXXYYZZ’ • WWXX’YYZZ + W’WXXYYZZ + WWXXYY’ZZ T8 = WXYZZ’ + WXX’YZ + WW’XYZ + WXYY’Z T3’ = 0 + 0 + 0 + 0 + 0 T5’ = 0 A4’ b. F = AB + ABC’D + ABDE’ + ABC’E + C’D = AB • (1 + C’D + DE’ + C’E) + C’D T8 = AB • (1 + DE’ + C’E) + C’D T6 = AB • (1) + C’D T2 = AB + C’D T1’ 3. a. F = X'Y + X'Y'Z’ XYZ X’ X’ • Y Y’ Z’ X’• Y’ • Z’ F 000 1 0 1 1 1 1 001 1 0 1 0 0 0 010 1 1 0 1 0 1 011 1 1 0 0 0 1 100 0 0 1 1 0 0 101 0 0 1 0 0 0 110 0 0 0 1 0 0 111 0 0 0 0 0 0 4 b. F = W’ + X' • (Y' + Z) WXYZ W’ Y’ Y’+Z X’ X’ • (Y’+Z) F 0000 1 1 1 1 1 1 0001 1 1 1 1 1 1 0010 1 0 0 1 0 1 0011 1 0 1 1 1 1 0100 1 1 1 0 0 1 0101 1 1 1 0 0 1 0110 1 0 0 0 0 1 0111 1 0 1 0 0 1 1000 0 1 1 1 1 1 1001 0 1 1 1 1 1 1010 0 0 0 1 0 0 1011 0 0 1 1 1 1 1100 0 1 1 0 0 0 1101 0 1 1 0 0 0 1110 0 0 0 0 0 0 1111 0 0 1 0 0 0 c. F = (A' + B' + CD) • (B + C' + D') ABCD A’ B’ CD A' + B' + CD C’ D’ B + C' + D’ F 0000 1 1 0 1 1 1 1 1 0001 1 1 0 1 1 0 1 1 0010 1 1 0 1 0 1 1 1 0011 1 1 1 1 0 0 0 0 0100 1 0 0 1 1 1 1 1 0101 1 0 0 1 1 0 1 1 0110 1 0 0 1 0 1 1 1 0111 1 0 1 1 0 0 1 1 1000 0 1 0 1 1 1 1 1 1001 0 1 0 1 1 0 1 1 1010 0 1 0 1 0 1 1 1 1011 0 1 1 1 0 0 0 0 1100 0 0 0 0 1 1 1 0 1101 0 0 0 0 1 0 1 0 1110 0 0 0 0 0 1 1 0 1111 0 0 1 1 0 0 1 1 5 d. F = (((A’ + B)' + C' )' + D)' = ((A’ + B)’ + C') • D’ DEMORGANS = (A • B’ + C’) • D’ DEMORGANS = AB’D’ + C’D’ T8 ABCD B’ D’ A•B’•D’ C’ C’•D’ F 0000 1 1 0 1 1 1 0001 1 0 0 1 0 0 0010 1 1 0 0 0 0 0011 1 0 0 0 0 0 0100 0 1 0 1 1 1 0101 0 0 0 1 0 0 0110 0 1 0 0 0 0 0111 0 0 0 0 0 0 1000 1 1 1 1 1 1 1001 1 0 0 1 0 0 1010 1 1 1 0 0 1 1011 1 0 0 0 0 0 1100 0 1 0 1 1 1 1101 0 0 0 1 0 0 1110 0 1 0 0 0 0 1111 0 0 0 0 0 0 4. a. F = XYZ (0,2,3) = m0 + m2 + m3 = X’Y’Z’ + X’YZ’ + X’YZ = XYZ (1,4,5,6,7) = M1 • M4 • M5 • M6 • M7 = (X+Y+Z’)•(X’+Y+Z)•(X’+Y+Z’)•(X’+Y’+Z)•(X’+Y’+Z’) b. F = ABC (1,2,4,6) = M1 • M2 • M4 • M6 = (A + B + C’) • (A + B’ + C) • (A’ + B + C) • (A’ + B’ + C) =  A,B,C (0,3,5,7) = m0 + m3 + m5 + m7 = A’B’C’ + A’BC + AB’C + ABC 6 c. F = ABCD(1,2,5,7) = m1 + m2 + m5 + m7 = A’B’C’D + A’B’CD’ + A’BC’D +A’BCD = ABCD (0,3,4,6,8,9,10,11,12,13,14,15) = M0 • M3 • M4 • M6 • M8 • M8• M9 • M10 • M11 • M12 • M13 • M14 • M15 = (A + B + C +D) • (A + B + C’ +D’) • (A + B’ + C +D) • (A + B’ + C’ +D) • (A’ + B + C +D) • (A’ + B + C +D’) • (A’ + B + C’ +D) • (A’ + B + C’ +D’) • (A’ + B’ + C +D) • (A’ + B’ + C +D’) • (A’ + B’ + C’ +D) • (A’ + B’ + C’ +D’) d. F = X' + YZ' +YZ' = X’ + YZ’ T3 = X’(Y + Y’)(Z + Z’) + (X+X’)(YZ’) T1’/T5 = X’Y’Z’ + X’Y’Z + X’YZ’ + X’YZ + XYZ’ + X’YZ’ T8 = m0 + m1 + m2 + m3 + m6 = XYZ(0,1,2,3,6) = X,Y,Z(4,5,7) = M4 • M5 • M7 = (X’ + Y + Z) • (X’ + Y + Z’) • (X’ + Y’ + Z’) e. F = A'B +B'C +AC’ = (A’BC + A’BC’) + (AB’C + A’B’C) + (ABC’ + AB’C’) T10 = m3 + m2 + m5 + m1 + m6 + m4 = ABC (1,2,3,4,5,6) = ABC(0,7) = M0 • M7 = (A + B + C) • (A’ + B’ + C’) 5. X + Y X + Y’ X = • x y x y x y 7 6. One possible solution: XY + X’Z + YZ XY + X’Z + YZ • 1 T1’ XY + X’Z + YZ • (X + X’) T5 XY + X’Z + XYZ + XYZ T8 (XY + XYZ) + (X’Z + X’ZY) T6 XY + X’Z T9 7. a. F = X’Z’ + (Y(X’+Z))’ = X’Z’ + Y’ + (X’+Z)’ DEMORGANS = X’Z’ + Y’ + XZ’ DEMORGANS = (X’ + X) • Z’ + Y’ T8 = 1 • Z’ + Y’ T5 POS = Z’ + Y’ T1’ b. G = XY + Y’Z’ = (XY + Y’)(XY + Z’) T8’ = (X + Y’)(Y + Y’)(X + Z’)(Y + Z’) T8’ = (X + Y’)(X + Z’)(Y + Z’) T3’ = (X + Y’)(Y + Z’) T11’ POS = (X + Y’) • (Y + Z’) c. H = AB • (CD)’ + A + D = (AB • (C’ + D’)) + A + D DEMORGANS = (AB + A + D) • (C’ + D’ + A + D) T8’ = (A(B+1) + D) • (1) = (A + D) • 1 T1’ POS = A + D 8. Z = AB + (C’ + A’B’)’ + A’(AB + AC’D’) = AB + (C’ + A’B’)’ + A’AB + A’AC’D’ T8 = AB + C• (A+B) + 0 + 0 DEMORGANS, T5’ SOP = AB + AC + BC T8 9. a. F = x’y’ + xy’z + z’ Let us simplify first = y’(x’ + xz) + z’ = y’(x’(z + z’) + xz) + z’ T1’, T5 = y’(x’z + x’z’ + xz) + z’ T8 8 = y’(x’z + x’z + x’z’ + xz) + z’ T3 (replicate terms) = y’(x’(z+z’) + z(x’+x)) + z’ T8 = y’(x’ +z) + z’ T5,T1’ Now let us convert to POS using T8’ = (z’+y’)(z’ + x’ + z) T8’ = (z’+y’) T5,T1’ b. G = (x’+y’)(y)(w’+y+z) Convert to SOP using T8 (simplifying as we go) = (x’y + y’y)(w’ + y + z) T8 = x’y(w’ + y + z) T5’/T1 = w’x’y + x’y + x’yz T8 = x’y(w’ + 1 + z) T8 = x’y T2 / T1’ 9 HW4 - Circuit Design w/ Karnaugh Maps 1. a. F = ABCD(1,2,4,5,9,10,12,13) FSOP = CD BC BCD + + b. F = MNOP(0,1,2,8,9) FSOP = N OP MO + + c. F = ABCD (0,1,2,5,8,10,11,15) FSOP = ACD ACD BD + + 1 1 0 1 1 1 1 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD 1 1 1 1 1 0 1 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD 1 1 1 1 1 1 1 0 1 1 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 MN OP 10 d. F = WXYZ (3,6,7,12,14) FSOP = WY WZ X Z + + 2. a. F = X’Y’ + X’Z’ + W’X + XYZ FPOS = (W’ + X’ + Y)(X + Y’ + Z’)(W’ + X’ + Z) b. The 4-bit prime numbers are: 2, 3, 5, 7, 11, 13 FPOS = (B’+D)(A’+D)(B+C) 1 1 1 1 1 0 1 1 1 1 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 WX YZ 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 WX YZ 1 1 1 0 0 1 0 0 0 0 0 0 0 1 1 0 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD 11 c. The 4-bit numbers that are not perfect squares or cubes are: 2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15. FPOS = (B + C)(A + C + D) d. The 4-bit numbers divisible by 3 or 5 are: 0, 3, 5, 6, 9, 10, 12, 15 FPOS = (A+B+C+D’)(A+B+C’+D)(A+B’+C+D)(A+B’+C’+D’)(A’+B+C+D)• (A’+B+C’+D’)(A’+B’+C+D’)(A’+B’+C’+D) 3. The K-Maps are below: FSOP = CD BC BCD + + 1 1 1 1 1 1 1 1 0 0 0 1 0 0 1 1 0 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD 1 0 0 1 1 0 0 1 1 0 0 1 0 1 0 1 0 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD 1 1 0 1 1 1 1 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD 12 FPOS = ( )( )( ) C D B C B C D + + + + a. AND – OR Implementation B C’ D C D’ B’ b. OR - AND Implementation D’ C’ B’ C D B 0 0 0 0 0 0 0 0 0 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD 13 c. NAND - NAND Implementation B C’ D C D’ B’ d. NOR - NOR Implementation D’ C’ B’ C D B 14 4. a. The values with an odd number of 1s are: 1, 2, 4, 7, 8, 11, 13, 14. The Don’t Care numbers are: 0, 4, 5, 8, 10, 12, and 15. SOP = A’C’ + BD + AC + B’D’ b. F = B’C’D’ + BCD’ + ABC’D, d = A’BC’D + A’B’CD’ F = (A+A’)B’C’D’ + (A+A’)BCD’ + ABC’D = AB’C’D’ + A’B’C’D’ + ABCD’ + A’BCD’ + ABC’D d = A’BC’D + A’B’CD’ FSOP = B’C’D’ + BC’D + BCD’ 1 1 d 1 1 d d 0 1 d 1 1 d 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD d 1 1 1 0 1 1 d 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD 15 5. a. F = A’C + ACE + CAB’ + B’A’DE + A’ D’ E’ FSOP = A’C + A’D’E’ + A’B’DE + B’C + CE b. F = (AB + A’B’) (C’D’ + CD) = ABC’D’ + A’B’C’D’ + ABCD + A’B’CD • K-Map: FSOP = ABC’D’ + A’B’C’D’ + ABCD + A’B’CD 1 1 1 0 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD 1 0 1 1 1 1 0 0 1 0 1 1 1 0 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 BC DE 0 0 1 1 1 0 0 0 0 0 1 0 0 0 1 1 0 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 BC DE A=0 A=1 16 • NAND – NAND Implementation B C’ D’ D C A A’ B’ c. F = A’B’C’D’ + A’B’CD’ + A’BC’D’ + A’BC’D + AB’CD’ • K-Map: FSOP = A’C’D’ + A’BC’ + B’CD’ • NAND – NAND Implementation A’ C’ D’ B B’ C 1 1 1 0 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD F F 17 d. The numbers are: 2, 3, 4, 5, 10, 12, 13, 14, 15. • K-Map: SOP = AB + BC’ + A’B’C + ACD’ • NAND – NAND Implementation A B C’ A’ B’ C D’ 1 1 1 1 1 0 1 1 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD 18 6. a. F = A’C + ACE + CAB’ + B’A’DE + A’ D’ E’ • K-Map: FPOS = (A’ + C’)(A’ + B’ + E)(B’+C+E’)(C + D + E’)(C + D’ + E) b. F = (AB + A’B’) (C’D’ + CD) • K-Map: FPOS = (A + B’)(A’ + B)(C + D’)(C’ + D) 0 0 0 0 0 0 0 0 0 0 0 0 0 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD 1 0 1 1 1 1 0 0 1 0 1 1 1 0 1 1 0 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 BC DE 0 0 1 1 1 0 0 0 0 0 1 0 0 0 1 1 0 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 BC DE A=0 A=1 19 NOR – NOR Implementation A B’ A’ B C D’ C’ D c. F = A’B’C’D’ + A’B’CD’ + A’BC’D’ + A’BC’D + AB’CD’ • K-Map: FPOS = (B + D’)(B’ + C’)(A’ + C) • NOR – NOR Implementation: B D’ B’ C’ A’ C 0 0 0 0 0 0 0 0 0 0 0 0 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD F F 20 d. The numbers are: 2, 3, 4, 5, 10, 12, 13, 14, 15. • K-Map: POS = (B + C)(A + B’ + C’)(A’ + B + D’) • NOR – NOR Implementation: B C A B’ C’ A’ D’ 7. A and B are both 2-bit signed numbers. They range from -2 to +1. Thus F = A + B is ranging from -4 to +2. We need 3 bits to represent the signed number F. a. The block diagram is below: A1 A0 B1 B0 F2 F1 F0 0 0 0 0 0 0 0 0 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 AB CD F 21 b. The truth table is below: A1 A0 B1 B0 F2 F1 F0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 1 1 0 0 0 1 1 1 1 1 0 1 0 0 0 0 1 0 1 0 1 0 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 1 0 0 0 1 1 0 1 0 0 1 1 1 1 1 0 1 0 1 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 1 1 1 0 1 0 0 0 1 1 1 0 1 0 1 1 1 1 1 1 1 0 c. K-Map of output “F2” K-Map of output “F1” F2SOP = A1A0’ + B1B0’ + A1B1 + A1B0’ + A0’B1 F1SOP = A1B1’B0’ + A1A0’B1’ + A1’A0’B1 + A1’B1B0’ + A1’A0B1’B0 + A1A0B1B0 1 1 1 1 1 1 1 0 1 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 A1A0 B1B0 1 1 1 1 0 1 1 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 A1A0 B1B0 22 K-Map of output “F0” F0SOP = A0B0’ + A0’B0 d. AND-OR implementation Implementation of F2 A1 A0' B1 B0' Implementation of F1 The implementation consists of four 3-input AND gates, two 4-input AND gates and one 6-input OR gate. Implementation of F0 A0 B0' A0' B0 1 1 1 1 0 1 1 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 A1A0 B1B0 23 NAND – NAND Implementation (1) Implementation of F2 A1 A0' B1 B0' (2) Implementation of F1 The implementation consists of four 3-input NAND gates, two 4-input NAND gates and one 6-input NAND gate. (3) Implementation of F0 A0 B0' A0' B0 24 8. X is a 3-bit unsigned number. It ranges from 0 to 7. Thus Y = 3X + 1 is ranging from 1 to 22. We need 5 bits to represent the unsigned number Y. a. The block diagram is below: X2 X1 X0 Y4 Y3 Y2 Y1 Y0 b. The truth table is below: X X2 X1 X0 Y4 Y3 Y2 Y1 Y0 Y 0 0 0 0 0 0 0 0 1 1 1 0 0 1 0 0 1 0 0 4 2 0 1 0 0 0 1 1 1 7 3 0 1 1 0 1 0 1 0 10 4 1 0 0 0 1 1 0 1 13 5 1 0 1 1 0 0 0 0 16 6 1 1 0 1 0 0 1 1 19 7 1 1 1 1 0 1 1 0 22 c. K-Map of output “Y4” K-Map of output “Y3” Y4POS = X2(X1 + X0) Y3POS = (X2 + X1)(X1’ + X0)(X2’ + X0’) K-Map of output “Y2” Y2POS = (X2 + X1 + X0)( X2 + X1’ + X0’)( X2’ + X1’ + X0)( X2’ + X1 + X0’) 0 0 0 0 0 0 6 4 5 7 3 1 00 01 11 10 0 1 2 X2X1 X0 0 0 0 0 0 0 0 6 4 5 7 3 1 00 01 11 10 0 1 2 X2X1 X0 0 0 0 0 0 6 4 5 7 3 1 00 01 11 10 0 1 2 X2X1 X0 25 K-Map of output “Y1” K-Map of output “Y0” Y1POS = X1 Y0POS = X0’ d. • OR – AND Implementation Implementation of Y4 X1 X0 X2 Implementation of Y3 X2 X1 X1' X0 X2' X0' 0 0 0 0 0 6 4 5 7 3 1 00 01 11 10 0 1 2 X2X1 X0 0 0 0 0 0 6 4 5 7 3 1 00 01 11 10 0 1 2 X2X1 X0 26 Implementation of Y2 X2 X1 X0 X1' X0' X2' Implementation of Y1 (straight wire connection) Y1 X1 Implementation of Y0 (straight wire connection) Y0 X0' • NOR – NOR Implementation Convert all gates in the above implementations to NOR gates. For Y1 and Y0 just use wires. 27 e. This will not be covered at this point of the class since we have not yet discussed adders. Ignore this question. 9. X is a 3-bit signed number. It ranges from -4 to +3. Thus, Z = X2 + 2X + 1 = (X + 1)2 is ranging from 0 to +16. We need 6 bits to represent signed number Z. a. The block diagram is below: Z5 Z4 Z3 Z2 Z1 Z0 X0 X1 X2 b. The truth table is below: X X2 X1 X0 Z5 Z4 Z3 Z2 Z1 Z0 Z +0 0 0 0 0 0 0 0 0 1 +1 +1 0 0 1 0 0 0 1 0 0 +4 +2 0 1 0 0 0 1 0 0 1 +9 +3 0 1 1 0 1 0 0 0 0 +16 -1 1 1 1 0 0 0 0 0 0 +0 -2 1 1 0 0 0 0 0 0 0 +1 -3 1 0 1 0 0 0 0 0 0 +4 -4 1 0 0 0 0 1 0 0 1 +9 c. No K-Map needed for Z5. “Z5” = 0 K-Map of output “Z4” K-Map of output “Z3” Z4SOP = X2’X1X0 Z3SOP = X2’X1X0’ + X2X1’X0’ 0 1 6 4 5 7 3 1 00 01 11 10 0 1 2 X2X1 X0 0 1 1 6 4 5 7 3 1 00 01 11 10 0 1 2 X2X1 X0 28 K-Map of output “Z2” K-Map of output “Z0” Z2SOP = X2’X1’X0 Z0SOP = X2’X0’ + X1’X0’ Note: Z1 = 0 (constant) 0 1 6 4 5 7 3 1 00 01 11 10 0 1 2 X2X1 X0 1 0 1 1 6 4 5 7 3 1 00 01 11 10 0 1 2 X2X1 X0 29 10. Design a 1-bit comparator that takes in a bit X and a bit Y and outputs XY, X=Y. Use a single 2-to-4 decoder and 1 single OR gate. Consider the truth table of each of these simple functions. We can implement each as a sum of minterms. And since a decoder implements the minterms of the input variables, we can arrive at the circuit above. X Y XY X==Y 0 0 0 0 1 0 1 1 0 0 1 0 0 1 0 1 1 0 0 1 Y0 Y1 Y2 Y3 A0 A1 2-to-4 Decoder X Y XY X=Y 30 HW5 - Signed Representations and Arithmetic 1. a. 11 1 10100 = 20 + 11101 = 29 Carry=1 110001 = 49 b. 11100 - 10100 01000 28 - 20 8 1 1 1 1 _ 11100 2’s comp 01011 + 1 01000 2. b. 1 1 1 1 _ 101.011 + 011.010 1000.101 3. For unsigned subtraction we still use take the 2's complement of the bottom number (but the check for overflow differs for unsigned vs. 2's complement) a. 01101102 – 10001002 cannot be performed as is because we cannot subtract a larger number from a smaller number in unsigned representation. For unsigned subtraction, after adding the 2's comp, check if the Cout=0, we can know there is overflow 0110110 - 1000100 +54 -68 0 1 1 1 1 1 1 _ 0110110 2’s comp 0111011 + 1 1110010 4. a. Unsigned binary: 28 < 28310 < 29. Therefore, 9 bits. 28310 = 1000110112 b. 2’s complement: with n-bits we can make –(2n-1) to +(2n-1-1). With n = 9, the range is -256 to +256. With n = 10, the range is -512 to +511. Thus 10 bits. 28310 = 1000110112 31 5. Consider the following decimal numbers +21, +55, +121, -32, -99, -128 a. What are the corresponding 8-bit signed-magnitude representation?. a. What are the corresponding 8-bit 2's complement representation?. b. Which of the above numbers can be represented in 6-bit signed-magnitude, 6-bit 1's complement, 6-bit 2's complement representations, explain ? 21 55 121 -32 -99 -128 A 16+4+1=21 00010101 00110111 01111001 10100000 11100011 NA B 00010101 00110111 01111001 1110000 10011101 10000000 C Y,Y,Y 010101 in all cases N,N,N N,N,N N,N,Y 100000 in 2’s comp N,N,N N,N,N 6. What are the corresponding decimal representations for the following binary numbers: 01011011 , 11010010, if a. The binary numbers are in 8-bit signed-magnitude format? b. The binary numbers are in 8-bit 2's complement format? 01011011 11010010 a) 64+16+8+2+1=91 -(64+16+2)=-82 b) 64+16+8+2+1=91 -128+64+16+2=-46 7. Perform the following addition problems for the following 2’s complement numbers. State whether overflow does or does not occur for each problem. Justify your answer for why overflow does or does not occur. Check your work by converting each number to decimal. a 1010 0111 b 1001 0110 c 0101 1100 d 0101 1101 +1110 0100 +1011 0011 +1011 0101 +0110 1001 11000 1011 10100 1001 10001 0001 1100 0110 n+n=n cin=cout=1 No Overflow n+n=p cin=0,cout=1 Overflow p+n=p cin=cout=1 NoOverflow p+p=n cin=1,cout=0 Overflow 32 8. This is an exercise to help you remember what gates are used to create a full adder. Using a full adder as shown below and no other gates, can you produce the function Z = A•B (Hint: Write out the logical equations for S and Cout and see if you can hook up the inputs to produce Z). ANSWER: Cin X Y S Cout 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 S X Y Cin Cout XY XCin YCin =   = + + We want Z=XY, and if we set Cin=0 then the formulas of S and Cout becomes as follows: S X Y Cout XY =  = Thus, Cout = Z = XY X Y S Cin Full Adder Cout A B 0 Z 33 9. Using 1 half-adder and a minimal number of 4-bit binary adders, design a circuit to calculate Y = 25X, where X is a 4-bit inputs. ANSWER: If a number is multiplied by a number that is a power of 2, then the number rewritten using the above formula: 2 0...0 n n X X = Y = 25X = 16X + 8X + X = X0000+X000+X Also adding the information that X is a 4-bit number above formula can be re-written X = X3X2X1X0 Y = X3X2X1X00000 + X3X2X1X0000 + X3X2X1X0 This can be written as addition by 4 bit adders: X3 X2 X1 X0 X3 X2 X1 X0 0 0 0 + X3 X2 X1 X0 0 0 0 0 Since one row in each addition is all zeros, they can be eliminated and the other two numbers can be added using a 4-bit adder. However, the lower bits require only a single bit adder. The carry from the first addition is transferred to the next one. B3 B2 B1 B0 A3 A2 A1 A0 S0 S1 S2 S3 C0 C4 4-bit Binary Adder X Y S Half Adder Cout X3 X0 X0 X1 X2 X3 X1 X2 X3 0 X0 X1 X2 Y8 Y7 Y6 Y5 Y4 Y3 Y0 Y1 Y2 34 10. Using a minimal number of 4-bit adders, design a circuit that implements Y=20X+107, where X is a 3-bit unsigned number. ANSWER: X = X2X1X0, and thus 20X= 16X+4X. These numbers are achieved by inserting zeros at the end. Once we add these there numbers, we see that not all bits require addition. We actually need only a single 4-bit adder and nothing more. 1 1 0 1 0 1 1 X2 X1 X0 0 0 + X2 X1 X0 0 0 0 0 Which can be collapsed to the following addition: 1 1 X2 X1 X0 0 0 + X2 X1 X0 1 0 1 1 B3 B2 B1 B0 A3 A2 A1 A0 S0 S1 S2 S3 C0 C4 4-bit Binary Adder X0 X1 X2 X1 X2 1 1 1 1 X0 Y7 Y6 Y5 Y4 Y3 Y0 Y1 Y2 0 1 35 11. Design a minimal circuit using AND/OR/NOT gates to implement the comparison A > 10 where A is a 4-bit number. a. Start by writing out the logical algorithm for when A > 10 then implement it using gates b. Check that your work is minimal by using a K-Map. ANSWER: Let’s assume that the number A = A3A2A1A0. Logical Algorithm: For any number A to be greater than a number B (in this case 1010 = 10102), a more significant bit of A must be greater than B. However, in the case of A compared with 1010, we realize A3 and A1 can never be greater than the 1’s in those places for 10. Similarly, A2 and A0 can never be less than the 0’s in those places. Thus for A > 10, there are only 2 cases that need to be checked: A3 = 1 and A2 > 0 (i.e. A2 = 1) OR A3=1 and A1 = 1 and A0 > 0 (i.e. A0 = 1). Thus A>10 = A3A2 + A3A2A0. Note that we don’t have to check A2 in the second case because it is DEFINITELY greater than or equal to the 0 in that place of 10102. A K-Map will also show the same equation SUM(11,12,13,14,15). A>10=A3A2+ A3A1A0 A3 A2 A3 A0 A1 A>10 1 1 1 0 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 A3A2 A1A0 36 HW 6 – Latches, and Flip-Flops 1. Complete the following waveforms for negative edge-triggered D Flip-Flops. D Q Q CK D Q Q CK CLK D1 D2 /D1 /D2 Z X CLK X D1 D2 Z 2. We should find the state diagram following the steps in the class notes a. Convert the diagram to a table. Current State Next State Output XY= 00 XY= 01 XY= 10 XY= 11 Q1 Q0 Q1 Q0 Q1 Q0 Q1 Q0 Q1 Q0 Z 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 1 1 1 0 0 0 0 1 0 0 0 1 0 1 1 0 0 0 1 0 1 0 1 0 37 Convert Q to D Current State Next State Output XY= 00 XY= 01 XY= 10 XY= 11 Q1 Q0 D1 D0 D1 D0 D1 D0 D1 D0 Z 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 1 1 1 0 0 0 0 1 0 0 0 1 0 1 1 0 0 0 1 0 1 0 1 0 b. Perform K-maps to find equations for D1 and D0. D0 = Q1·Y + Q0·X D1 = Q1’·Q0’·X c. Draw the circuit and implement the initial state (reset) condition. D Q Q CK D Q Q CK Z X Clock Q0 Q1 Y PRE PRE CLR CLR Reset Reset GND GND 38 3. Because we have 2 states in the state diagram, we use one D flip-flop to implement state memory. We design the combinational logic for the next-state logic and the output function logic by building up the state/transition table. Current State Next State / Output XY = 00 XY = 01 XY = 10 XY = 11 Symbol/Code Symbol/Code Symbol/Code Symbol/Code Symbol/Code H F / 0 F / 0 G / 1 G / 1 G / 1 0 G / 1 F / 0 F / 0 F / 0 G / 1 1 Q0(t) Q0 D0 Q0 D0 Q0 D0 Q0 D0 After we get the state/transition table, we build the K-Maps to simplify expressions for D0, and H. • D0 K-Map: D0 = YQ0’ + XQ0’ + XY H = Q0 0 1 1 1 1 6 4 5 7 3 1 00 01 11 10 0 1 2 XY Q0 39 4. Current State Next State & Flop-Flop Inputs Outputs Next State Next State X = 0 X = 1 Symbol/Code Symbol/Code Symbol/Code Z A/00 B/01 D/11 0 B/01 C/10 B/01 0 C/10 B/01 A/00 1 D/11 B/01 C/10 0 Q1(t),Q0(t) Q1(t+1),Q0(t+1) Q1(t+1),Q0(t+1) After we get the state/transition table, we build the K-Maps to simplify expressions for D1, D0, and Z. • D1 K-Map: D0 K-Map D1 = X’Q1'Q0 + XQ1'Q0’+ XQ1Q0 D0 = X’Q1+X'Q0' + XQ1’ Z K-Map: Z = Q1Q0’ 1 1 0 0 1 1 0 1 00 01 11 10 0 Q1Q0 X 1 1 0 1 0 0 0 1 0 00 01 11 10 0 Q1Q0 X 1 1 0 0 0 0 1 0 Q0 Q1 1 40 a. The waveform is below: CLK X Q1 Q0 State Z A D B C A B 41 5. a. We need 4 states and each of them stores one 2-bit binary number ranging from 00 to 11. State S0 S1 S2 S3 Assignment (Q1 Q0) 00 01 10 11 The state diagram is below: b. Because we have 4 states in the state diagram, we use two D flip-flops to implement state memory. We design the combinational logic for the next-state logic and the output function logic by building up the state/transition table. U – UP, D – DOWN Current State Next State U·D = 00 U·D = 01 U·D = 10 U·D = 11 Symbol/Code Symbol/Code Symbol/Code Symbol/Code Symbol/Code S0/00 S0/00 S3/11 S1/01 S0/00 S1/01 S1/01 S0/00 S2/10 S1/01 S2/10 S2/10 S1/01 S3/11 S2/10 S3/11 S3/11 S2/10 S0/00 S3/11 Q1(t),Q0(t) Q1(t+1),Q0(t+1) D1,D0 Q1(t+1),Q0(t+1) D1,D0 Q1(t+1),Q0(t+1) D1,D0 Q1(t+1),Q0(t+1) D1,D0 After we get the state/transition table, we build the K-Maps to simplify expressions for D1, and D0. S0 S1 S2 S3 UP=DOWN=1 or UP=DOWN=0 UP=DOWN=1 or UP=DOWN=0 UP=DOWN=1 or UP=DOWN=0 UP=DOWN=1 or UP=DOWN=0 UP=1 and DOWN=0 UP=1 and DOWN=0 UP=1 and DOWN=0 UP=1 and DOWN=0 UP=0 and DOWN=1 UP=0 and DOWN=1 UP=0 and DOWN=1 UP=0 and DOWN=1 ~Reset 42 • D1 K-Map: D1 = U’·D’Q1 + D·Q1Q0 + U·Q1Q0’ + U’·DQ1’Q0’ + U·D’Q1’Q0 • D0 K-Map: D0 = U’·D’Q0 + U·DQ0 + U’·DQ0’ + U·D’Q0’ c. The waveform is below: CLK UP DOWN Q1 Q0 1 1 1 1 1 1 0 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 UD Q1Q0 1 1 1 1 0 1 1 1 1 3 12 8 9 13 5 1 00 01 11 10 00 01 2 6 7 15 14 10 11 4 11 10 UD Q1Q0 43 6. Complete the waveform for the 4-bit counter presented in class. 1111 Clock SR CEP CET PE P3-P0 Q3-Q0 1101 1101 1110 0000 0001 0000 1110 1111 7. Complete the waveform for a 4-bit D-Register with Data (Load) Enable and asynchronous reset. CLK AR EN 0110 D[3:0] 0001 1011 1010 1101 1001 1000 1111 0111 Q[3:0] 1010 1111 1101 1000 0000 1011 44 8.1. Unit 7 – Datapath Design 1. Implement a circuit that takes in a 4-bit number X[3:0] and produces a 4-bit value Z[3:0] according to the following function: if X < 8 then Z = X + 5; else Z = X – 2; Using the building blocks below and at most 1 inverter (no other gates), implement this function. Note: To check X<8, subtract X-8 by taking the 2’s complement of 8. The 2’s comp. of 8 is 0111+1. Recall, a carry of 0 when subtracting unsigned numbers (A-B) indicates overflow which can only occur in unsigned subtraction if the result needed to be negative (i.e. A < B). So C4 will only be 1 if X is not less-than 8. We can invert it to get a signal that is true when X < 8. Use those signals to either add 5 (0101) or subtract 2 (add 1110). B3 B2 B1 B0 A3 A2 A1 A0 S0 S1 S2 S3 C0 C4 4-bit Binary Adder X !< 8 X<8 X0 X1 X2 X3 Z0 Z1 Z2 Z3 B3 B2 B1 B0 A3 A2 A1 A0 S0 S1 S2 S3 C0 C4 4-bit Binary Adder X3 X2 X1 X0 1 1 1 0 1 0 1 45 2. Design a circuit that takes in an 4-bit number X[3:0] and outputs a number Y, and performs the operation below. Assume these numbers are unsigned. Use a 4-bit adder and 2-to-1 muxes. Hint: Multiplying by 4 (i.e. 22) in binary can be done with no gates just as multiplying by 100 (i.e. 102) in decimal can be done simply and easily. if(X < 00112) Y = 4X else Y = X 4X would require 6 outputs and be X3 X2 X1 X0 0 0, thus even if we pass X we need to produce 6 outputs. Recall subtracting 3=0011 bin. is the same as adding the 2's complement. Next, a carry of 0 when subtracting unsigned numbers (A-B) indicates overflow which can only occur in unsigned subtraction if the result needed to be negative (i.e. A < B) I1 Y S I0 I1 Y S I0 I1 Y S I0 I1 Y S I0 I1 Y S I0 I1 Y S I0 X0 X1 X2 X3 0 0 0 0 X0 X1 X2 X3 Y5 Y4 Y3 Y2 Y0 Y1 X !< 3 B3 B2 B1 B0 A3 A2 A1 A0 S0 S1 S2 S3 C0 C4 4-bit Binary Adder X3 X2 X1 X0 0 0 1 1 1 46 3. Design a circuit that takes in two 4-bit numbers, X[3:0] and Y[3:0] along with two function select bits, FS1 and FS0, and produces an output, Z[3:0], according to the following table: FS1,FS0 Z 0,0 X+Y 0,1 X-Y 1,0 Y-X 1,1 don’t care Use (1) 4-bit adder, 4-bit wide 2-to-1 muxes, and any basic gates you desire. Approach: Take a 4-bit adder and for each operation listed in the table above identify what should be passed to the A and B inputs of the adder. Then use muxes to perform that function. Design logic for the select bits of the muxes and the carry-in of the adder that use FS1 and FS0 as input. Note: In the solution below the A and B inputs of the muxes are like the 0th and 1st inputs of 4 2-to-1 muxes, respectively. Ignore the G input. B3 B2 B1 B0 A3 A2 A1 A0 S3 S2 S1 S0 C0 C4 S G 1A 1Y 2Y 2-to-1 4-bit wide mux 2A 3A 4A 3Y 4Y 1B 2B 3B 4B 0 S G 1A 1Y 2Y 2-to-1 4-bit wide mux 2A 3A 4A 3Y 4Y 1B 2B 3B 4B 0 X3 X2 X1 X0 X3 X2 X1 X0 Y3 Y2 Y1 Y0 Y3 Y2 Y1 Y0 Z3 Z2 Z1 Z0 FS1 FS0 FS1 FS0 74LS 283 47 4. Design a circuit that takes in two 4-bit signed magnitude numbers, X[3:0] and Y[3:0] and produces two 4-bit outputs, A[3:0] and B[3:0] according to the following algorithm: if X >= Y then A = X and B = Y else A = Y and B = X Use a 4-bit adder to subtract and perform the comparison, 4-bit wide 2-to-1 muxes, and basic logic gates. Assume the input 1000 = -0 will never occur on X or Y. Approach: First think about how to compare signed magnitude numbers by using unsigned comparison techniques and how to precondition the inputs to allow the use of unsigned comparison. For example, 1001 = -1 in signed magnitude and 0100 = +4 in signed magnitude. So 1001 is really less than 0100. Use the comparison result to control the selects of the muxes. 48 5. Design a circuit that adds two 4-bit, unsigned numbers: X[3:0] and Y[3:0] and outputs either the sum if there is no overflow or 1111 (=1510, the maximum unsigned 4-bit number) if there is overflow. Unsigned overflow is found by looking at Cout. You could use a mux to select the sum or 1111 but realize you can just use OR gates for this function. B3 B2 B1 B0 A3 A2 A1 A0 S3 S2 S1 S0 C0 C4 74LS 283 X3 X2 X1 X0 UNS_OV 0 Y3 Y2 Y1 Y0 G0 G1 G2 G3 49 1. We would like to build an adder whose inputs are in units of pennies and produces sum in nickels and pennies. Essentially, this is a base-5 adder. The 2 inputs of this adder are A={A2A1A0} and B={B2B1B0}. Assume that the maximum input number that A and B can be is 4 pennies. Design an adder to produce the correct nickels/pennies sum, C={ [CN] ; [CP2CP1CP0] }. To implement this design you may assume you can use as many 4-bit adders as you like. if X+Y > 4 dec. then CN = 1, CP[2:0] = X+Y-5 else CN = 0, CP[2:0] = X+Y A2 A1 A0 B2 B1 B0 CP2 CP1 CP0 To be designed by you CN Block diagram for the nickel/penny adder you are going to design 50 See below for some alternate implementations and simplifications. 51 Now, noting that the final 4-bit adder really only produces a 3-bit output, we can substitute and use a 3-bit adder in its place…OR…simplify the 4-bit adder shown. Since the 4-bit adder does not utilize the S3 nor C4 output, we can simply and not connect A3, B3 since they will only affect S3 and C4 to arrive at the design below (which uses B0 rather than C0 for the +1 when taking the 2's complement. B3 B2 B1 B0 A3 A2 A1 A0 S0 S1 S2 S3 C0 C4 4-bit Binary Adder Z4 Z2 Z1 CP2 0 0 B3 B2 B1 B0 A3 A2 A1 A0 S0 S1 S2 S3 C0 C4 4-bit Binary Adder 0 X4 X2 X1 0 Y4 Y2 Y1 Z8 Z4 Z2 Z1 0 B3 B2 B1 B0 A3 A2 A1 A0 S0 S1 S2 S3 C0 C4 4-bit Binary Adder 0 1 0 0 Z8 Z4 Z2 Z1 1 Need actual inverters here, just didn t draw them. CN (4 < X+Y) CP1 CP0 52 Another approach using a mux rather than a 3rd adder is also shown (and also compares (X+Y) >= 5 rather than 4 < (X+Y)
188004	https://answers.everydaycalculation.com/percent-of/80-600	What is 80 percent of 600? 80% of 600 Answers Solutions by everydaycalculation.com Answers.everydaycalculation.com » What is A% of B What is 80 percent of 600? 80% of 600 is 480 Working out 80% of 600 Write 80% as 80/100 Since, finding the fraction of a number is same as multiplying the fraction with the number, we have 80/100 of 600 = 80/100 × 600 3. Therefore, the answer is 480 If you are using a calculator, simply enter 80÷100×600 which will give you 480 as the answer. MathStep (Works offline) Download our mobile app and learn how to work with percentages in your own time: Android and iPhone/ iPad More percentage problems: 160% of 60080% of 1200240% of 60080% of 1800400% of 60080% of 3000560% of 60080% of 420080% of what number is 60080 is what percent of 600 Find another What is % of © everydaycalculation.com
188005	https://examobjective.com/ped621-most-sensitive-indicator-of-intravascular-volume-depletion-in-infant-is/?srsltid=AfmBOopZAXWoD-2Cwqn-1G427SwF9smuJ7D7ssqwGgufdctKyhiiTQxr	Most sensitive indicator of intravascular volume depletion in Skip to content Exam Objective by Asha D. Keep learning with us. Download App Search for:Search Search Home Posts Nursing MCQ E-Book Shop Other MCQs Exam Objective by Asha D. Keep learning with us. Search for:Search Search Download App Main Menu Home Posts Nursing MCQ E-Book Shop Other MCQs Most sensitive indicator of intravascular volume depletion in infant is: mcq given below: Q. Most sensitive indicator of intravascular volume depletion in infant is: (a) Stroke volume (b) Heart rate (c) Cardiac output (d) Blood pressure Show Answer Ans: (d) Blood pressure / Next Question Back Do you Agree with Answer comment below 🚀 Daily Nursing MCQ Go Nursing Exam Mock Test-1 Take Test Now 🚀 Top 100 Nursing MCQ Read Now 📂 Subjects 🩺 Fundamentals of Nursing 💊 Pharmacology 👶 Pediatric Nursing 🫁 Medical & Surgical 🌍 Community Health 📕 PDFs / E-Books 📗 Nursing MCQ PDF [8000+] 🚗 Automobile Engineering PDF 🏆 Quizzes / Mock Tests 🫀 Anatomy & Physiology – Mock Test 🩺 Fundamentals Quiz 📱 Follow Us FacebookInstagramWhatsAppLinkedInPinterest 🩺 Fundamentals of Nursing Set-1Set-2Set-3Set-4Set-5Next ➤ 💊 Pharmacology Set-1Set-2Set-3Set-4Set-5Next ➤ 👶 Pediatric Nursing Set-1Set-2Set-3Set-4Set-5Next ➤ 🫁 Medical & Surgical Nursing Set-1Set-2Set-3Set-4Set-5Next ➤ 🌍 Community Health Nursing Set-1Set-2Set-3Set-4Set-5Next ➤ 👩‍🍼 Obstetric & Gynecological Nursing Set-1Set-2Set-3Set-4Set-5Next ➤ 🫀 Anatomy & Physiology Set-1Set-2Set-3Set-4Set-5Next ➤ 📘 Nursing Education Set-1Set-2Set-3Set-4Set-5Next ➤ 🧬 Microbiology Set-1Set-2Set-3Set-4Set-5Next ➤ Shopping Cart Download AppDownload App Copyright © 2025 www.examobjective.com Email:examobjective@gmail.com About Us \| Terms & Conditions \| Private Policy \| Refund Policy \| Disclaimer \|Contact Us\| Pricing \| Shipping Policy Scroll to Top Hi, Ask me anything regarding your exam 1 Scan the code Help Desk Hi, Ask me anything regarding your exam preparation. Talk to Me
188006	https://phys.libretexts.org/Bookshelves/Conceptual_Physics/Introduction_to_Physics_(Park)/03%3A_Unit_2-_Mechanics_II_-_Energy_and_Momentum_Oscillations_and_Waves_Rotation_and_Fluids/03%3A_Work_and_Energy/3.06%3A_Spring_Potential_Energy	Search x Text Color Text Size Margin Size Font Type selected template will load here Error This action is not available. 3.6: Spring Potential Energy ( \newcommand{\vecs}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } ) ( \newcommand{\vecd}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash {#1}}} ) ( \newcommand{\id}{\mathrm{id}}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\kernel}{\mathrm{null}\,}) ( \newcommand{\range}{\mathrm{range}\,}) ( \newcommand{\RealPart}{\mathrm{Re}}) ( \newcommand{\ImaginaryPart}{\mathrm{Im}}) ( \newcommand{\Argument}{\mathrm{Arg}}) ( \newcommand{\norm}{\\| #1 \\|}) ( \newcommand{\inner}{\langle #1, #2 \rangle}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\id}{\mathrm{id}}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\kernel}{\mathrm{null}\,}) ( \newcommand{\range}{\mathrm{range}\,}) ( \newcommand{\RealPart}{\mathrm{Re}}) ( \newcommand{\ImaginaryPart}{\mathrm{Im}}) ( \newcommand{\Argument}{\mathrm{Arg}}) ( \newcommand{\norm}{\\| #1 \\|}) ( \newcommand{\inner}{\langle #1, #2 \rangle}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\AA}{\unicode[.8,0]{x212B}}) ( \newcommand{\vectorA}{\vec{#1}} % arrow) ( \newcommand{\vectorAt}{\vec{\text{#1}}} % arrow) ( \newcommand{\vectorB}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } ) ( \newcommand{\vectorC}{\textbf{#1}} ) ( \newcommand{\vectorD}{\overrightarrow{#1}} ) ( \newcommand{\vectorDt}{\overrightarrow{\text{#1}}} ) ( \newcommand{\vectE}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} ) ( \newcommand{\vecs}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } ) ( \newcommand{\vecd}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash {#1}}} ) Learning Objectives Hooke's Law, (F=-k x), describes force exerted by a spring being deformed. Here, (F) is the restoring force, (x) is the displacement from equilibrium or deformation, and (k) is a constant related to the difficulty in deforming the system. The minus sign indicates the restoring force is in the direction opposite to the displacement. In order to produce a deformation, work must be done. That is, a force must be exerted through a distance, whether you pluck a guitar string or compress a car spring. If the only result is deformation, and no work goes into thermal, sound, or kinetic energy, then all the work is initially stored in the deformed object as some form of potential energy. The potential energy stored in a spring is (\mathrm{PE}_{\mathrm{el}}=\frac{1}{2} k x^{2}). Here, we generalize the idea to elastic potential energy for a deformation of any system that can be described by Hooke’s law. Hence, [\mathrm{PE}_{\mathrm{el}}=\frac{1}{2} k x^{2}, \nonumber ] where (\mathrm{PE}_{\mathrm{el}}) is the elastic potential energy stored in any deformed system that obeys Hooke’s law and has a displacement (x) from equilibrium and a force constant (k). It is possible to find the work done in deforming a system in order to find the energy stored. This work is performed by an applied force (F_{\text {app }}). The applied force is exactly opposite to the restoring force (action-reaction), and so (F_{\text {app }}=k x). Figure (\PageIndex{1}) shows a graph of the applied force versus deformation (x) for a system that can be described by Hooke’s law. Work done on the system is force multiplied by distance, which equals the area under the curve or ((1 / 2) k x^{2}) (Method A in the figure). Another way to determine the work is to note that the force increases linearly from 0 to (kx), so that the average force is ((1 / 2) k x), the distance moved is (x), and thus (W=F_{\mathrm{app}} d=(1 / 2) k x=(1 / 2) k x^{2}) (Method B in the figure). Example (\PageIndex{1}): Calculating Stored Energy: A Tranquilizer Gun Spring We can use a toy gun’s spring mechanism to ask and answer two simple questions: (a) How much energy is stored in the spring of a tranquilizer gun that has a force constant of 50.0 N/m and is compressed 0.150 m? (b) If you neglect friction and the mass of the spring, at what speed will a 2.00-g projectile be ejected from the gun? Strategy for a (a): The energy stored in the spring can be found directly from elastic potential energy equation, because (k) and (x) are given. Solution for a Entering the given values for (k) and (x) yields [\begin{aligned} \mathrm{PE}_{\mathrm{el}} &=\frac{1}{2} k x^{2}=\frac{1}{2}(50.0 \mathrm{~N} / \mathrm{m})(0.150 \mathrm{~m})^{2}=0.563 \mathrm{~N} \cdot \mathrm{m} \ &=0.563 \mathrm{~J} \end{aligned} \nonumber] Strategy for b Because there is no friction, the potential energy is converted entirely into kinetic energy. The expression for kinetic energy can be solved for the projectile’s speed. Solution for b [\mathrm{KE}{\mathrm{f}}=\mathrm{PE}{\mathrm{el}} \text { or } 1 / 2 m v^{2}=(1 / 2) k x^{2}=\mathrm{PE}_{\mathrm{el}}=0.563 \mathrm{~J} \nonumber] [v=\left[\frac{2 \mathrm{PE}_{\mathrm{el}}}{m}\right]^{1 / 2}=\left[\frac{2(0.563 \mathrm{~J})}{0.002 \mathrm{~kg}}\right]^{1 / 2}=23.7(\mathrm{~J} / \mathrm{kg})^{1 / 2} \nonumber] Discussion (a) and (b): This projectile speed is impressive for a tranquilizer gun (more than 80 km/h). The numbers in this problem seem reasonable. The force needed to compress the spring is small enough for an adult to manage, and the energy imparted to the dart is small enough to limit the damage it might do. Yet, the speed of the dart is great enough for it to travel an acceptable distance. Exercise (\PageIndex{1}) Envision holding the end of a ruler with one hand and deforming it with the other. When you let go, you can see the oscillations of the ruler. In what way could you modify this simple experiment to increase the rigidity of the system? You could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment. Exercise (\PageIndex{2}) If you apply a deforming force on an object and let it come to equilibrium, what happened to the work you did on the system? It was stored in the object as potential energy. Section Summary [F=-k x, \nonumber] where (F) is the restoring force, (x) is the displacement from equilibrium or deformation, and (k) is the force constant of the system. [\mathrm{PE}_{\mathrm{el}}=(1 / 2) k x^{2}. \nonumber] Glossary This page titled 3.6: Spring Potential Energy is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax. Recommended articles The LibreTexts libraries are Powered by NICE CXone Expert and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement. For more information contact us at info@libretexts.org.
188007	https://healthy.kaiserpermanente.org/health-wellness/health-encyclopedia/he.herpangina-in-children-care-instructions.uh3752	Herpangina in Children: Care Instructions \| Kaiser Permanente Language English Englishselected Español Chinese Vietnamese Support Center Support Center Brokers Employers Help paying your bills My Account Sign inMenu Search Search Beginning of navigation menu RegisterSign in Search Search Contextual Menu Learn Shop Plans Doctors & Locations Health & Wellness Get Care Pay Bills Sign in New Member?Register Additional resources Brokers Employers Help paying your bills Support Center Language English Englishselected Español Chinese Vietnamese End of navigation menu Error There's a chat in progress. Back Print Herpangina in Children: Care Instructions Skip Navigation Home Click to begin search. Topic Contents Top of the page Your Care Instructions Herpangina (say "HUR-pann-JY-nuh") is an illness that is caused by a virus. It causes sores inside the mouth, a sore throat, and a high fever. Adults usually do not get it. Herpangina easily spreads to other children through exposure to a sick child's runny nose or saliva. While herpangina can make your child feel very ill for a few days, this illness usually clears up within a week. The most common concern is that your child may get dehydrated because it is painful to swallow. You can use home treatment to reduce your child's pain and discomfort. Since this illness is caused by a virus, antibiotic medicine is not used to treat it. Follow-up care is a key part of your child's treatment and safety. Be sure to make and go to all appointments, and call your doctor if your child is having problems. It's also a good idea to know your child's test results and keep a list of the medicines your child takes. How can you care for your child at home? Give acetaminophen (Tylenol) or ibuprofen (Advil, Motrin) for fever, pain, or fussiness. Read and follow all instructions on the label. Do not give aspirin to anyone younger than 20. It has been linked to Reye syndrome, a serious illness. Do not give your child over-the-counter antidiarrhea or upset-stomach medicines without talking to your doctor first. Do not give Pepto-Bismol or other medicines that contain salicylates, a form of aspirin, or aspirin. Aspirin has been linked to Reye syndrome, a serious illness. Make sure your child rests. Keep your child home as long as your child has a fever. Have your child drink plenty of fluids. Warm fluids such as soup, warm water, or warm lemonade may ease throat pain. Ice cream, gelatin dessert, and sherbet can also soothe the throat. If your child is eating solids, try offering bland foods, such as yogurt and warm cereal. Watch for and treat signs of dehydration, which means that the body has lost too much water. Your child's mouth may feel very dry. Your child may have sunken eyes with few tears when crying. Your child may lack energy and want to be held a lot. Your child may not urinate as often as usual. Give your child lots of fluids. This is very important if your child is vomiting or has diarrhea. Give your child sips of water or drinks such as Pedialyte or Infalyte. These drinks contain a mix of salt, sugar, and minerals. You can buy them at drugstores or grocery stores. Give these drinks as long as your child is throwing up or has diarrhea. Do not use them as the only source of liquids or food for more than 12 to 24 hours. Wash your hands after changing diapers and before you touch food. Have your child wash his or her hands after using the toilet and before eating. When should you call for help? Call 911 anytime you think your child may need emergency care. For example, call if: Your child has severe trouble breathing. Signs may include the chest sinking in, using belly muscles to breathe, or nostrils flaring while your child is struggling to breathe. Your child is confused, does not know where he or she is, or is extremely sleepy or hard to wake up. Your child passes out (loses consciousness). Your child has a seizure. Call your doctor now or seek immediate medical care if: Your child has a fever with a stiff neck or a severe headache. Your child still has a fever after 5 days of home treatment. Your child has signs of needing more fluids. These signs include sunken eyes with few tears, a dry mouth with little or no spit, and little or no urine for 6 hours. Watch closely for changes in your child's health, and be sure to contact your doctor if: Your child's mouth sores and sore throat get worse or are not improving. Your child does not get better as expected. Current as of: October 27, 2024 Author: Ignite Healthwise, LLC Staff Clinical Review Board All Ignite Healthwise, LLC education is reviewed by a team that includes physicians, nurses, advanced practitioners, registered dieticians, and other healthcare professionals. Topic Contents Go to top Your Care Instructions How can you care for your child at home? When should you call for help? Current as of: October 27, 2024 Author: Ignite Healthwise, LLC Staff Clinical Review Board All Ignite Healthwise, LLC education is reviewed by a team that includes physicians, nurses, advanced practitioners, registered dieticians, and other healthcare professionals. ©2024-2025 Ignite Healthwise, LLC. Healthwise, Healthwise for every health decision, and the Healthwise logo are trademarks of Ignite Healthwise, LLC. This information does not replace the advice of a doctor. Ignite Healthwise, LLC disclaims any warranty or liability for your use of this information. Your use of this information means that you agree to the Terms of Use and Privacy Policy. The Health Encyclopedia contains general health information. Not all treatments or services described are covered benefits for Kaiser Permanente members or offered as services by Kaiser Permanente. For a list of covered benefits, please refer to your Evidence of Coverage or Summary Plan Description. For recommended treatments, please consult with your health care provider. 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188008	https://www.asha.org/practice-portal/clinical-topics/selective-mutism/?srsltid=AfmBOorViVXu4P-e5vb7TpRuMANGWBkVNborhJkbyV1RuJ4IJNKzVJS-	Selective Mutism about Join ASHA ASHA Store My Account Login LOGOUT Toggle navigation American Speech-Language-Hearing Association Making effective communication, a human right, accessible and achievable for all. Type your search query here Careers Certification Publications Events Advocacy Continuing Education Practice Management Research Audiologists Speech-Language Pathologists Academic & Faculty Audiology & SLP Assistants Students Public ASHA/Practice Portal/Clinical Topics/ Selective Mutism View All Portal Topics Overview The scope of this page includes information about selective mutism occurring during preschool age through adolescence. Considerations for selective mutism as it extends into adulthood are briefly discussed. Selective mutism is a complex anxiety disorder that affects pragmatic language. Despite the term “selective,” individuals with selective mutism do not elect where to speak but are more comfortable speaking in select situations. According to the Diagnostic and Statistical Manual of Mental Disorders, Fifth Edition, Text Revision (American Psychiatric Association, 2022, p. 222), selective mutism is an anxiety disorder, and the diagnostic criteria for selective mutism are as follows: The child shows consistent failure to speak in specific social situations in which there is an expectation for speaking (e.g., at school), despite speaking in other situations. The lack of verbal communication interferes with educational or occupational achievement or with social communication. The duration of the mutism is at least 1 month (not limited to the first month of school). The failure to speak is not attributable to a lack of knowledge of, or comfort with, the spoken language required in the social situation. The mutism is not better explained by a communication disorder (e.g., childhood-onset fluency disorder) or exclusively due to the presence of autism spectrum disorder, schizophrenia, or another psychotic disorder. The onset of selective mutism typically occurs between 3 and 6 years of age, with diagnosis often occurring when the child enters school (Sharp et al., 2007). Different characteristics of the three primary factors (i.e., person, place, activity) can trigger a child’s mutism and influence the child’s ability to socially engage and communicate (Schwenck et al., 2022). Some examples are as follows: The child is generally able to speak to familiar people who they are comfortable with in familiar settings. With the same familiar person, the child may be verbal in one setting but mute in another setting. Within the same setting, the child may be verbal with some people but mute with others or may be mute during specific anxiety-producing activities (e.g., reading out loud, music class). Performance is most difficult when there is an expectation for speaking (mostly at school). Patterns of selective mutism can vary greatly and can interfere with academic, educational, and/or social performance. Speech-language pathologists are integral members of an interprofessional team and often collaborate with school-based teams (e.g., teachers, guidance counselor, school staff) and behavioral health professionals (e.g., school or clinical psychologist, psychiatrist, school social worker). Collaboration between the speech-language pathologist and assigned team members is particularly important for appropriate assessment and treatment because selective mutism is an anxiety-based disorder that can significantly impact the ability to access speech and language skills. Incidence and Prevalence The incidence of selective mutism refers to the number of new cases identified in a specified time period. Prevalence is the number of individuals who are living with selective mutism in a given time period. Accurate population estimates of selective mutism are difficult to ascertain due to the relative rarity of the condition, differences in sampled populations, variations in diagnostic procedures (e.g., chart review, standardized assessment), and the use of different diagnostic criteria (Busse & Downey, 2011; Sharkey & McNicholas, 2008; Viana et al., 2009). Most prevalence estimates for selective mutism range between 0.2% and 1.6% (Bergman et al., 2002; Chavira et al., 2004; Elizur & Perednik, 2003; Sharkey & McNicholas, 2012). Prevalence can be somewhat higher among immigrant children, language-minority children, and children with speech and language delays (Elizur & Perednik, 2003; Kristensen, 2000; Manassis et al., 2003; Steinhausen & Juzi, 1996). However, it is important to note that selective mutism must exist in all languages to confirm an accurate diagnosis in these populations (Toppelberg et al., 2005). There is currently a lack of consensus regarding the incidence and prevalence of selective mutism and gender assigned at birth. While most studies report that selective mutism affects more females than males by a ratio of about 1.5–2.5:1.0 (Cohan et al., 2008; Cunningham et al., 2004; Dummit et al., 1997; Kumpulainen et al., 1998), some studies report that it affects more males than females with a ratio of about 1.3:1.0 (Karakaya et al., 2008) or that there is no difference between genders (Bergman et al., 2002; Elizur & Perednik, 2003). Signs and Symptoms As with many anxiety disorders, children with selective mutism attempt to protect themselves from the discomfort they experience by avoiding the unpleasant activity (i.e., speaking and/or communicating). Varied characteristics and behaviors associated with selective mutism are a method of self-protection but may be interpreted as deliberately oppositional (e.g., “difficult” or “rude”; Kotrba, 2015). Children with selective mutism are often anxious about communication demands. This anxiety may impair the child’s ability to attend to class instruction and participate fully in school or social expectations (Klein et al., 2019). Misunderstanding such behaviors may complicate the identification of selective mutism. Individuals with selective mutism may demonstrate the following characteristics and behaviors in specific environments; however, they are not required for a diagnosis (Beidel et al., 1999; Doll, 2022; Kearney, 2010). Home Able to speak to one or more immediate family members. Exhibit difficulty speaking to extended family members or close family friends. May not be able to speak to immediate family members when visitors are present. May refuse to leave home to avoid social communication demands (e.g., school, birthday parties). May have an emotional–behavioral response (e.g., tantrum, withdrawal) when the child has an awareness of social and expressive communication expectations. School and Community Exhibit physical manifestations of anxiety: Fight, flight, or freeze response; rigid or restricted body movement; or minimal to no facial expression or eye contact. May display emotional–behavioral responses (e.g., clinging to the parent, behavioral meltdowns, school refusal). May be perceived as withdrawn, inattentive, or aloof. May have difficulty with language processing in specific situations due to a heightened level of anxiety. Unable to speak with adults or children in social or educational settings. Unable to respond nonverbally or verbally when spoken to; unable to initiate speech to provide information or comment. May use nonverbal methods of communication (e.g., body posture, eye gaze, facial expression, gesture) to respond to or initiate with people in settings where they are more comfortable and less anxious. Unable to initiate using any mode of communication to request help. Unable to speak at school, which impacts both educational performance and social development. Unable to speak to immediate family outside the home or when other people are present. Unable to speak with unfamiliar communication partners; may be able to use nonverbal modes of expression (e.g., eye gaze, head nod, pointing) over time as they become more comfortable in the social environment. Additional conditions that may be associated with selective mutism are as follows (Capozzi et al., 2018): enuresis (i.e., urine accidents) and encopresis (i.e., bowel accidents) eating challenges (e.g., eating with or in front of others, food selectivity) sleep disturbance Causes No single cause of selective mutism has been identified, and its causes may be multifactorial (Cohan, Price, & Stein, 2006). The following factors may coexist and play a role in selective mutism: Psychological factors, such as social phobia, separation anxiety, and obsessive-compulsive disorder (Beidel & Turner, 2007; Black & Uhde, 1995; Manassis et al., 2003). Hereditary or genetic predisposition of selective mutism and social anxiety disorder (Black & Uhde, 1995; Cohan, Price, & Stein, 2006; Viana et al., 2009). Family and environmental factors, such as reduced opportunities for social contact, parenting style, or reinforced avoidance behaviors (Viana et al., 2009). Neurological/neurodevelopmental vulnerabilities, such as delays in achieving speech, language, or fine and gross motor milestones (Viana et al., 2009). Overactive autonomic nervous system response that impacts physiological, sensory, and emotional–behavioral responses (e.g., Melfsen et al., 2021). Other factors, such as shy or timid temperament (American Psychiatric Association, 2022; Steinhausen & Juzi, 1996). Roles and Responsibilities Speech-language pathologists (SLPs) play an integral role in the screening, assessment, diagnosis, and treatment of individuals with selective mutism. The professional roles and activities in speech-language pathology include clinical services (diagnosis, assessment, planning, and treatment); prevention and advocacy; and education, administration, and research. See ASHA’s Scope of Practice in Speech-Language Pathology. The following roles are appropriate for SLPs: Educate other professionals on the needs of individuals with selective mutism and the role of the SLP in diagnosing and managing selective mutism. Screen individuals who present with language and communication difficulties to determine the need for further assessment and/or referral for other services. Conduct a comprehensive, culturally and linguistically appropriate assessment of speech, language, and communication. Aid in diagnosing the presence or absence of selective mutism with an interdisciplinary team. Refer to other professionals to rule out other conditions, determine etiology, and facilitate access to comprehensive services. Make decisions about the management of selective mutism. Develop treatment plans, provide treatment, document progress, and determine appropriate dismissal criteria. Counsel individuals with selective mutism and their care partners regarding communication-related issues and provide education aimed at preventing further complications relating to selective mutism. Consult and collaborate with other professionals, family members, care partners, and others to facilitate program development and to provide supervision, evaluation, and/or expert testimony, as appropriate. Remain informed of research in selective mutism and help advance the knowledge base related to the nature and treatment of selective mutism. Advocate for individuals with selective mutism and their families/care partners at the local, state, and national levels. Serve as an integral member of an interdisciplinary team working with individuals with selective mutism and their families/care partners. As indicated in the Code of Ethics (ASHA, 2023), clinicians who serve this population should be specifically educated and appropriately trained to do so. SLPs take part in the aspects of the profession that are within the scope of their professional practice and competence. If an SLP has advanced training in and knowledge of selective mutism, diagnosis is possible in accordance with existing state credentialing laws. However, a diagnosis made by an interdisciplinary team ensures that a full differential diagnosis was completed. Assessment Screening Screening for selective mutism is conducted whenever selective mutism is suspected or as part of a comprehensive speech and language evaluation for a child with communication concerns. If a parent or care partner reports that a child is communicating successfully at home but not in one or more settings, the speech-language pathologist (SLP) may want to consider screening for selective mutism. Screening typically includes norm-referenced parent/care partner and teacher report measures, competency-based tools such as interviews and observations, and hearing screening to rule out hearing loss as a possible contributing factor. See ASHA’s Practice Portal pages on Hearing Loss in Children and Hearing Loss in Adults for more information. Comprehensive Assessment See theSelective Mutism Evidence Map for pertinent scientific evidence, expert opinion, and client/ care partner perspective. Please see ASHA’s resource, Assessment Tools, Techniques, and Data Sources, for information on the elements of a comprehensive assessment, considerations, and best practices. Information specific to these practices in the comprehensive assessment of individuals with selective mutism is discussed below. Assessment of children with selective mutism involves a collaborative approach with an interdisciplinary team, which may consist of a pediatrician, a psychologist or psychiatrist, an SLP, a teacher, a school social worker or guidance counselor, and family/care partners. During the evaluation, parents/care partners may need to help elicit verbal output. The SLP can also involve parents/care partners by requesting a video recording of the child’s communicative behavior at home and then compare the child’s behavior in a clinical or school setting. Video recordings may also be used for subsequent language sample analysis. Several techniques can be used throughout assessment to reduce stress on the child, increase participation, and improve the quality of assessment findings. See “Meeting the Child” section below for more details. Case History A diagnostic interview with parents/care partners and teachers is conducted without the child present to help gather information about the following: Any suspected co-occurring disorders (e.g., schizophrenia, autism spectrum disorder). Environmental factors (e.g., amount of language stimulation). Circumstances of communication (Kotrba, 2015): With whom does the child communicate? In what circumstances is the child most likely to communicate? Where and in what settings is the child able to communicate? How does the child communicate—gestures? writing? sounds? whispering? short responses? The child’s symptom history (e.g., onset and behavior). Family history (e.g., psychiatric, personality, and/or physical problems). Speech and language development (e.g., how well does the child express themself and understand others?). Educational history, such as information on academic reports, parent/care partner and teacher comments, previous testing (e.g., psychological), and standardized testing. If the child is multilingual, the SLP will need to obtain the following information (Mayworm et al., 2015; Toppelberg et al., 2005): What languages does the child use now, and with whom? How well does the child understand the different languages to which they are exposed? Does the child use their primary language successfully outside the home environment? If so, in what settings and with whom? Speech and Language Evaluation During the speech and language evaluation, the SLP gathers information on the child’s language comprehension; expressive language ability; nonverbal communication (e.g., pretend play, drawing); pragmatic language, including situations, speakers, and contexts that encourage or discourage speech (Hungerford, 2017); functional communication ability across various circumstances and settings (Kotrba, 2015; Selective Mutism Anxiety Research and Treatment Center & Shipon-Blum, 2012); and oral–motor functioning, including strength, coordination, and range of motion of the lips, jaw, and tongue. A child with selective mutism might not be able to participate in formal evaluation activities, and they may lack verbal responses and use nonverbal responses (e.g., pointing or gesturing). These behaviors provide diagnostic information regarding the child’s response to social communication. The SLP can also use audio or video recordings from home to supplement parent/care partner descriptions. Any discrepancy between the child’s communication at home and their communication in public may suggest an overarching problem of difficulty with social language. In some situations, it may be feasible to train parents/care partners, or other familiar adults with whom the child is able to speak, to administer standardized tests (Klein et al., 2013). In these cases, parents can administer test items with the SLP in or out of the room to promote verbal responses from the child; however, the SLP is still responsible for scoring and interpreting test performance. It is the responsibility of the SLP to review the examiner manual to see if parents are listed as a potential assessor based on the prescribed educational and expertise requirements. Using standardized tools in a nonstandardized way may invalidate standardized scores; however, information gleaned from the assessment can still be reported. Speech Sound Production Speech sound disorders may occur in children with selective mutism and may magnify the child’s anxiety of interacting with others (Anstendig, 1999). These children may benefit from direct assessment and treatment with parental involvement and support. See ASHA’s Practice Portal page on Speech Sound Disorders – Articulation and Phonology for more information related to speech assessment and treatment. Voice Some children with selective mutism have reported that they do not like their voice, they don’t want their voice to be heard, or their voice “sounds funny” (Henkin & Bar-Haim, 2015; Vogel et al., 2019). Voicing requires control and coordination of airflow and the vocal mechanism that may be disrupted by their level of anxiety and may present a challenge (e.g., increased laryngeal tension) for an individual with selective mutism (Ruiz & Klein, 2018). Even in cases where a child verbalizes in front of the clinician, this speech may be produced in a whisper, at a decreased vocal intensity, or in an altered vocal quality. The SLP documents vocal quality at the time of the initial evaluation and then reassesses during intervention. The altered vocal quality can lessen as anxiety decreases. Clinicians may also want to evaluate the level of vocal tension during the assessment. Language Ability Receptive and expressive language skills may vary in children with selective mutism. For example, expressive–receptive and receptive language disorders may coexist with selective mutism (e.g., Viana et al., 2009). Some children with selective mutism with average receptive language abilities may demonstrate shorter, less detailed, and more linguistically simplistic narratives (McInnes et al., 2004). These subtle deficits in expressive language are theorized to be a compilation of anxiety, mild language deficits, and lack of experience with high-level language skills. It may be beneficial to use low-stress tasks, such as a picture-pointing task when assessing language ability. If the child is unable to speak, SLPs acknowledge and respond to the child’s gestures or written/typed responses, assess the effectiveness of the child’s attempts at nonverbal communication, and assess the child’s behaviors when engaged in communication. There may be cultural differences within nonverbal communication that the SLP needs to consider when assessing communication. See ASHA’s Practice Portal pages on Cultural Responsiveness and Social Communication Disorder for further information as well as ASHA’s Practice Portal page on Spoken Language Disorders for more information related to language assessment and treatment. Cognitive Abilities While children with selective mutism may demonstrate average cognitive and academic abilities (Manassis et al., 2003; McInnes et al., 2004), some children with selective mutism may have impaired visual memory or auditory–verbal memory (Kristensen & Oerbeck, 2006; Manassis et al., 2007). Difficulty responding using verbal and nonverbal responses, avoidance of interacting with unfamiliar adults, and slowness to respond can lead to lower test scores and misinterpretation of the child’s ability (Kotrba, 2015). Social Communication Skills Social communication skills for children with selective mutism typically appear limited outside the home and other familiar environments and, at times, may appear limited in the home as well. Research is not clear as to whether children with selective mutism have pragmatic language challenges beyond avoiding communicating in certain circumstances outside the home setting (McInnes et al., 2004). Social immaturity is not uncommon because the child with selective mutism has fewer social interactions and may lack social awareness (Kotrba, 2015). Children with selective mutism can display decreased nonverbal and verbal indicators of social engagement, such as proxemics, facial expressions, gestures, eye contact, turn-taking, participation in joint activity routines, and joint attention (Hungerford et al., 2003). Home video samples may be helpful in assessing social communication variations across settings. Please see ASHA’s Practice Portal page on Social Communication Disorder for more information related to assessment and treatment. Assessment Considerations Meeting the Child Prior to initiating speech and language services, the SLP can provide parental or teacher questionnaires regarding selective mutism or conduct a diagnostic interview with parents, care partners, and teachers to prepare for the initial meeting. Clinicians may consider meeting the child one-on-one or with the parent/care partner present prior to formal assessment. Conditions of meeting the child with selective mutism may vary based on the school, home, or private practice setting. The clinician can reassure the parents/care partners and child of the expectations for the first meeting, such as the child will not be pressured to speak, there will be no interruptions, and no one else will be present in the meeting setting (Doll, 2022). First sessions may be informal and flexible. The SLP may develop a relationship with the child prior to the evaluation by scheduling two to three sessions for age-appropriate recreational or play-based interactions without the expectation for speech. Clinicians may play at the child’s level and follow their lead with open-ended, creative play involving arts and crafts, building blocks, and/or board games (Kotrba, 2015). The child and parent/care partner may benefit from playing in the assessment room for 5–10 minutes without the SLP in the room to increase comfort and familiarity with the setting. During this time, parents are encouraged to actively engage with their child or ask their child questions to promote verbal output. The SLP can observe if an observation room or video is available. This allows for comparison of the child’s communication with and without an unfamiliar person in the area. Then, the SLP can enter the room, allow the child and parent/care partner to continue playing for several minutes, and then enter the child’s circle of play (Middendorf & Buringrud, 2009). The following defocused communication strategies can help build a positive rapport and establish trust (Oerbeck et al., 2014): Minimize eye contact. Sustaining eye contact from unfamiliar people can make children with selective mutism uncomfortable. Maintain a calm demeanor. Make environmental modifications. For example, some children may prefer that the SLP sit by their side rather than face-to-face, whereas this may be too close for others. Create opportunities for joint attention using an activity that the child enjoys. Think aloud by providing behavioral descriptions of what the child is doing, rather than by asking direct questions (e.g., “I see that you’re playing with the truck!” instead of “What are you doing?”). Use phrases and terms that encourage the child to communicate, including using the terms “words” or “voice” rather than “talk” or “speak .” The latter two words may have negative connotations for the child (Kotrba, 2015). Also, encourage the child to show, gesture, write, or draw if they are not able to speak (Schum, 2006). Reflect back language that the child shares. Offer choices or options to respond instead of open-ended or yes/no questions. Allow plenty of time for the child to process and respond rather than talking for the child. Continue the conversation, even when the child does not respond verbally. Receive the child’s responses in a neutral way. Within an evaluation process, it is also important to be mindful of the communication demands required for specific tasks completed in the evaluation. An SLP may need to modify the order in which they present materials, starting with tasks with no verbal communication demands and moving to verbal communication tasks based on the child’s presentation and responsiveness. Visit the Selective Mutism Association’s Educator Toolkit for more information. Interprofessional Collaboration and Referrals During evaluation and treatment, the SLP may collaborate with and refer to the following professionals: audiologist behavior analyst/behavioral specialist extended family and/or care partners family guidance counselor pediatrician psychiatrist school or clinical psychologist social worker teacher The SLP’s role on the evaluation team is to identify and describe (a) the child’s communication skills and coexisting communication disorders and (b) the impact of those skills on the child’s ability to consistently participate in various settings (Kotrba, 2015). If the SLP is the first professional that a family encounters, the SLP can initiate the collaborative process and provide referrals to behavioral health professionals with training and experience in working with children with anxiety disorders (e.g., behavioral therapists, cognitive therapists). A collaborative interprofessional team that develops a treatment plan and communicates regularly can optimize treatment outcomes and promote generalization of effective communication skills across people, settings, and situations. See ASHA’s webpage on Interprofessional Education/Interprofessional Practice (IPE/IPP). Differential Diagnosis The major difference between selective mutism and other disorders is that the child with selective mutism can talk in certain situations but not others due to anxiety (Kotrba, 2015). SLPs consider whether a child’s absence of speech may be due to a communication disorder, a developmental disorder, or other psychiatric disorders (Kearney, 2010). Diagnosis by an interdisciplinary team, including behavioral health care professionals, ensures a complete differential diagnosis process. Although selective mutism is not better explained by a communication disorder or psychological disorder, selective mutism may occur simultaneously with the following (Driessen et al., 2020; Steffenburg et al., 2018): social anxiety generalized anxiety separation anxiety autism specific phobia obsessive-compulsive disorder speech and/or language disorder (Viana et al., 2009) SLPs also consider if the child is immersed in a new language environment because acquiring another language is a complex process. When children are exposed to a new language, they may experience a brief silent period in which they are quiet and speak little. Although children may not speak in situations in which the new language is used, children with typical second-language acquisition demonstrate appropriate social communication skills in settings and with people who speak the child’s primary language (Doll, 2022). When working with a multilingual child, diagnosing selective mutism depends on understanding typical multilingual child development. Multilingual children with true selective mutism present with mutism in both languages, in several unfamiliar settings, and for significant periods of time (Toppelberg et al., 2005). Interviewing parents/care partners to determine if the child speaks in their primary language successfully outside of the home environment is important information for the SLP to gather to inform differential diagnosis (Mayworm et al., 2015). Please see ASHA’s Practice Portal page on Multilingual Service Delivery in Audiology and Speech-Language Pathologyfor further information. It is necessary to collaborate with an interpreter or a translator if the SLP does not speak the language(s) of the child. The SLP should be mindful of the number of people in the room and consider how the introduction of an additional person may impact performance. The SLP may need to consider asking a family member to act as an interpreter in this circumstance so as not to create additional anxiety or stress for the child. See ASHA’s Practice Portal page on Collaborating With Interpreters, Transliterators, and Translators for more information. Some children will not speak after a traumatic event or ongoing social–emotional difficulties, such as parental divorce. Children who do not speak following trauma are mute in all settings (Manassis et al., 2003). If the child spoke well prior to these events, then a diagnosis of selective mutism may not be appropriate. Instead, the child may require assistance in adjusting to the trauma or other life challenges (Kearney, 2010); in which case, referral to a behavioral health professional is appropriate. See ASHA’s resource on trauma-informed care. Determining Educational Eligibility Interprofessional practice and family involvement are essential in assessing and diagnosing selective mutism; the SLP is a key member of a multidisciplinary team. The multidisciplinary team reaches a consensus that assessment results are consistent with the diagnostic characteristics of selective mutism. Within school settings, children can be supported through informal services, Section 504 plans, or individualized education programs. There is no single, preferred, consistent diagnostic category for children and youth with selective mutism in the school setting. Eligibility for special education services under the Individuals with Disabilities Education Improvement Act of 2004 could be determined to fall within the disability categories of Other Health Impairment, Speech-Language Impairment, or Emotional Disturbance/Disability. The level of accommodations will depend on the functional impact of selective mutism in the school setting. For example, a newly identified student may need regular access to a “buddy,” someone who the child can speak to throughout the day, versus a student farther in the treatment process may need opportunities to work in small groups with less familiar peers (Doll, 2022). Some children with selective mutism may benefit from the classroom accommodations offered through a Section 504 plan, whereas others may need more direct services within special education to address the communication concerns. Treatment See the Selective Mutism Evidence Map for pertinent scientific evidence, expert opinion, and client/ care partner perspective. Early intervention for selective mutism is key to remediation. Communication partners sometimes speak for the child with selective mutism when the child demonstrates distress. This “rescuing” behavior may discourage the child’s future speech attempts and results in negatively reinforcing the child’s avoidance of speaking. Treatment works to break the cycle of negative reinforcement. Consistency in the intervention and expectations, at home and in school, of everyone on the team is important. Speech-language pathologists (SLPs) work to provide predictability and control for children with selective mutism, which may decrease anxiety and improve self-image based on mastery of skills in a variety of settings (Kotrba, 2015). Pharmacological treatment may be prescribed by the individual’s treating pediatrician or psychiatrist (Manassis et al., 2016). Clinicians consider the behavioral influences and side effects of medications (e.g., selective serotonin reuptake inhibitors) on speech and language interventions and collaborate with behavioral health professionals, as appropriate. Monitoring the individual’s success at each level of the treatment plan through ongoing assessment will determine the overall success for consistent communication with a variety of people in different settings. Anxiety and avoidance behaviors will indicate the need to break down communication steps, locations, or audience size into more manageable steps of facing a fear (Kotrba, 2015). Behavioral and Cognitive-Behavioral Strategies and Definitions The behavioral perspective views selective mutism as a learned behavior that the individual has developed as a coping mechanism for anxiety. The purpose of treatment is to decrease anxiety and increase verbal communication in a variety of settings, incorporating practice and reinforcement for speaking in subtle, nonthreatening ways (Camposano, 2011; Cohan, Chavira, & Stein, 2006; Kotrba, 2015). Reinforcements may be verbal (e.g., praise); tangible (e.g., toys, special outings, belongings); and/or privileges (e.g., staying up later, having additional time to play a video game, choosing a movie or board game to enjoy with a parent/care partner). The trained behavioral health professional, SLP, and school staff collaborate to incorporate behavioral and cognitive-behavioral strategies into interventions across settings for children with selective mutism. These strategies include the following. Exposure-based practice involves the child saying words in gradually but increasingly difficult or anxiety-provoking situations. Exposure-based practice aims to (a) replace anxious feelings/behaviors with more relaxed feelings and (b) increase the child’s feelings of independence by gradually improving their ability to speak in different situations (Kearney, 2010; Middendorf & Buringrud, 2009). Systematic desensitization involves the use of relaxation techniques along with gradual exposure to subsequently more anxiety-provoking situations (Cohan, Chavira, & Stein, 2006; Kearney, 2010). Stimulus fading involves gradually increasing exposure to a fear-evoking stimulus (e.g., the number of people present or the presence of an unfamiliar person in the room while the child is speaking). For example, if a child does not speak in school, then a child’s parent would be brought into the child’s classroom. When the child speaks to the parent, the clinician slowly brings a new person into the room (e.g., a teacher). This process usually includes rewarding the child when they are speaking in the presence of someone to whom they do not typically speak (Middendorf & Buringrud, 2009; Viana et al., 2009). Contingency management, positive reinforcement, and shaping includes (a) providing positive reinforcement contingent upon verbalization and (b) reinforcing attempts and approximations to communicate (i.e., shaping) until such attempts are shaped into verbalizations, with the goal of making verbalizing more rewarding than not responding. Shaping is commonly used in combination with contingency management and positive reinforcement. Treatment Options and Techniques The treatment options below include approaches that are within the scope of an SLP, may involve an SLP in an interprofessional team, or may require additional training. Augmentative and Alternative Communication Augmentative and alternative communication (AAC) involves supplementing or replacing natural speech with aided symbols (e.g., pictures, line drawings, tangible objects, and writing) and/or unaided symbols (e.g., gestures). Some children who have been diagnosed with selective mutism may adapt an AAC system to facilitate classroom communication. Some individuals may use AAC only in the initial stages of intervention. Some individuals may use AAC only in the initial stages of intervention, with AAC faded over time as an individual with selective mutism finds more success with verbal communication. Other clients and their care partners may have long-term preferences for AAC as their primary communication method. In such cases, language and communication treatment goals incorporating the client’s preferred communication modality may be appropriate. Please see ASHA’s Practice Portal page on Augmentative and Alternative Communication for further information. Augmented Self-Modeling In augmented self-modeling, the individual with selective mutism watches a video segment or listens to an audio segment in which they are engaging in a positive verbal interaction in a comfortable setting (typically, at home). This approach may be paired with additional behavioral and cognitive-behavioral strategies, such as positive reinforcement and stimulus fading (Kehle et al., 2011). This process may also involve making a video recording of the child and editing it so that the video shows the child speaking in settings where the child does not speak, such as the classroom. The child watches and listens to themself speaking to learn to think positively about speaking in front of others. DIR Floortime® DIR (Developmental, Individual Differences, Relationship-Based) Floortime is a developmental and interdisciplinary framework based on functional emotional developmental capacities (FEDCs). It utilizes the concepts of self-regulation, attention, engagement, intentional communication, and purposeful problem-solving communication. Goals are based on evaluating the child’s FEDC (i.e., moving from nonresponsive to using gestures, to making sounds, and then to being verbal) and supporting individual differences (sensory processing, praxis, speech and language challenges, visual–spatial processing, postural stability) to move the child up the FEDC ladder. It incorporates sensorimotor and play-based activities (often having co-treatments with an occupational therapist) and instruction regarding antianxiety strategies from a social worker or other behavioral and mental health professionals (Fernald, 2011). ECHO Program ECHO: A Vocal Language Program for Easing Anxiety in Conversation (Ruiz et al., 2022) aims to support individuals, who are of late elementary age through adolescence, who may experience social anxiety related to speaking in certain situations or with certain individuals. This program, which can be implemented by SLPs, bridges the gap from vocalization to conversation. The following three modules include both face-to-face and computer-based interactive activities: Module 1—Voice Control: The individual learns how to initiate voice, modulate intonation and volume, and produce speech sounds in words and sentences. Module 2—Social Pragmatic Language: The individual learns to use language for different purposes, change language for the listeners or situation, and follow rules for conversation and storytelling. Module 3—Role Play: The individual uses the skills learned in the previous two modules to participate in conversational role plays that simulate real life (e.g., school, home, social, public). A cognitive behavioral therapy framework is used to help reduce cognitive distortions (e.g., “Everyone will laugh at me if I talk because my voice sounds funny”). EXPRESS Program EXPanding Receptive and Expressive Skills through Stories (EXPRESS): Language Formulation in Children With Selective Mutism and Other Communication Needs (Klein et al., 2018) aims to expand receptive and expressive language skills with five levels of communication (i.e., nonvocal communication through spontaneous vocalization). The EXPRESS approach, which supports the Common Core State Standards for English Language Arts, uses classic children’s stories to correspond with each module to help expand vocabulary and grammar, engage in question–answer routines, improve sentence formulation, and generate narrative language. Integrated Behavioral Therapy for Selective Mutism Integrated behavioral therapy for selective mutism, originally developed for children ages 4–8 years, aims to increase successful speaking behaviors in anxiety-provoking situations, habituate speaking-related anxiety, and positively reinforce speaking (Bergman, 2013). Using a combination of behavioral techniques (e.g., stimulus fading, shaping, desensitization) and exposure-based interventions, the clinician systematically and gradually exposes the child to increasingly difficult speaking situations. This program takes place over 24 weeks during the school year. Intensive Group Behavioral Treatment Intensive Group Behavioral Treatment focuses on providing a full course of intervention for selective mutism in a condensed period, such as a 1-week summer camp program (Cornacchio et al., 2019). In a 1:1 child–staff ratio, trained counselors and at least one clinical psychologist incorporate aspects of the parent–child interaction therapy and cognitive behavioral therapy in a group setting. Components of the Intensive Group Behavioral Treatment may also include parent training and coaching. Parent–Child Interaction Therapy for Children With Selective Mutism Parent–child interaction therapy for children with selective mutism aims to increase verbal interactions in social settings and decrease avoidance behaviors (Cotter et al., 2018). Intervention includes the following two phases that involve specific techniques, procedures, and tasks to promote verbalization: Child-directed interaction (CDI)—This phase focuses on building the child’s comfort with the communication partner and environment (Doll, 2022). The communication partner uses strategies (e.g., labeled praise, reflection, enthusiasm) to provide verbal models of communication and to take away the pressure of the child speaking. Verbal-directed interaction (VDI)—Once rapport is established, VDI is introduced to prompt the child’s speech. Exposure tasks are used to begin generalizing speech to new environments and people (Cotter et al., 2018). Clinicians continue to address CDI skills during the VDI phase. Social Communication Anxiety Treatment® Social Communication Anxiety Treatment (S-CAT) uses a multimodal approach to increase the social engagement, verbal communication, and confidence of the person with selective mutism (SMart Center, n.d.). S-CAT focuses on reducing the child’s anxiety about speaking and the parent/care partner’s rescuing behaviors that enable the child’s avoidance behaviors (Klein et al., 2016). Using behavioral and cognitive-behavioral strategies, the clinician helps the individual move through the four stages of communication (i.e., noncommunicative, nonverbal, transition to verbal, and verbal). The clinician can incorporate the Ritual Sound Approach® into the S-CAT program to systematically increase the child’s comfort with making sounds and words (Shipon-Blum, n.d.). In the Ritual Sound Approach, the clinician teaches and models how sounds are made through a mechanical perspective. Once the child with selective mutism is comfortable with making nonspeech sounds, the clinician can gradually introduce different phonemes. Eventually, the clinician can help the child blend the phonemes into simple words. Involvement of the child, parent/care partner, and school staff is integral to establishing skills across all environments and communication partners. Social–Pragmatic Approach This integrated approach emphasizes participation in social engagement (nonverbal and verbal) at increasingly difficult levels. Shaping and reinforcement, in the context of interactive routines, are used to move the child with selective mutism through building acceptance of joining social activities (e.g., games, art, social play); using nonverbal communication during social activities (reaching, pointing, gesturing “yes” or “no,” facial expression); and using a hierarchy of sound production (i.e., from nonspeech sounds to speech sounds to using words). The clinician considers the hierarchy of language functions at the word level and beyond. For example, the child may begin with answering noninvasive questions (e.g., “What color is your shirt?”) and progress to answering increasingly more personal questions (likes/dislikes, family and friends) before eventually being able to ask noninvasive personal questions and participate in conversation over multiple turns. Tasks may need to be simplified when the child changes communication partners or contexts. The approach considers different variables of the communication context, as follows: who the child is communicating with (familiar vs. unfamiliar) where the child is communicating (e.g., treatment room, school library, classroom before school starts, in small group inside the classroom) the purpose of communication (e.g., regulating another’s behavior, social interaction, joint attention) the ability to manage conversation (i.e., multiple turns, repair conversation, select/maintain/terminate conversation, take another’s perspective; Hungerford et al., 2003) Generalization/Carryover Several of the treatment programs described above incorporate ways to generalize speaking in new environments and with new communication partners. Overall, generalizing spontaneous speech to different settings and communication partners may involve (Kotrba, 2015; Middendorf & Buringrud, 2009) having the individual with selective mutism rate situations and people from “most difficult” to “least difficult” in a hierarchy; preparing and reassuring the individual with selective mutism of their abilities by thoroughly explaining the plan to generalize their skills; establishing a keyworker, an adult trained in behavioral and cognitive-behavioral strategies, in the school setting; changing only one variable at a time (e.g., either the location or the people present); moving from structured and carefully planned occurrences to spontaneous and unplanned situations; and/or practicing frequently and repetitively. Interprofessional Collaboration Continued collaboration between the SLP and behavioral health professionals, classroom teachers, and the family is necessary for treatment continuity, clear delineation of roles and responsibilities, and appropriate hierarchical goal setting. Having the SLP on the team helps the child with selective mutism gain confidence in what they may perceive as decreased communication skills (Dow et al., 1995). The SLP can work with the child’s teacher and school staff to use the following strategies: Form small, cooperative learning groups that include the child’s preferred peers. Help the child communicate with peers in a group by first using nonverbal methods (e.g., signals, gestures, pictures, writing) and then gradually working toward verbal participation. Watch for opportunities to reinforce small improvements. Reassure others that the child is still comprehending even if they are not talking. Try to minimize symptoms—the child may not want to talk, but they can point, show, gesture, or draw. Avoid speaking forthe child, justifying the child’s silences, or pressuring the child to speak, all of which may reinforce mutism. Support peer acceptance of nonverbal participation in classroom and recreational activities. Find nonverbal jobs that the child with selective mutism can perform to build confidence. Maintain the classroom routine and try making the same request of the child at the same point in the schedule to decrease anxiety. Strategize speaking assignments that the individual agrees to complete with the teacher prompting or reminding the student as necessary (Bergman, 2013). Arrange one-on-one time with the teacher and student so that they can seek assistance quietly rather than in front of peers (Richard, 2011; Schum, 2002, 2006). See also ASHA’s webpage on Interprofessional Education/Interprofessional Practice (IPE/IPP). Special Considerations Structuring Treatment Initially, children may require individual treatment sessions to establish rapport and practice relaxation techniques and pragmatic skills in a comfortable setting. Typically, treatment progresses from CDI to VDI. During CDI, the adult observes the child performing an activity that the child chooses, and then, the adult joins in by imitating, describing, and demonstrating enjoyment without asking questions, giving commands, or using negative talk. VDI allows adults and peers to ask questions, direct play, and give instructions (Kurtz, 2015; Mac, 2015). A trained keyworker could also provide behavioral interventions, CDI, and VDI throughout the school environment (Kotrba, 2015). English Language Learners The SLP and the interprofessional team incorporate the following considerations when an English language learner is suspected or confirmed of having selective mutism (Mayworm et al., 2015): early identification and intervention the language(s) that providers use to implement intervention and the role of language development in the type of treatment willingness and training of team members to implement interventions in multiple contexts When treating an English language learner with selective mutism, the SLP is aware of possible stressors within the child’s school setting that will need to be addressed through staff development, interventions, and accommodations (Toppelberg et al., 2005). These may include lack of class support for learning another language, the potential for negative views of the child’s culture or language used at home, and/or limited communication between the parent/care partner and the school. See Multilingual Service Delivery in Audiology and Speech-Language Pathologyand Cultural Responsivenessfor more information related to providing culturally and linguistically appropriate services. Treating Concomitant Speech and Language Problems Children with selective mutism can also have a concomitant communication delay, disorder, or weakness (Richard, 2011). Children with selective mutism may avoid speaking out of fear of being teased regarding speech sound production or vocal quality (Anstendig, 1999). Evidence of a concomitant communication disorder is not restricted to specific settings or social situations, even when co-occurring with selective mutism. Before addressing specific speech and language deficits, the child may benefit from addressing only selective mutism goals to increase their confidence in communicating and to establish rapport with the SLP. Adolescents and Adults Selective mutism may be resolved in childhood; however, selective mutism in childhood may persist into adolescence and adulthood, or it may develop into another anxiety disorder or phobia (Steinhausen et al., 2006). Adolescents and adults with selective mutism may report not wanting to talk because they do not see the benefits of speaking. At times, young adults may have the desire to speak but are unable to speak because of significant anxiety or lack of strategies (Walker & Tobbell, 2015). The inability to speak may bring about feelings of shame, isolation, frustration, and hopelessness because they have difficulty fulfilling expected social roles. Older individuals often develop strategies to avoid talking and may have defined themselves as being primarily nonverbal. Motivational interviewing is a client-centered counseling technique that helps the adolescent or adult explore and resolve ambivalence through discussion and aims to increase internal motivation for behavioral change. A motivational interview for someone with selective mutism could include asking about the positive and negative aspects of selective mutism, exploring life goals and values, and then determining goals (Kotrba, 2015; Rollnick & Miller, 1995). The client may be more comfortable with sharing their experiences and concerns through online interview methods (Walker & Tobbell, 2015). Coding and Reimbursement Payment and coverage of services related to the evaluation and treatment of selective mutism varies based on factors such as the patient’s diagnosis(es), the payer (e.g., Medicare, Medicaid, or commercial insurance), and the patient’s specific health insurance plan. It is important for clinicians to understand coverage policies for the payers they commonly bill, to verify coverage for each patient prior to initiating services, and to be familiar with correct diagnosis and procedure coding for accurate claims submission. Clinicians use International Classification of Diseases and Related Health Problems, 10th Revision, Clinical Modification (ICD-10-CM) codes to describe the patient’s diagnosis and Current Procedural Terminology codes to describe related evaluation and treatment services. The term “selective mutism” is used to classify this diagnosis within the ICD-10-CM family of codes for behavioral and emotional disorders with onset usually occurring in childhood and adolescence. Clinicians may also report specific diagnosis codes for speech, language, and communication disorders, as needed. Payer policies may outline specific guidelines based on this diagnosis, such as who may assign a diagnosis for behavioral and emotional disorders and what types of services are covered. For example, some payers may only cover services related to this diagnosis under a mental health benefit; this could require an initial evaluation by a physician or mental health professional and limit coverage of evaluation and treatment by an SLP. For more information about coding, see the following ASHA resources: ICD-10-CM Diagnosis Codes Related to Speech, Language, and Swallowing Disorders [PDF] Current Procedural Terminology (CPT) Codes for Speech-Language Pathology Services Service Delivery Format Format refers to the structure of the treatment session (e.g., group vs. individual) provided. Children may require individual treatment sessions initially, depending on the strategies and techniques being applied, to establish rapport and to practice relaxation techniques and pragmatic skills in a safe, comfortable setting. Small-group treatment can facilitate communication with peers, beginning with nonverbal play using scripted interactions involving single words and phrases and moving toward the goal of speaking spontaneously (Klein & Armstrong, 2013). Another format involves forming groups of individuals with selective mutism who are of similar age, cognitive functioning, and speech-language skills. Groups may need to be adjusted based on each individual’s progress (Kearney, 2010). Some individuals may prefer telepractice to receive treatment. Many of the treatment strategies noted above can be implemented through virtual means (Busman et al., 2020; Hong et al., 2022). The clinician can use technology (e.g., mobile device) to coach the child or care partner through in vivo exposure activities in school or in the community. Telepractice provides increased access to services for children who may not otherwise have access to trained professionals with experience treating selective mutism. It can also allow for increased collaboration between professionals and family members in different settings. See ASHA’s Practice Portal page on Telepractice for more information. Provider Provider refers to the person offering the treatment (e.g., SLP, speech-language pathology assistant, care partner). In treating selective mutism in a school setting, an established and trained keyworker may be the provider of interventions (Kotrba, 2015). A keyworker is an adult in the school setting who is trained to provide consistent behavioral interventions to the student. The keyworker can help the student generalize skills throughout the school environment and communicate with the treatment team. Dosage Dosagerefers to the frequency, intensity, and duration of service. Intensive treatment sessions for selective mutism may be helpful for some individuals and can take place in a variety of settings. For example, in the school setting, using stimulus fading and/or shaping can take place over the course of a week; however, an intensive treatment can disrupt the child’s schedule for the duration of a week. With this type of treatment schedule, school staff receive training in the intervention approach to continue with appropriate treatment and provide accommodations after the intensive treatment ends (Kotrba, 2015). Intensive group treatment in a summer camp simulates a school setting, and the child with selective mutism can receive intensive practice in a safe setting without interruption to their school schedule. Families also receive the benefit of meeting other families who have a child with selective mutism (Cornacchio et al., 2019; Kotrba, 2015). Setting Setting refers to the location of treatment (e.g., home, school, community-based). Generalization of skills to new environments is an important aspect to selective mutism treatment. Treatment may occur within the clinical office, school, and community to reinforce the individual’s speaking skills. Resources ASHA Resources Assessment Tools, Techniques, and Data Sources Consumer Information: Selective Mutism Conquering Challenges of Interprofessional Treatment for Selective Mutism Current Procedural Terminology (CPT) Codes for Speech Language Pathology Services ICD-10-CM Diagnosis Codes Related to Speech, Language, and Swallowing Disorders [PDF] Interprofessional Education/Interprofessional Practice (IPE/IPP) Selective Mutism: An Integrated Approach Selective Mutism in Elementary School: Multidisciplinary Interventions Teaching the Language of Feelings to Students With Severe Emotional and Behavioral Handicaps Trauma-Informed Care Other Resources This list of resources is not exhaustive, and the inclusion of any specific resource does not imply endorsement from ASHA. Bergman, R. L., Keller, M. L., Piacentini, J., & Bergman, A. J. (2008). The development and psychometric properties of the Selective Mutism Questionnaire. Journal of Clinical Child & Adolescent Psychology, 37(2), 456–464. Child Mind Institute: Complete Guide to Selective Mutism Kurtz Psychology: Selective Mutism Learning University RCSLT: New Long COVID Guidance and Patient Handbook Striving to Speak: Children’s Books About Selective Mutism Selective Mutism Association Selective Mutism Association: Educator Toolkit [PDF] Selective Mutism Association: Speech Language Therapy and Selective Mutism References American Psychiatric Association. (2022). Anxiety disorders. In Diagnostic and statistical manual of mental disorders(5th ed., text rev.). American Speech-Language-Hearing Association. (2023). Code of ethics [Ethics]. www.asha.org/policy/ Anstendig, K. D. (1999). Is selective mutism an anxiety disorder? Rethinking its DSM-IV classification. Journal of Anxiety Disorders, 13(4), 417–434. Beidel, D. C., & Turner, S. M. (2007). Shy children, phobic adults: Nature and treatment of social anxiety disorders (2nd ed.). 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Neuropsychiatric Disease and Treatment, 14, 1163–1169. Steinhausen, H.-C., & Juzi, C. (1996). Elective mutism: An analysis of 100 cases. Journal of the American Academy of Child & Adolescent Psychiatry, 35(5), 606–614. Steinhausen, H.C., Wachter, M., Laimböck, K., & Metzke, C. W. (2006). A long-term outcome study of selective mutism in childhood. The Journal of Child Psychology and Psychiatry , 47(7), 751–756. Toppelberg, C. O., Tabors, P., Coggins, A., Lum, K., & Burger, C. (2005). Differential diagnosis of selective mutism in bilingual children. Journal of the American Academy of Child & Adolescent Psychiatry, 44(6), 592–595. Viana, A. G., Beidel, D. C., & Rabian, B. (2009). Selective mutism: A review and integration of the last 15 years. Clinical Psychology Review, 29(1), 57–67. Vogel, F., Gensthaler, A., Stahl, J., & Schwenck, C. (2019). Fears and fear-related cognitions in children with selective mutism. European Child & Adolescent Psychiatry, 28(9), 1169–1181. Walker, A. S., & Tobbell, J. (2015). Lost voices and unlived lives: Exploring adults’ experiences of selective mutism using interpretative phenomenological analysis. Qualitative Research in Psychology, 12(4), 453–471. About This Content Acknowledgments Content for ASHA’s Practice Portal is developed and updated through a comprehensive process that includes multiple rounds of subject matter expert input and review. ASHA extends its gratitude to the following subject matter experts who were involved in the development of the Selective Mutism page: Sharon Lee Armstrong, PhD Brittany Bice-Urbach, PhD Rachel Cortese, MS, CCC-SLP Emily R. Doll, MA, MS, CCC-SLP Joleen Fernald, PhD, CCC-SLP Suzanne Hungerford, PhD, CCC-SLP Evelyn Klein, PhD, CCC-SLP Janet Middendorf, MA, CCC-SLP Gail Richard, PhD, CCC-SLP Robert Schum, PhD Donna Spillman-Kennedy, MS, CCC-SLP Robert Thompson, PhD, CCC-SLP ASHA seeks input from subject matter experts representing differing perspectives and backgrounds. At times a subject matter expert may request to have their name removed from our acknowledgment. We continue to appreciate their work. Citing Practice Portal Pages The recommended citation for this Practice Portal page is: American Speech-Language-Hearing Association. (n.d.). Selective mutism [Practice Portal]. In This Section Practice Portal Home Clinical Topics Professional Topics Advertising Disclaimer Advertise with us Evidence Maps ASHA Evidence Maps Peer Connections Connect with your colleagues in the ASHA Community ASHA Special Interest Groups ASHA Related Content Find related products in ASHA's Store Search for articles on ASHAWire ASHA Stream Content Disclaimer: The Practice Portal, ASHA policy documents, and guidelines contain information for use in all settings; however, members must consider all applicable local, state and federal requirements when applying the information in their specific work setting. 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188009	https://www.youtube.com/watch?v=UCJ8AzThSpI	Cylinder Lateral Area, Surface Area and Volume Mario's Math Tutoring 458000 subscribers 108 likes Description 9773 views Posted: 19 May 2020 Learn how to find the lateral area, surface area and volume of a cylinder in this math tutorial by Mario's Math Tutoring. We go through where the formulas come from as well as applying them to an example problem. Related Videos to Help You Succeed!: Surface Area and Volume Review: Organized List of My Video Lessons to Help You Raise Your Scores & Pass Your Class. Videos Arranged by Math Subject as well as by Chapter/Topic. (Bookmark the Link Below) ➡️JOIN the channel as a CHANNEL MEMBER at the "ADDITIONAL VIDEOS" level to get access to my math video courses(Algebra 1, Algebra 2/College Algebra, Geometry, PreCalculus), midterm & final exam reviews, ACT and SAT prep videos and more! (Over 390+ videos) 9 comments Transcript: and this last thing you're gonna learn how to find the lateral area surface area and volume of a cylinder and the first thing we want to talk about are the formulas and what's interesting if you're familiar with prisms you can think of a cylinder kind of like a circular prism in a sense because you've got a these two parallel and congruent bases which in this case their circles separated by the height so what you can do is you could take the perimeter of the base okay the perimeter of this circle and multiply it by the height and the perimeter of a circle is just the circumference that's 2 PI R times the height so instead of memorizing this formula you can memorize the formula for a prism and then just go ahead and substitute in the circle for the perimeter of the circumference for surface area are taking the lateral area which is the area of the sides but then you're adding on the area of the base times 2 since we have a top and a bottom so surface area is like the total area but of course when you find the area of a circle that's PI R squared and we have two of those so 2 PI R squared and the lateral area we already talked about that that's 2 pi RH so a lot of times students like to memorize this formula but I kind of like to use the general formula so there's little bit less for me to memorize and then the last thing is to find the volume that's the area of the base but of course we said the base is a circle so that's PI R squared times the height so we could think of PI R squared H for volume let's go ahead and do this problem for number 1 here we want to find the lateral area which is the perimeter of the base 2 pi R that's the circumference times the height now keep in mind if I was to cut this and unroll it you would have like a rectangle that makes up the sides and then at the top you have like a circle and the bottom you'd have a circle but the lateral area is just the area of this rectangle this right here represents this circumference of the circle which is 2 PI R times the height ok and that's the area of a rectangle right so that's where that form that comes from so we have 2 pi times the radius which is 4 times the height which is 7 and that comes out to 56 pi inches squared so inches squared since its area it's like you're covering the surface here with little by one square inches you know little squares so for surface area now what we're going to do is we're going to add on these two bases in addition to the lateral area so we've got area of a circle is PI R squared so that's going to be pi times 4 squared but then you have two of them so we have to double that so that's going to be 16 times 2 which is 32 pi plus the lateral area which that's the area of the side so that's going to give us 88 pi inches squared for the surface area or the total area last part is finding the volume that's like if you were going to fill this up with water like this is a glass how much would that contain well we're gonna do the area of the base which is a circle PI R squared times the height the radius we said was 4 squared and the height is 7 so let's see that it's gonna be 16 times 7 which is a hundred and 12 PI inches cubed so it's like you're filling this up with like little one by one by one ice cubes and you're that's how many we fill up the inside of that glass now if you want to see more about like cones and pyramids and prisms and all the different shapes together I put together a comprehensive video right there where we go through all the different shapes when we talk about how the formulas are related and easy way to memorize them follow me over that video we'll dive into some more examples
188010	https://math.stackexchange.com/questions/3563427/using-markov-inequality-to-prove-chebyshev-inequality	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Using Markov inequality to prove Chebyshev inequality Ask Question Asked Modified 5 years, 7 months ago Viewed 2k times 2 $\begingroup$ I was introduced to the Markov inequality: For any r.v. $X$ and constant $a > 0$, $$P(\vert X \vert \ge a) \le \dfrac{E \vert X \vert}{a}.$$ I was then introduced to the Chebyshev inequality: Let $X$ have mean $\mu$ and variance $\sigma^2$. Then for any $a > 0$, $$P(\vert X - \mu \vert \ge a) \le \dfrac{\sigma^2}{a^2}$$ The proof for the Chebyshev inequality was given as follows: By Markov's inequality, $$P(\vert X - \mu \vert \ge a) = P((X - \mu)^2 \ge a^2) \le \dfrac{E(X - \mu)^2}{a^2} = \dfrac{\sigma^2}{a^2}.$$ Substituting $c \sigma$ for $a$, for $c > 0$, we have the following equivalent form of Chebyshev's inequality: $$P(\vert X - \mu \vert \ge c \sigma) \le \dfrac{1}{c^2}.$$ This gives us the upper bound on the probability of an r.v. being more than $c$ standard deviations away from its mean, e.g., there can't be more than a 25% chance of being 2 or more standard deviations from the mean. I have two questions. First, I'm wondering how the authors went from $P(\vert X - \mu \vert \ge a)$ to $P((X - \mu)^2 \ge a^2)$? I'm unsure of the algebra that $\vert X - \mu \vert = (X - \mu)^2$. And then, how did the authors go from "being more than $c$ standard deviations away from its mean" to the specific claim that "there can't be more than a 25% chance of being 2 or more standard deviations from the mean"? Thank you. probability inequality proof-explanation standard-deviation Share edited Jun 12, 2020 at 10:38 CommunityBot 1 asked Feb 28, 2020 at 17:02 Dom FomelloDom Fomello 47933 silver badges1717 bronze badges $\endgroup$ Add a comment \| 1 Answer 1 Reset to default 0 $\begingroup$ For the first question, just see that the event $\|X-\mu\|\ge a$ is the same event as $(\|X-\mu\|)^2\ge a^2$ (and we can omit the absolute value). For the second question , it just an example of a usage of this inequality with $c=2$. Share edited Feb 28, 2020 at 17:07 answered Feb 28, 2020 at 17:05 infinityinfinity 96655 silver badges1010 bronze badges $\endgroup$ 2 $\begingroup$ But $\vert X - \mu \vert^2$ is not the same as $(X - \mu)^2$? $\endgroup$ Dom Fomello – Dom Fomello 2020-02-28 17:07:43 +00:00 Commented Feb 28, 2020 at 17:07 1 $\begingroup$ @DomFomello just see that $\|x\|^2 = x^2$ , i think that's what confusing you. write $x = X-\mu$ after that. $\endgroup$ infinity – infinity 2020-02-28 17:14:25 +00:00 Commented Feb 28, 2020 at 17:14 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions probability inequality proof-explanation standard-deviation See similar questions with these tags. 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188011	https://www.cancerresearchuk.org/about-cancer/hodgkin-lymphoma/treatment/treatment-decisions	Skip to main content Together we will beat cancer Home About cancer Hodgkin lymphoma Treatment for Hodgkin lymphoma Treatment options for Hodgkin lymphoma Treatment options for Hodgkin lymphoma When you are diagnosed with Hodgkin lymphoma, a team of healthcare professionals decide if you need treatment straight away. They also look at what treatment options you have and will discuss those with you. This depends on factors such as the type and stage of Hodgkin lymphoma, as well as your general health. You might have more than one treatment. The most common treatments for Hodgkin lymphoma are chemotherapy and radiotherapy. A team of healthcare professionals will discuss the best treatment for you. This team is called a multidisciplinary team (MDT). The team members specialise in different areas of Hodgkin lymphoma treatment and support. The team usually includes: a haematologist (a doctor who treats blood conditions) a clinical oncologist (a specialist in radiotherapy) a radiologist who looks at your scans a specialist cancer nurse (also called a clinical nurse specialist) a pathologist (a specialist who looks at tissue samples) a stem cell transplant specialist You might not start treatment straightway if you are well. This depends on the type of Hodgkin lymphoma you have and the results of your blood tests and scans. Find out more about the tests you might have for Hodgkin lymphoma If you need treatment straight away, your team plan it depending on: what stage your Hodgkin lymphoma is what type of Hodgkin lymphoma you have if you have fevers, sweating or unexpected weight loss, these are called B symptoms your general health and level of fitness your personal wishes Your doctor will talk to you about your treatment options. They will discuss the benefits and the possible side effects with you. Treatment overview The treatment of Hodgkin lymphoma is usually very successful in many cases and most people are cured. But sometimes Hodgkin lymphoma can come back (relapse). Some relapsed Hodgkin lymphomas can be difficult to treat. The most common treatments for Hodgkin lymphoma are: chemotherapy radiotherapy targeted cancer drugs immunotherapy a stem cell transplant Some people only need one type of treatment. Others need a combination of treatments. Surgery is not usually used as a treatment for Hodgkin lymphoma. But you might have a small operation to remove a lymph node to check for lymphoma cells. This is called a lymph node biopsy. Chemotherapy Chemotherapy uses anti cancer drugs called cytotoxic drugs to destroy cancer cells. The chemotherapy drugs circulate throughout the body in the bloodstream. Your treatment usually includes a combination of different chemotherapy drugs and a steroid. This is called a chemotherapy regimen. Steroids are substances made naturally in the body that affect many functions. They can be made artificially and used in cancer treatment as tablets or injections. Read about chemotherapy for Hodgkin lymphoma Radiotherapy Radiotherapy is a type of cancer treatment that uses high energy waves similar to x-rays to kill cancer cells. You have treatment from a radiotherapy machine in the hospital radiotherapy department. You might have radiotherapy to the lymph nodes where the lymphoma is and the areas around it. This is called involved site radiotherapy (ISRT). You usually have radiotherapy after chemotherapy. Read about radiotherapy for Hodgkin lymphoma Targeted cancer drugs Cancer cells have changes in their genes (DNA) that make them different from normal cells. These changes mean that the cancer cells might grow faster and sometimes spread. Targeted cancer drugs work by targeting those differences that a cancer cell has. There are different types of targeted cancer drugs. For Hodgkin lymphoma you might have brentuximab or rituximab. Immunotherapy Immunotherapy is treatment that stimulates the body's immune system to fight cancer. You might have immunotherapy if your Hodgkin lymphoma has come back. You may have different immunotherapy drugs such as pembrolizumab or nivolumab. Read about immunotherapy and targeted cancer drugs for Hodgkin lymphoma Stem cell transplant Stem cell transplants are a possible treatment for some types of blood cancers such as lymphoma. You might have a stem cell transplant if you have relapsed Hodgkin lymphoma. Or if your treatment hasn’t worked as well as your doctor would like. Before you have a stem cell transplant, you have high doses of chemotherapy. This kills the cancer cells and the stem cells in your bone marrow. The bone marrow is the spongy tissue inside your bones that makes blood cells. After the chemotherapy, you have the new stem cells into your bloodstream. You usually have a stem cell transplant using your own cells. Read about stem cell transplants for Hodgkin lymphoma Treatment by stage Doctors simplify the staging of Hodgkin lymphoma into: early (limited) stage intermediate stage advanced stage They will consider different things when deciding what stage your Hodgkin lymphoma is. This includes: where your Hodgkin lymphoma is the size what symptoms you have other risk factors such as having lots of lymphoma in your chest or being older than 50 Read more about the stages of Hodgkin lymphoma, B symptoms and risk factors Early (limited) stage Hodgkin lymphoma Early stage generally means stage 1 or 2 lymphoma with no risk factors. It is also called early stage favourable Hodgkin lymphoma. For early stage Hodgkin lymphoma, you usually have 2 cycles of chemotherapy followed by radiotherapy. A cycle of chemotherapy means that you have the drugs and then a rest to allow your body to recover. Each cycle is usually 4 weeks. You might have a PET-CT scan after 2 cycles of chemotherapy. Depending on the results of your PET-CT scan, you might have another 1 or 2 cycles of chemotherapy before radiotherapy. Some people don’t have radiotherapy. The treatment for nodular lymphocyte predominant Hodgkin lymphoma (NLPHL) is slightly different. You usually have radiotherapy on its own if you have early stage NLPHL. If you have very early stage lymphoma that is removed when you have a biopsy, your doctor might prefer to wait and monitor you before starting any treatment. Intermediate stage Hodgkin lymphoma Intermediate stage usually means stage 1 or 2 with one or more risk factors. It is also called early stage unfavourable Hodgkin lymphoma. There are a few treatment options your doctor may discuss with you for intermediate stage Hodgkin lymphoma. You might have 4 to 6 cycles of chemotherapy, sometimes followed by radiotherapy. You may have a PET-CT scan midway through your treatment so your doctor can see how well your treatment is working. Intermediate stage NLPHL is usually treated in the same way as classical Hodgkin lymphoma but you might have a targeted cancer drug called rituximab as well. In some cases your doctor may decide to wait for a little while before starting treatment if it is not causing any problems for you. Advanced stage Hodgkin lymphoma Advanced stage generally means that you have stage 3 or 4 Hodgkin lymphoma. However stage 2 with B symptoms and bulky disease or extranodal sites is usually treated as advanced stage. You usually have 4 to 6 cycles of chemotherapy if you have advanced stage Hodgkin lymphoma. You usually have a PET-CT scan after 2 cycles of chemotherapy so your doctor can decide how many more cycles of treatment you need and whether it needs to be intensified or reduced. You may have another PET-CT scan when you finish your chemotherapy to decide if you need radiotherapy or not. For advanced stage NLPHL, you usually have targeted therapy called rituximab with chemotherapy. Some people also have radiotherapy. Lymphoma that has come back Lymphoma that comes back after treatment is called relapsed or recurrent Hodgkin lymphoma. You usually have another lymph node biopsy and PET-CT scan if your doctors think that your Hodgkin lymphoma has relapsed. If you have Hodgkin lymphoma that comes back, it may still be possible to get rid of your lymphoma again. This will depend on: how well further treatment works for your type of Hodgkin lymphoma the treatment you have already had your general health and level of fitness For some people, treatment can cure Hodgkin lymphoma that comes back. Even if your lymphoma can't be cured, you will be able to have treatment to shrink the lymphoma. You may then be very well for some time. Treatment for relapsed Hodgkin lymphoma includes: high dose chemotherapy a stem cell transplant radiotherapy targeted therapy immunotherapy more chemotherapy For relapsed NLPHL, you usually have targeted cancer drugs and chemotherapy. If your lymphoma is early stage when it comes back, you might have targeted cancer drugs on their own. For some people NLPHL can change (transform) into a different type of lymphoma called non Hodgkin lymphoma . Your doctor will check if your lymphoma has transformed when they do a lymph node biopsy. You will have a different treatment if your lymphoma transforms into non Hodgkin lymphoma. Sometimes treatment for Hodgkin lymphoma doesn’t work as well as doctors would hope. If this happens your doctor might change the type of treatment you have. Your feelings It can be hard to cope with the news that your cancer has come back or that it isn’t responding to treatment. You may feel shocked and anxious even if treatment is still likely to get rid of your lymphoma again. Talking to friends and family about your feelings can help. You may also want to think about having counselling. A trained counsellor can help you to talk about your fears and worries. And sometimes it feels easier to talk to someone outside your family and friends. Many people with cancer find that counselling can help them to deal with their emotions and to discuss any difficulties that they have. Find a counsellor Fertility and cancer treatment Chemotherapy and radiotherapy for Hodgkin lymphoma can affect your ability to have children (fertility). Talk to your doctor about this before treatment if this important for you. They can help you look at fertility treatments or getting support to help you cope. Read more on fertility and chemotherapy Clinical trials Your doctor might ask if you’d like to take part in a clinical trial. Doctors and researchers do trials to make existing treatments better and develop new treatments. Research and clinical trials for Hodgkin lymphoma Getting a second opinion Some people like to get an opinion from a second doctor. This is before they decide on their treatment. Most doctors are happy to refer you to another NHS specialist if you would find this helpful. Go to more information about getting a second opinion Vaccinations Your doctor or specialist nurse may recommend that you have flu and pneumonia vaccinations. The vaccinations help to protect you from these infections if you have low immunity during treatment. Read more about the flu vaccine and cancer treatment Related links Stages of Hodgkin lymphoma The stage shows whether the lymphoma is in one area of your body (localised) or has spread to other areas.There are 4 stages for Hodgkin lymphoma. ## Types of Hodgkin lymphoma The two types of Hodgkin lymphoma are classical Hodgkin lymphoma and nodular lymphocyte predominant Hodgkin lymphoma (NLPHL). There are also subtypes. Knowing the type helps your doctor choose the right treatment. ## Chemotherapy for Hodgkin lymphoma Chemotherapy uses cytotoxic drugs to kill cancer cells. You normally have several chemotherapy drugs together during a course of treatment. Find out about the chemotherapy drugs used to treat Hodgkin lymphoma. ## Radiotherapy for Hodgkin lymphoma Radiotherapy uses high energy x-rays to kill Hodgkin lymphoma cells. You may have radiotherapy after chemotherapy or on its own. ## Living with Hodgkin lymphoma Get practical and emotional support to help you cope with a diagnosis of Hodgkin lymphoma, and life during and after treatment. ## Hodgkin lymphoma home page Hodgkin lymphoma is a cancer of a type of white blood cell called lymphocytes. Find out more about the tests, treatments and support available if you have Hodgkin lymphoma. It’s a worrying time for many people and we want to be there for you whenever - and wherever - you need us. Cancer Chat is our fully moderated forum where you can talk to others affected by cancer, share experiences, and get support. Cancer Chat is free to join and available 24 hours a day. 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188012	https://fiveable.me/key-terms/intro-philosophy/sufficient-condition	printables 🤔intro to philosophy review key term - Sufficient Condition Citation: MLA Definition A sufficient condition is a condition that, if true, guarantees the truth of a given statement or conclusion. In other words, if the sufficient condition is met, the statement or conclusion must also be true, regardless of any other circumstances. 5 Must Know Facts For Your Next Test A sufficient condition is denoted by the symbol '⇒', which can be read as 'implies' or 'if-then'. The statement 'If P, then Q' is logically equivalent to 'P is a sufficient condition for Q'. Sufficient conditions are often used in logical reasoning and mathematical proofs to establish the truth of a conclusion. In the context of logical statements, a sufficient condition is a statement that, if true, guarantees the truth of another statement. Sufficient conditions are important in understanding the relationships between different logical statements and their implications. Review Questions Explain the relationship between a sufficient condition and a necessary condition. A sufficient condition is a condition that, if true, guarantees the truth of a given statement or conclusion. In contrast, a necessary condition is a condition that must be true for a given statement or conclusion to be true. While a sufficient condition is enough to ensure the truth of a statement, a necessary condition is the minimum requirement for the statement to be true. A condition can be both necessary and sufficient, but a necessary condition alone is not enough to guarantee the truth of a statement. Describe how a sufficient condition is used in logical reasoning and mathematical proofs. In logical reasoning and mathematical proofs, sufficient conditions are used to establish the truth of a conclusion. If a statement P is a sufficient condition for a statement Q, then proving that P is true is enough to conclude that Q is also true, regardless of any other circumstances. This is a powerful tool in deductive reasoning, as it allows one to draw definite conclusions from known premises. Sufficient conditions are often used to construct logical arguments and to prove theorems in mathematics, where establishing the truth of a statement is crucial. Analyze the differences and similarities between a sufficient condition and logical implication. A sufficient condition and logical implication are closely related concepts. Both involve a relationship where if one statement is true, another statement must also be true. However, the key difference is that a sufficient condition is a more specific relationship, where the first statement (the sufficient condition) is enough to guarantee the truth of the second statement. In contrast, logical implication is a more general relationship, where the truth of the first statement implies the truth of the second statement, but the first statement may not be the only way to ensure the truth of the second statement. In other words, a sufficient condition is a type of logical implication, but a logical implication does not necessarily imply a sufficient condition. Related terms Necessary Condition: A necessary condition is a condition that must be true for a given statement or conclusion to be true. If the necessary condition is not met, the statement or conclusion cannot be true. Logical Implication: Logical implication is the relationship between two statements where if the first statement is true, the second statement must also be true. This is the same as a sufficient condition. Logical Equivalence: Logical equivalence is a relationship between two statements where if one statement is true, the other statement must also be true, and vice versa. This is a stronger relationship than a sufficient condition. "Sufficient Condition" also found in: Subjects (1) Formal Logic I
188013	https://www.khanacademy.org/math/grade-5-math-fl-best/x7a7f452d9f51baa9:properties-of-shapes/x7a7f452d9f51baa9:triangles/v/recog-triangles-example	Worked example: Classifying triangles (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. 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Course: 5th grade math (FL B.E.S.T.)>Unit 14 Lesson 2: Classifying triangles Classifying triangles Classifying triangles by angles Worked example: Classifying triangles Classify triangles by angles Classify triangles by side lengths Classify triangles by both sides and angles Types of triangles review Math> 5th grade math (FL B.E.S.T.)> Properties of shapes> Classifying triangles © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Worked example: Classifying triangles CCSS.Math: 4.G.A.2 Google Classroom Microsoft Teams About About this video Transcript Work through sample problems categorizing triangles as scalene, isosceles, equilateral, acute, right, or obtuse.Created by Sal Khan. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted albrei.duncan a year ago Posted a year ago. Direct link to albrei.duncan's post “you are pretty we Love yo...” more you are pretty we Love you no matter what you pretty like the stars and we love God and we are 4th grade we are pretty we can read we can write and do anything. Answer Button navigates to signup page •2 comments Comment on albrei.duncan's post “you are pretty we Love yo...” (8 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer mala a year ago Posted a year ago. Direct link to mala's post “correct that is true we ...” more correct that is true we love that Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Fried Rice 10 years ago Posted 10 years ago. Direct link to Fried Rice's post “Can a equilateral triangl...” more Can a equilateral triangle be a scalene triangle? Answer Button navigates to signup page •1 comment Comment on Fried Rice's post “Can a equilateral triangl...” (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer LeafSwirl222🍃 10 years ago Posted 10 years ago. Direct link to LeafSwirl222🍃's post “A scalene triangle means ...” more A scalene triangle means all sides have a different length. An equilateral triangle means all sides are the same length. Comment Button navigates to signup page (11 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... AbrahamC 2 years ago Posted 2 years ago. Direct link to AbrahamC's post “what the heck is this guy...” more what the heck is this guys it look I got a bunch of answer Answer Button navigates to signup page •2 comments Comment on AbrahamC's post “what the heck is this guy...” (8 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer annmanacker a year ago Posted a year ago. Direct link to annmanacker's post “How would you describe ob...” more How would you describe obtuse but also describe other angles with it in one sentence. Answer Button navigates to signup page •2 comments Comment on annmanacker's post “How would you describe ob...” (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer gabyswaggy a year ago Posted a year ago. Direct link to gabyswaggy's post “Obtuse triangles have one...” more Obtuse triangles have one angle larger than 90 degrees while acute angles have all angles smaller then 90, and right triangles have exactly one 90 degree angle. Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Tyler 9 years ago Posted 9 years ago. Direct link to Tyler's post “at 0:39 I didn't hear wha...” more at 0:39 I didn't hear what he was saying Answer Button navigates to signup page •Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer ༼ つ ◕_◕ ༽つEli 6 months ago Posted 6 months ago. Direct link to ༼ つ ◕_◕ ༽つEli's post “"and we can assume these ...” more "and we can assume these are drawn to scale" 3 comments Comment on ༼ つ ◕_◕ ༽つEli's post “"and we can assume these ...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more fey 4 years ago Posted 4 years ago. Direct link to fey's post “Can someone explain why a...” more Can someone explain why all equilateral triangles are also Isosceles triangle? Shouldn’t it be the other way around? Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Ian Pulizzotto 4 years ago Posted 4 years ago. Direct link to Ian Pulizzotto's post “Compare the following def...” more Compare the following definitions: An isosceles triangle is a triangle with two or three equal sides. An equilateral triangle is a triangle with three equal sides. It is clear from these definitions that an equilateral triangle is a special type of isosceles triangle (but not the other way around). So all equilateral triangles are isosceles, but not the other way around. Have a blessed, wonderful day! 1 comment Comment on Ian Pulizzotto's post “Compare the following def...” (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Alan 8 years ago Posted 8 years ago. Direct link to Alan's post “I'm going to 4th grade th...” more I'm going to 4th grade this year and I want to improve my reading. And also, I'm looking forward to getting mostly 4’s and 3's on my report card (We grade by 1,2,3,4 with 1 being the worst and 4 being the best.) Any Tips? Thank you very much. Answer Button navigates to signup page •3 comments Comment on Alan's post “I'm going to 4th grade th...” (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer X marks the spot 5 months ago Posted 5 months ago. Direct link to X marks the spot's post “Do math that is higher th...” more Do math that is higher than your level. I do that and I can do collage math. 4 comments Comment on X marks the spot's post “Do math that is higher th...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... jedhughe a year ago Posted a year ago. Direct link to jedhughe's post “tell me the type of angle...” more tell me the type of angles Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Jerry Nilsson a year ago Posted a year ago. Direct link to Jerry Nilsson's post “A zero angle is 0°. An ac...” more A zero angle is 0°. An acute angle is more than 0° but less than 90°. A right angle is 90°. An obtuse angle is more than 90° but less than 180°. A straight angle is 180°. A reflex angle is more than 180° but less than 360°. A full angle is 360°. Complementary angles are two angles that together form a right angle. Supplementary angles are two angles that together form a straight angle. Explementary angles are two angles that together form a full angle. Comment Button navigates to signup page (8 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more augh 2 years ago Posted 2 years ago. Direct link to augh's post “@it was a gift and a curs...” more @it was a gift and a curse Answer Button navigates to signup page •Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer xk-class insomniac a year ago Posted a year ago. Direct link to xk-class insomniac's post “also why is an equilatera...” more also why is an equilateral triangle an isosceles triangle, isn't it just supposed to be one or the other? Answer Button navigates to signup page •Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Patrick Wood a year ago Posted a year ago. Direct link to Patrick Wood's post “Because a equilateral tri...” more Because a equilateral triangle has three sides that are equal and three is greater than 2 2 comments Comment on Patrick Wood's post “Because a equilateral tri...” (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Video transcript Categorize the following triangles according to whether or not they are obtuse triangles. So an obtuse triangle is a triangle that has an obtuse angle in it, or an angle that is larger than 90 degrees. So it's pretty clear that this one does not have any obtuse angles. This is a 90 degree angle, and these are going to have to be less than 90 degrees. So this is not obtuse. This one right over here, just by looking at it, you see that all of them are less than 90 degrees, so I'll put this not obtuse. Now, these two are interesting. Angle JKL looks like it is larger than 90 degrees, and we can assume that these are drawn to scale, so we can base it on our visual judgment. So I would say that this is obtuse, and angle BAC also looks like it is larger than 90 degrees. So that's an obtuse angle, so I would throw that in the obtuse bucket as well. So we had two not obtuse, two obtuse. Check our answer. Let's do a couple more of these. Which of the following are correct descriptions of triangle PIG? Or I guess triangle pig? Select all that apply. PIG is equilateral. Well, that's not true. To be equilateral, all the sides have to be the same length. And we see here two sides are seven, but one side has length 4. So that's not true. Angle PIG has two equal angles. Well, we see that right over here, these two angles that are 74 degrees. So that's true. Triangle PIG, I guess, has an obtuse angle. Well, an obtuse angle is one that's larger than 90 degrees. None of these are larger than 90. That's not true. Triangle PIG has three acute angles. Well, that is true. All of these angles are less than 90 degrees. Triangle PIG has a right angle. No, none of these angles are exactly 90 degrees, so it does not have a right angle. So let's check our answer. Let's do one more of this. This is actually kind of fun. Categorize the following triangles according to whether or not they are equilateral. So to be equilateral, all of the sides have to have the same length. So this is equilateral. This one, the sides are definitely not the same length. In fact, not even two of the sides are the same length. That's really a scalene triangle. Here we have two sides that are the same length, but the third is different. This would be an isosceles triangle, but it's not equilateral. And here they're all the same length, so we have an equilateral triangle. We got it right. Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: exercise Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. 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188014	https://clp.math.uky.edu/clp_3_mc_text.pdf	CLP-3 MULTIVARIABLE CALCULUS Joel FELDMAN Andrew RECHNITZER Elyse YEAGER THIS DOCUMENT WAS TYPESET ON FRIDAY 16TH AUGUST, 2024. § § Legal stuff • Copyright © 2017–2021 Joel Feldman, Andrew Rechnitzer and Elyse Yeager. • This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. You can view a copy of the license at • Links to the source files can be found at the text webpage 2 CONTENTS 1 Vectors and Geometry in Two and Three Dimensions 1 1.1 Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.2.1 Addition of Vectors and Multiplication of a Vector by a Scalar . . . . 9 1.2.2 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.2.3 (Optional) Using Dot Products to Resolve Forces — The Pendulum . 19 1.2.4 (Optional) Areas of Parallelograms . . . . . . . . . . . . . . . . . . . . 23 1.2.5 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 1.2.6 (Optional) Some Vector Identities . . . . . . . . . . . . . . . . . . . . . 34 1.2.7 (Optional) Application of Cross Products to Rotational Motion . . . 36 1.2.8 (Optional) Application of Cross Products to Rotating Reference Frames 37 1.3 Equations of Lines in 2d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 1.4 Equations of Planes in 3d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 1.5 Equations of Lines in 3d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 1.6 Curves and their Tangent Vectors . . . . . . . . . . . . . . . . . . . . . . . . . 56 1.6.1 Derivatives and Tangent Vectors . . . . . . . . . . . . . . . . . . . . . 62 1.7 Sketching Surfaces in 3d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 1.7.1 Level Curves and Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . 79 1.8 Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 1.9 Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 2 Partial Derivatives 87 2.1 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 2.1.1 Optional — A Nasty Limit That Doesn’t Exist . . . . . . . . . . . . . 94 2.2 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 2.3 Higher Order Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 2.3.1 Optional — The Proof of Theorem 2.3.4 . . . . . . . . . . . . . . . . . 112 2.3.2 Optional — An Example of B2 f BxBy(x0, y0) ‰ B2 f ByBx(x0, y0) . . . . . . . . . 115 2.4 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 2.4.1 Memory Aids for the Chain Rule . . . . . . . . . . . . . . . . . . . . . 117 2.4.2 Chain Rule Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 i CONTENTS CONTENTS 2.4.3 Review of the Proof of d dt f (x(t)) = df dx(x(t)) dx dt (t) . . . . . . . . . . . 125 2.4.4 Proof of Theorem 2.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 2.5 Tangent Planes and Normal Lines . . . . . . . . . . . . . . . . . . . . . . . . 127 2.5.1 Surfaces of the Form z = f (x, y). . . . . . . . . . . . . . . . . . . . . . 128 2.5.2 Surfaces of the Form G(x, y, z) = 0. . . . . . . . . . . . . . . . . . . . . 134 2.6 Linear Approximations and Error . . . . . . . . . . . . . . . . . . . . . . . . . 147 2.6.1 Quadratic Approximation and Error Bounds . . . . . . . . . . . . . . 155 2.6.2 Optional — Taylor Polynomials . . . . . . . . . . . . . . . . . . . . . . 160 2.7 Directional Derivatives and the Gradient . . . . . . . . . . . . . . . . . . . . 162 2.8 A First Look at Partial Differential Equations . . . . . . . . . . . . . . . . . . 172 2.8.1 Optional — Solving the Advection and Wave Equations . . . . . . . 174 2.8.2 Really Optional — Derivation of the Wave Equation . . . . . . . . . . 179 2.9 Maximum and Minimum Values . . . . . . . . . . . . . . . . . . . . . . . . . 181 2.9.1 Absolute Minima and Maxima . . . . . . . . . . . . . . . . . . . . . . 205 2.10 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 2.10.1 (Optional) An Example with Two Lagrange Multipliers . . . . . . . . 223 3 Multiple Integrals 228 3.1 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 3.1.1 Vertical Slices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 3.1.2 Horizontal Slices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 3.1.3 Volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 3.1.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 3.1.5 Optional — More about the Definition of ť R f (x, y) dxdy . . . . . . 257 3.1.6 Even and Odd Functions . . . . . . . . . . . . . . . . . . . . . . . . . 262 3.2 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . 273 3.2.1 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 3.2.2 Polar Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 3.2.3 Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . 280 3.2.4 Optional— Error Control for the Polar Area Formula . . . . . . . . . 294 3.3 Applications of Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . 296 3.4 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314 3.5 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322 3.6 Triple Integrals in Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . 330 3.6.1 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . 330 3.6.2 The Volume Element in Cylindrical Coordinates . . . . . . . . . . . . 332 3.6.3 Sample Integrals in Cylindrical Coordinates . . . . . . . . . . . . . . 333 3.7 Triple Integrals in Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . 340 3.7.1 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 340 3.7.2 The Volume Element in Spherical Coordinates . . . . . . . . . . . . . 341 3.7.3 Sample Integrals in Spherical Coordinates . . . . . . . . . . . . . . . 345 3.8 Optional— Integrals in General Coordinates . . . . . . . . . . . . . . . . . . 352 3.8.1 Optional — Dropping Higher Order Terms in du, dv . . . . . . . . . 361 ii CONTENTS CONTENTS A Trigonometry 363 A.1 Trigonometry — Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 A.2 Trigonometry — Special Triangles . . . . . . . . . . . . . . . . . . . . . . . . 363 A.3 Trigonometry — Simple Identities . . . . . . . . . . . . . . . . . . . . . . . . 364 A.4 Trigonometry — Add and Subtract Angles . . . . . . . . . . . . . . . . . . . 365 A.5 Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . 366 B Powers and Logarithms 368 B.1 Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 B.2 Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369 C Table of Derivatives 371 D Table of Integrals 373 E Table of Taylor Expansions 375 F 3d Coordinate Systems 377 F.1 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 F.2 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378 F.3 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379 G ISO Coordinate System Notation 382 G.1 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382 G.2 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 G.3 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385 H Conic Sections and Quadric Surfaces 388 iii CONTENTS CONTENTS iv VECTORS AND GEOMETRY IN TWO AND THREE DIMENSIONS Chapter 1 Before we get started doing calculus in two and three dimensions we need to brush up on some basic geometry, that we will use a lot. We are already familiar with the Cartesian plane1, but we’ll start from the beginning. 1.1Ĳ Points Each point in two dimensions may be labeled by two coordinates2 (x, y) which specify the position of the point in some units with respect to some axes as in the figure below. x y x y (x, y) The set of all points in two dimensions is denoted3 R2. Observe that 1 Ren´ e Descartes (1596–1650) was a French scientist and philosopher, who lived in the Dutch Republic for roughly twenty years after serving in the (mercenary) Dutch States Army. He is viewed as the father of analytic geometry, which uses numbers to study geometry. 2 This is why the xy-plane is called “two dimensional” — the name of each point consists of two real numbers. 3 Not surprisingly, the 2 in R2 signifies that each point is labelled by two numbers and the R in R2 signifies that the numbers in question are real numbers. There are more advanced applications (for example in signal analysis and in quantum mechanics) where complex numbers are used. The space of all pairs (z1, z2), with z1 and z2 complex numbers is denoted C2. 1 VECTORS AND GEOMETRY 1.1 POINTS • the distance from the point (x, y) to the x-axis is \|y\| • if y ą 0, then (x, y) is above the x-axis and if y ă 0, then (x, y) is below the x-axis • the distance from the point (x, y) to the y-axis is \|x\| • if x ą 0, then (x, y) is the right of the y-axis and if x ă 0, then (x, y) is to the left of the y-axis • the distance from the point (x, y) to the origin (0, 0) is a x2 + y2 Similarly, each point in three dimensions may be labeled by three coordinates (x, y, z), as in the two figures below. (x, y, z) x y z x y z (x, y, z) x y z x y z The set of all points in three dimensions is denoted R3. The plane that contains, for exam-ple, the x- and y-axes is called the xy-plane. • The xy-plane is the set of all points (x, y, z) that satisfy z = 0. • The xz-plane is the set of all points (x, y, z) that satisfy y = 0. • The yz-plane is the set of all points (x, y, z) that satisfy x = 0. More generally, • The set of all points (x, y, z) that obey z = c is a plane that is parallel to the xy-plane and is a distance \|c\| from it. If c ą 0, the plane z = c is above the xy-plane. If c ă 0, the plane z = c is below the xy-plane. We say that the plane z = c is a signed distance c from the xy-plane. • The set of all points (x, y, z) that obey y = b is a plane that is parallel to the xz-plane and is a signed distance b from it. • The set of all points (x, y, z) that obey x = a is a plane that is parallel to the yz-plane and is a signed distance a from it. 2 VECTORS AND GEOMETRY 1.1 POINTS z “ c x y z y “ b x y z x “ a x y z Observe that our 2d distances extend quite easily to 3d. • the distance from the point (x, y, z) to the xy-plane is \|z\| • the distance from the point (x, y, z) to the xz-plane is \|y\| • the distance from the point (x, y, z) to the yz-plane is \|x\| • the distance from the point (x, y, z) to the origin (0, 0, 0) is a x2 + y2 + z2 To see that the distance from the point (x, y, z) to the origin (0, 0, 0) is indeed a x2 + y2 + z2, • apply Pythagoras to the right-angled triangle with vertices (0, 0, 0), (x, 0, 0) and (x, y, 0) to see that the distance from (0, 0, 0) to (x, y, 0) is a x2 + y2 and then • apply Pythagoras to the right-angled triangle with vertices (0, 0, 0), (x, y, 0) and (x, y, z) to see that the distance from (0, 0, 0) to (x, y, z) is ba x2 + y22 + z2 = a x2 + y2 + z2. px, 0, 0q px, y, 0q px, y, zq x y z x y z More generally, the distance from the point (x, y, z) to the point (x1, y1, z1) is b (x ´ x1)2 + (y ´ y1)2 + (z ´ z1)2 Notice that this gives us the equation for a sphere quite directly. All the points on a sphere are equidistant from the centre of the sphere. So, for example, the equation of the sphere centered on (1, 2, 3) with radius 4, that is, the set of all points (x, y, z) whose distance from (1, 2, 3) is 4, is (x ´ 1)2 + (y ´ 2)2 + (z ´ 3)2 = 16 3 VECTORS AND GEOMETRY 1.1 POINTS Here is an example in which we sketch a region in the xy-plane that is specified using inequalities. Example 1.1.1 In this example, we sketch the region ␣(x, y) ˇ ˇ ´ 12 ď x2 ´ 6x + y2 ´ 4y ď ´9, y ě 1 ( in the xy-plane. We do so in two steps. In the first step, we sketch the curves x2 ´ 6x + y2 ´ 4y = ´12, x2 ´ 6x + y2 ´ 4y = ´9, and y = 1. • By completing squares, we see that the equation x2 ´ 6x + y2 ´ 4y = ´12 is equiva-lent to (x ´ 3)2 + (y ´ 2)2 = 1, which is the circle of radius 1 centred on (3, 2). It is sketched in the figure below. • By completing squares, we see that the equation x2 ´ 6x + y2 ´ 4y = ´9 is equivalent to (x ´ 3)2 + (y ´ 2)2 = 4, which is the circle of radius 2 centred on (3, 2). It is sketched in the figure below. • The point (x, y) obeys y = 1 if and only if it is a distance 1 vertically above the x-axis. So y = 1 is the line that is parallel to the x-axis and is one unit above it. This line is also sketched in the figure below. px ´ 3q2 py ´ 2q2 “ 1 px ´ 3q2 py ´ 2q2 “ 4 y “ 1 x y In the second step we determine the impact that the inequalities have. • The inequality x2 ´ 6x + y2 ´ 4y ě ´12 is equivalent to (x ´ 3)2 + (y ´ 2)2 ě 1 and hence is equivalent to a (x ´ 3)2 + (y ´ 2)2 ě 1. So the point (x, y) satisfies x2 ´ 6x + y2 ´ 4y ě ´12 if and only if the distance from (x, y) to (3, 2) is at least 1, i.e. if and only if (x, y) is outside (or on) the circle (x ´ 3)2 + (y ´ 2)2 = 1. • The inequality x2 ´ 6x + y2 ´ 4y ď ´9 is equivalent to (x ´ 3)2 + (y ´ 2)2 ď 4 and hence is equivalent to a (x ´ 3)2 + (y ´ 2)2 ď 2. So the point (x, y) satisfies the inequality x2 ´ 6x + y2 ´ 4y ď ´9 if and only if the distance from (x, y) to (3, 2) is at most 2, i.e. if and only if (x, y) is inside (or on) the circle (x ´ 3)2 + (y ´ 2)2 = 4. • The point (x, y) obeys y ě 1 if and only if (x, y) is a vertical distance at least 1 above the x-axis, i.e. is above (or on) the line y = 1. 4 VECTORS AND GEOMETRY 1.1 POINTS • So the region ␣(x, y) ˇ ˇ ´ 12 ď x2 ´ 6x + y2 ´ 4y ď ´9, y ě 1 ( consists of all points (x, y) that – are inside or on the circle (x ´ 3)2 + (y ´ 2)2 = 4 and – are also outside or on the circle (x ´ 3)2 + (y ´ 2)2 = 1 and – are also above or on the line y = 1. It is the shaded region in the figure below. px ´ 3q2 py ´ 2q2 “ 1 px ´ 3q2 py ´ 2q2 “ 4 y “ 1 x y Example 1.1.1 Here are a couple of examples that involve spheres. Example 1.1.2 In this example, we are going to find the curve formed by the intersection of the xy-plane and the sphere of radius 5 centred on (0, 0, 4). The point (x, y, z) lies on the xy-plane if and only if z = 0, and lies on the sphere of radius 5 centred on (0, 0, 4) if and only if x2 + y2 + (z ´ 4)2 = 25. So the point (x, y, z) lies on the curve of intersection if and only if both z = 0 and x2 + y2 + (z ´ 4)2 = 25, or equivalently z = 0, x2 + y2 + (0 ´ 4)2 = 25 ð ñ z = 0, x2 + y2 = 9 This is the circle in the xy-plane that is centred on the origin and has radius 3. Here is a sketch that show the parts of the sphere and the circle of intersection that are in the first octant. That is, that have x ě 0, y ě 0 and z ě 0. 5 VECTORS AND GEOMETRY 1.2 VECTORS z y x Example 1.1.2 Example 1.1.3 In this example, we are going to find all points (x, y, z) for which the distance from (x, y, z) to (9, ´12, 15) is twice the distance from (x, y, z) to the origin (0, 0, 0). The distance from (x, y, z) to (9, ´12, 15) is a (x ´ 9)2 + (y + 12)2 + (z ´ 15)2. The dis-tance from (x, y, z) to (0, 0, 0) is a x2 + y2 + z2. So we want to find all points (x, y, z) for which b (x ´ 9)2 + (y + 12)2 + (z ´ 15)2 = 2 b x2 + y2 + z2 Squaring both sides of this equation gives x2 ´ 18x + 81 + y2 + 24y + 144 + z2 ´ 30z + 225 = 4 x2 + y2 + z2) Collecting up terms gives 3x2 + 18x + 3y2 ´ 24y + 3z2 + 30z = 450 and then, dividing by 3, x2 + 6x + y2 ´ 8y + z2 + 10z = 150 and then, completing squares, x2 + 6x + 9 + y2 ´ 8y + 16 + z2 + 10z + 25 = 200 or (x + 3)2 + (y ´ 4)2 + (z + 5)2 = 200 This is the sphere of radius 10 ? 2 centred on (´3, 4, ´5). Example 1.1.3 1.2Ĳ Vectors In many of our applications in 2d and 3d, we will encounter quantities that have both a magnitude (like a distance) and also a direction. Such quantities are called vectors. That is, a vector is a quantity which has both a direction and a magnitude, like a velocity. If you are moving, the magnitude (length) of your velocity vector is your speed (distance travelled 6 VECTORS AND GEOMETRY 1.2 VECTORS per unit time) and the direction of your velocity vector is your direction of motion. To specify a vector in three dimensions you have to give three components, just as for a point. To draw the vector with components a, b, c you can draw an arrow from the point (0, 0, 0) to the point (a, b, c). Similarly, to specify a vector in two dimensions you have to x y a b (a, b) pa, b, cq a b c x y z give two components and to draw the vector with components a, b you can draw an arrow from the point (0, 0) to the point (a, b). There are many situations in which it is preferable to draw a vector with its tail at some point other than the origin. For example, it is natural to draw the velocity vector of a moving particle with the tail of the velocity vector at the position of the particle, whether or not the particle is at the origin. The sketch below shows a moving particle and its velocity vector at two different times. v v x y As a second example, suppose that you are analyzing the motion of a pendulum. There are three forces acting on the pendulum bob: gravity g, which is pulling the bob straight down, tension t in the rod, which is pulling the bob in the direction of the rod, and air resistance r, which is pulling the bob in a direction opposite to its direction of motion. All three forces are acting on the bob. So it is natural to draw all three arrows representing the forces with their tails at the bob. 7 VECTORS AND GEOMETRY 1.2 VECTORS g t r In this text, we will used bold faced letters, like v, t, g, to designate vectors. In hand-writing, it is clearer to use a small overhead arrow4, as in ⃗ v,⃗ t, ⃗ g, instead. Also, when we want to emphasise that some quantity is a number, rather than a vector, we will call the number a scalar. Both points and vectors in 2d are specified by two numbers. Until you get used to this, it might confuse you sometimes — does a given pair of numbers represent a point or a vector? To distinguish5 between the components of a vector and the coordinates of the point at its head, when its tail is at some point other than the origin, we shall use angle brackets rather than round brackets around the components of a vector. For example, the figure below shows the two-dimensional vector ⟨2, 1⟩drawn in three different positions. In each case, when the tail is at the point (u, v) the head is at (2 + u, 1 + v). We warn you that, out in the real world6, no one uses notation that distinguishes between components of a vector and the coordinates of its head — usually round brackets are used for both. It is up to you to keep straight which is being referred to. x y (0, 0) (2, 1) (8, 0) (10, 1) (4, 2) (6, 3) ⟨2, 1⟩ ⟨2, 1⟩ By way of summary, we use • bold faced letters, like v, t, g, to designate vectors, and • angle brackets, like ⟨2, 1⟩, around the components of a vector, but use • round brackets, like (2, 1), around the coordinates of a point, and use • “scalar” to emphasise that some quantity is a number, rather than a vector. Notation 1.2.1. 4 Some people use an underline, as in v, rather than an arrow. 5 Or, in the Wikipedia jargon, disambiguate. 6 OK. OK. Out in that (admittedly very small) part of the real world that actually knows what a vector is. 8 VECTORS AND GEOMETRY 1.2 VECTORS 1.2.1 § § Addition of Vectors and Multiplication of a Vector by a Scalar Just as we have done many times in the CLP texts, when we define a new type of object, we want to understand how it interacts with the basic operations of addition and multipli-cation. Vectors are no different, and we shall shortly see a natural way to define addition of vectors. Multiplication will be more subtle, and we shall start with multiplication of a vector by a number (rather than with multiplication of a vector by another vector). By way of motivation for the definitions of addition and multiplication by a number, imagine that we are out for a walk on the xy-plane. • Suppose that we take a step and, in doing so, we move a1 units parallel to the x-axis and a2 units parallel to the y-axis. Then we say that ⟨a1, a2⟩is the displacement vector for the step. Suppose now that we take a second step which moves us an additional b1 units parallel to the x-axis and an additional b2 units parallel to the y-axis, as in the figure on the left below. So the displacement vector for the second step is ⟨b1, b2⟩. All together, we have moved a1 + b1 units parallel to the x-axis and a2 + b2 units parallel to the y-axis. The displacement vector for the two steps com-bined is ⟨a1 + b1, a2 + b2⟩. We shall define the sum of ⟨a1, a2⟩and ⟨b1, b2⟩, denoted by ⟨a1, a2⟩+ ⟨b1, b2⟩, to be ⟨a1 + b1, a2 + b2⟩. • Suppose now that, instead, we decide to step in the same direction as the first step above, but to move twice as far, as in the figure on the right below. That is, our step will move us 2a1 units in the direction of the x-axis and 2a2 units in the direction of the y-axis and the corresponding displacement vector will be ⟨2a1, 2a2⟩. We shall define the product of the number 2 and the vector ⟨a1, a2⟩, denoted by 2 ⟨a1, a2⟩, to be ⟨2a1, 2a2⟩. a2 a2 b2 b2 ⟨a1, a2⟩ ⟨b1, b2⟩ ⟨a1 b1, a2 b2⟩ a2 2a2 ⟨a1, a2⟩ ⟨2a1, 2a2⟩ Here are the formal definitions. These two operations have the obvious definitions a = ⟨a1, a2⟩, b = ⟨b1, b2⟩ ù ñ a + b = ⟨a1 + b1, a2 + b2⟩ a = ⟨a1, a2⟩, s a number ù ñ sa = ⟨sa1, sa2⟩ and similarly in three dimensions. Definition 1.2.2 (Adding Vectors and Multiplying a Vector by a Number). 9 VECTORS AND GEOMETRY 1.2 VECTORS Pictorially, you add the vector b to the vector a by drawing b with its tail at the head of a and then drawing a vector from the tail of a to the head of b, as in the figure on the left below. For a number s, we can draw the vector sa, by just • changing the vector a’s length by the factor \|s\|, and, • if s ă 0, reversing the arrow’s direction, as in the other two figures below. a2 a2 b2 b2 a b a b a2 2a2 a 2a a ´2a The special case of multiplication by s = ´1 appears so frequently that (´1)a is given the shorter notation ´a. That is, ´ ⟨a1, a2⟩= ⟨´a1, ´a2⟩ Of course a + (´a) is 0, the vector all of whose components are zero. To subtract b from a pictorially, you may add ´b (which is drawn by reversing the direction of b) to a. Alternatively, if you draw a and b with their tails at a common point, then a ´ b is the vector from the head of b to the head of a. That is, a ´ b is the vector you must add to b in order to get a. a −b a b −b a −b The operations of addition and multiplication by a scalar that we have just defined are quite natural and rarely cause any problems, because they inherit from the real numbers the properties of addition and multiplication that you are used to. 10 VECTORS AND GEOMETRY 1.2 VECTORS Let a, b and c be vectors and s and t be scalars. Then (1) a + b = b + a (2) a + (b + c) = (a + b) + c (3) a + 0 = a (4) a + (´a) = 0 (5) s(a + b) = sa + sb (6) (s + t)a = sa + ta (7) (st)a = s(ta) (8) 1a = a Theorem 1.2.3 (Properties of Addition and Scalar Multiplication). We have just been introduced to many definitions. Let’s see some of them in action. Example 1.2.4 For example, if a = ⟨1, 2, 3⟩ b = ⟨3, 2, 1⟩ c = ⟨1, 0, 1⟩ then 2a = 2 ⟨1, 2, 3⟩= ⟨2, 4, 6⟩ ´b = ´ ⟨3, 2, 1⟩= ⟨´3, ´2, ´1⟩ 3c = 3 ⟨1, 0, 1⟩= ⟨3, 0, 3⟩ and 2a ´ b + 3c = ⟨2, 4, 6⟩+ ⟨´3, ´2, ´1⟩+ ⟨3, 0, 3⟩ = ⟨2 ´ 3 + 3 , 4 ´ 2 + 0 , 6 ´ 1 + 3⟩ = ⟨2, 2, 8⟩ Example 1.2.4 Two vectors a and b • are said to be parallel if a = s b for some nonzero real number s and • are said to have the same direction if a = s b for some number s ą 0. Definition 1.2.5. There are some vectors that occur sufficiently commonly that they are given special names. One is the vector 0. Some others are the “standard basis vectors”. 11 VECTORS AND GEOMETRY 1.2 VECTORS (a) The standard basis vectors in two dimensions are ˆ ı ı ı = ⟨1, 0⟩ ˆ ȷ ȷ ȷ = ⟨0, 1⟩ x y ˆ ı ı ı ˆ    (b) The standard basis vectors in three dimensions are ˆ ı ı ı = ⟨1, 0, 0⟩ ˆ ȷ ȷ ȷ = ⟨0, 1, 0⟩ ˆ k = ⟨0, 0, 1⟩ x y z ˆ ı ı ı ˆ    ˆ k Definition 1.2.6. We’ll explain the little hats in the notation ˆ ı ı ı, ˆ ȷ ȷ ȷ, ˆ k shortly. Some people rename ˆ ı ı ı, ˆ ȷ ȷ ȷ and ˆ k to e1, e2 and e3 respectively. Using the above properties we have, for all vectors, ⟨a1, a2⟩= a1 ˆ ı ı ı + a2 ˆ ȷ ȷ ȷ ⟨a1, a2, a3⟩= a1 ˆ ı ı ı + a2 ˆ ȷ ȷ ȷ + a3 ˆ k A sum of numbers times vectors, like a1ˆ ı ı ı + a2ˆ ȷ ȷ ȷ is called a linear combination of the vectors. Thus all vectors can be expressed as linear combinations of the standard basis vectors. This makes basis vectors very helpful in computations. The standard basis vectors are unit vectors, meaning that they are of length one, where the length of a vector a is denoted7 \|a\| and is defined by a = ⟨a1, a2⟩ ù ñ \|a\| = b a2 1 + a2 2 a = ⟨a1, a2, a3⟩ ù ñ \|a\| = b a2 1 + a2 2 + a2 3 A unit vector is a vector of length one. We’ll sometimes use the accent ˆ to em-phasise that the vector ˆ a is a unit vector. That is, \|ˆ a\| = 1. Definition 1.2.7 (Length of a Vector). Example 1.2.8 Recall that multiplying a vector a by a positive number s, changes the length of the vector by a factor s without changing the direction of the vector. So (assuming that \|a\| ‰ 0) a \|a\| is a unit vector that has the same direction as a. For example, ⟨1,1,1⟩ ? 3 is a unit vector that points in the same direction as ⟨1, 1, 1⟩. Example 1.2.8 7 The notation }a} is also used for the length of a. 12 VECTORS AND GEOMETRY 1.2 VECTORS Example 1.2.9 We go for a walk on a flat Earth. We use a coordinate system with the positive x-axis point-ing due east and the positive y-axis pointing due north. We • start at the origin and • walk due east for 4 units and then • walk northeast for 5 ? 2 units and then • head towards the point (0, 11), but we only go • one third of the way. p4, 0q p9, 5q p6, 7q p0, 11q x y E N W S ⟨4, 0⟩ ⟨5, 5⟩ ⟨´9, 6⟩ ⟨´3, 2⟩ 45˝ 1 1 ⟨1, 1⟩ We will now use vectors to figure out our final location. • On the first leg of our walk, we go 4 units in the positive x-direction. So our dis-placement vector — the vector whose tail is at our starting point and whose head is at the end point of the first leg — is ⟨4, 0⟩. As we started at (0, 0) we finish the first leg of the walk at (4, 0). • On the second leg of our walk, our direction of motion is northeast, i.e. is 45˝ above the direction of the positive x-axis. Looking at the figure on the right above, we see that our displacement vector, for the second leg of the walk, has to be in the same direction as the vector ⟨1, 1⟩. So our displacement vector is the vector of length 5 ? 2 with the same direction as ⟨1, 1⟩. The vector ⟨1, 1⟩has length ? 12 + 12 = ? 2 and so ⟨1,1⟩ ? 2 has length one and our displacement vector is 5 ? 2 ⟨1, 1⟩ ? 2 = 5 ⟨1, 1⟩= ⟨5, 5⟩ If we draw this displacement vector, ⟨5, 5⟩with its tail at (4, 0), the starting point of the second leg of the walk, then its head will be at (4 + 5, 0 + 5) = (9, 5) and that is the end point of the second leg of the walk. • On the final leg of our walk, we start at (9, 5) and walk towards (0, 11). The vector from (9, 5) to (0, 11) is ⟨0 ´ 9 , 11 ´ 5⟩= ⟨´9, 6⟩. As we go only one third of the way, our final displacement vector is 1 3 ⟨´9, 6⟩= ⟨´3, 2⟩ 13 VECTORS AND GEOMETRY 1.2 VECTORS If we draw this displacement vector with its tail at (9, 5), the starting point of the final leg, then its head will be at (9 ´ 3, 5 + 2) = (6, 7) and that is the end point of the final leg of the walk, and our final location. Example 1.2.9 1.2.2 § § The Dot Product Let’s get back to the arithmetic operations of addition and multiplication. We will be using both scalars and vectors. So, for each operation there are three possibilities that we need to explore: • “scalar plus scalar”, “scalar plus vector” and “vector plus vector” • “scalar times scalar”, “scalar times vector” and “vector times vector” We have been using “scalar plus scalar” and “scalar times scalar” since childhood. “vector plus vector” and “scalar times vector” were just defined above. There is no sensible way to define “scalar plus vector”, so we won’t. This leaves “vector times vector”. There are actually two widely used such products. The first is the dot product, which is the topic of this section, and which is used to easily determine the angle θ (or more precisely, cos θ) between two vectors. We’ll get to the second, the cross product, later. Here is preview of what we will do in this dot product subsection §1.2.2. We are going to give two formulae for the dot product, a ¨ b, of the pair of vectors a = ⟨a1, a2, a3⟩and b = ⟨b1, b2, b3⟩. • The first formula is a ¨ b = a1b1 + a2b2 + a3b3. We will take it as our official definition of a ¨ b. This formula provides us with an easy way to compute dot products. • The second formula is a ¨ b = \|a\| \|b\| cos θ, where θ is the angle between a and b. a b θ We will show, in Theorem 1.2.11 below, that this second formula always gives the same answer as the first formula. The second formula provides us with an easy way to determine the angle between two vectors. In particular, it provides us with an easy way to test whether or not two vectors are perpendicular to each other. For example, the vectors ⟨1, 2, 3⟩and ⟨´1, ´1, 1⟩have dot product ⟨1, 2, 3⟩¨ ⟨´1, ´1, 1⟩= 1 ˆ (´1) + 2 ˆ (´1) + 3 ˆ 1 = 0 This tell us as the angle θ between the two vectors obeys cos θ = 0, so that θ = π 2 . That is, the two vectors are perpendicular to each other. 14 VECTORS AND GEOMETRY 1.2 VECTORS After we give our official definition of the dot product in Definition 1.2.10, and give the important properties of the dot product, including the formula a ¨ b = \|a\| \|b\| cos θ, in Theorem 1.2.11, we’ll give some examples. Finally, to see the dot product in action, we’ll define what it means to project one vector on another vector and give an example. The dot product of the vectors a and b is denoted a ¨ b and is defined by a = ⟨a1, a2⟩, b = ⟨b1, b2⟩ ù ñ a ¨ b = a1b1 + a2b2 a = ⟨a1, a2, a3⟩, b = ⟨b1, b2, b3⟩ ù ñ a ¨ b = a1b1 + a2b2 + a3b3 in two and three dimensions respectively. Definition 1.2.10 (Dot Product). The properties of the dot product are as follows: Let a, b and c be vectors and let s be a scalar. Then (0) a, b are vectors and a ¨ b is a scalar (1) a ¨ a = \|a\|2 (2) a ¨ b = b ¨ a (3) a ¨ (b + c) = a ¨ b + a ¨ c, (a + b) ¨ c = a ¨ c + b ¨ c (4) (sa) ¨ b = s(a ¨ b) (5) 0 ¨ a = 0 (6) a ¨ b = \|a\| \|b\| cos θ where θ is the angle between a and b (7) a ¨ b = 0 ð ñ a = 0 or b = 0 or a K b Theorem 1.2.11 (Properties of the Dot Product). Proof. Properties 0 through 5 are almost immediate consequences of the definition. For example, for property 3 (which is called the distributive law) in dimension 2, a ¨ (b + c) = ⟨a1, a2⟩¨ ⟨b1 + c1, b2 + c2⟩ = a1(b1 + c1) + a2(b2 + c2) = a1b1 + a1c1 + a2b2 + a2c2 a ¨ b + a ¨ c = ⟨a1, a2⟩¨ ⟨b1, b2⟩+ ⟨a1, a2⟩¨ ⟨c1, c2⟩ = a1b1 + a2b2 + a1c1 + a2c2 Property 6 is sufficiently important that it is often used as the definition of dot product. It is not at all an obvious consequence of the definition. To verify it, we just write \|a ´ b\|2 in two different ways. The first expresses \|a ´ b\|2 in terms of a ¨ b. It is \|a ´ b\|2 1 = (a ´ b ) ¨ (a ´ b ) 3 = a ¨ a ´ a ¨ b ´ b ¨ a + b ¨ b 1,2 = \|a\|2 + \|b\|2 ´ 2a ¨ b 15 VECTORS AND GEOMETRY 1.2 VECTORS Here, 1 =, for example, means that the equality is a consequence of property 1. The second way we write \|a ´ b\|2 involves cos θ and follows from the cosine law for triangles. Just in case you don’t remember the cosine law, we’ll derive it right now! Start by applying Pythagoras to the shaded triangle in the right hand figure of b θ a a ´ b \|b\| \|a\| cos θ \|a\| sin θ θ \|a\| \|a ´ b\| That triangle is a right triangle whose hypotenuse has length \|a ´ b\| and whose other two sides have lengths \|b\| ´ \|a\| cos θ and \|a\| sin θ. So Pythagoras gives \|a ´ b\|2 = \|b\| ´ \|a\| cos θ 2 + \|a\| sin θ 2 = \|b\|2 ´ 2\|a\| \|b\| cos θ + \|a\|2 cos2 θ + \|a\|2 sin2 θ = \|b\|2 ´ 2\|a\| \|b\| cos θ + \|a\|2 This is precisely the cosine law8. Observe that, when θ = π 2 , this reduces to, (surprise!) Pythagoras’ theorem. Setting our two expressions for \|a ´ b\|2 equal to each other, \|a ´ b\|2 = \|a\|2 + \|b\|2 ´ 2a ¨ b = \|b\|2 ´ 2\|a\| \|b\| cos θ + \|a\|2 cancelling the \|a\|2 and \|b\|2 common to both sides ´2a ¨ b = ´2\|a\| \|b\| cos θ and dividing by ´2 gives a ¨ b = \|a\| \|b\| cos θ which is exactly property 6. Property 7 follows directly from property 6. First note that the dot product a ¨ b = \|a\| \|b\| cos θ is zero if and only if at least one of the three factors \|a\|, \|b\|, cos θ is zero. The first factor is zero if and only if a = 0. The second factor is zero if and only if b = 0. The third factor is zero if and only if θ = ˘π 2 + 2kπ, for some integer k, which in turn is true if and only if a and b are mutually perpendicular. Because of Property 7 of Theorem 1.2.11, the dot product can be used to test whether or not two vectors are perpendicular to each other. That is, whether or not the angle between 8 You may be used to seeing it written as c2 = a2 + b2 ´ 2ab cos C, where a, b and c are the lengths of the three sides of the triangle and C is the angle opposite the side of length c 16 VECTORS AND GEOMETRY 1.2 VECTORS the two vectors is 90˝. Another name9 for “perpendicular” is “orthogonal”. Testing for orthogonality is one of the main uses of the dot product. Example 1.2.12 Consider the three vectors a = ⟨1, 1, 0⟩ b = ⟨1, 0, 1⟩ c = ⟨´1, 1, 1⟩ Their dot products a ¨ b = ⟨1, 1, 0⟩¨ ⟨1, 0, 1⟩ = 1 ˆ 1 + 1 ˆ 0 + 0 ˆ 1 = 1 a ¨ c = ⟨1, 1, 0⟩¨ ⟨´1, 1, 1⟩= 1 ˆ (´1) + 1 ˆ 1 + 0 ˆ 1 = 0 b ¨ c = ⟨1, 0, 1⟩¨ ⟨´1, 1, 1⟩= 1 ˆ (´1) + 0 ˆ 1 + 1 ˆ 1 = 0 tell us that c is perpendicular to both a and b. Since both \|a\| = \|b\| = ? 12 + 12 + 02 = ? 2 the first dot product tells us that the angle, θ, between a and b obeys cos θ = a ¨ b \|a\| \|b\| = 1 2 ù ñ θ = π 3 z y x ⟨1, 1, 0⟩ ⟨1, 0, 1⟩ ⟨´1, 1, 1⟩ Example 1.2.12 Dot products are also used to compute projections. First, here’s the definition. 9 The concepts of the dot product and perpendicularity have been generalized a lot in mathematics (for example, from 2d and 3d vectors to functions). The generalization of the dot product is called the “inner product” and the generalization of perpendicularity is called “orthogonality”. 17 VECTORS AND GEOMETRY 1.2 VECTORS Draw two vectors, a and b, with their tails at a common point and drop a per-pendicular from the head of a to the line that passes through both the head and tail of b. By definition, the projection of the vector a on the vector b is the vector from the tail of b to the point on the line where the perpendicular hits. a b projb a θ a b projb a θ Definition 1.2.13 (Projection). Think of the projection of a on b as the part of a that is in the direction of b. Now let’s develop a formula for the projection of a on b. Denote by θ the angle between a and b. If \|θ\| is no more than 90˝, as in the figure on the left above, the length of the projection of a on b is \|a\| cos θ. By Property 6 of Theorem 1.2.11, \|a\| cos θ = a ¨ b/\|b\|, so the projection is a vector whose length is a ¨ b/\|b\| and whose direction is given by the unit vector b/\|b\|. Hence projection of a on b = projb a = a ¨ b \|b\| b \|b\| = a ¨ b \|b\|2 b If \|θ\| is larger than 90˝, as in the figure on the right above, the projection has length \|a\| cos(π ´ θ) = ´\|a\| cos θ = ´a ¨ b/\|b\| and direction ´b/\|b\|. In this case projb a = ´a ¨ b \|b\| ´b \|b\| = a ¨ b \|b\|2 b So the formula projb a = a ¨ b \|b\|2 b Equation 1.2.14. is applicable whenever b ‰ 0. We may rewrite projb a = a¨b \|b\| b \|b\|. The coefficient, a¨b \|b\| , of the unit vector b \|b\|, is called the component of a in the direction b. As a special case, if b happens to be a unit vector, which, for emphasis, we’ll now write has ˆ b, the projection formula simplifies to proj ˆ b a = (a ¨ ˆ b) ˆ b Equation 1.2.15. 18 VECTORS AND GEOMETRY 1.2 VECTORS Example 1.2.16 In this example, we will find the projection of the vector ⟨0, 3⟩on the vector ⟨1, 1⟩, as in the figure x y ⟨0, 3⟩ ⟨1, 1⟩ proj⟨1,1⟩⟨0, 3⟩ By Equation 1.2.14 with a = ⟨0, 3⟩and b = ⟨1, 1⟩, that projection is proj⟨1,1⟩⟨0, 3⟩= ⟨0, 3⟩¨ ⟨1, 1⟩ \| ⟨1, 1⟩\|2 ⟨1, 1⟩ = 0 ˆ 1 + 3 ˆ 1 12 + 12 ⟨1, 1⟩= 3 2, 3 2 Example 1.2.16 One use of projections is to “resolve forces”. There is an example in the next (optional) section. 1.2.3 § § (Optional) Using Dot Products to Resolve Forces — The Pendulum Model a pendulum by a mass m that is connected to a hinge by an idealized rod that is massless and of fixed length ℓ. Denote by θ the angle between the rod and vertical. The forces acting on the mass are • gravity, which has magnitude mg and direction ⟨0, ´1⟩, • tension in the rod, whose magnitude τ(t) automatically adjusts itself so that the distance between the mass and the hinge is fixed at ℓ(so that the rod does not stretch or contract) and whose direction is always parallel to the rod, • and possibly some frictional forces, like friction in the hinge and air resistance. As-sume that the total frictional force has magnitude proportional10 to the speed of the mass and has direction opposite to the direction of motion of the mass. We’ll call the constant of proportionality β. 10 The behaviour of air resistance (sometimes called drag) is pretty complicated. We’re using a reasonable low speed approximation. At high speeds drag is typically proportional to the square of the speed. 19 VECTORS AND GEOMETRY 1.2 VECTORS θ ℓ mg (gravity) (tension) τ ´βℓdθ dt (friction) If we use a coordinate system centered on the hinge, the (x, y) coordinates of the mass at time t are x(t) = ℓsin θ(t) y(t) = ´ℓcos θ(t) where θ(t) is the angle between the rod and vertical at time t. We are now going to use Newton’s law of motion mass ˆ acceleration = total applied force to determine now θ evolves in time. By definition, the velocity and acceleration vectors11 for the position vector ⟨x(t), y(t)⟩are d dt ⟨x(t), y(t)⟩= dx dt (t), dy dt (t) d2 dt2 ⟨x(t), y(t)⟩= d2x dt2 (t), d2y dt2 (t) So, the velocity and acceleration vectors of our mass are v(t) = d dt ⟨x(t), y(t)⟩= ℓd dt sin θ(t), ´ℓd dt cos θ(t) = ℓcos θ(t) dθ dt (t) , ℓsin θ(t) dθ dt (t) = ℓdθ dt (t) ⟨cos θ(t), sin θ(t)⟩ a(t) = d2 dt2 ⟨x(t), y(t)⟩= d dt " ℓdθ dt (t) ⟨cos θ(t), sin θ(t)⟩ = ℓd2θ dt2 (t) ⟨cos θ(t), sin θ(t)⟩+ ℓdθ dt (t) d dt cos θ(t), d dt sin θ(t) = ℓd2θ dt2 (t) ⟨cos θ(t), sin θ(t)⟩+ ℓ dθ dt (t) 2 ⟨´ sin θ(t), cos θ(t)⟩ 11 For a more comprehensive treatment of derivatives of vector-valued functions r(t), and in particular of velocity and acceleration, see Section 1.6 in this text and Section 1.1 in the CLP-4 text. 20 VECTORS AND GEOMETRY 1.2 VECTORS The negative of the velocity vector is ´ℓdθ dt ⟨cos θ, sin θ⟩, so the total frictional force is ´βℓdθ dt ⟨cos θ, sin θ⟩ with β our constant of proportionality. The vector τ(t) ⟨´ sin θ(t), cos θ(t)⟩ has magnitude τ(t) and direction parallel to the rod pointing from the mass towards the hinge and so is the force due to tension in the rod. Hence, for this physical system, Newton’s law of motion is massˆacceleration hkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkikkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkj mℓd2θ dt2 ⟨cos θ, sin θ⟩+ mℓ dθ dt 2 ⟨´ sin θ, cos θ⟩ = gravity hkkkkkikkkkkj mg ⟨0, ´1⟩+ tension hkkkkkkkkkikkkkkkkkkj τ ⟨´ sin θ, cos θ⟩´ friction hkkkkkkkkkkikkkkkkkkkkj βℓdθ dt ⟨cos θ, sin θ⟩ (˚) This is a rather complicated looking equation. Writing out its x- and y-components doesn’t help. They also look complicated. Instead, the equation can be considerably simplified (and consequently better understood) by “taking its components parallel to and perpen-dicular to the direction of motion”. From the velocity vector v(t), we see that ⟨cos θ(t), sin θ(t)⟩ is a unit vector parallel to the direction of motion at time t. Recall, from (1.2.15), that the projection of any vector b on any unit vector ˆ d (with the “hat” on ˆ d reminding ourselves that the vector is a unit vector) is b ¨ ˆ d ˆ d The coefficient b ¨ ˆ d is, by definition, the component of b in the direction ˆ d. So, by dot-ting both sides of the equation of motion (˚) with ˆ d = ⟨cos θ(t), sin θ(t)⟩, we extract the component parallel to the direction of motion. Since ⟨cos θ, sin θ⟩¨ ⟨cos θ, sin θ⟩= 1 ⟨cos θ, sin θ⟩¨ ⟨´ sin θ, cos θ⟩= 0 ⟨cos θ, sin θ⟩¨ ⟨0, ´1⟩= ´ sin θ this gives mℓd2θ dt2 = ´mg sin θ ´ βℓdθ dt which is much cleaner than (˚)! When θ is small, we can approximate sin θ « θ and get the equation d2θ dt2 + β m dθ dt + g ℓθ = 0 which is easily solved. There are systematic procedures for finding the solution, but we’ll just guess. 21 VECTORS AND GEOMETRY 1.2 VECTORS When there is no friction (so that β = 0), we would expect the pendulum to just oscil-late. So it is natural to guess θ(t) = A sin(ωt ´ δ) which is an oscillation with (unknown) amplitude A, frequency ω (radians per unit time) and phase δ. Substituting this guess into the left hand side, θ2 + g ℓθ, yields ´Aω2 sin(ωt ´ δ) + A g ℓsin(ωt ´ δ) which is zero if ω = a g/ℓ. So θ(t) = A sin(ωt ´ δ) is a solution for any amplitude A and phase δ, provided the frequency ω = a g/ℓ. When there is some, but not too much, friction, so that β ą 0 is relatively small, we would expect “oscillation with decaying amplitude”. So we guess θ(t) = Ae´γt sin(ωt ´ δ) for some constant decay rate γ, to be determined. With this guess, θ(t) = Ae´γtsin(ωt ´ δ) θ1(t) = ´ γAe´γtsin(ωt ´ δ) + ωAe´γtcos(ωt ´ δ) θ2(t) = (γ2 ´ ω2)Ae´γtsin(ωt ´ δ) ´ 2γωAe´γtcos(ωt ´ δ) and the left hand side d2θ dt2 + β m dθ dt + g ℓθ = γ2 ´ ω2 ´ β mγ + g ℓ Ae´γt sin(ωt ´ δ) + ´2γω + β mω Ae´γt cos(ωt ´ δ) vanishes if γ2 ´ ω2 ´ β mγ + g ℓ= 0 and ´2γω + β mω = 0. The second equation tells us the decay rate γ = β 2m and then the first tells us the frequency ω = b γ2 ´ β mγ + g ℓ= c g ℓ´ β2 4m2 When there is a lot of friction (namely when β2 4m2 ą g ℓ, so that the frequency ω is not a real number), we would expect damping without oscillation and so would guess θ(t) = Ae´γt. You can determine the allowed values of γ by substituting this guess in. To extract the components perpendicular to the direction of motion, we dot with ⟨´ sin θ, cos θ⟩ rather than ⟨cos θ, sin θ⟩. Note that, because ⟨´ sin θ, cos θ⟩¨ ⟨cos θ, sin θ⟩= 0, the vector ⟨´ sin θ, cos θ⟩really is perpendicular to the direction of motion. Since ⟨´ sin θ, cos θ⟩¨ ⟨cos θ, sin θ⟩= 0 ⟨´ sin θ, cos θ⟩¨ ⟨´ sin θ, cosθ⟩= 1 ⟨´ sin θ, cos θ⟩¨ ⟨0, ´1⟩= ´ cos θ dotting both sides of the equation of motion (˚) with ⟨´ sin θ, cos θ⟩gives mℓ dθ dt 2 = ´mg cos θ + τ This equation just determines the tension τ = mℓ dθ dt 2 + mg cos θ in the rod, once you know θ(t). 22 VECTORS AND GEOMETRY 1.2 VECTORS 1.2.4 § § (Optional) Areas of Parallelograms A parallelogram is naturally determined by the two vectors that define its sides. We’ll now develop a formula for the area of a parallelogram in terms of these two vectors. Construct a parallelogram as follows. Pick two vectors ⟨a, b⟩and ⟨c, d⟩. Draw them with their tails at a common point. Then draw ⟨a, b⟩a second time with its tail at the head of ⟨c, d⟩and draw ⟨c, d⟩a second time with its tail at the head of ⟨a, b⟩. If the com-mon point is the origin, you get a picture like the figure below. Any parallelogram can pa c, b dq pa, bq pc, dq a c d b c a b d be constructed like this if you pick the common point and two vectors appropriately. Let’s compute the area of the parallelogram. The area of the large rectangle with ver-tices (0, 0), (0, b + d), (a + c, 0) and (a + c, b + d) is (a + c)(b + d). The parallelogram we want can be extracted from the large rectangle by deleting the two small rectangles (each of area bc), and the two lightly shaded triangles (each of area 1 2cd), and the two darkly shaded triangles (each of area 1 2ab). So the desired area = (a + c)(b + d) ´ (2 ˆ bc) ´ 2 ˆ 1 2cd ´ 2 ˆ 1 2ab = ad ´ bc In the above figure, we have implicitly assumed that a, b, c, d ě 0 and d/c ě b/a. In words, we have assumed that both vectors ⟨a, b⟩, ⟨c, d⟩lie in the first quadrant and that ⟨c, d⟩lies above ⟨a, b⟩. By simply interchanging a Ø c and b Ø d in the picture and throughout the argument, we see that when a, b, c, d ě 0 and b/a ě d/c, so that the vector ⟨c, d⟩lies below ⟨a, b⟩, the area of the parallelogram is bc ´ ad. In fact, all cases are covered by the formula area of parallelogram with sides ⟨a, b⟩and ⟨c, d⟩= \|ad ´ bc\| Equation 1.2.17. Given two vectors ⟨a, b⟩and ⟨c, d⟩, the expression ad ´ bc is generally written det a b c d = ad ´ bc 23 VECTORS AND GEOMETRY 1.2 VECTORS and is called the determinant of the matrix12 a b c d with rows ⟨a, b⟩and ⟨c, d⟩. The determinant of a 2 ˆ 2 matrix is the product of the diagonal entries minus the product of the off-diagonal entries. There is a similar formula in three dimensions. Any three vectors a = ⟨a1, a2, a3⟩, b = ⟨b1, b2, b3⟩and c = ⟨c1, c2, c3⟩in three dimensions determine a parallelepiped (three di-a b c mensional parallelogram). Its volume is given by the formula volume of parallelepiped with edges a, b, c = ˇ ˇ ˇ ˇ ˇ ˇ det   a1 a2 a3 b1 b2 b3 c1 c2 c3   ˇ ˇ ˇ ˇ ˇ ˇ Equation 1.2.18. The determinant of a 3 ˆ 3 matrix can be defined in terms of some 2 ˆ 2 determinants by det   a1 a2 a3 b1 b2 b3 c1 c2 c3  = a1 det   a1 a2 a3 b1 b2 b3 c1 c2 c3  −a2 det   a1 a2 a3 b1 b2 b3 c1 c2 c3  + a3 det   a1 a2 a3 b1 b2 b3 c1 c2 c3   = a1 (b2c3 −b3c2) −a2 (b1c3 −b3c1) + a3 (b1c2 −b2c1) This formula is called “expansion along the top row”. There is one term in the formula for each entry in the top row of the 3 ˆ 3 matrix. The term is a sign times the entry itself times the determinant of the 2 ˆ 2 matrix gotten by deleting the row and column that contains the entry. The sign alternates, starting with a +. We shall not prove this formula completely here13. It gets a little tedious. But, there is one case in which we can easily verify that the volume of the parallelepiped is really 12 The topics of matrices and determinants appear prominently in linear algebra courses. We are only going to use them as notation, and we will explicitly explain that notation. A linear algebra course is not a prerequisite for this text. 13 For a full derivation, see Example 1.2.25 24 VECTORS AND GEOMETRY 1.2 VECTORS given by the absolute value of the claimed determinant. If the vectors b and c happen to lie in the xy plane, so that b3 = c3 = 0, then det   a1 a2 a3 b1 b2 0 c1 c2 0  = a1(b20 ´ 0c2) ´ a2(b10 ´ 0c1) + a3(b1c2 ´ b2c1) = a3(b1c2 ´ b2c1) The first factor, a3, is the z-coordinate of the one vector not contained in the xy-plane. It is (up to a sign) the height of the parallelepiped. The second factor is, up to a sign, the area of the parallelogram determined by b and c. This parallelogram forms the base of the parallelepiped. The product is indeed, up to a sign, the volume of the parallelepiped. That the formula is true in general is a consequence of the fact (that we will not prove) that the value of a determinant does not change when one rotates the coordinate system and that one can always rotate our coordinate axes around so that b and c both lie in the xy-plane. 1.2.5 § § The Cross Product We have already seen two different products involving vectors — the multiplication of a vector by a scalar and the dot product of two vectors. The dot product of two vectors yields a scalar. We now introduce another product of two vectors, called the cross product. The cross product of two vectors will give a vector. There are applications which have two vectors as inputs and produce one vector as an output, and which are related to the cross product. Here is a very brief mention of two such applications. We will look at them in much more detail later. • Consider a parallelogram in three dimensions. A parallelogram is naturally deter-mined by the two vectors that define its sides. One measure of the size of a parallel-ogram is its area. One way to specify the orientation of the parallelogram is to give a vector that is perpendicular to it. A very compact way to encode both the area and the orientation of the parallelogram is to give a vector whose direction is perpen-dicular to the plane in which it lies and whose magnitude is its area. We shall see that such a vector can be easily constructed by taking the cross product (definition coming shortly) of the two vectors that give the sides of the parallelogram. • Imagine a rigid body which is rotating at a rate Ωradians per second about an axis whose direction is given by the unit vector ˆ a. Let P be any point on the body. We shall see, in the (optional) §1.2.7, that the velocity, v, of the point P is the cross prod-uct (again, definition coming shortly) of the vector Ωˆ a with the vector r from any point on the axis of rotation to P. 25 VECTORS AND GEOMETRY 1.2 VECTORS P r v ˆ a Ω Finally, here is the definition of the cross product. Note that it applies only to vectors in three dimensions. The cross product of the vectors a = ⟨a1, a2, a3⟩and b = ⟨b1, b2, b3⟩is denoted a ˆ b and is defined by a ˆ b = ⟨a2b3 ´ a3b2 , a3b1 ´ a1b3 , a1b2 ´ a2b1⟩ Definition 1.2.19 (Cross Product). Note that each component has the form aibj ´ ajbi. The index i of the first a in compo-nent number k of a ˆ b is just after k in the list 1, 2, 3, 1, 2, 3, 1, 2, 3, ¨ ¨ ¨ . The index j of the first b is just before k in the list. (a ˆ b)k = ajust after k bjust before k ´ ajust before k bjust after k For example, for component number k = 3, “just after 3” is 1 “just before 3” is 2 ù ñ (a ˆ b)3 = a1b2 ´ a2b1 There is a much better way to remember this definition. Recall that a 2 ˆ 2 matrix is an array of numbers having two rows and two columns and that the determinant of a 2 ˆ 2 matrix is defined by det a b c d = ad ´ bc It is the product of the entries on the diagonal minus the product of the entries not on the diagonal. A 3 ˆ 3 matrix is an array of numbers having three rows and three columns.   i j k a1 a2 a3 b1 b2 b3   You will shortly see why the entries in the top row have been given the rather peculiar names i, j and k. The determinant of a 3 ˆ 3 matrix can be defined in terms of some 2 ˆ 2 determinants by 26 VECTORS AND GEOMETRY 1.2 VECTORS det   i j k a1 a2 a3 b1 b2 b3  = i det   i j k a1 a2 a3 b1 b2 b3  −j det   i j k a1 a2 a3 b1 b2 b3  + k det   i j k a1 a2 a3 b1 b2 b3   = i (a2b3 −a3b2) −j (a1b3 −a3b1) + k (a1b2 −a2b1) This formula is called “expansion of the determinant along the top row”. There is one term in the formula for each entry in the top row. The term is a sign times the entry itself times the determinant of the 2 ˆ 2 matrix gotten by deleting the row and column that contains the entry. The sign alternates, starting with a +. If we now replace i by ˆ ı ı ı, j by ˆ ȷ ȷ ȷ and k by ˆ k, we get exactly the formula for a ˆ b of Definition 1.2.19. That is the reason for the peculiar choice of names for the matrix entries. So a ˆ b = det   ˆ ı ı ı ˆ ȷ ȷ ȷ ˆ k a1 a2 a3 b1 b2 b3   = ˆ ı ı ı a2b3 ´ a3b2) ´ ˆ ȷ ȷ ȷ(a1b3 ´ a3b1) + ˆ k(a1b2 ´ a2b1) is a mnemonic device for remembering the definition of a ˆ b. It is also good from the point of view of evaluating a ˆ b. Here are several examples in which we use the deter-minant mnemonic device to evaluate cross products. Example 1.2.20 In this example, we’ll use the mnemonic device to compute two very simple cross prod-ucts. First ˆ ı ı ı ˆ ˆ   “ det » – ˆ ı ı ı ˆ    ˆ k 1 0 0 0 1 0 ﬁ ﬂ“ ˆ ı ı ı det » – ˆ ı ı ı ˆ    ˆ k 1 0 0 0 1 0 ﬁ ﬂ´ ˆ   det » – ˆ ı ı ı ˆ    ˆ k 1 0 0 0 1 0 ﬁ ﬂ ˆ k det » – ˆ ı ı ı ˆ    ˆ k 1 0 0 0 1 0 ﬁ ﬂ “ ˆ ı ı ı p0 ˆ 0 ´ 0 ˆ 1q´ˆ   p1 ˆ 0 ´ 0 ˆ 0q ˆ k p1 ˆ 1 ´ 0 ˆ 0q “ ˆ k Second ˆ   ˆˆ ı ı ı “ det » – ˆ ı ı ı ˆ    ˆ k 0 1 0 1 0 0 ﬁ ﬂ“ ˆ ı ı ı det » – ˆ ı ı ı ˆ    ˆ k 0 1 0 1 0 0 ﬁ ﬂ´ ˆ   det » – ˆ ı ı ı ˆ    ˆ k 0 1 0 1 0 0 ﬁ ﬂ ˆ k det » – ˆ ı ı ı ˆ    ˆ k 0 1 0 1 0 0 ﬁ ﬂ “ ˆ ı ı ı p1 ˆ 0 ´ 0 ˆ 0q´ˆ   p0 ˆ 0 ´ 0 ˆ 1q ˆ k p0 ˆ 0 ´ 1 ˆ 1q “ ´ˆ k Note that, unlike most (or maybe even all) products that you have seen before, ˆ ı ı ı ˆ ˆ ȷ ȷ ȷ is not the same as ˆ ȷ ȷ ȷ ˆ ˆ ı ı ı! Example 1.2.20 Example 1.2.21 In this example, we’ll use the mnemonic device to compute two more complicated cross products. Let a = ⟨1, 2, 3⟩and b = ⟨1, ´1, 2⟩. First 27 VECTORS AND GEOMETRY 1.2 VECTORS a ˆ b “ det » – ˆ ı ı ı ˆ    ˆ k 1 2 3 1 ´1 2 ﬁ ﬂ“ ˆ ı ı ı det » – ˆ ı ı ı ˆ    ˆ k 1 2 3 1 ´1 2 ﬁ ﬂ´ ˆ   det » – ˆ ı ı ı ˆ    ˆ k 1 2 3 1 ´1 2 ﬁ ﬂ ˆ k det » – ˆ ı ı ı ˆ    ˆ k 1 2 3 1 ´1 2 ﬁ ﬂ “ ˆ ı ı ı t2 ˆ 2 ´ 3 ˆ p´1qu ´ˆ   t1 ˆ 2 ´ 3 ˆ 1uˆ k t1 ˆ p´1q ´ 2 ˆ 1u “ 7ˆ ı ı ı ˆ   ´ 3 ˆ k Second b ˆ a “ det » – ˆ ı ı ı ˆ    ˆ k 1 ´1 2 1 2 3 ﬁ ﬂ“ ˆ ı ı ı det » – ˆ ı ı ı ˆ    ˆ k 1 ´1 2 1 2 3 ﬁ ﬂ´ ˆ   det » – ˆ ı ı ı ˆ    ˆ k 1 ´1 2 1 2 3 ﬁ ﬂˆ k det » – ˆ ı ı ı ˆ    ˆ k 1 ´1 2 1 2 3 ﬁ ﬂ “ ˆ ı ı ı tp´1q ˆ 3 ´ 2 ˆ 2u ´ˆ   t1 ˆ 3 ´ 2 ˆ 1u ˆ k t1 ˆ 2 ´ p´1q ˆ 1u “ ´7ˆ ı ı ı ´ ˆ   3 ˆ k Here are some important observations. • The vectors a ˆ b and b ˆ a are not the same! In fact b ˆ a = ´a ˆ b. We shall see in Theorem 1.2.23 below that this was not a fluke. • The vector a ˆ b has dot product zero with both a and b. So the vector a ˆ b is perpendicular to both a and b. We shall see in Theorem 1.2.23 below that this was also not a fluke. Example 1.2.21 Example 1.2.22 Yet again we use the mnemonic device to compute a more complicated cross product. This time let a = ⟨3, 2, 1⟩and b = ⟨6, 4, 2⟩. Then a ˆ b “ det » – ˆ ı ı ı ˆ    ˆ k 3 2 1 6 4 2 ﬁ ﬂ “ ˆ ı ı ı det » – ˆ ı ı ı ˆ    ˆ k 3 2 1 6 4 2 ﬁ ﬂ´ ˆ   det » – ˆ ı ı ı ˆ    ˆ k 3 2 1 6 4 2 ﬁ ﬂ ˆ k det » – ˆ ı ı ı ˆ    ˆ k 3 2 1 6 4 2 ﬁ ﬂ “ ˆ ı ı ı p2 ˆ 2 ´ 1 ˆ 4q´ˆ   p3 ˆ 2 ´ 1 ˆ 6q ˆ k p3 ˆ 4 ´ 2 ˆ 6q “ 0 We shall see in Theorem 1.2.23 below that it is not a fluke that the cross product is 0. It is a consequence of the fact that a and b = 2a are parallel. Example 1.2.22 We now move on to learning about the properties of the cross product. Our first prop-erties lead up to a more intuitive geometric definition of a ˆ b, which is better for inter-preting a ˆ b. These properties of the cross product, which state that a ˆ b is a vector and then determine its direction and length, are as follows. We will collect these properties, and a few others, into a theorem shortly. (0) a, b are vectors in three dimensions and a ˆ b is a vector in three dimensions. 28 VECTORS AND GEOMETRY 1.2 VECTORS (1) a ˆ b is perpendicular to both a and b. Proof: To check that a and a ˆ b are perpendicular, one just has to check that the dot product a ¨ (a ˆ b) = 0. The six terms in a ¨ (a ˆ b) = a1(a2b3 ´ a3b2) + a2(a3b1 ´ a1b3) + a3(a1b2 ´ a2b1) cancel pairwise. The computation showing that b ¨ (a ˆ b) = 0 is similar. (2) \|a ˆ b\| = \|a\| \|b\| sin θ where 0 ď θ ď π is the angle between a and b = the area of the parallelogram with sides a and b a a b b θ Proof: The formula \|a ˆ b\| = \|a\| \|b\| sin θ is gotten by verifying that \|a ˆ b\|2 = a ˆ b ¨ a ˆ b = (a2b3 ´ a3b2)2 + (a3b1 ´ a1b3)2 + (a1b2 ´ a2b1)2 = a2 2b2 3 ´ 2a2b3a3b2 + a2 3b2 2 + a2 3b2 1 ´ 2a3b1a1b3 + a2 1b2 3 + a2 1b2 2 ´ 2a1b2a2b1 + a2 2b2 1 is equal to \|a\|2 \|b\|2 sin2 θ = \|a\|2\|b\|2(1 ´ cos2 θ) = \|a\|2\|b\|2 ´ (a ¨ b)2 = a2 1 + a2 2 + a2 3 b2 1 + b2 2 + b2 3 ´ a1b1 + a2b2 + a3b3 2 = a2 1b2 2 + a2 1b2 3 + a2 2b2 1 + a2 2b2 3 + a2 3b2 1 + a2 3b2 2 ´ 2a1b1a2b2 + 2a1b1a3b3 + 2a2b2a3b3 To see that \|a\| \|b\| sin θ is the area of the parallelogram with sides a and b, just recall that the area of any parallelogram is given by the length of its base times its height. Think of a as the base of the parallelogram. Then \|a\| is the length of the base and \|b\| sin θ is the height. a a b b θ These properties almost determine a ˆ b. Property 1 forces the vector a ˆ b to lie on the line perpendicular to the plane containing a and b. There are precisely two vectors on this line that have the length given by property 2. In the left hand figure of 29 VECTORS AND GEOMETRY 1.2 VECTORS a b c d a b a ˆ b the two vectors are labeled c and d. Which of these two candidates is correct is determined by the right hand rule14, which says that if you form your right hand into a fist with your fingers curling from a to b, then when you stick your thumb straight out from the fist, it points in the direction of a ˆ b. This is illustrated in the figure on the right above 15. The important special cases (3) ˆ ı ı ı ˆ ˆ ȷ ȷ ȷ = ˆ k, ˆ ȷ ȷ ȷ ˆ ˆ k = ˆ ı ı ı, ˆ k ˆ ˆ ı ı ı = ˆ ȷ ȷ ȷ ˆ ȷ ȷ ȷ ˆ ˆ ı ı ı = ´ ˆ k, ˆ k ˆ ˆ ȷ ȷ ȷ = ´ˆ ı ı ı, ˆ ı ı ı ˆ ˆ k = ´ˆ ȷ ȷ ȷ all follow directly from the definition of the cross product (see, for example, Example 1.2.20) and all obey the right hand rule. Combining properties 1, 2 and the right hand rule give the geometric definition of a ˆ b. To remember these three special cases, just remember this figure. The product of any two standard basis vectors, taken in the order of the arrows in the figure, is the third standard basis vector. Going against the arrows introduces a minus sign. (4) a ˆ b = \|a\| \|b\| sin θ ˆ n where θ is the angle between a and b, \| ˆ n\| = 1, ˆ n K a, b and (a, b, ˆ n) obey the right hand rule. Outline of Proof: We have already seen that the right hand side has the correct length and, except possibly for a sign, direction. To check that the right hand rule holds in general, rotate your coordinate system around16 so that a points along the positive x axis and b lies in the xy-plane with positive y component. That is a = αˆ ı ı ı and b = βˆ ı ı ı + γˆ ȷ ȷ ȷ with α, γ ě 0. Then a ˆ b = αˆ ı ı ı ˆ (βˆ ı ı ı + γˆ ȷ ȷ ȷ) = αβ ˆ ı ı ı ˆ ˆ ı ı ı + αγ ˆ ı ı ı ˆ ˆ ȷ ȷ ȷ. The first term vanishes by property 2, because the angle θ between ˆ ı ı ı and ˆ ı ı ı is zero. So, by property 3, a ˆ b = αγ ˆ k points along the positive z axis, which is consistent with the right hand rule. 14 That the cross product uses the right hand rule, rather than the left hand rule, is an example of the tyranny of the masses — only roughly 10% of humans are left-handed. 15 This figure is a variant of hand rule simple.png 16 Note that as you translate or rotate the coordinate system, the right hand rule is preserved. If (a, b, ˆ n) obey the right hand rule so do their rotated and translated versions. 30 VECTORS AND GEOMETRY 1.2 VECTORS The analog of property 7 of the dot product (which says that a ¨ b is zero if and only if a = 0 or b = 0 or a K b) follows immediately from property 2. (5) a ˆ b = 0 ð ñ a = 0 or b = 0 or a ∥b The remaining properties are all tools for helping do computations with cross products. Here is a theorem which summarizes the properties of the cross product. We have already seen the first five. The other properties are all tools for helping do computations with cross products. (0) a, b are vectors in three dimensions and a ˆ b is a vector in three dimensions. (1) a ˆ b is perpendicular to both a and b. (2) \|a ˆ b\| = \|a\| \|b\| sin θ where θ is the angle between a and b = the area of the parallelogram with sides a and b a a b b θ (3) ˆ ı ı ı ˆ ˆ ȷ ȷ ȷ = ˆ k, ˆ ȷ ȷ ȷ ˆ ˆ k = ˆ ı ı ı, ˆ k ˆ ˆ ı ı ı = ˆ ȷ ȷ ȷ (4) a ˆ b = \|a\| \|b\| sin θ n where θ is the angle between a and b, \|n\| = 1, n K a, b and (a, b, n) obey the right hand rule. (5) a ˆ b = 0 ð ñ a = 0 or b = 0 or a ∥b (6) a ˆ b = ´b ˆ a (7) (sa) ˆ b = a ˆ (sb) = s(a ˆ b) for any scalar (i.e. number) s. (8) a ˆ (b + c) = a ˆ b + a ˆ c (9) a ¨ (b ˆ c) = (a ˆ b) ¨ c (10) a ˆ (b ˆ c) = (c ¨ a)b ´ (b ¨ a)c Theorem 1.2.23 (Properties of the Cross Product). Proof. We have already seen the proofs up to number 5. Numbers 6, 7 and 8 follow im-mediately from the definition, using a little algebra. To prove numbers 9 and 10 we just write out the definitions of the left hand sides and the right hand sides and observe that they are equal. (9) The left hand side is a ¨ (b ˆ c) = ⟨a1, a2, a3⟩¨ ⟨b2c3 ´ b3c2 , b3c1 ´ b1c3 , b1c2 ´ b2c1⟩ = a1b2c3 ´ a1b3c2 + a2b3c1 ´ a2b1c3 + a3b1c2 ´ a3b2c1 The right hand side is (a ˆ b) ¨ c = ⟨a2b3 ´ a3b2 , a3b1 ´ a1b3 , a1b2 ´ a2b1⟩¨ ⟨c1, c2, c3⟩ = a2b3c1 ´ a3b2c1 + a3b1c2 ´ a1b3c2 + a1b2c3 ´ a2b1c3 31 VECTORS AND GEOMETRY 1.2 VECTORS The left and right hand sides are the same. (10) We will give the straightforward, but slightly tedious, computations in (the optional) §1.2.6. Take particular care with properties 6 and 10. They are counterintuitive and are a frequent source of errors. In particular, for general vectors a, b, c, the cross product is neither commutative nor associative, meaning that a ˆ b ‰ b ˆ a a ˆ (b ˆ c) ‰ (a ˆ b) ˆ c For example ˆ ı ı ı ˆ (ˆ ı ı ı ˆ ˆ ȷ ȷ ȷ) = ˆ ı ı ı ˆ ˆ k = ´ ˆ k ˆ ˆ ı ı ı = ´ˆ ȷ ȷ ȷ (ˆ ı ı ı ˆ ˆ ı ı ı) ˆ ˆ ȷ ȷ ȷ = 0 ˆ ˆ ȷ ȷ ȷ = 0 Warning 1.2.24. Example 1.2.25 As an illustration of the properties of the dot and cross product, we now derive the for-mula for the volume of the parallelepiped with edges a = ⟨a1, a2, a3⟩, b = ⟨b1, b2, b3⟩, c = ⟨c1, c2, c3⟩that was mentioned in §1.2.4. The volume of the parallelepiped is the area a c b b ˆ c θ ϕ of its base times its height17. The base is the parallelogram with sides b and c. Its area is the length of its base, which is \|b\|, times its height, which is \|c\| sin θ. (Drop a perpendic-ular from the head of c to the line containing b). Here θ is the angle between b and c. So the area of the base is \|b\| \|c\| sin θ = \|b ˆ c\|, by property 2 of the cross product. To get the height of the parallelepiped, we drop a perpendicular from the head of a to the line that passes through the tail of a and is perpendicular to the base of the parallelepiped. In other words, from the head of a to the line that contains both the head and the tail of b ˆ c. So 17 This is a simple integral calculus exercise. 32 VECTORS AND GEOMETRY 1.2 VECTORS the height of the parallelepiped is \|a\| \| cos φ\|. (The absolute values have been included be-cause if the angle between b ˆ c and a happens to be greater than 90˝, the cos φ produced by taking the dot product of a and (b ˆ c) will be negative.) All together volume of parallelepiped = (area of base) (height) = \|b ˆ c\| \|a\| \| cos φ\| = ˇ ˇa ¨ (b ˆ c) ˇ ˇ = \|a1(b ˆ c)1 + a2(b ˆ c)2 + a3(b ˆ c)3\| = ˇ ˇ ˇ ˇa1 det b2 b3 c2 c3 ´ a2 det b1 b3 c1 c3 + a3 det b1 b2 c1 c2 ˇ ˇ ˇ ˇ = ˇ ˇ ˇ ˇ ˇ ˇ det   a1 a2 a3 b1 b2 b3 c1 c2 c3   ˇ ˇ ˇ ˇ ˇ ˇ Example 1.2.25 Example 1.2.26 As a concrete example of the computation of the volume of a parallelepiped, we consider the parallelepiped with edges a = ⟨0, 1, 2⟩ b = ⟨1, 1, 0⟩ c = ⟨0, 1, 0⟩ Here is a sketch. The base of the parallelepiped is the parallelogram with sides b and c. It x y z a b c b ˆ c 33 VECTORS AND GEOMETRY 1.2 VECTORS is the shaded parallelogram in the sketch above. As b ˆ c = det   ˆ ı ı ı ˆ ȷ ȷ ȷ ˆ k 1 1 0 0 1 0   = ˆ ı ı ı det 1 0 1 0 ´ ˆ ȷ ȷ ȷ det 1 0 0 0 + ˆ k det 1 1 0 1 = ˆ ı ı ı 1 ˆ 0 ´ 0 ˆ 1) ´ ˆ ȷ ȷ ȷ(1 ˆ 0 ´ 0 ˆ 0) + ˆ k(1 ˆ 1 ´ 1 ˆ 0) = ˆ k We should not be surprised that b ˆ c has direction ˆ k. ˝ b ˆ c has to be perpendicular to both b and c and ˝ both b and c lie in the xy-plane, ˝ so that b ˆ c has to the perpendicular to the xy-plane, ˝ so that b ˆ c has to the parallel to the z-axis. The area of the base, i.e. of the shaded parallelogram in the figure above, is \|b ˆ c\| = \| ˆ k\| = 1 and the volume of the parallelepiped is \|a ¨ (b ˆ c)\| = \| ⟨0, 1, 2⟩¨ ⟨0, 0, 1⟩\| = 2 Example 1.2.26 1.2.6 § § (Optional) Some Vector Identities Here are a few identities involving dot and cross products. (a) a ¨ (b ˆ c) = (a ˆ b) ¨ c (b) a ˆ (b ˆ c) = (c ¨ a)b ´ (b ¨ a)c (c) a ˆ (b ˆ c) + b ˆ (c ˆ a) + c ˆ (a ˆ b) = 0 Lemma 1.2.27. Proof. (a) We proved this in Theorem 1.2.23, by evaluating the left and right hand sides, and observing that they are the same. Here is a second proof, in which we again write out 34 VECTORS AND GEOMETRY 1.2 VECTORS both sides, but this time we express them in terms of determinants. a ¨ b ˆ c = (a1, a2, a3) ¨ det   ˆ ı ı ı ˆ ȷ ȷ ȷ ˆ k b1 b2 b3 c1 c2 c3   = a1 det b2 b3 c2 c3 ´ a2 det b1 b3 c1 c3 + a3 det b1 b2 c1 c2 = det   a1 a2 a3 b1 b2 b3 c1 c2 c3   a ˆ b ¨ c = det   ˆ ı ı ı ˆ ȷ ȷ ȷ ˆ k a1 a2 a3 b1 b2 b3  ¨ (c1, c2, c3) = c1 det a2 a3 b2 b3 ´ c2 det a1 a3 b1 b3 + c3 det a1 a2 b1 b2 = det   c1 c2 c3 a1 a2 a3 b1 b2 b3   Exchanging two rows in a determinant changes the sign of the determinant. Moving the top row of a 3 ˆ 3 determinant to the bottom row requires two exchanges of rows. So the two 3 ˆ 3 determinants are equal. (b) The proof is not exceptionally difficult — just write out both sides and grind. Substi-tuting in b ˆ c = (b2c3 ´ b3c2)ˆ ı ı ı ´ (b1c3 ´ b3c1)ˆ ȷ ȷ ȷ + (b1c2 ´ b2c1) ˆ k gives, for the left hand side, a ˆ (b ˆ c) = det   ˆ ı ı ı ˆ ȷ ȷ ȷ ˆ k a1 a2 a3 b2c3 ´ b3c2 ´b1c3 + b3c1 b1c2 ´ b2c1   = ˆ ı ı ı a2(b1c2 ´ b2c1) ´ a3(´b1c3 + b3c1) ´ˆ ȷ ȷ ȷ a1(b1c2 ´ b2c1) ´ a3(b2c3 ´ b3c2) + ˆ k a1(´b1c3 + b3c1) ´ a2(b2c3 ´ b3c2) On the other hand, the right hand side (a ¨ c)b ´ (a ¨ b)c = (a1c1 + a2c2 + a3c3)(b1ˆ ı ı ı + b2ˆ ȷ ȷ ȷ + b3 ˆ k) ´ (a1b1 + a2b2 + a3b3)(c1ˆ ı ı ı + c2ˆ ȷ ȷ ȷ + c3 ˆ k) = ˆ ı ı ı a1b1c1 + a2b1c2 + a3b1c3 ´ a1b1c1 ´ a2b2c1 ´ a3b3c1 + ˆ ȷ ȷ ȷ a1b2c1 + a2b2c2 + a3b2c3 ´ a1b1c2 ´ a2b2c2 ´ a3b3c2 + ˆ k a1b3c1 + a2b3c2 + a3b3c3 ´ a1b1c3 ´ a2b2c3 ´ a3b3c3 = ˆ ı ı ı [a2b1c2 + a3b1c3 ´ a2b2c1 ´ a3b3c1] + ˆ ȷ ȷ ȷ [a1b2c1 + a3b2c3 ´ a1b1c2 ´ a3b3c2] + ˆ k [a1b3c1 + a2b3c2 ´ a1b1c3 ´ a2b2c3] 35 VECTORS AND GEOMETRY 1.2 VECTORS The last formula that we had for the left hand side is the same as the last formula we had for the right hand side. Oof! This is a little tedious to do by hand. But any computer algebra system will do it for you in a flash. (c) We just apply part (b) three times a ˆ (b ˆ c) + b ˆ (c ˆ a) + c ˆ (a ˆ b) = (c ¨ a)b ´ (b ¨ a)c + (a ¨ b)c ´ (c ¨ b)a + (b ¨ c)a ´ (a ¨ c)b = 0 1.2.7 § § (Optional) Application of Cross Products to Rotational Motion In most computations involving rotational motion, the cross product shows up in one form or another. This is one of the main applications of the cross product. Consider, for example, a rigid body which is rotating at a constant rate of Ωradians per second about an axis whose direction is given by the unit vector ˆ a. Let P be any point on the body. Let’s figure out its velocity. Pick any point on the axis of rotation and designate it as the origin of our coordinate system. Denote by r the vector from the origin to the point P. Let θ denote the angle between ˆ a and r. As time progresses the point P sweeps out a circle of radius R = \|r \| sin θ. P r v 0 ˆ a θ Ω In one second P travels along an arc that subtends an angle of Ωradians, which is the fraction Ω 2π of a full circle. The length of this arc is Ω 2π ˆ 2πR = ΩR = Ω\|r \| sin θ so P travels the distance Ω\|r \| sin θ in one second and its speed, which is also the length of its velocity vector, is Ω\|r \| sin θ. Now we just need to figure out the direction of the velocity vector. That is, the direction of motion of the point P. Imagine that both ˆ a and r lie in the plane of a piece of paper, as in the figure above. Then v points either straight into or straight out of the page and consequently is perpendicular to both ˆ a and r. To distinguish between the “into the page” and “out of the page” cases, let’s impose the conventions that Ωą 0 and the axis of rotation ˆ a is chosen to obey the right hand rule, meaning that if the thumb of your right hand is pointing in the direction ˆ a, then your fingers are pointing in the direction of motion of the rigid body. Under these conventions, the velocity vector v obeys 36 VECTORS AND GEOMETRY 1.2 VECTORS • \|v\| = Ω\|r\|\|ˆ a\| sin θ • v K ˆ a, r • (ˆ a, r, v) obey the right hand rule That is, v is exactly Ωˆ a ˆ r. It is conventional to define the “angular velocity” of a rigid body to be vector Ω= Ωˆ a. That is, the vector with length given by the rate of rotation and direction given by the axis of rotation of the rigid body. In particular, the bigger the rate of rotation, the longer the angular velocity vector. In terms of this angular velocity vector, the velocity of the point P is v = Ωˆ r 1.2.8 § § (Optional) Application of Cross Products to Rotating Reference Frames Imagine a moving particle that is being tracked by two observers. (a) One observer is fixed (out in space) and measures the position of the particle to be X(t), Y(t), Z(t) . (b) The other observer is tied to a merry-go-round (the Earth) and measures the position of the particle to be x(t), y(t), z(t) . The merry-go-round is sketched in the figure on the left below. It is rotating about the Z-axis at a (constant) rate of Ωradians per second. The vector Ω= Ωˆ k, whose length is the rate of rotation and whose direction is the axis of rotation, is called the angular velocity. The x- and y-axes of the moving observer are painted in red on the merry-go-round. The Ω x y moving observer Ωt x y top view X Y figure on the right above shows a top view of the merry-go-round. The x- and y-axes of the moving observer are again red. The X- and Y-axes of the fixed observer are blue. We are assuming that at time 0, the x-axis of the moving observer and the X-axis of the fixed observer coincide. As the merry-go-round is rotating at Ωradians per second, the angle between the X-axis and x-axis after t seconds is Ωt. As an example, suppose that the moving particle is tied to the tip of the moving ob-server’s unit x vector. Then x(t) = 1 y(t) = 0 z(t) = 0 X(t) = cos(Ωt) Y(t) = sin(Ωt) Z(t) = 0 37 VECTORS AND GEOMETRY 1.2 VECTORS or, if we write r(t) = x(t), y(t), z(t) and R(t) = X(t), Y(t), Z(t) , then r(t) = (1 , 0 , 0) R(t) = cos(Ωt) , sin(Ωt) , 0 In general, denote by ˆ ı ı ı(t) the coordinates of the unit x-vector of the moving observer at time t, as measured by the fixed observer. Similarly ˆ ȷ ȷ ȷ(t) for the unit y-vector, and ˆ k(t) for the unit z-vector. As the merry-go-round is rotating about the Z-axis at a rate of Ω radians per second, the angle between the X-axis and x-axis after t seconds is Ωt, and ˆ ı ı ı(t) = cos(Ωt) , sin(Ωt) , 0 ˆ ȷ ȷ ȷ(t) = ´ sin(Ωt) , cos(Ωt) , 0 Ωt Ωt ˆ ı ı ıptq ˆ   ptq X Y ˆ k(t) = 0 , 0 , 1 The position of the moving particle, as seen by the fixed observer is R(t) = x(t) ˆ ı ı ı(t) + y(t) ˆ ȷ ȷ ȷ(t) + z(t) ˆ k(t) Differentiating, the velocity of the moving particle, as measured by the fixed observer is V(t) = dR dt = dx dt(t) ˆ ı ı ı(t) + dy dt(t) ˆ ȷ ȷ ȷ(t) + dz dt(t) ˆ k(t) + x(t)d dtˆ ı ı ı(t) + y(t)d dtˆ ȷ ȷ ȷ(t) + z(t)d dt ˆ k(t) We saw, in the last (optional) §1.2.7, that d dtˆ ı ı ı(t) = Ωˆ ˆ ı ı ı(t) d dtˆ ȷ ȷ ȷ(t) = Ωˆ ˆ ȷ ȷ ȷ(t) d dt ˆ k(t) = Ωˆ ˆ k(t) (You could also verify that these are correct by putting in Ω= (0, 0, Ω) and explicitly computing the cross products.) So V(t) = dx dt(t) ˆ ı ı ı(t) + dy dt(t) ˆ ȷ ȷ ȷ(t) + dz dt(t) ˆ k(t) + Ωˆ x(t) ˆ ı ı ı(t) + y(t) ˆ ȷ ȷ ȷ(t) + z(t) ˆ k(t) Differentiating a second time, the acceleration of the moving particle (which is also F m, where F is the net force being applied to the particle and m is the mass of the particle) as measured by the fixed observer is F m = A(t) = d2x dt2(t) ˆ ı ı ı(t) + d2y dt2(t) ˆ ȷ ȷ ȷ(t) + d2z dt2(t) ˆ k(t) + 2Ωˆ dx dt(t) ˆ ı ı ı(t) + dy dt(t) ˆ ȷ ȷ ȷ(t) + dz dt(t) ˆ k(t) + Ωˆ Ωˆ x(t) ˆ ı ı ı(t) + y(t) ˆ ȷ ȷ ȷ(t) + z(t) ˆ k(t) Recall that the angular velocity Ω= (0, 0, Ω) does not depend on time. The rotating observer sees ˆ ı ı ı(t) as ˆ ı ı ı = (1, 0, 0), sees ˆ ȷ ȷ ȷ(t) as ˆ ȷ ȷ ȷ = (0, 1, 0), and sees ˆ k(t) as ˆ k = (0, 0, 1) and so sees F m = a(t) + 2Ωˆ v(t) + Ωˆ Ωˆ r(t) 38 VECTORS AND GEOMETRY 1.2 VECTORS where, as usual, v(t) = d dtr(t) = dx dt (t) , dy dt (t) , dz dt (t) a(t) = d2 dt2 r(t) = d2x dt2 (t) , d2y dt2 (t) , d2z dt2 (t) So the acceleration of the particle seen by the moving observer is a(t) = F m ´ 2Ωˆ v(t) ´ Ωˆ Ωˆ r(t) Here • F is the sum of all external forces acting on the moving particle, • Fcor = ´2Ωˆ v(t) is called the Coriolis force and • ´Ωˆ Ωˆ r(t) is called the centrifugal force. As an example, suppose that you are the moving particle and that you are at the edge of the merry-go-round. Let’s say t = 0 and you are at ˆ ı ı ı. Then F is the friction that the surface of the merry-go-round applies to the soles of your shoes. If you are just standing there, v(t) = 0, so that Fcor = 0, and the friction F exactly cancels the centrifugal force ´Ωˆ Ωˆ r(t) so that you remain at ˆ ı ı ı(t). Assume that Ωą 0. Now suppose that you start walking around the edge of the merry-go-round. Then, at t = 0, r = ˆ ı ı ı and • if you walk in the direction of rotation (with speed one), as in the figure on the left below, v = ˆ ȷ ȷ ȷ and the Coriolis force Fcor = ´2Ωˆ k ˆ ˆ ȷ ȷ ȷ = 2Ωˆ ı ı ı tries to push you off of the merry-go-round, while • if you walk opposite to the direction of rotation (with speed one), as in the figure on the right below, v = ´ˆ ȷ ȷ ȷ so that the Coriolis force Fcor = ´2Ωˆ k ˆ (´ˆ ȷ ȷ ȷ) = ´2Ωˆ ı ı ı tries to pull you into the centre of the merry-go-round. Ω v Fcor Ω v Fcor On a rotating ball, such as the Earth, the Coriolis force deflects wind to the right (coun-terclockwise) in the northern hemisphere and to the left (clockwise) is the southern hemi-sphere. In particular, hurricanes/cyclones/typhoons rotate counterclockwise in the north-ern hemisphere and clockwise in the southern hemisphere. On the other hand, when it comes to water draining out of, for example, a toilet, Coriolis force effects are dominated by other factors like asymmetry of the toilet. 39 VECTORS AND GEOMETRY 1.3 EQUATIONS OF LINES IN 2D 1.3Ĳ Equations of Lines in 2d A line in two dimensions can be specified by giving one point (x0, y0) on the line and one vector d = dx, dy whose direction is parallel to the line. If (x, y) is any point on the line (x0, y0) (x, y) d then the vector ⟨x ´ x0, y ´ y0⟩, whose tail is at (x0, y0) and whose head is at (x, y), must be parallel to d and hence must be a scalar multiple of d. So ⟨x ´ x0, y ´ y0⟩= td or, writing out in components, x ´ x0 = tdx y ´ y0 = tdy Equation 1.3.1 (Parametric Equations). These are called the parametric equations of the line, because they contain a free parame-ter, namely t. As t varies from ´8 to 8, the point (x0 + tdx, y0 + tdy) traverses the entire line. It is easy to eliminate the parameter t from the equations. Just multiply x ´ x0 = tdx by dy, multiply y ´ y0 = tdy by dx and subtract to give (x ´ x0)dy ´ (y ´ y0)dx = 0 In the event that dx and dy are both nonzero, we can rewrite this as x ´ x0 dx = y ´ y0 dy Equation 1.3.2 (Symmetric Equation). which is called the symmetric equation for the line. A second way to specify a line in two dimensions is to give one point (x0, y0) on the line and one vector n = nx, ny whose direction is perpendicular to that of the line. If 40 VECTORS AND GEOMETRY 1.3 EQUATIONS OF LINES IN 2D (x0, y0) (x, y) n (x, y) is any point on the line then the vector ⟨x ´ x0, y ´ y0⟩, whose tail is at (x0, y0) and whose head is at (x, y), must be perpendicular to n so that n ¨ ⟨x ´ x0, y ´ y0⟩= 0 Writing out in components nx(x ´ x0) + ny(y ´ y0) = 0 or nxx + nyy = nxx0 + nyy0 Equation 1.3.3. Observe that the coefficients nx, ny of x and y in the equation of the line are the compo-nents of a vector nx, ny perpendicular to the line. This enables us to read off a vector perpendicular to any given line directly from the equation of the line. Such a vector is called a normal vector for the line. Example 1.3.4 Consider, for example, the line y = 3x + 7. To rewrite this equation in the form nxx + nyy = nxx0 + nyy0 we have to move terms around so that x and y are on one side of the equation and 7 is on the other side: 3x ´ y = ´7. Then nx is the coefficient of x, namely 3, and ny is the coefficient of y, namely ´1. One normal vector for y = 3x + 7 is ⟨3, ´1⟩. Of course, if ⟨3, ´1⟩is perpendicular to y = 3x + 7, so is ´5 ⟨3, ´1⟩= ⟨´15, 5⟩. In fact, if we first multiply the equation 3x ´ y = ´7 by ´5 to get ´15x + 5y = 35 and then set nx and ny to the coefficients of x and y respectively, we get n = ⟨´15, 5⟩. Example 1.3.4 Example 1.3.5 In this example, we find the point on the line y = 6 ´ 3x (call the line L) that is closest to the point (7, 5). We’ll start by sketching the line. To do so, we guess two points on L and then draw the line that passes through the two points. • If (x, y) is on L and x = 0, then y = 6. So (0, 6) is on L. • If (x, y) is on L and y = 0, then x = 2. So (2, 0) is on L. 41 VECTORS AND GEOMETRY 1.3 EQUATIONS OF LINES IN 2D L p0, 6q p2, 0q p7, 5q x y L p0, 6q p2, 0q p7, 5q P Q x y Denote by P the point on L that is closest to (7, 5). It is characterized by the property that the line from (7, 5) to P is perpendicular to L. This is the case just because if Q is any other point on L, then, by Pythagoras, the distance from (7, 5) to Q is larger than the distance from (7, 5) to P. See the figure on the right above. Let’s use N to denote the line which passes through (7, 5) and which is perpendicular to L. Since L has the equation 3x + y = 6, one vector perpendicular to L, and hence L (0, 6) (2, 0) (7, 5) P N (x, y) ⟨3, 1⟩ x y parallel to N, is ⟨3, 1⟩. So if (x, y) is any point on N, the vector ⟨x ´ 7, y ´ 5⟩must be of the form t ⟨3, 1⟩. So the parametric equations of N are ⟨x ´ 7, y ´ 5⟩= t ⟨3, 1⟩ or x = 7 + 3t, y = 5 + t Now let (x, y) be the coordinates of P. Since P is on N, we have x = 7 + 3t, y = 5 + t for some t. Since P is also on L, we also have 3x + y = 6. So 3(7 + 3t) + (5 + t) = 6 ð ñ 10t + 26 = 6 ð ñ t = ´2 ù ñ x = 7 + 3 ˆ (´2) = 1, y = 5 + (´2) = 3 and P is (1, 3). Example 1.3.5 42 VECTORS AND GEOMETRY 1.4 EQUATIONS OF PLANES IN 3D 1.4Ĳ Equations of Planes in 3d Specifying one point (x0, y0, z0) on a plane and a vector d parallel to the plane does not uniquely determine the plane, because it is free to rotate about d. On the other hand, giving one point on the plane and one vector n = nx, ny, nz with direction perpendicular (x0, y0, z0) d (x0, y0, z0) (x, y, z) n to that of the plane does uniquely determine the plane. If (x, y, z) is any point on the line then the vector ⟨x ´ x0, y ´ y0, z ´ z0⟩, whose tail is at (x0, y0, z0) and whose head is at (x, y, z), lies entirely inside the plane and so must be perpendicular to n. That is, n ¨ ⟨x ´ x0, y ´ y0, z ´ z0⟩= 0 Writing out in components nx(x ´ x0) + ny(y ´ y0) + nz(z ´ z0) = 0 or nxx + nyy + nzz = d where d = nxx0 + nyy0 + nzz0. Equation 1.4.1 (The Equation of a Plane). Again, the coefficients nx, ny, nz of x, y and z in the equation of the plane are the com-ponents of a vector nx, ny, nz perpendicular to the plane. The vector n is often called a normal vector for the plane. Any nonzero multiple of n will also be perpendicular to the plane and is also called a normal vector. Example 1.4.2 We have just seen that if we write the equation of a plane in the standard form ax + by + cz = d then it is easy to read off a normal vector for the plane. It is just ⟨a, b, c⟩. So for example the planes P : x + 2y + 3z = 4 P1 : 3x + 6y + 9z = 7 have normal vectors n = ⟨1, 2, 3⟩and n1 = ⟨3, 6, 9⟩, respectively. Since n1 = 3n, the two normal vectors n and n1 are parallel to each other. This tells us that the planes P and P1 are parallel to each other. 43 VECTORS AND GEOMETRY 1.4 EQUATIONS OF PLANES IN 3D When the normal vectors of two planes are perpendicular to each other, we say that the planes are perpendicular to each other. For example the planes P : x + 2y + 3z = 4 P2 : 2x ´ y = 7 have normal vectors n = ⟨1, 2, 3⟩and n2 = ⟨2, ´1, 0⟩, respectively. Since n ¨ n2 = 1 ˆ 2 + 2 ˆ (´1) + 3 ˆ 0 = 0 the normal vectors n and n2 are mutually perpendicular, so the corresponding planes P and P2 are perpendicular to each other. Example 1.4.2 Example 1.4.3 In this example, we’ll sketch the plane P : 4x + 3y + 2z = 12 A good way to prepare for sketching a plane is to find the intersection points of the plane with the x-, y- and z-axes, just as you are used to doing when sketching lines in the xy-plane. For example, any point on the x axis must be of the form (x, 0, 0). For (x, 0, 0) to also be on P we need x = 12/4 = 3. So P intersects the x-axis at (3, 0, 0). Similarly, P intersects the y-axis at (0, 4, 0) and the z-axis at (0, 0, 6). Now plot the points (3, 0, 0), (0, 4, 0) and (0, 0, 6). P is the plane containing these three points. Often a visually effective way to sketch a surface in three dimensions is to • only sketch the part of the surface in the first octant. That is, the part with x ě 0, y ě 0 and z ě 0. • To do so, sketch the curve of intersection of the surface with the part of the xy-plane in the first octant and, • similarly, sketch the curve of intersection of the surface with the part of the xz-plane in the first octant and the curve of intersection of the surface with the part of the yz-plane in the first octant. That’s what we’ll do. The intersection of the plane P with the xy-plane is the straight line through the two points (3, 0, 0) and (0, 4, 0). So the part of that intersection in the first octant is the line segment from (3, 0, 0) to (0, 4, 0). Similarly the part of the intersection of P with the xz-plane that is in the first octant is the line segment from (3, 0, 0) to (0, 0, 6) and the part of the intersection of P with the yz-plane that is in the first octant is the line segment from (0, 4, 0) to (0, 0, 6). So we just have to sketch the three line segments joining the three axis intercepts (3, 0, 0), (0, 4, 0) and (0, 0, 6). That’s it. 44 VECTORS AND GEOMETRY 1.4 EQUATIONS OF PLANES IN 3D z y x p3, 0, 0q p0, 4, 0q p0, 0, 6q Example 1.4.3 Example 1.4.4 In this example, we’ll compute the distance between the point x = (1, ´1, ´3) and the plane P : x + 2y + 3z = 18 By the “distance between x and the plane P” we mean the shortest distance between x and any point y on P. In fact, we’ll evaluate the distance in two different ways. In the next Example 1.4.5, we’ll use projection. In this example, our strategy for finding the distance will be to • first observe that the vector n = ⟨1, 2, 3⟩is normal to P and then • start walking18 away from x in the direction of the normal vector n and • keep walking until we hit P. Call the point on P where we hit, y. Then the desired distance is the distance between x and y. From the figure below it does indeed look like distance between x and y is the shortest distance between x and any point on P. This is in fact true, though we won’t prove it. P x y n n x tn 18 To see why heading in the normal direction gives the shortest walk, revisit Example 1.3.5. 45 VECTORS AND GEOMETRY 1.4 EQUATIONS OF PLANES IN 3D So imagine that we start walking, and that we start at time t = 0 at x and walk in the direction n. Then at time t we might be at x + tn = (1, ´1, ´3) + t ⟨1, 2, 3⟩= (1 + t, ´1 + 2t, ´3 + 3t) We hit the plane P at exactly the time t for which (1 + t, ´1 + 2t, ´3 + 3t) satisfies the equation for P, which is x + 2y + 3z = 18. So we are on P at the unique time t obeying (1 + t) + 2(´1 + 2t) + 3(´3 + 3t) = 18 ð ñ 14t = 28 ð ñ t = 2 So the point on P which is closest to x is y = x + tn t=2 = (1 + t, ´1 + 2t, ´3 + 3t) ˇ ˇ t=2 = (3, 3, 3) and the distance from x to P is the distance from x to y, which is \|y ´ x\| = 2\|n\| = 2 a 12 + 22 + 32 = 2 ? 14 Example 1.4.4 Example 1.4.5 (Example 1.4.4, revisited) We are again going to find the distance from the point x = (1, ´1, ´3) to the plane P : x + 2y + 3z = 18 But this time we will use the following strategy. • We’ll first find any point z on P and then • we’ll denote by y the point on P nearest x, and we’ll denote by v the vector from x to z (see the figure below) and then • we’ll realize, by looking at the figure, that the vector from x to y is exactly the pro-jection19 of the vector v on n so that • the distance from x to P, i.e. the length of the vector from x to y, is exactly ˇ ˇprojnv ˇ ˇ. x y z projn v v P n n 19 Now might be a good time to review the Definition 1.2.13 of projection. 46 VECTORS AND GEOMETRY 1.4 EQUATIONS OF PLANES IN 3D Now let’s find a point on P. The plane P is given by a single equation, namely x + 2y + 3z = 18 in the three unknowns, x, y, z. The easiest way to find one solution to this equation is to assign two of the unknowns the value zero and then solve for the third unknown. For example, if we set x = y = 0, then the equation reduces to 3z = 18. So we may take z = (0, 0, 6). Then v, the vector from x = (1, ´1, ´3) to z = (0, 0, 6) is ⟨0 ´ 1 , 0 ´ (´1) , 6 ´ (´3)⟩= ⟨´1, 1, 9⟩so that, by Equation (1.2.14), projn v = v ¨ n \|n\|2 n = ⟨´1, 1, 9⟩¨ ⟨1, 2, 3⟩ \| ⟨1, 2, 3⟩\|2 ⟨1, 2, 3⟩ = 28 14 ⟨1, 2, 3⟩ = 2 ⟨1, 2, 3⟩ and the distance from x to P is ˇ ˇprojn v ˇ ˇ = ˇ ˇ2 ⟨1, 2, 3⟩ ˇ ˇ = 2 ? 14 just as we found in Example 1.4.4. Example 1.4.5 Example 1.4.6 Now we’ll increase the degree of difficulty a tiny bit, and compute the distance between the planes P : x + 2y + 2z = 1 and P1 : 2x + 4y + 4z = 11 By the “distance between the planes P and P1” we mean the shortest distance between any pair of points x and x1 with x in P and x1 in P1. First observe that the normal vectors n = ⟨1, 2, 2⟩ and n1 = ⟨2, 4, 4⟩= 2n to P and P1 are parallel to each other. So the planes P and P1 are parallel to each other. If they had not been parallel, they would have crossed and the distance between them would have been zero. Our strategy for finding the distance will be to • first find a point x on P and then, like we did in Example 1.4.4, • start walking away from P in the direction of the normal vector n and • keep walking until we hit P1. Call the point on P1 that we hit x1. Then the desired distance is the distance between x and x1. From the figure below it does indeed look like distance between x and x1 is the shortest distance between any pair of points with one point on P and one point on P1. Again, this is in fact true, though we won’t prove it. 47 VECTORS AND GEOMETRY 1.4 EQUATIONS OF PLANES IN 3D P P ′ x x′ n x + tn We can find a point on P just as we did on Example 1.4.5. The plane P is given by the single equation x + 2y + 2z = 1 in the three unknowns, x, y, z. We can find one solution to this equation by assigning two of the unknowns the value zero and then solving for the third unknown. For example, if we set y = z = 0, then the equation reduces to x = 1. So we may take x = (1, 0, 0). Now imagine that we start walking, and that we start at time t = 0 at x and walk in the direction n. Then at time t we might be at x + tn = (1, 0, 0) + t ⟨1, 2, 2⟩= (1 + t, 2t, 2t) We hit the second plane P1 at exactly the time t for which (1+ t, 2t, 2t) satisfies the equation for P1, which is 2x + 4y + 4z = 11. So we are on P1 at the unique time t obeying 2(1 + t) + 4(2t) + 4(2t) = 11 ð ñ 18t = 9 ð ñ t = 1 2 So the point on P1 which is closest to x is x1 = x + tn t=1/2 = (1 + t, 2t, 2t) ˇ ˇ t=1/2 = (3/2, 1, 1) and the distance from P to P1 is the distance from x to x1 which is b (1 ´ 3/2)2 + (0 ´ 1)2 + (0 ´ 1)2 = a 9/4 = 3/2 Example 1.4.6 Example 1.4.7 The orientation (i.e. direction) of a plane is determined by its normal vector. So, by defini-tion, the angle between two planes is the angle between their normal vectors. For example, the normal vectors of the two planes P1 : 2x + y ´ z = 3 P2 : x + y + z = 4 48 VECTORS AND GEOMETRY 1.5 EQUATIONS OF LINES IN 3D are n1 = ⟨2, 1, ´1⟩ n2 = ⟨1, 1, 1⟩ If we use θ to denote the angle between n1 and n2, then cos θ = n1 ¨ n2 \|n1\| \|n2\| = ⟨2, 1, ´1⟩¨ ⟨1, 1, 1⟩ \| ⟨2, 1, ´1⟩\| \| ⟨1, 1, 1⟩\| = 2 ? 6 ? 3 so that θ = arccos 2 ? 18 = 1.0799 to four decimal places. That’s in radians. In degrees, it is 1.0799180 π = 61.87˝ to two decimal places. Example 1.4.7 1.5Ĳ Equations of Lines in 3d Just as in two dimensions, a line in three dimensions can be specified by giving one point (x0, y0, z0) on the line and one vector d = dx, dy, dz whose direction is parallel to that of the line. If (x, y, z) is any point on the line then the vector ⟨x ´ x0, y ´ y0, z ´ z0⟩, whose tail is at (x0, y0, z0) and whose arrow is at (x, y, z), must be parallel to d and hence a scalar multiple of d. By translating this statement into a vector equation we get ⟨x ´ x0, y ´ y0, z ´ z0⟩= td or the three corresponding scalar equations x ´ x0 = tdx y ´ y0 = tdy z ´ z0 = tdz Equation 1.5.1 (Parametric Equations of a Line). These are called the parametric equations of the line. Solving all three equations for the parameter t (assuming that dx, dy and dz are all nonzero) t = x ´ x0 dx = y ´ y0 dy = z ´ z0 dz and erasing the “t =” again gives the (so called) symmetric equations for the line. 49 VECTORS AND GEOMETRY 1.5 EQUATIONS OF LINES IN 3D Example 1.5.2 The set of points (x, y, z) that obey x + y + z = 2 form a plane. The set of points (x, y, z) that obey x ´ y = 0 form a second plane. The set of points (x, y, z) that obey both x + y + z = 2 and x ´ y = 0 lie on the intersection of these two planes and hence form a line. We shall find the parametric equations for that line. To sketch x + y + z = 2 we observe that if any two of x, y, z are zero, then the third is 2. So all of (0, 0, 2), (0, 2, 0) and (2, 0, 0) are on x + y + z = 2. The plane x ´ y = 0 contains all of the z-axis, since (0, 0, z) obeys x ´ y = 0 for all z. Here are separate sketches of (parts of) the two planes. p0, 2, 0q p0, 0, 2q p2, 0, 0q x y z “ 2 z y x x ´ y “ 0 And here is a sketch of their intersection p0, 2, 0q p0, 0, 2q p2, 0, 0q p1, 1, 0q x y z “ 2 x ´ y “ 0 d Method 1. Each point on the line has a different value of z. We’ll use z as the parameter. (We could just as well use x or y.) There is no law that requires us to use the parameter name t, but that’s what we have done so far, so set t = z. If (x, y, z) is on the line then z = t and x + y + t = 2 x ´ y = 0 The second equation forces y = x. Substituting this into the first equation gives 2x + t = 2 ù ñ x = y = 1 ´ t 2 50 VECTORS AND GEOMETRY 1.5 EQUATIONS OF LINES IN 3D So the parametric equations are x = 1 ´ t 2, y = 1 ´ t 2, z = t or ⟨x ´ 1, y ´ 1, z⟩= t ´1 2, ´1 2, 1 Method 2. We first find one point on the line. There are lots of them. We’ll find the point with z = 0. (We could just as well use z=123.4, but arguably z = 0 is a little easier.) If (x, y, z) is on the line and z = 0, then x + y = 2 x ´ y = 0 The second equation again forces y = x. Substituting this into the first equation gives 2x = 2 ù ñ x = y = 1 So (1, 1, 0) is on the line. Now we’ll find a direction vector, d, for the line. • Since the line is contained in the plane x + y + z = 2, any vector lying on the line, like d, is also completely contained in that plane. So d must be perpendicular to the normal vector of x + y + z = 2, which is ⟨1, 1, 1⟩. • Similarly, since the line is contained in the plane x ´ y = 0, any vector lying on the line, like d, is also completely contained in that plane. So d must be perpendicular to the normal vector of x ´ y = 0, which is ⟨1, ´1, 0⟩. So we may choose for d any vector which is perpendicular to both ⟨1, 1, 1⟩and ⟨1, ´1, 0⟩, like, for example, d = ⟨1, ´1, 0⟩ˆ ⟨1, 1, 1⟩ = det   ˆ ı ı ı ˆ ȷ ȷ ȷ ˆ k 1 ´1 0 1 1 1  = ˆ ı ı ı det ´1 0 1 1 ´ ˆ ȷ ȷ ȷ det 1 0 1 1 + ˆ k det 1 ´1 1 1 = ´ˆ ı ı ı ´ ˆ ȷ ȷ ȷ + 2 ˆ k We now have both a point on the line (namely (1, 1, 0)) and a direction vector for the line (namely ⟨´1, ´1, 2⟩), so, as usual, the parametric equations for the line are ⟨x ´ 1, y ´ 1, z⟩= t ⟨´1, ´1, 2⟩ or x = 1 ´ t, y = 1 ´ t, z = 2t This looks a little different than the solution from method 1, but we’ll see in a moment that they are really the same. Before that, let’s do one more method. Method 3. We’ll find two points on the line. We have already found that (1, 1, 0) is on the line. From the picture above, it looks like (0, 0, 2) is also on the line. This is indeed the case since (0, 0, 2) obeys both x + y + z = 2 and x ´ y = 0. Notice that we could also have guessed (0, 0, 2) by setting x = 0 and then solving y + z = x + y + z = 2, ´y = x ´ y = 0 for x and y. As both (1, 1, 0) and (0, 0, 2) are on the line, the vector with head at (1, 1, 0) and tail at (0, 0, 2), which is ⟨1 ´ 0, 1 ´ 0, 0 ´ 2⟩= ⟨1, 1, ´2⟩, is a direction vector for the 51 VECTORS AND GEOMETRY 1.5 EQUATIONS OF LINES IN 3D line. As (0, 0, 2) is a point on the line and ⟨1, 1, ´2⟩is a direction vector for the line, the parametric equations for the line are ⟨x ´ 0, y ´ 0, z ´ 2⟩= t ⟨1, 1, ´2⟩ or x = t, y = t, z = 2 ´ 2t This also looks similar, but not quite identical, to our previous answers. Time for a com-parison. Comparing the answers. The parametric equations given by the three methods are different. That’s just because we have really used different parameters in the three methods, even though we have called the parameter t in each case. To clarify the relation between the three answers, rename the parameter of method 1 to t1, the parameter of method 2 to t2 and the parameter of method 3 to t3. The parametric equations then become Method 1: x = 1 ´ t1 2 y = 1 ´ t1 2 z = t1 Method 2: x = 1 ´ t2 y = 1 ´ t2 z = 2t2 Method 3: x = t3 y = t3 z = 2 ´ 2t3 Substituting t1 = 2t2 into the Method 1 equations gives the Method 2 equations, and substituting t3 = 1 ´ t2 into the Method 3 equations gives the Method 2 equations. So all three really give the same line, just parametrized a little differently. Example 1.5.2 A line in three dimensions has infinitely many normal vectors. For example, the line ⟨x ´ 1, y ´ 1, z⟩= t ⟨1, 2, ´2⟩ has direction vector ⟨1, 2, ´2⟩. Any vector perpendicular to ⟨1, 2, ´2⟩is perpen-dicular to the line. The vector ⟨n1, n2, n3⟩is perpendicular to ⟨1, 2, ´2⟩if and only if 0 = ⟨1, 2, ´2⟩¨ ⟨n1, n2, n3⟩= n1 + 2n2 ´ 2n3 There is whole plane of ⟨n1, n2, n3⟩’s obeying this condition, of which ⟨2, ´1, 0⟩, ⟨0, 1, 1⟩and ⟨2, 0, 1⟩are only three examples. Warning 1.5.3. The next two examples illustrate two different methods for finding the distance be-tween a point and a line. Example 1.5.4 In this example, we find the distance between the point (2, 3, ´1) and the line L : ⟨x ´ 1, y ´ 2, z ´ 3⟩= t ⟨1, 1, 2⟩ or, equivalently, x = 1 + t, y = 2 + t, z = 3 + 2t 52 VECTORS AND GEOMETRY 1.5 EQUATIONS OF LINES IN 3D The vector from (2, 3, ´1) to the point (1 + t , 2 + t , 3 + 2t) on L is ⟨t ´ 1 , t ´ 1 , 2t + 4⟩. The square of the distance between (2, 3, ´1) and the point (1 + t , 2 + t , 3 + 2t) on L is the square of the length of that vector, namely d(t)2 = (t ´ 1)2 + (t ´ 1)2 + (2t + 4)2 The point on L that is closest to (2, 3, ´1) is that whose value of t obeys 0 = d dtd(t)2 = 2(t ´ 1) + 2(t ´ 1) + 2(2)(2t + 4) (˚) Before we solve this equation for t and finish of our computation, observe that this equation (divided by 2) says that ⟨1 , 1 , 2⟩¨ ⟨t ´ 1 , t ´ 1 , 2t + 4⟩= 0 That is, the vector from (2, 3, ´1) to the point on L nearest (2, 3, ´1) is perpendicular to L’s direction vector. Now back to our computation. The equation (˚) simplifies to 12t + 12 = 0. So the optimal t = ´1 and the distance is d(´1) = b (´1 ´ 1)2 + (´1 ´ 1)2 + (´2 + 4)2 = ? 12 Example 1.5.4 Example 1.5.5 (Example 1.5.4 revisited) In this example, we again find the distance between the point (2, 3, ´1) and the line L : ⟨x ´ 1, y ´ 2, z ´ 3⟩= t ⟨1, 1, 2⟩ but we use a different method. In the figure below, Q is the point (2, 3, ´1). Q N P projd v v w d L If we drop a perpendicular from Q to the line L, it hits the line L at the point N, which is the point on L that is nearest Q. So the distance from Q to L is exactly the distance from Q to N, which is exactly the length of the vector from Q to N. In the figure above, w is the vector from Q to N. Now the vector w has to be perpendicular to the direction vector for L. That is, w has to be perpendicular to d = ⟨1, 1, 2⟩. However, as we saw in Warning 1.5.3, there are a huge number of vectors in different directions that are perpendicular to d. So you might think that it is very hard to even determine the direction of w. Fortunately, it isn’t. Here is the strategy. 53 VECTORS AND GEOMETRY 1.5 EQUATIONS OF LINES IN 3D • Pick any point on L and call it P. • It is very easy to find the vector from P to N — it is just the projection of the vector from P to Q (called v in the figure above) on d. • Once we know projd v, we will be able to compute w = projd v ´ v • and then the distance from Q to the line L is just \|w\|. Here is the computation. We’ll choose P to be the point on L that has t = 0, which is (1, 2, 3). So the vector from P = (1, 2, 3) to Q = (2, 3, ´1) is v = ⟨2 ´ 1, 3 ´ 2, ´1 ´ 3⟩= ⟨1, 1, ´4⟩ The projection of v = ⟨1, 1, ´4⟩on d = ⟨1, 1, 2⟩is projd v = ⟨1, 1, ´4⟩¨ ⟨1, 1, 2⟩ \| ⟨1, 1, 2⟩\|2 ⟨1, 1, 2⟩= ´6 6 ⟨1, 1, 2⟩= ⟨´1, ´1, ´2⟩ and then w = projd v ´ v = ⟨´1, ´1, ´2⟩´ ⟨1, 1, ´4⟩= ⟨´2, ´2, 2⟩ and finally the distance from Q to the line L is \|w\| = \| ⟨´2, ´2, 2⟩\| = \|2 ⟨´1, ´1, 1⟩\| = 2 ? 3 Example 1.5.5 The next two (optional) examples illustrate two different methods for finding the dis-tance between two lines. Example 1.5.6 (Optional) In this example, we find the distance between the lines L : ⟨x ´ 1, y ´ 2, z ´ 3⟩= t ⟨1, 0, ´1⟩ L1 : ⟨x ´ 1, y ´ 2, z ´ 1⟩= t ⟨1, ´2, 1⟩ We can rewrite the equations of the lines as L : x = 1 + t, y = 2, z = 3 ´ t L1 : x = 1 + t, y = 2 ´ 2t, z = 1 + t Of course the value of t in the parametric equation for L need not be the same as the value of t in the parametric equation for L1. So let us denote by x(s) = (1 + s , 2 , 3 ´ s) and y(t) = (1 + t , 2 ´ 2t , 1 + t) the points on L and L1, respectively, that are closest together. Note that the vector from x(s) to y(t) is ⟨t ´ s , ´2t , ´2 + s + t⟩. Then, in particular, • x(s) is the point on L that is closest to the point y(t), and • y(t) is the point on L1 that is closest to the point x(s). 54 VECTORS AND GEOMETRY 1.5 EQUATIONS OF LINES IN 3D So, as we saw in Example 1.5.4, the vector, ⟨t ´ s , ´2t , ´2 + s + t⟩, that joins x(s) and y(t), must be perpendicular to both the direction vector of L and the direction vector of L1. Consequently 0 = ⟨1, 0, ´1⟩¨ ⟨t ´ s , ´2t , ´2 + s + t⟩= 2 ´ 2s 0 = ⟨1, ´2, 1⟩¨ ⟨t ´ s , ´2t , ´2 + s + t⟩= ´2 + 6t So s = 1 and t = 1/3 and the distance between L and L1 is ˇ ˇ ⟨t ´ s , ´2t , ´2 + s + t⟩ ˇ ˇ s=1, t=1/3 = ˇ ˇ ⟨´2/3 , ´2/3 , ´2/3⟩ ˇ ˇ = 2 ? 3 Example 1.5.6 Example 1.5.7 (Example 1.5.6 revisited, again optional) In this example, we again find the distance between the lines L : ⟨x ´ 1, y ´ 2, z ´ 3⟩= t ⟨1, 0, ´1⟩ L1 : ⟨x ´ 1, y ´ 2, z ´ 1⟩= t ⟨1, ´2, 1⟩ this time using a projection, much as in Example 1.4.5. The procedure, which will be justified below, is • first form a vector n that is perpendicular to the direction vectors of both lines by taking the cross product of the two direction vectors. In this example, ⟨1, 0, ´1⟩ˆ ⟨1, ´2, 1⟩= det   ˆ ı ı ı ˆ ȷ ȷ ȷ ˆ k 1 0 ´1 1 ´2 1  = ´2ˆ ı ı ı ´ 2ˆ ȷ ȷ ȷ ´ 2 ˆ k Since we just want ˆ n to be perpendicular to both direction vectors, we may simplify our computations by dividing this vector by ´2, and take n = ⟨1, 1, 1⟩. • Next find one point on L and one point on L1 and subtract to form a vector v whose tail is at one point and whose head is at the other point. This vector goes from one line to the other line. In this example, the point (1, 2, 3) is on L (just set t = 0 in the equation for L) and the point (1, 2, 1) is on L1 (just set t = 0 in the equation for L1), so that we may take v = ⟨1 ´ 1 , 2 ´ 2 , 3 ´ 1⟩= ⟨0, 0, 2⟩ • The distance between the two lines is the length of the projection of v on n. In this example, by (1.2.14), the distance is ˇ ˇprojn v ˇ ˇ = ˇ ˇ ˇ ˇ v ¨ n \|n\|2 n ˇ ˇ ˇ ˇ = \|v ¨ n\| \|n\| = \| ⟨0, 0, 2⟩¨ ⟨1, 1, 1⟩\| \| ⟨1, 1, 1⟩\| = 2 ? 3 just as we found in Example 1.5.6 55 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS Now, here is the justification for the procedure. • As we did in Example 1.5.6, denote by x(s) and y(t) the points on L and L1, re-spectively, that are closest together. Note that, as we observed in Example 1.5.6, the vector from x(s) to y(t) is perpendicular to the direction vectors of both lines, and so is parallel to n. • Denote by P the plane through x(s) that is perpendicular to n. As x(s) is on L and the direction vector of L is perpendicular to n, the line L is contained in P. • Denote by P1 the plane through y(t) that is perpendicular to n. As y(t) is on L1 and the direction vector of L1 is perpendicular to n, the line L1 is contained in P1. • The planes P and P1 are parallel to each other. As x(s) is on P and y(t) is on P1, and the vector from x(s) to y(t) is perpendicular to both P and P1, the distance from P to P1 is exactly the length of the vector from x(s) to y(t). That is also the distance from L to L1. • The vector v constructed in the procedure above is a vector between L and L1 and so is also a vector between P and P1. Looking at the figure below20, we see that the vector from x(s) to y(t) is (up to a sign) the projection of v on n. projnv P P 1 xpsq yptq n v • So the distance from P to P1, and hence the distance from L to L1, is exactly the length of projnv. Example 1.5.7 1.6Ĳ Curves and their Tangent Vectors The right hand side of the parametric equation (x, y, z) = (1, 1, 0) + t ⟨1, 2, ´2⟩that we just saw in Warning 1.5.3 is a vector-valued function of the one real variable t. We are now going to study more general vector-valued functions of one real variable. That is, we are going to study functions that assign to each real number t (typically in some interval) a vector r(t). For example r(t) = x(t), y(t), z(t) 20 and possibly reviewing the Definition 1.2.13 of projection 56 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS might be the position21 of a particle at time t. As t varies r(t) sweeps out a curve. r(0) r(1) r(2) While in some applications t will indeed be “time”, it does not have to be. It can be simply a parameter that is used to label the different points on the curve that r(t) sweeps out. We then say that r(t) provides a parametrization of the curve. Example 1.6.1 Parametrization of x2 + y2 = a2 While we will often use t as the parameter in a parametrized curve r(t), there is no need to call it t. Sometimes it is natural to use a different name for the parameter. For example, consider the circle22 x2 + y2 = a2. It is natural to use the angle θ in the sketch below to label the point a cos θ , a sin θ on the circle. x y x2 y2 “ a2 a cos θ , a sin θ ˘ θ That is, r(θ) = a cos θ , a sin θ 0 ď θ ă 2π is a parametrization of the circle x2 + y2 = a2. Just looking at the figure above, it is clear that, as θ runs from 0 to 2π, r(θ) traces out the full circle. However beware that just knowing that r(t) lies on a specified curve does not guaran-tee that, as t varies, r(t) covers the entire curve. For example, as t runs over the whole real line, 2 π arctan(t) runs over the interval (´1, 1). For all t, r(t) = x(t), y(t) = a 2 π arctan(t) , c 1 ´ 4 π2 arctan2(t) ! is well-defined and obeys x(t)2 + y(t)2 = a2. But this r(t) does not cover the entire circle because y(t) is always positive. 21 When we say r(t) = x(t), y(t), z(t) , we mean that x(t), y(t), z(t) is the point at the head of the vector r(t) when its tail is at the origin. 22 We of course assume that the constant a ą 0. 57 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS Example 1.6.1 Example 1.6.2 Parametrization of (x ´ h)2 + (y ´ k)2 = a2 We can tweak the parametrization of Example 1.6.1 to get a parametrization of the circle of radius a that is centred on (h, k). One way to do so is to redraw the sketch of Example 1.6.1 with the circle translated so that its centre is at (h, k). x y px ´ hq2 py ´ kq2 “ a2 a sin θ a ph, kq h a cos θ , k a sin θ ˘ θ We see from the sketch that r(θ) = h + a cos θ , k + a sin θ 0 ď θ ă 2π is a parametrization of the circle (x ´ h)2 + (y ´ k)2 = a2. A second way to come up with this parametrization is to observe that we can turn the trig identity cos2 t + sin2 t = 1 into the equation (x ´ h)2 + (y ´ k)2 = a2 of the circle by • multiplying the trig identity by a2 to get (a cos t)2 + (a sin t)2 = a2 and then • setting a cos t = x ´ h and a sin t = y ´ k , which turns (a cos t)2 + (a sin t)2 = a2 into (x ´ h)2 + (y ´ k)2 = a2. Example 1.6.2 Example 1.6.3 Parametrization of x2 a2 + y2 b2 = 1 and of x2/3 + y2/3 = a2/3 We can build parametrizations of the curves x2 a2 + y2 b2 = 1 and x2/3 + y2/3 = a2/3 from the trig identity cos2 t + sin2 t = 1, like we did in the second part of the last example. • Setting cos t = x a and sin t = y b turns cos2 t + sin2 t = 1 into x2 a2 + y2 b2 = 1. • Setting cos t = x a 1 3 and sin t = y a 1 3 turns cos2 t + sin2 t = 1 into x2/3 a2/3 + y2/3 a2/3 = 1. 58 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS So r(t) = a cos t , b sin t 0 ď t ă 2π r(t) = a cos3 t , a sin3 t 0 ď t ă 2π give parametrizations of x2 a2 + y2 b2 = 1 and x2/3 + y2/3 = a2/3, respectively. To see that running t from 0 to 2π runs r(t) once around the curve, look at the figures below. x y x2 a2 y2 b2 “ 1 t “ 0 t “ π t “ π{2 t “ 3π{2 x y x2/3 + y2/3 = a2/3 t = 0 t = π t = π/2 t = 3π/2 The curve x2/3 + y2/3 = a2/3 is called an astroid. From its equation, we would expect its sketch to look like a deformed circle. But it is probably not so obvious that it would have the pointy bits of the right hand figure. We will not explain here why they arise. The astroid is studied in some detail in Example 1.1.9 of the CLP-4 text. In particular, the above sketch is carefully developed there. Example 1.6.3 Example 1.6.4 Parametrization of ey = 1 + x2 A very easy method that can often create parametrizations for a curve is to use x or y as a parameter. Because we can solve ey = 1 + x2 for y as a function of x, namely y = ln 1 + x2 , we can use x as the parameter simply by setting t = x. This gives the parametrization r(t) = t , ln(1 + t2) ´ 8 ă t ă 8 Example 1.6.4 Example 1.6.5 Parametrization of x2 + y2 = a2, again It is also quite common that one can use either x or y to parametrize part of, but not all of, a curve. A simple example is the circle x2 + y2 = a2. For each ´a ă x ă a, there are two points on the circle with that value of x. So one cannot use x to parametrize the whole 59 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS circle. Similarly, for each ´a ă y ă a, there are two points on the circle with that value of y. So one cannot use y to parametrize the whole circle. On the other hand r(t) = t , a a2 ´ t2 ´a ă t ă a r(t) = t , ´ a a2 ´ t2 ´a ă t ă a provide parametrizations of the top half and bottom half, respectively, of the circle using x as the parameter, and r(t) = a a2 ´ t2 , t ´a ă t ă a r(t) = ´ a a2 ´ t2 , t ´a ă t ă a provide parametrizations of the right half and left half, respectively, of the circle using y as the parameter. Example 1.6.5 Example 1.6.6 (Unparametrization of r(t) = (cos t, 7 ´ t)) In this example, we will undo the parametrization r(t) = (cos t, 7 ´ t) and find the Carte-sian equation of the curve in question. We may rewrite the parametrization as x = cos t y = 7 ´ t Note that we can eliminate the parameter t simply by using the second equation to solve for t as a function of y. Namely t = 7 ´ y. Substituting this into the first equation gives us the Cartesian equation x = cos(7 ´ y) Example 1.6.6 Curves often arise as the intersection of two surfaces. For example, the intersection of the sphere x2 + y2 + z2 = 1 with the plane y = x is a circle. The part of that circle that is in the first octant is the red curve in the figure below. One way to parametrize such curves z y x y “ x x2 y2 z2 “ 1 60 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS is to choose one of the three coordinates x, y, z as the parameter, and solve the two given equations for the remaining two coordinates, as functions of the parameter. Here are two examples. Example 1.6.7 The set of all (x, y, z) obeying x ´ y = 0 x2 + y2 + z2 = 1 is the circle sketched above. We can choose to use y as the parameter and think of x = y x2 + z2 = 1 ´ y2 as a system of two equations for the two unknowns x and z, with y being treated as a given constant, rather than as an unknown. We can now (trivially) solve the first equation for x, substitute the result into the second equation, and finally solve for z. x = y, x2 + z2= 1 ´ y2 ù ñ z2 = 1 ´ 2y2 If, for example, we are interested in points (x, y, z) on the curve with z ě 0, we have z = a 1 ´ 2y2 and r(y) = y , y , b 1 ´ 2y2 , ´ 1 ? 2 ď y ď 1 ? 2 is a parametrization for the part of the circle above the xy-plane. If, on the other hand, we are interested in points (x, y, z) on the curve with z ď 0, we have z = ´ a 1 ´ 2y2 and r(y) = y , y , ´ b 1 ´ 2y2 , ´ 1 ? 2 ď y ď 1 ? 2 is a parametrization for the part of the circle below the xy-plane. Example 1.6.7 Example 1.6.8 The previous example was rigged so that it was easy to solve for x and z as functions of y. In practice it is not always easy, or even possible, to do so. A more realistic example is the set of all (x, y, z) obeying x2 + y2 2 + z2 3 = 1 x2 + 2y2 = z which is the blue curve in the figure 61 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS x2 y2 2 z2 3 “ 1 z “ x2 2y2 (Don’t worry about how we make sketches like this. We’ll develop some surface sketching technique in §1.7 below.) Substituting x2 = z ´ 2y2 (from the second equation) into the first equation gives ´3 2y2 + z + z2 3 = 1 or, completing the square, ´3 2y2 + 1 3 z + 3 2 2 = 7 4 If, for example, we are interested in points (x, y, z) on the curve with y ě 0, this can be solved to give y as a function of z. y = c 2 9 z + 3 2 2 ´ 14 12 Then x2 = z ´ 2y2 also gives x as a function of z. If x ě 0, x = c z ´ 4 9 z + 3 2 2 + 14 6 = c 4 3 ´ 4 9z2 ´ 1 3z The other signs of x and y can be gotten by using the appropriate square roots. In this example, (x, y, z) is on the curve, i.e. satisfies the two original equations, if and only if all of (˘x, ˘y, z) are also on the curve. Example 1.6.8 1.6.1 § § Derivatives and Tangent Vectors This being a Calculus text, one of our main operations is differentiation. We are now interested in parametrizations r(t). It is very easy and natural to extend our definition of derivative to r(t) as follows. 62 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS The derivative of the vector-valued function r(t) is defined to be r1(t) = dr dt(t) = lim hÑ0 r(t + h) ´ r(t) h rptq rpt hq rpt hq ´ rptq when the limit exists. In particular, if r(t) = x(t)ˆ ı ı ı + y(t)ˆ ȷ ȷ ȷ + z(t) ˆ k, then r1(t) = x1(t)ˆ ı ı ı + y1(t)ˆ ȷ ȷ ȷ + z1(t) ˆ k That is, to differentiate a vector-valued function of t, just differentiate each of its components. Definition 1.6.9. And of course differentiation interacts with arithmetic operations, like addition, in the obvious way. Only a little more thought is required to see that differentiation interacts quite nicely with dot and cross products too. Here are some examples. Example 1.6.10 Let a(t) = t2 ˆ ı ı ı + t4 ˆ ȷ ȷ ȷ + t6 ˆ k b(t) = e´t ˆ ı ı ı + e´3t ˆ ȷ ȷ ȷ + e´5t ˆ k γ(t) = t2 s(t) = sin t We are about to compute some derivatives. To make it easier to follow what is going on, we’ll use some colour. When we apply the product rule d dt f (t) g(t) = f 1(t) g(t) + f (t) g1(t) we’ll use blue to highlight the factors f 1(t) and g1(t). Here we go. γ(t) b(t) = t2e´t ˆ ı ı ı + t2e´3t ˆ ȷ ȷ ȷ + t2e´5t ˆ k ù ñ d dt γ(t)b(t) = 2te´t´t2e´tˆ ı ı ı + 2te´3t´3t2e´3tˆ ȷ ȷ ȷ + 2te´5t´5t2e´5t ˆ k = 2t ␣ e´t ˆ ı ı ı + e´3t ˆ ȷ ȷ ȷ + e´5t ˆ k ( + t2␣ ´ e´t ˆ ı ı ı ´ 3e´3t ˆ ȷ ȷ ȷ ´ 5e´5t ˆ k ( = γ1(t)b(t) + γ(t)b1(t) 63 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS and a(t) ¨ b(t) = t2e´t + t4e´3t + t6e´5t ù ñ d dt a(t) ¨ b(t) = 2te´t´t2e´t + 4t3e´3t´3t4e´3t + 6t5e´5t´5t6e´5t = 2te´t + 4t3e´3t + 6t5e´5t + ´t2e´t´3t4e´3t´5t6e´5t = ␣ 2t ˆ ı ı ı + 4t3 ˆ ȷ ȷ ȷ + 6t5 ˆ k ( ¨ ␣ e´t ˆ ı ı ı + e´3t ˆ ȷ ȷ ȷ + e´5t ˆ k ( + ␣ t2 ˆ ı ı ı + t4 ˆ ȷ ȷ ȷ + t6 ˆ k ( ¨ ␣ ´ e´t ˆ ı ı ı ´ 3e´3t ˆ ȷ ȷ ȷ ´ 5e´5t ˆ k ( = a1(t) ¨ b(t) + a(t) ¨ b1(t) and a(t) ˆ b(t) = det   ˆ ı ı ı ˆ ȷ ȷ ȷ ˆ k t2 t4 t6 e´t e´3t e´5t   = ˆ ı ı ı t4e´5t ´ t6e´3t) ´ ˆ ȷ ȷ ȷ(t2e´5t ´ t6e´t) + ˆ k(t2e´3t ´ t4e´t) ù ñ d dt a(t) ˆ b(t) = ˆ ı ı ı 4t3e´5t ´ 6t5e´3t) ´ ˆ ȷ ȷ ȷ( 2te´5t ´ 6t5e´t) + ˆ k( 2te´3t ´ 4t3e´t) + ˆ ı ı ı ´5t4e´5t+3t6e´3t) ´ ˆ ȷ ȷ ȷ(´5t2e´5t+t6e´t) + ˆ k(´3t2e´3t+t4e´t) = ␣ 2t ˆ ı ı ı + 4t3 ˆ ȷ ȷ ȷ + 6t5 ˆ k ( ˆ ␣ e´t ˆ ı ı ı + e´3t ˆ ȷ ȷ ȷ + e´5t ˆ k ( + ␣ t2 ˆ ı ı ı + t4 ˆ ȷ ȷ ȷ + t6 ˆ k ( ˆ ␣ ´ e´t ˆ ı ı ı ´ 3e´3t ˆ ȷ ȷ ȷ ´ 5e´5t ˆ k ( = a1(t) ˆ b(t) + a(t) ˆ b1(t) and a s(t) = (sin t)2 ˆ ı ı ı + (sin t)4 ˆ ȷ ȷ ȷ + (sin t)6 ˆ k ù ñ d dt a s(t) = 2(sin t) cos t ˆ ı ı ı + 4(sin t)3 cos t ˆ ȷ ȷ ȷ + 6(sin t)5 cos t ˆ k = ␣ 2(sin t) ˆ ı ı ı + 4(sin t)3ˆ ȷ ȷ ȷ + 6(sin t)5 ˆ k ( cos t = a1s(t) s1(t) Example 1.6.10 Of course these examples extend to general (differentiable) a(t), b(t), γ(t) and s(t) and give us (most of) the following theorem. 64 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS Let ˝ a(t), b(t) be vector-valued differentiable functions of t P R that take values in Rn and ˝ α, β P R be constants and ˝ γ(t) and s(t) be real-valued differentiable functions of t P R Then (a) d dt α a(t) + β b(t) = α a1(t) + β b1(t) (linear combination) (b) d dt γ(t)b(t) = γ1(t)b(t) + γ(t)b1(t) (multiplication by scalar function) (c) d dt a(t) ¨ b(t) = a1(t) ¨ b(t) + a(t) ¨ b1(t) (dot product) (d) d dt a(t) ˆ b(t) = a1(t) ˆ b(t) + a(t) ˆ b1(t) (cross product) (e) d dt a s(t) = a1s(t) s1(t) (composition) Theorem 1.6.11 (Arithmetic of differentiation). Let’s think about the geometric significance of r1(t). In particular, let’s think about the relationship between r1(t) and distances along the curve. The derivative r1(t) is the limit of r(t+h)´r(t) h as h Ñ 0. The numerator, r(t + h) ´ r(t), is the vector with head at r(t + h) and tail at r(t). r(t) r(t + h) r(t + h) −r(t) ≈r′(t) h When h is very small this vector ˝ has the essentially the same direction as the tangent vector to the curve at r(t) and ˝ has length being essentially the length of the part of the curve between r(t) and r(t + h). Taking the limit as h Ñ 0 yields that ˝ r1(t) is a tangent vector to the curve at r(t) that points in the direction of increasing t and ˝ if s(t) is the length of the part of the curve between r(0) and r(t), then ds dt(t) = ˇ ˇdr dt(t) ˇ ˇ. This is worth stating formally. 65 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS Let r(t) be a parametrized curve. (a) Denote by ˆ T(t) the unit tangent vector to the curve at r(t) pointing in the direction of increasing t. If r1(t) ‰ 0 then ˆ T(t) = r1(t) \|r1(t)\| (b) Denote by s(t) the length of the part of the curve between r(0) and r(t). Then ds dt(t) = ˇ ˇ ˇ ˇ dr dt(t) ˇ ˇ ˇ ˇ s(T) ´ s(T0) = ż T T0 ˇ ˇ ˇ ˇ dr dt(t) ˇ ˇ ˇ ˇ dt rp0q rptq ˆ Tptq sptq (c) In particular, if the parameter happens to be arc length, i.e. if t = s, so that ds ds = 1, then ˇ ˇ ˇ ˇ dr ds(s) ˇ ˇ ˇ ˇ = 1 ˆ T(s) = r1(s) Lemma 1.6.12. As an application, we have the If r(t) = x(t) , y(t) , z(t) is the position of a particle at time t, then velocity at time t = v(t) = r1(t) = x1(t)ˆ ı ı ı + y1(t)ˆ ȷ ȷ ȷ + z1(t) ˆ k = ds dt(t) ˆ T(t) speed at time t = ds dt(t) = \|v(t)\| = \|r1(t)\| = b (x1(t)2 + y1(t)2 + z1(t)2 acceleration at time t = a(t) = r2(t) = v1(t) = x2(t)ˆ ı ı ı + y2(t)ˆ ȷ ȷ ȷ + z2(t) ˆ k and the distance travelled between times T0 and T is s(T) ´ s(T0) = ż T T0 ˇ ˇ ˇdr dt(t) ˇ ˇ ˇ dt = ż T T0 b (x1(t)2 + y1(t)2 + z1(t)2 dt Lemma 1.6.13. Note that the velocity v(t) = r1(t) is a vector quantity while the speed ds dt(t) = \|r1(t)\| is a scalar quantity. 66 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS Example 1.6.14 (Circumference of a circle) In general it can be quite difficult to compute arc lengths. So, as an easy warmup exam-ple, we will compute the circumference of the circle23 x2 + y2 = a2. We’ll also find a unit tangent to the circle at any point on the circle. We’ll use the parametrization r(θ) = a cos θ , a sin θ 0 ď θ ď 2π of Example 1.6.1. Using Lemma 1.6.12, but with the parameter t renamed to θ r1(θ) = ´a sin θˆ ı ı ı + a cos θˆ ȷ ȷ ȷ ˆ T(θ) = r1(θ) \|r1(θ)\| = ´ sin θˆ ı ı ı + cos θˆ ȷ ȷ ȷ ds dθ(θ) = ˇ ˇr1(θ) ˇ ˇ = a s(Θ) ´ s(0) = ż Θ 0 ˇ ˇr1(θ) ˇ ˇ dθ = aΘ As24 s(Θ) is the arc length of the part of the circle with 0 ď θ ď Θ, the circumference of the whole circle is s(2π) = 2πa which is reassuring, since this formula has been known25 for thousands of years. The x y x2 y2 “ a2 a cos θ , a sin θ ˘ θ ˆ Tpθq formula s(Θ) ´ s(0) = aΘ also makes sense — the part of the circle with 0 ď θ ď Θ is the fraction Θ 2π of the whole circle, and so should have length Θ 2π ˆ 2πa. Also note that r(θ) ¨ ˆ T(θ) = a cos θ , a sin θ ¨ ´ sin θ , cos θ = 0 23 We of course assume that the constant a ą 0. 24 You might guess that Θ is a capital Greek theta. You’d be right. 25 The earliest known written approximations of π, in Egypt and Babylon, date from 1900–1600BC. The first recorded algorithm for rigorously evaluating π was developed by Archimedes around 250 BC. The first use of the symbol π, for the ratio between the circumference of a circle and its diameter, in print was in 1706 by William Jones. 67 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS so that the tangent to the circle at any point is perpendicular to the radius vector of the circle at that point. This is another geometric fact that has been known26 for thousands of years. Example 1.6.14 Example 1.6.15 (Arc length of a helix) Consider the curve r(t) = 6 sin(2t)ˆ ı ı ı + 6 cos(2t)ˆ ȷ ȷ ȷ + 5t ˆ k where the standard basis vectors ˆ ı ı ı = (1, 0, 0), ˆ ȷ ȷ ȷ = (0, 1, 0) and ˆ k = (0, 0, 1). We’ll first sketch it, by observing that ˝ x(t) = 6 sin(2t) and y(t) = 6 cos(2t) obey x(t)2 + y(t)2 = 36 sin2(2t) + 36 cos2(2t) = 36 So all points of the curve lie on the cylinder x2 + y2 = 36 and ˝ as t increases, x(t), y(t) runs clockwise around the circle x2 + y2 = 36 and at the same time z(t) = 5t just increases linearly. Our curve is the helix y z x t “ 0 t “ π 2 t “ π We have marked three points of the curve on the above sketch. The first has t = 0 and is 0ˆ ı ı ı + 6ˆ ȷ ȷ ȷ + 0 ˆ k. The second has t = π 2 and is 0ˆ ı ı ı ´ 6ˆ ȷ ȷ ȷ + 5π 2 ˆ k, and the third has t = π and is 0ˆ ı ı ı + 6ˆ ȷ ȷ ȷ + 5π ˆ k. We’ll now use Lemma 1.6.12 to find a unit tangent ˆ T(t) to the curve at r(t) 26 It is Proposition 18 in Book 3 of Euclid’s Elements. It was published around 300BC. 68 VECTORS AND GEOMETRY 1.6 CURVES AND THEIR TANGENT VECTORS and also the arclength of the part of curve between t = 0 and t = π. r(t) = 6 sin(2t)ˆ ı ı ı + 6 cos(2t)ˆ ȷ ȷ ȷ + 5t ˆ k r1(t) = 12 cos(2t)ˆ ı ı ı ´ 12 sin(2t)ˆ ȷ ȷ ȷ + 5 ˆ k ds dt(t) = ˇ ˇr1(t) ˇ ˇ = b 122 cos2(2t) + 122 sin2(2t) + 52 = a 122 + 52 = 13 ˆ T(t) = r1(t) \|r1(t))\| = 12 13 cos(2t)ˆ ı ı ı ´ 12 13 sin(2t)ˆ ȷ ȷ ȷ + 5 13 ˆ k s(π) ´ s(0) = ż π 0 ˇ ˇr1(t) ˇ ˇ dt = 13π Example 1.6.15 Example 1.6.16 (Velocity and acceleration) Imagine that, at time t, a particle is at r(t) = h + a cos 2π t T ˆ ı ı ı + k + a sin 2π t T ˆ ȷ ȷ ȷ As \|r(t) ´ h ˆ ı ı ı ´ k ˆ ȷ ȷ ȷ\| = a, the particle is running around the circle of radius a centred on (h, k). When t increases by T, the argument, 2π t T, of cos 2π t T and sin 2π t T increases by exactly 2π and the particle runs exactly once around the circle. In particular, it travels a distance 2πa. So it is moving at speed 2πa T . According to Lemma 1.6.13, it has velocity = r1(t) = ´2πa T sin 2π t T ˆ ı ı ı + 2πa T cos 2π t T ˆ ȷ ȷ ȷ speed = ds dt(t) = \|r1(t)\| = 2πa T acceleration = r2(t) = ´4π2a T2 cos 2π t T ˆ ı ı ı ´ 4π2a T2 sin 2π t T ˆ ȷ ȷ ȷ = ´4π2 T2 r(t) ´ h ˆ ı ı ı ´ k ˆ ȷ ȷ ȷ Here are some observations. • The velocity r1(t) has dot product zero with r(t) ´ h ˆ ı ı ı ´ k ˆ ȷ ȷ ȷ, which is the radius vector from the centre of the circle to the particle. So the velocity is perpendicular to the radius vector, and hence parallel to the tangent vector of the circle at r(t). • The speed given by Lemma 1.6.13 is exactly the speed we found above, just before we started applying Lemma 1.6.13. • The acceleration r2(t) points in the direction opposite to the radius vector. 69 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D x y rptq r1ptq r2ptq ph, kq Example 1.6.16 1.7Ĳ Sketching Surfaces in 3d In practice students taking multivariable calculus regularly have great difficulty visual-ising surfaces in three dimensions, despite the fact that we all live in three dimensions. We’ll now develop some technique to help us sketch surfaces in three dimensions27. We all have a fair bit of experience drawing curves in two dimensions. Typically the intersection of a surface (in three dimensions) with a plane is a curve lying in the (two dimensional) plane. Such an intersection is usually called a cross-section. In the special case that the plane is one of the coordinate planes, the intersection is sometimes called a trace. One can often get a pretty good idea of what a surface looks like by sketching a bunch of cross-sections. Here are some examples. Example 1.7.1 4x2 + y2 ´ z2 = 1 Sketch the surface that satisfies 4x2 + y2 ´ z2 = 1. Solution. We’ll start by fixing any number z0 and sketching the part of the surface that lies in the horizontal plane z = z0. z y x z “ z0 27 Of course you could instead use some fancy graphing software, but part of the point is to build intuition. Not to mention that you can’t use fancy graphing software on your exam. 70 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D The intersection of our surface with that horizontal plane is a horizontal cross-section. Any point (x, y, z) lying on that horizontal cross-section satisfies both z = z0 and 4x2 + y2 ´ z2 = 1 ð ñ z = z0 and 4x2 + y2 = 1 + z2 0 Think of z0 as a constant. Then 4x2 + y2 = 1 + z2 0 is a curve in the xy-plane. As 1 + z2 0 is a constant, the curve is an ellipse. To determine its semi-axes28, we observe that when y = 0, we have x = ˘1 2 b 1 + z2 0 and when x = 0, we have y = ˘ b 1 + z2 0. So the curve is just an ellipse with x semi-axis 1 2 b 1 + z2 0 and y semi-axis b 1 + z2 0. It’s easy to sketch. x y ( 1 2 p 1 + z2 0 , 0) (0 , p 1 + z2 0 ) Remember that this ellipse is the part of our surface that lies in the plane z = z0. Imagine that the sketch of the ellipse is on a single sheet of paper. Lift the sheet of paper up, move it around so that the x- and y-axes point in the directions of the three dimensional x- and y-axes and place the sheet of paper into the three dimensional sketch at height z0. This gives a single horizontal ellipse in 3d, as in the figure below. z y x z “ z0 We can build up the full surface by stacking many of these horizontal ellipses — one for each possible height z0. So we now draw a few of them as in the figure below. To reduce the amount of clutter in the sketch, we have only drawn the first octant (i.e. the part of three dimensions that has x ě 0, y ě 0 and z ě 0). 28 The semi-axes of an ellipse are the line segments from the centre of the ellipse to the farthest points on the ellipse and to the nearest points on the ellipse. For a circle the lengths of all of these line segments are just the radius. 71 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D z y x z = 1 z = 2 z = 3 Here is why it is OK, in this case, to just sketch the first octant. Replacing x by ´x in the equation 4x2 + y2 ´ z2 = 1 does not change the equation. That means that a point (x, y, z) is on the surface if and only if the point (´x, y, z) is on the surface. So the surface is invariant under reflection in the yz-plane. Similarly, the equation 4x2 + y2 ´ z2 = 1 does not change when y is replaced by ´y or z is replaced by ´z. Our surface is also invariant under reflection in the xz- and xy-planes. Once we have the part in the first octant, the remaining octants can be gotten simply by reflecting about the coordinate planes. We can get a more visually meaningful sketch by adding in some vertical cross-sections. The x = 0 and y = 0 cross-sections (also called traces — they are the parts of our surface that are in the yz- and xz-planes, respectively) are x = 0, y2 ´ z2 = 1 and y = 0, 4x2 ´ z2 = 1 These equations describe hyperbolae29. If you don’t remember how to sketch them, don’t worry. We’ll do it now. We’ll first sketch them in 2d. Since y2 = 1 + z2 ù ñ \|y\| ě 1 and y = ˘1 when z = 0 and for large z, y « ˘z 4x2 = 1 + z2 ù ñ \|x\| ě 1 2 and x = ˘1 2 when z = 0 and for large z, x « ˘1 2z the sketches are y z z = y y2 −z2 = 1 x z 4x2 −z2 = 1 Now we’ll incorporate them into the 3d sketch. Once again imagine that each is a single sheet of paper. Pick each up and move it into the 3d sketch, carefully matching up the axes. The red (blue) parts of the hyperbolas above become the red (blue) parts of the 3d sketch below (assuming of course that you are looking at this on a colour screen). 29 It’s not just a figure of speech! 72 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D z y x z = 1 z = 2 z = 3 Now that we have a pretty good idea of what the surface looks like we can clean up and simplify the sketch. Here are a couple of possibilities. z y x Here are two figures created by graphing software. -2 -1 0 1 2 2 1 -1 0 -1 0 1 -2 -2 2 This type of surface is called a hyperboloid of one sheet. 73 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D There are also hyperboloids of two sheets. For example, replacing the +1 on the right hand side of x2 + y2 ´ z2 = 1 gives x2 + y2 ´ z2 = ´1, which is a hyperboloid of two sheets. We’ll sketch it quickly in the next example. Example 1.7.1 Example 1.7.2 4x2 + y2 ´ z2 = ´1 Sketch the surface that satisfies 4x2 + y2 ´ z2 = ´1. Solution. As in the last example, we’ll start by fixing any number z0 and sketching the part of the surface that lies in the horizontal plane z = z0. The intersection of our surface with that horizontal plane is z = z0 and 4x2 + y2 = z2 0 ´ 1 Think of z0 as a constant. • If \|z0\| ă 1, then z2 0 ´ 1 ă 0 and there are no solutions to x2 + y2 = z2 0 ´ 1. • If \|z0\| = 1 there is exactly one solution, namely x = y = 0. • If \|z0\| ą 1 then 4x2 + y2 = z2 0 ´ 1 is an ellipse with x semi-axis 1 2 b z2 0 ´ 1 and y semi-axis b z2 0 ´ 1. These semi-axes are small when \|z0\| is close to 1 and grow as \|z0\| increases. The first octant parts of a few of these horizontal cross-sections are drawn in the figure below. z y x z “ 1.02 z “ 2 z “ 3 Next we add in the x = 0 and y = 0 cross-sections (i.e. the parts of our surface that are in the yz- and xz-planes, respectively) x = 0, z2 = 1 + y2 and y = 0, z2 = 1 + 4x2 74 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D z y x z “ 1.05 z “ 2 z “ 3 Now that we have a pretty good idea of what the surface looks like we clean up and simplify the sketch. z y x Here are two figures created by graphing software. -2 -1 0 1 2 2 1 0 -1 -2 -2 -1 0 1 2 This type of surface is called a hyperboloid of two sheets. Example 1.7.2 75 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D Example 1.7.3 (yz = 1) Sketch the surface yz = 1. Solution. This surface has a special property that makes it relatively easy to sketch. There are no x’s in the equation yz = 1. That means that if some y0 and z0 obey y0z0 = 1, then the point (x, y0, z0) lies on the surface yz = 1 for all values of x. As x runs from ´8 to 8, the point (x, y0, z0) sweeps out a straight line parallel to the x-axis. So the surface yz = 1 is a union of lines parallel to the x-axis. It is invariant under translations parallel to the x-axis. To sketch yz = 1, we just need to sketch its intersection with the yz-plane and then translate the resulting curve parallel to the x-axis to sweep out the surface. We’ll start with a sketch of the hyperbola yz = 1 in two dimensions. y z yz = 1 Next we’ll move this 2d sketch into the yz-plane, i.e. the plane x = 0, in 3d, except that we’ll only draw in the part in the first octant. z y x The we’ll draw in x = x0 cross-sections for a couple of more values of x0 76 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D z y x and clean up the sketch a bit z y x Here are two figures created by graphing software. 0 1 2 3 4 5 0 1 5 4 3 2 0 1 2 3 4 5 Example 1.7.3 Example 1.7.4 (xyz = 4) Sketch the surface xyz = 4. 77 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D Solution. We’ll sketch this surface using much the same procedure as we used in Examples 1.7.1 and 1.7.2. We’ll only sketch the part of the surface in the first octant. The remaining parts (in the octants with x, y ă 0, z ě 0, with x, z ă 0, y ě 0 and with y, z ă 0, x ě 0) are just reflections of the first octant part. As usual, we start by fixing any number z0 and sketching the part of the surface that lies in the horizontal plane z = z0. The intersection of our surface with that horizontal plane is the hyperbola z = z0 and xy = 4 z0 Note that x Ñ 8 as y Ñ 0 and that y Ñ 8 as x Ñ 0. So the hyperbola has both the x-axis and the y-axis as asymptotes, when drawn in the xy-plane. The first octant parts of a few of these horizontal cross-sections (namely, z0 = 4, z0 = 2 and z0 = 1 2) are drawn in the figure below. z y x z “ 1{2 z “ 2 z “ 4 Next we add some vertical cross-sections. We can’t use x = 0 or y = 0 because any point on xyz = 4 must have all of x, y, z nonzero. So we use x = 4, yz = 1 and y = 4, xz = 1 instead. They are again hyperbolae. z y x x “ 4 y “ 4 78 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D Finally, we clean up and simplify the sketch. z y x Here are two figures created by graphing software. 0 1 2 3 4 0 1 2 0 1 2 4 3 3 4 5 5 Example 1.7.4 1.7.1 § § Level Curves and Surfaces Often the reason you are interested in a surface in 3d is that it is the graph z = f (x, y) of a function of two variables f (x, y). Another good way to visualize the behaviour of a function f (x, y) is to sketch what are called its level curves. By definition, a level curve of f (x, y) is a curve whose equation is f (x, y) = C, for some constant C. It is the set of points in the xy-plane where f takes the value C. Because it is a curve in 2d, it is usually easier to sketch than the graph of f. Here are a couple of examples. Example 1.7.5 f (x, y) = x2 + 4y2 ´ 2x + 2 Sketch the level curves of f (x, y) = x2 + 4y2 ´ 2x + 2. 79 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D Solution. Fix any real number C. Then, for the specified function f, the level curve f (x, y) = C is the set of points (x, y) that obey x2 + 4y2 ´ 2x + 2 = C ð ñ x2 ´ 2x + 1 + 4y2 + 1 = C ð ñ (x ´ 1)2 + 4y2 = C ´ 1 Now (x ´1)2 + 4y2 is the sum of two squares, and so is always at least zero. So if C ´1 ă 0, i.e. if C ă 1, there is no curve f (x, y) = C. If C ´ 1 = 0, i.e. if C = 1, then f (x, y) = C ´ 1 = 0 if and only if both (x ´ 1)2 = 0 and 4y2 = 0 and so the level curve consists of the single point (1, 0). If C ą 1, then f (x, y) = C become (x ´ 1)2 + 4y2 = C ´ 1 ą 0 which describes an ellipse centred on (1, 0). It intersects the x-axis when y = 0 and (x ´ 1)2 = C ´ 1 ð ñ x ´ 1 = ˘ ? C ´ 1 ð ñ x = 1 ˘ ? C ´ 1 and it intersects the line x = 1 (i.e. the vertical line through the centre) when 4y2 = C ´ 1 ð ñ 2y = ˘ ? C ´ 1 ð ñ y = ˘1 2 ? C ´ 1 So, when C ą 1, f (x, y) = C is the ellipse centred on (1, 0) with x semi-axis ? C ´ 1 and y semi-axis 1 2 ? C ´ 1. Here is a sketch of some representative level curves of f (x, y) = x2 + 4y2 ´ 2x + 2. x y x “ 1 1 1 f“1 f“2 f“5 f“10 f“17 It is often easier to develop an understanding of the behaviour of a function f (x, y) by looking at a sketch of its level curves, than it is by looking at a sketch of its graph. On the other hand, you can also use a sketch of the level curves of f (x, y) as the first step in building a sketch of the graph z = f (x, y). The next step would be to redraw, for each C, the level curve f (x, y) = C, in the plane z = C, as we did in Example 1.7.1. Example 1.7.5 80 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D Example 1.7.6 (ex+y+z = 1) The function f (x, y) is given implicitly by the equation ex+y+z = 1. Sketch the level curves of f. Solution. This one is not as nasty as it appears. That “f (x, y) is given implicitly by the equation ex+y+z = 1” means that, for each x, y, the solution z of ex+y+z = 1 is f (x, y). So, for the specified function f and any fixed real number C, the level curve f (x, y) = C is the set of points (x, y) that obey ex+y+C = 1 ð ñ x + y + C = 0 (by taking the logarithm of both sides) ð ñ x + y = ´C This is of course a straight line. It intersects the x-axis when y = 0 and x = ´C and it intersects the y-axis when x = 0 and y = ´C. Here is a sketch of some level curves. x y 1 1 f=0 f=1 f=−1 f=2 f=−2 f=3 f=−3 Example 1.7.6 We have just seen that sketching the level curves of a function f (x, y) can help us understand the behaviour of f. We can generalise this to functions F(x, y, z) of three vari-ables. A level surface of F(x, y, z) is a surface whose equation is of the form F(x, y, z) = C for some constant C. It is the set of points (x, y, z) at which F takes the value C. Example 1.7.7 F(x, y, z) = x2 + y2 + z2 Let F(x, y, z) = x2 + y2 + z2. If C ą 0, then the level surface F(x, y, z) = C is the sphere of radius ? C centred on the origin. Here is a sketch of the parts of the level surfaces F = 1 (radius 1), F = 4 (radius 2) and F = 9 (radius 3) that are in the first octant. 81 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D z y x F “ 9 F “ 1 F “ 4 Example 1.7.7 Example 1.7.8 F(x, y, z) = x2 + z2 Let F(x, y, z) = x2 + z2 and C ą 0. Consider the level surface x2 + z2 = C. The variable y does not appear in this equation. So for any fixed y0, the intersection of our surface x2 + z2 = C with the plane y = y0 is the circle of radius ? C centred on x = z = 0. Here is a sketch of the first quadrant part of one such circle. z y x F “ C y “ y0 The full surface is the horizontal stack of all of those circles with y0 running over R. It is the cylinder of radius ? C centred on the y-axis. Here is a sketch of the parts of the level surfaces F = 1 (radius 1), F = 4 (radius 2) and F = 9 (radius 3) that are in the first octant. 82 VECTORS AND GEOMETRY 1.7 SKETCHING SURFACES IN 3D z y x F “ 9 F “ 4 F “ 1 Example 1.7.8 Example 1.7.9 (F(x, y, z) = ex+y+z) Let F(x, y, z) = ex+y+z and C ą 0. Consider the level surface ex+y+z = C, or equivalently, x + y + z = ln C. It is the plane that contains the intercepts (ln C, 0, 0), (0, ln C, 0) and (0, 0, ln C). Here is a sketch of the parts of the level surfaces • F = e (intercepts (1, 0, 0), (0, 1, 0), (0, 0, 1)), • F = e2 (intercepts (2, 0, 0), (0, 2, 0), (0, 0, 2)) and • F = e3 (intercepts (3, 0, 0), (0, 3, 0), (0, 0, 3)) that are in the first octant. z y x F “e3 F “e F “e2 Example 1.7.9 83 VECTORS AND GEOMETRY 1.8 CYLINDERS 1.8Ĳ Cylinders There are some classes of relatively simple, but commonly occurring, surfaces that are given their own names. One such class is cylindrical surfaces. You are probably used to thinking of a cylinder as being something that looks like x2 + y2 = 1. x2 y2 “ 1 In Mathematics, the word “cylinder” is given a more general meaning. A cylinder is a surface that consists of all points that are on all lines that are • parallel to a given line and • pass through a given fixed curve, that lies in a fixed plane that is not parallel to the given line. Definition 1.8.1 (Cylinder). Example 1.8.2 Here are sketches of three cylinders. The familiar cylinder on the left below x2 y2 “ 1 x2 py ´ zq2 “ 1 is called a right circular cylinder, because the given fixed curve (x2 + y2 = 1, z = 0) is a circle and the given line (the z-axis) is perpendicular (i.e. at right angles) to the fixed curve. The cylinder on the left above can be thought of as a vertical stack of circles. The cylinder on the right above can also be thought of as a stack of circles, but the centre of the circle at height z has been shifted rightward to (0, z, z). For that cylinder, the given fixed curve is once again the circle x2 + y2 = 1, z = 0, but the given line is y = z, x = 0. We have already seen the third cylinder 84 VECTORS AND GEOMETRY 1.9 QUADRIC SURFACES z y x yz “ 1 x, y, z ą 0 in Example 1.7.3. It is called a hyperbolic cylinder. In this example, the given fixed curve is the hyperbola yz = 1, x = 0 and the given line is the x-axis. Example 1.8.2 1.9Ĳ Quadric Surfaces Another named class of relatively simple, but commonly occurring, surfaces is the quadric surfaces. A quadric surface is surface that consists of all points that obey Q(x, y, z) = 0, with Q being a polynomial of degree two30. Definition 1.9.1 (Quadrics). For Q(x, y, z) to be a polynomial of degree two, it must be of the form Q(x, y, z) = Ax2 + By2 + Cz2 + Dxy + Eyz + Fxz + Gx + Hy + Iz + J for some constants A, B, ¨ ¨ ¨ , J. Each constant z cross section of a quadric surface has an equation of the form Ax2 + Dxy + By2 + gx + hy + j = 0, z = z0 If A = B = D = 0 but g and h are not both zero, this is a straight line. If A, B, and D are not all zero, then by rotating and translating our coordinate system the equation of the cross section can be brought into one of the forms31 • αx2 + βy2 = γ with α, β ą 0, which, if γ ą 0, is an ellipse (or a circle), 30 Technically, we should also require that the polynomial can’t be factored into the product of two poly-nomials of degree one. 31 This statement can be justified using a linear algebra eigenvalue/eigenvector analysis. It is beyond what we can cover here, but is not too difficult for a standard linear algebra course. 85 VECTORS AND GEOMETRY 1.9 QUADRIC SURFACES • αx2 ´ βy2 = γ with α, β ą 0, which, if γ ‰ 0, is a hyperbola, and if γ = 0 is two lines, • x2 = δy, which, if δ ‰ 0 is a parabola, and if δ = 0 is a straight line. There are similar statements for the constant x cross sections and the constant y cross sections. Hence quadratic surfaces are built by stacking these three types of curves. We have already seen a number of quadric surfaces in the last couple of sections. • We saw the quadric surface 4x2 + y2 ´ z2 = 1 in Example 1.7.1. Its constant z cross sections are ellipses and its x = 0 and y = 0 cross sections are hyperbolae. It is called a hyperboloid of one sheet. • We saw the quadric surface x2 + y2 = 1 in Example 1.8.2. Its constant z cross sections are circles and its x = 0 and y = 0 cross sections are straight lines. It is called a right circular cylinder. Appendix H contains other quadric surfaces. 86 PARTIAL DERIVATIVES Chapter 2 In this chapter we are going to generalize the definition of “derivative” to functions of more than one variable and then we are going to use those derivatives. We will parallel the development in Chapters 1 and 2 of the CLP-1 text. We shall • define limits and continuity of functions of more than one variable (Definitions 2.1.2 and 2.1.3) and then • study the properties of limits in more than one dimension (Theorem 2.1.5) and then • define derivatives of functions of more than one variable (Definition 2.2.1). We are going to be able to speed things up considerably by recycling what we have already learned in the CLP-1 text. We start by generalizing the definition of “limit” to functions of more than one vari-able. 2.1Ĳ Limits Before we really start, let’s recall some useful notation. • N is the set t1, 2, 3, ¨ ¨ ¨ u of all natural numbers. • R is the set of all real numbers. • P is read “is an element of”. • R is read “is not an element of”. • ␣ A ˇ ˇ B ( is read “the set of all A such that B” Notation 2.1.1. 87 PARTIAL DERIVATIVES 2.1 LIMITS • If S is a set and T is a subset of S, then SzT is ␣ x P S ˇ ˇ x R T ( , the set S with the elements of T removed. In particular, if S is a set and a is an element of S, then Sztau = ␣ x P S ˇ ˇ x ‰ a ( is the set S with the element a removed. • If n is a natural number, Rn is used for both the set of n-component vectors ⟨x1, x2, ¨ ¨ ¨ , xn⟩and the set of points (x1, x2, ¨ ¨ ¨ , xn) with n coordinates. • If S and T are sets, then f : S Ñ T means that f is a function which assigns to each element of S an element of T. The set S is called the domain of f. • [a, b] = ␣ x P R ˇ ˇ a ď x ď b ( (a, b] = ␣ x P R ˇ ˇ a ă x ď b ( [a, b) = ␣ x P R ˇ ˇ a ď x ă b ( (a, b) = ␣ x P R ˇ ˇ a ă x ă b ( Notation 2.1.1 (continued). The definition of the limit of a function of more than one variable looks just like the definition1 of the limit of a function of one variable. Very roughly speaking lim xÑa f (x) = L if f (x) approaches L whenever x approaches a. Here is a more careful definition of limit. Let ˝ m and n be natural numbers2 ˝ a P Rm ˝ the function f (x) be defined for all x near3 a and take values in Rn ˝ L P Rn We write lim xÑa f (x) = L if4 the value of the function f (x) is sure to be arbitrarily close to L whenever the value of x is close enough to a, without5 being exactly a. Definition 2.1.2 (Limit). 1 Definition 1.3.3 in the CLP-1 text. 2 In this text, we will interested in m, n P ␣ 1, 2, 3 ( , but the definition works for all natural numbers m, n. 3 To be precise, there is a number r ą 0 such that f (x) is defined for all x obeying \|x ´ a\| ă r. 4 There is a precise, formal version of this definition that looks just like Definition 1.7.1 of the CLP-1 text. 5 You may find the condition “without being exactly a” a little strange, but there is a good reason for it, which we have already seen in Calculus I. In the definition f 1(x) = lim xÑa f (x)´f (a) x´a , the function whose limit is being taken, namely f (x)´f (a) x´a , is not defined at all at x = a. This will again happen when we define derivatives of functions of more than one variable. 88 PARTIAL DERIVATIVES 2.1 LIMITS Now that we have extended the definition of limit, we can extend the definition of conti-nuity. Let ˝ m and n be natural numbers ˝ a P Rm ˝ the function f (x) be defined for all x near a and take values in Rn (a) The function f is continuous at a point a if lim xÑa f (x) = f (a) (b) The function f is continuous on a set D if it is continuous at every point of D. Definition 2.1.3 (Continuity). Here are a few very simple examples. There will be some more substantial examples later — after, as we did in the CLP-1 text, we build some tools that can be used to build complicated limits from simpler ones. Example 2.1.4 (a) If f (x, y) is the constant function which always takes the value L, then lim (x,y)Ñ(a,b) f (x, y) = L (b) If f : R2 Ñ R2 is defined by f (x, y) = (x, y), then lim (x,y)Ñ(a,b) f (x, y) = (a, b) (c) By definition, as (x, y) approaches (a, b), x approaches a and y approaches b, so that if f : R2 Ñ R is defined by f (x, y) = x, then lim (x,y)Ñ(a,b) f (x, y) = a Similarly, if g : R2 Ñ R is defined by g(x, y) = y, then lim (x,y)Ñ(a,b) g(x, y) = b Example 2.1.4 89 PARTIAL DERIVATIVES 2.1 LIMITS Limits of multivariable functions have much the same computational properties as limits of functions of one variable. The following theorem summarizes a bunch of them. For simplicity, it concerns primarily real-valued functions. That is, functions that output real numbers as opposed to vectors. However it does contain one vector-valued function. The function X in the theorem takes as input an n-component vector and returns an m-component vector. We will not deal with many vector-valued functions here in CLP-3, but we will see a lot in CLP-4. Let ˝ m and n be natural numbers ˝ a P Rm and b P Rn ˝ D be a subset of Rm that contains all x P Rm that are near a ˝ c, F, G P R and f, g : Dztau Ñ R X : Rnztbu Ñ Dztau γ : R Ñ R Assume that lim xÑa f (x) = F lim xÑa g(x) = G lim yÑb X(y) = a lim tÑF γ(t) = γ(F) Then (a) lim xÑa f (x) + g(x) = F + G lim xÑa f (x) ´ g(x) = F ´ G (b) lim xÑa f (x) g(x) = FG lim xÑa c f (x) = cF (c) lim xÑa f (x) g(x) = F G if G ‰ 0 (d) lim yÑb f X(y) = F (e) lim xÑa γ f (x) = γ(F) Theorem 2.1.5 (Arithmetic, and Other, Properties of Limits). This shows that multivariable limits interact very nicely with arithmetic, just as single variable limits did. Also recall, from Theorem 1.6.8 in the CLP-1 text, 90 PARTIAL DERIVATIVES 2.1 LIMITS The following functions are continuous everywhere in their domains • polynomials, rational functions • roots and powers • trig functions and their inverses • exponential and the logarithm Theorem 2.1.6. Example 2.1.7 In this example we evaluate lim (x,y)Ñ(2,3) x + sin y x2y2 + 1 as a typical application of Theorem 2.1.5. Here “ a = ” means that part (a) of Theorem 2.1.5 justifies that equality. Start by computing separately the limits of the numerator and denominator. lim (x,y)Ñ(2,3) x + sin y a = lim (x,y)Ñ(2,3) x + lim (x,y)Ñ(2,3) sin y e = lim (x,y)Ñ(2,3) x + sin lim (x,y)Ñ(2,3) y = 2 + sin 3 lim (x,y)Ñ(2,3) x2y2 + 1 a = lim (x,y)Ñ(2,3) x2y2 + lim (x,y)Ñ(2,3) 1 b = lim (x,y)Ñ(2,3) x lim (x,y)Ñ(2,3) x lim (x,y)Ñ(2,3) y lim (x,y)Ñ(2,3) y + 1 = 2232 + 1 Since the limit of the denominator is nonzero, we can simply divide. lim (x,y)Ñ(2,3) x + sin y x2y2 + 1 c = lim (x,y)Ñ(2,3)(x + sin y) lim (x,y)Ñ(2,3)(x2y2 + 1) = 2 + sin 3 37 Here we have used that sin x is a continuous function. Example 2.1.7 While the CLP-1 text’s Definition 1.3.3 of the limit of a function of one variable, and our Definition 2.1.2 of the limit of a multivariable function look virtually identical, there is 91 PARTIAL DERIVATIVES 2.1 LIMITS a substantial practical difference between the two. In dimension one, you can approach a point from the left or from the right and that’s it. There are only two possible directions of approach. In two or more dimensions there is “much more room” and there are infinitely many possible types of approach. One can even spiral in to a point. See the middle and right hand figures below. The next few examples illustrate the impact that the“extra room” in dimensions greater than one has on limits. Example 2.1.8 As a second example, we consider lim (x,y)Ñ(0,0) x2y x2+y2. In this example, both the numerator, x2y, and the denominator, x2 + y2, tend to zero as (x, y) approaches (0, 0), so we have to be more careful. A good way to see the behaviour of a function f (x, y) when (x, y) is close to (0, 0) is to switch to the polar coordinates, r, θ, that are defined by x = r cos θ x y px, yq r θ y = r sin θ The points (x, y) that are close to (0, 0) are those with small r, regardless of what θ is. Recall that lim (x,y)Ñ(0,0) f (x, y) = L when f (x, y) approaches L as (x, y) approaches (0, 0). Substituting x = r cos θ, y = r sin θ into that statement turns it into the statement that lim (x,y)Ñ(0,0) f (x, y) = L when f (r cos θ, r sin θ) approaches L as r approaches 0. For our current example x2y x2 + y2 = (r cos θ)2(r sin θ) r2 = r cos2 θ sin θ As ˇ ˇr cos2 θ sin θ ˇ ˇ ď r tends to 0 as r tends to 0 (regardless of what θ does as r tends to 0) we have lim (x,y)Ñ(0,0) x2y x2 + y2 = 0 92 PARTIAL DERIVATIVES 2.1 LIMITS Example 2.1.8 Example 2.1.9 As a third example, we consider lim (x,y)Ñ(0,0) x2´y2 x2+y2. Once again, the best way to see the be-haviour of f (x, y) = x2´y2 x2+y2 for (x, y) close to (0, 0) is to switch to polar coordinates. f (x, y) = x2 ´ y2 x2 + y2 = (r cos θ)2 ´ (r sin θ)2 r2 = cos2 θ ´ sin2 θ = cos(2θ) Note that, this time, f is independent of r but does depend on θ. Here is a greatly magni-fied sketch of a number of level curves for f (x, y). f“cosp90˝q“0 f“cosp0q“1 f“cosp30˝q“ ? 3{2 f“cosp60˝q“1{2 f“cosp135˝q“´1{ ? 2 f“cosp180˝q“´1 r “ 10´137 Observe that ˝ as (x, y) approaches (0, 0) along the ray with 2θ = 30˝, f (x, y) approaches the value ? 3 2 (and in fact f (x, y) takes the value cos(30˝) = ? 3 2 at every point of that ray) ˝ as (x, y) approaches (0, 0) along the ray with 2θ = 60˝, f (x, y) approaches the value 1 2 (and in fact f (x, y) takes the value cos(60˝) = 1 2 at every point of that ray) ˝ as (x, y) approaches (0, 0) along the ray with 2θ = 90˝, f (x, y) approaches the value 0 (and in fact f (x, y) takes the value cos(90˝) = 0 at every point of that ray) ˝ and so on So there is not a single number L such that f (x, y) approaches L as r = \|(x, y)\| Ñ 0, no matter what the direction of approach is. The limit lim (x,y)Ñ(0,0) x2´y2 x2+y2 does not exist. Here is another way to come to the same conclusion. ˝ Pick any really small positive number. We’ll use 10´137 as an example. ˝ Pick any real number F between ´1 and 1. We’ll use F = ? 3 2 as an example. 93 PARTIAL DERIVATIVES 2.1 LIMITS ˝ Looking at the sketch above, we see that f (x, y) takes the value F along an entire ray θ = const, r ą 0. In the case F = ? 3 2 , the ray is 2θ = 30˝, r ą 0. In particular, because the ray extends all the way to (0, 0), f takes the value F for some (x, y) obeying \|(x, y)\| ă 10´137. ˝ That is true regardless of which really small number you picked. So f (x, y) = x2´y2 x2+y2 does not approach any single value as r = \|(x, y)\| approaches 0 and we conclude that lim (x,y)Ñ(0,0) x2´y2 x2+y2 does not exist. Example 2.1.9 2.1.1 § § Optional — A Nasty Limit That Doesn’t Exist Example 2.1.10 In this example we study the behaviour of the function f (x, y) = $ & % (2x´y)2 x´y if x ‰ y 0 if x = y as (x, y) Ñ (0, 0). Here is a graph of the level curve, f (x, y) = ´3, for this function. y x ´3 f “ ´3 Here is a larger graph of level curves, f (x, y) = c, for various values of the constant c. 94 PARTIAL DERIVATIVES 2.1 LIMITS y x 1 ´1 2 f “ 2 ´2 3 f “ ´3 1 2 1 2 ´ 1 2 ´ 1 2 As before, it helps to convert to polar coordinates — it is a good approach6. In polar coordinates f (r cos θ, r sin θ) = $ & % r (2 cos θ´sin θ)2 cos θ´sin θ if cos θ ‰ sin θ 0 if cos θ = sin θ If we approach the origin along any fixed ray θ = const, then f (r cos θ, r sin θ) is the constant (2 cos θ´sin θ)2 cos θ´sin θ (or 0 if cos θ = sin θ) times r and so approaches zero as r approaches zero. You can see this in the figure below, which shows the level curves again, with the rays θ = 1 8π and θ = 3 16π superimposed. If you move towards the origin on either of y x 1 2 f “ 3 1 2 θ “ 3 16π θ “ 1 8π those rays, you first cross the f = 3 level curve, then the f = 2 level curve, then the f = 1 level curve, then the f = 1 2 level curve, and so on. 6 Not just a pun. 95 PARTIAL DERIVATIVES 2.1 LIMITS That f (x, y) Ñ 0 as (x, y) Ñ (0, 0) along any fixed ray is suggestive, but does not imply that the limit exists and is zero. Recall that to have lim(x,y)Ñ(0,0) f (x, y) = 0, we need f (x, y) Ñ 0 no matter how (x, y) Ñ (0, 0). It is not sufficient to check only straight line approaches. In fact, the limit of f (x, y) as (x, y) Ñ (0, 0) does not exist. A good way to see this is to observe that if you fix any r ą 0, no matter how small, f (x, y) takes all values from ´8 to +8 on the circle x2 + y2 = r2. You can see this in the figure below, which shows the level curves yet again, with a circle x2 + y2 = r2 superimposed. For every single ´8 ă c ă 8, the level curve f (x, y) = c crosses the circle. Consequently there is no one y x 1 ´1 2 ´2 f “ 3 ´3 1 2 ´ 1 2 number L such that f (x, y) is close to L whenever (x, y) is sufficiently close to (0, 0). The limit lim(x,y)Ñ(0,0) f (x, y) does not exist. Another way to see that f (x, y) does not have any limit as (x, y) Ñ (0, 0) is to show that f (x, y) does not have a limit as (x, y) approaches (0, 0) along some specific curve. This can be done by picking a curve that makes the denominator, x ´ y, tend to zero very quickly. One such curve is x ´ y = x3 or, equivalently, y = x ´ x3. Along this curve, for x ‰ 0, f (x, x ´ x3) = (2x ´ x + x3)2 x ´ x + x3 = (x + x3)2 x3 = (1 + x2)2 x Ý Ñ $ & % +8 as x Ñ 0 with x ą 0 ´8 as x Ñ 0 with x ă 0 The choice of the specific power x3 is not important. Any power xp with p ą 2 will have the same effect. If we send (x, y) to (0, 0) along the curve x ´ y = ax2 or, equivalently, y = x ´ ax2, where a is a nonzero constant, lim xÑ0 f (x, x ´ ax2) = lim xÑ0 (2x ´ x + ax2)2 x ´ x + ax2 = lim xÑ0 (x + ax2)2 ax2 = lim xÑ0 (1 + ax)2 a = 1 a 96 PARTIAL DERIVATIVES 2.2 PARTIAL DERIVATIVES This limit depends on the choice of the constant a. Once again, this proves that f (x, y) does not have a limit as (x, y) Ñ (0, 0). Example 2.1.10 2.2Ĳ Partial Derivatives We are now ready to define derivatives of functions of more than one variable. First, recall how we defined the derivative, f 1(a), of a function of one variable, f (x). We imagined that we were walking along the x-axis, in the positive direction, measuring, for example, the temperature along the way. We denoted by f (x) the temperature at x. The instantaneous rate of change of temperature that we observed as we passed through x = a was df dx(a) = lim hÑ0 f (a + h) ´ f (a) h = lim xÑa f (x) ´ f (a) x ´ a Next suppose that we are walking in the xy-plane and that the temperature at (x, y) is f (x, y). We can pass through the point (x, y) = (a, b) moving in many different directions, and we cannot expect the measured rate of change of temperature if we walk parallel to the x-axis, in the direction of increasing x, to be the same as the measured rate of change of temperature if we walk parallel to the y-axis in the direction of increasing y. We’ll start by considering just those two directions. We’ll consider other directions (like walking parallel to the line y = x) later. Suppose that we are passing through the point (x, y) = (a, b) and that we are walking parallel to the x-axis (in the positive direction). Then our y-coordinate will be constant, al-ways taking the value y = b. So we can think of the measured temperature as the function of one variable B(x) = f (x, b) and we will observe the rate of change of temperature dB dx (a) = lim hÑ0 B(a + h) ´ B(a) h = lim hÑ0 f (a + h, b) ´ f (a, b) h This is called the “partial derivative f with respect to x at (a, b)” and is denoted B f Bx(a, b). Here ˝ the symbol B, which is read “partial”, indicates that we are dealing with a function of more than one variable, and ˝ the x in B f Bx indicates that we are differentiating with respect to x, while y is being held fixed, i.e. being treated as a constant. ˝ B f Bx is read “ partial dee f dee x”. Do not write d dx when B Bx is appropriate. We shall later encounter situations when d dx f and B Bx f are both defined and have different meanings. If, instead, we are passing through the point (x, y) = (a, b) and are walking parallel to the y-axis (in the positive direction), then our x-coordinate will be constant, always taking the value x = a. So we can think of the measured temperature as the function of one variable A(y) = f (a, y) and we will observe the rate of change of temperature dA dy (b) = lim hÑ0 A(b + h) ´ A(b) h = lim hÑ0 f (a, b + h) ´ f (a, b) h 97 PARTIAL DERIVATIVES 2.2 PARTIAL DERIVATIVES This is called the “partial derivative f with respect to y at (a, b)” and is denoted B f By(a, b). Just as was the case for the ordinary derivative df dx(x) (see Definition 2.2.6 in the CLP-1 text), it is common to treat the partial derivatives of f (x, y) as functions of (x, y) simply by evaluating the partial derivatives at (x, y) rather than at (a, b). The x- and y-partial derivatives of the function f (x, y) are B f Bx(x, y) = lim hÑ0 f (x + h, y) ´ f (x, y) h B f By(x, y) = lim hÑ0 f (x, y + h) ´ f (x, y) h respectively. The partial derivatives of functions of more than two variables are defined analogously. Definition 2.2.1 (Partial Derivatives). Partial derivatives are used a lot. And there many notations for them. The partial derivative B f Bx(x, y) of a function f (x, y) is also denoted B f Bx fx(x, y) fx Dx f (x, y) Dx f D1 f (x, y) D1 f The subscript 1 on D1 f indicates that f is being differentiated with respect to its first variable. The partial derivative B f Bx(a, b) is also denoted B f Bx ˇ ˇ ˇ ˇ (a,b) with the subscript (a, b) indicating that B f Bx is being evaluated at (x, y) = (a, b). The notation B f Bx y is used to make explicit that the variable y is being held.7 Notation 2.2.2. Remark 2.2.3 (The Geometric Interpretation of Partial Derivatives). We’ll now develop a geometric interpretation of the partial derivative B f Bx(a, b) = lim hÑ0 f (a + h, b) ´ f (a, b) h 7 There are applications in which there are several variables that cannot be varied independently. For example, the pressure, volume and temperature of an ideal gas are related by the equation of state PV = (constant)T. In those applications, it may not be clear from the context which variables are being held fixed. 98 PARTIAL DERIVATIVES 2.2 PARTIAL DERIVATIVES in terms of the shape of the graph z = f (x, y) of the function f (x, y). That graph appears in the figure below. It looks like the part of a deformed sphere that is in the first octant. The definition of B f Bx(a, b) concerns only points on the graph that have y = b. In other words, the curve of intersection of the surface z = f (x, y) with the plane y = b. That is the red curve in the figure. The two blue vertical line segments in the figure have heights f (a, b) and f (a + h, b), which are the two numbers in the numerator of f (a+h,b)´f (a,b) h . z y x z “ fpx, yq y “ b h fpa, bq fpa h, bq pa, b, 0q pa h, b, 0q fpa h, bq ´ fpa, bq A side view of the curve (looking from the left side of the y-axis) is sketched in the figure below. Again, the two blue vertical line segments in the figure have heights f (a, b) z x z “ fpx, bq, y “ b fpa, bq fpa h, bq pa, b, 0q pa h, b, 0q fpa h, bq ´ fpa, bq 99 PARTIAL DERIVATIVES 2.2 PARTIAL DERIVATIVES and f (a + h, b), which are the two numbers in the numerator of f (a+h,b)´f (a,b) h . So the numerator f (a + h, b) ´ f (a, b) and denominator h are the rise and run, respectively, of the curve z = f (x, b) from x = a to x = a + h. Thus B f Bx(a, b) is exactly the slope of (the tangent to) the curve of intersection of the surface z = f (x, y) and the plane y = b at the point a, b, f (a, b) . In the same way B f By(a, b) is exactly the slope of (the tangent to) the curve of intersection of the surface z = f (x, y) and the plane x = a at the point a, b, f (a, b) . §§§ Evaluation of Partial Derivatives From the above discussion, we see that we can readily compute partial derivatives B Bx by using what we already know about ordinary derivatives d dx. More precisely, • to evaluate B f Bx(x, y), treat the y in f (x, y) as a constant and differentiate the resulting function of x with respect to x. • To evaluate B f By(x, y), treat the x in f (x, y) as a constant and differentiate the resulting function of y with respect to y. • To evaluate B f Bx(a, b), treat the y in f (x, y) as a constant and differentiate the resulting function of x with respect to x. Then evaluate the result at x = a, y = b. • To evaluate B f By(a, b), treat the x in f (x, y) as a constant and differentiate the resulting function of y with respect to y. Then evaluate the result at x = a, y = b. Now for some examples. Example 2.2.4 Let f (x, y) = x3 + y2 + 4xy2 Then, since B Bx treats y as a constant, B f Bx = B Bx(x3) + B Bx(y2) + B Bx(4xy2) = 3x2 + 0 + 4y2 B Bx(x) = 3x2 + 4y2 and, since B By treats x as a constant, B f By = B By(x3) + B By(y2) + B By(4xy2) = 0 + 2y + 4x B By(y2) = 2y + 8xy 100 PARTIAL DERIVATIVES 2.2 PARTIAL DERIVATIVES In particular, at (x, y) = (1, 0) these partial derivatives take the values B f Bx(1, 0) = 3(1)2 + 4(0)2 = 3 B f By(1, 0) = 2(0) + 8(1)(0) = 0 Example 2.2.4 Example 2.2.5 Let f (x, y) = y cos x + xexy Then, since B Bx treats y as a constant, B Bxeyx = yeyx and B f Bx(x, y) = y B Bx(cos x) + exy B Bx(x) + x B Bx exy (by the product rule) = ´y sin x + exy + xyexy B f By(x, y) = cos x B By(y) + x B By exy = cos x + x2exy Example 2.2.5 Let’s move up to a function of four variables. Things generalize in a quite straight forward way. Example 2.2.6 Let f (x, y, z, t) = x sin(y + 2z) + t2e3y ln z Then B f Bx(x, y, z, t) = sin(y + 2z) B f By(x, y, z, t) = x cos(y + 2z) + 3t2e3y ln z B f Bz (x, y, z, t) = 2x cos(y + 2z) + t2e3y/z B f Bt (x, y, z, t) = 2te3y ln z Example 2.2.6 Now here is a more complicated example — our function takes a special value at (0, 0). To compute derivatives there we revert to the definition. 101 PARTIAL DERIVATIVES 2.2 PARTIAL DERIVATIVES Example 2.2.7 Set f (x, y) = #cos x´cos y x´y if x ‰ y 0 if x = y If b ‰ a, then for all (x, y) sufficiently close to (a, b), f (x, y) = cos x´cos y x´y and we can compute the partial derivatives of f at (a, b) using the familiar rules of differentiation. However that is not the case for (a, b) = (0, 0). To evaluate fx(0, 0), we need to set y = 0 and find the derivative of f (x, 0) = # cos x´1 x if x ‰ 0 0 if x = 0 with respect to x at x = 0. As we cannot use the usual differentiation rules, we evaluate the derivative8 by applying the definition fx(0, 0) = lim hÑ0 f (h, 0) ´ f (0, 0) h = lim hÑ0 cos h´1 h ´ 0 h (Recall that h ‰ 0 in the limit.) = lim hÑ0 cos h ´ 1 h2 = lim hÑ0 ´ sin h 2h (By l’Hˆ opital’s rule.) = lim hÑ0 ´ cos h 2 (By l’Hˆ opital again.) = ´1 2 We could also evaluate the limit of cos h´1 h2 by substituting in the Taylor expansion cos h = 1 ´ h2 2 + h4 4! ´ ¨ ¨ ¨ We can also use Taylor expansions to understand the behaviour of f (x, y) for (x, y) near (0, 0). For x ‰ y, cos x ´ cos y x ´ y = h 1 ´ x2 2! + x4 4! ´ ¨ ¨ ¨ i ´ h 1 ´ y2 2! + y4 4! ´ ¨ ¨ ¨ i x ´ y = ´ x2´y2 2! + x4´y4 4! ´ ¨ ¨ ¨ x ´ y = ´ 1 2! x2 ´ y2 x ´ y + 1 4! x4 ´ y4 x ´ y ´ ¨ ¨ ¨ = ´x + y 2! + x3 + x2y + xy2 + y3 4! ´ ¨ ¨ ¨ 8 It is also possible to evaluate the derivative by using the technique of the optional §2.15 in the CLP-1 text. 102 PARTIAL DERIVATIVES 2.2 PARTIAL DERIVATIVES So for (x, y) near (0, 0), f (x, y) « # ´ x+y 2 if x ‰ y 0 if x = y So it sure looks like (and in fact it is true that) • f (x, y) is continuous at (0, 0) and • f (x, y) is not continuous at (a, a) for small a ‰ 0 and • fx(0, 0) = fy(0, 0) = ´1 2 Example 2.2.7 Example 2.2.8 Again set f (x, y) = #cos x´cos y x´y if x ‰ y 0 if x = y We’ll now compute fy(x, y) for all (x, y). The case y ‰ x: When y ‰ x, fy(x, y) = B By cos x ´ cos y x ´ y = (x ´ y) B By(cos x ´ cos y) ´ (cos x ´ cos y) B By(x ´ y) (x ´ y)2 by the quotient rule = (x ´ y) sin y + cos x ´ cos y (x ´ y)2 The case y = x: When y = x, fy(x, y) = lim hÑ0 f (x, y + h) ´ f (x, y) h = lim hÑ0 f (x, x + h) ´ f (x, x) h = lim hÑ0 cos x´cos(x+h) x´(x+h) ´ 0 h (Recall that h ‰ 0 in the limit.) = lim hÑ0 cos(x + h) ´ cos x h2 Now we apply L’Hˆ opital’s rule, remembering that, in this limit, x is a constant and h is the variable — so we differentiate with respect to h. fy(x, y) = lim hÑ0 ´ sin(x + h) 2h 103 PARTIAL DERIVATIVES 2.2 PARTIAL DERIVATIVES Note that if x is not an integer multiple of π, then the numerator ´ sin(x + h) does not tend to zero as h tends to zero, and the limit giving fy(x, y) does not exist. On the other hand, if x is an integer multiple of π, both the numerator and denominator tend to zero as h tends to zero, and we can apply L’Hˆ opital’s rule a second time. Then fy(x, y) = lim hÑ0 ´ cos(x + h) 2 = ´cos x 2 The conclusion: fy(x, y) = $ ’ ’ & ’ ’ % (x´y) sin y+cos x´cos y (x´y)2 if x ‰ y ´cos x 2 if x = y with x an integer multiple of π DNE if x = y with x not an integer multiple of π Example 2.2.8 Example 2.2.9 (Optional — A Little Weirdness) In this example, we will see that the function f (x, y) = # x2 x´y if x ‰ y 0 if x = y is not continuous at (0, 0) and yet has both partial derivatives fx(0, 0) and fy(0, 0) perfectly well defined. We’ll also see how that is possible. First let’s compute the partial derivatives. By definition, fx(0, 0) = lim hÑ0 f (0 + h, 0) ´ f (0, 0) h = lim hÑ0 h hk kik kj h2 h´0 ´ 0 h = lim hÑ0 1 = 1 fy(0, 0) = lim hÑ0 f (0, 0 + h) ´ f (0, 0) h = lim hÑ0 02 0´h ´ 0 h = lim hÑ0 0 = 0 So the first order partial derivatives fx(0, 0) and fy(0, 0) are perfectly well defined. To see that, nonetheless, f (x, y) is not continuous at (0, 0), we take the limit of f (x, y) as (x, y) approaches (0, 0) along the curve y = x ´ x3. The limit is lim xÑ0 f x, x ´ x3 = lim xÑ0 x2 x ´ (x ´ x3) = lim xÑ0 1 x which does not exist. Indeed as x approaches 0 through positive numbers, 1 x approaches +8, and as x approaches 0 through negative numbers, 1 x approaches ´8. 104 PARTIAL DERIVATIVES 2.2 PARTIAL DERIVATIVES So how is this possible? The answer is that fx(0, 0) only involves values of f (x, y) with y = 0. As f (x, 0) = x, for all values of x, we have that f (x, 0) is a continuous, and indeed a differentiable, function. Similarly, fy(0, 0) only involves values of f (x, y) with x = 0. As f (0, y) = 0, for all values of y, we have that f (0, y) is a continuous, and indeed a differentiable, function. On the other hand, the bad behaviour of f (x, y) for (x, y) near (0, 0) only happens for x and y both nonzero. Example 2.2.9 Our next example uses implicit differentiation. Example 2.2.10 The equation z5 + y2ez + e2x = 0 implicitly determines z as a function of x and y. That is, the function z(x, y) obeys z(x, y)5 + y2ez(x,y) + e2x = 0 For example, when x = y = 0, the equation reduces to z(0, 0)5 = ´1 which forces9 z(0, 0) = ´1. Let’s find the partial derivative Bz Bx(0, 0). We are not going to be able to explicitly solve the equation for z(x, y). All we know is that z(x, y)5 + y2ez(x,y) + e2x = 0 for all x and y. We can turn this into an equation for Bz Bx(0, 0) by differentiating10 the whole equation with respect to x, giving 5z(x, y)4 Bz Bx(x, y) + y2ez(x,y) Bz Bx(x, y) + 2e2x = 0 and then setting x = y = 0, giving 5z(0, 0)4 Bz Bx(0, 0) + 2 = 0 As we already know that z(0, 0) = ´1, Bz Bx(0, 0) = ´ 2 5z(0, 0)4 = ´2 5 Example 2.2.10 Next we have a partial derivative disguised as a limit. 9 The only real number z which obeys z5 = ´1 is z = ´1. However there are four other complex numbers which also obey z5 = ´1. 10 You should have already seen this technique, called implicit differentiation, in your first Calculus course. It is covered in Section 2.11 in the CLP-1 text. 105 PARTIAL DERIVATIVES 2.2 PARTIAL DERIVATIVES Example 2.2.11 In this example we are going to evaluate the limit lim zÑ0 (x + y + z)3 ´ (x + y)3 (x + y)z The critical observation is that, in taking the limit z Ñ 0, x and y are fixed. They do not change as z is getting smaller and smaller. Furthermore this limit is exactly of the form of the limits in the Definition 2.2.1 of partial derivative, disguised by some obfuscating changes of notation. Set f (x, y, z) = (x + y + z)3 (x + y) Then lim zÑ0 (x + y + z)3 ´ (x + y)3 (x + y)z = lim zÑ0 f (x, y, z) ´ f (x, y, 0) z = lim hÑ0 f (x, y, 0 + h) ´ f (x, y, 0) h = B f Bz (x, y, 0) = B Bz (x + y + z)3 x + y z=0 Recalling that B Bz treats x and y as constants, we are evaluating the derivative of a function of the form (const+z)3 const . So lim zÑ0 (x + y + z)3 ´ (x + y)3 (x + y)z = 3(x + y + z)2 x + y ˇ ˇ ˇ ˇ z=0 = 3(x + y) Example 2.2.11 The next example highlights a potentially dangerous difference between ordinary and partial derivatives. Example 2.2.12 In this example we are going to see that, in contrast to the ordinary derivative case, Br Bx is not, in general, the same as Bx Br ´1. Recall that Cartesian and polar coordinates11 (for (x, y) ‰ (0, 0) and r ą 0) are related 11 If you are not familiar with polar coordinates, don’t worry about it. There will be an introduction to them in §3.2.1. 106 PARTIAL DERIVATIVES 2.2 PARTIAL DERIVATIVES by x = r cos θ x y px, yq r θ y = r sin θ r = b x2 + y2 tan θ = y x We will use the functions x(r, θ) = r cos θ and r(x, y) = b x2 + y2 Fix any point (x0, y0) ‰ (0, 0) and let (r0, θ0), 0 ď θ0 ă 2π, be the corresponding polar coordinates. Then Bx Br (r, θ) = cos θ Br Bx(x, y) = x a x2 + y2 so that Bx Br (r0, θ0) = Br Bx(x0, y0) ´1 ð ñ cos θ0 =   x0 b x2 0 + y2 0   ´1 = (cos θ0)´1 ð ñ cos2 θ0 = 1 ð ñ θ0 = 0, π We can also see pictorially why this happens. By definition, the partial derivatives Bx Br (r0, θ0) = lim drÑ0 x(r0 + dr, θ0) ´ x(r0, θ0) dr Br Bx(x0, y0) = lim dxÑ0 r(x0 + dx, y0) ´ r(x0, y0) dx Here we have just renamed the h of Definition 2.2.1 to dr and to dx in the two definitions. In computing Bx Br (r0, θ0), θ0 is held fixed, r is changed by a small amount dr and the resulting dx = x(r0 + dr, θ0) ´ x(r0, θ0) is computed. In the figure on the left below, dr is the length of the orange line segment and dx is the length of the blue line segment. dr θ0 dx px0,y0q dr dx px0,y0q 107 PARTIAL DERIVATIVES 2.2 PARTIAL DERIVATIVES On the other hand, in computing Br Bx, y is held fixed, x is changed by a small amount dx and the resulting dr = r(x0 + dx, y0) ´ r(x0, y0) is computed. In the figure on the right above, dx is the length of the pink line segment and dr is the length of the orange line segment. Here are the two figures combined together. We have arranged that the same dr is used in both computations. In order for the dr’s to be the same in both computations, the two dx’s have to be different (unless θ0 = 0, π). So, in general, Bx Br (r0, θ0) ‰ Br Bx(x0, y0) ´1. dr θ0 dx dx px0,y0q Example 2.2.12 Example 2.2.13 (Optional — Example 2.2.12, continued) The inverse function theorem, for functions of one variable, says that, if y(x) and x(y) are inverse functions, meaning that y x(y) = y and x y(x) = x, and are differentiable with dy dx ‰ 0, then dx dy(y) = 1 dy dx x(y) To see this, just apply d dy to both sides of y x(y) = y to get dy dx x(y) dx dy(y) = 1, by the chain rule (see Theorem 2.9.3 in the CLP-1 text). In the CLP-1 text, we used this to compute the derivatives of the logarithm (see Theorem 2.10.1 in the CLP-1 text) and of the inverse trig functions (see Theorem 2.12.7 in the CLP-1 text). We have just seen, in Example 2.2.12, that we can’t be too naive in extending the single variable inverse function theorem to functions of two (or more) variables. On the other hand, there is such an extension, which we will now illustrate, using Cartesian and polar coordinates. For simplicity, we’ll restrict our attention to x ą 0, y ą 0, or equivalently, r ą 0, 0 ă θ ă π 2 . The functions which convert between Cartesian and polar coordinates are x(r, θ) = r cos θ r(x, y) = b x2 + y2 x y px, yq r θ y(r, θ) = r sin θ θ(x, y) = arctan y x The two functions on the left convert from polar to Cartesian coordinates and the two functions on the right convert from Cartesian to polar coordinates. The inverse function theorem (for functions of two variables) says that, 108 PARTIAL DERIVATIVES 2.2 PARTIAL DERIVATIVES • if you form the first order partial derivatives of the left hand functions into the matrix " Bx Br (r, θ) Bx Bθ (r, θ) By Br (r, θ) By Bθ(r, θ) # = " cos θ ´r sin θ sin θ r cos θ # • and you form the first order partial derivatives of the right hand functions into the matrix " Br Bx(x, y) Br By(x, y) Bθ Bx(x, y) Bθ By(x, y) # =    x ? x2+y2 y ? x2+y2 ´ y x2 1+( y x )2 1 x 1+( y x )2   =   x ? x2+y2 y ? x2+y2 ´y x2+y2 x x2+y2   • and if you evaluate the second matrix at x = x(r, θ), y = y(r, θ), " Br Bx x(r, θ), y(r, θ) Br By x(r, θ), y(r, θ) Bθ Bx x(r, θ), y(r, θ) Bθ By x(r, θ), y(r, θ) # = " cos θ sin θ ´sin θ r cos θ r # • and if you multiply12 the two matrices together " Br Bx x(r, θ), y(r, θ) Br By x(r, θ), y(r, θ) Bθ Bx x(r, θ), y(r, θ) Bθ By x(r, θ), y(r, θ) # " Bx Br (r, θ) Bx Bθ (r, θ) By Br (r, θ) By Bθ(r, θ) # = " cos θ sin θ ´sin θ r cos θ r # " cos θ ´r sin θ sin θ r cos θ # = " (cos θ)(cos θ) + (sin θ)(sin θ) (cos θ)(´r sin θ) + (sin θ)(r cos θ) (´sin θ r )(cos θ) + (cos θ r )(sin θ) (´sin θ r )(´r sin θ) + (cos θ r )(r cos θ) # • then the result is the identity matrix 1 0 0 1 and indeed it is! This two variable version of the inverse function theorem can be derived by applying the derivatives B Br and B Bθ to the equations r x(r, θ), y(r, θ) = r θ x(r, θ), y(r, θ) = θ and using the two variable version of the chain rule, which we will see in §2.4. Example 2.2.13 12 Matrix multiplication is usually covered in courses on linear algebra, which you may or may not have taken. That’s why this example is optional. 109 PARTIAL DERIVATIVES 2.3 HIGHER ORDER DERIVATIVES 2.3Ĳ Higher Order Derivatives You have already observed, in your first Calculus course, that if f (x) is a function of x, then its derivative, df dx(x), is also a function of x, and can be differentiated to give the second order derivative d2 f dx2 (x), which can in turn be differentiated yet again to give the third order derivative, f (3)(x), and so on. We can do the same for functions of more than one variable. If f (x, y) is a function of x and y, then both of its partial derivatives, B f Bx(x, y) and B f By(x, y) are also functions of x and y. They can both be differentiated with respect to x and they can both be differentiated with respect to y. So there are four possible second order derivatives. Here they are, together with various alternate notations. B Bx B f Bx (x, y) = B2 f Bx2 (x, y) = fxx(x, y) B By B f Bx (x, y) = B2 f ByBx(x, y)= fxy(x, y) B Bx B f By (x, y) = B2 f BxBy(x, y)= fyx(x, y) B By B f By (x, y) = B2 f By2 (x, y) = fyy(x, y) In B2 f By Bx = B2 By Bx f, the derivative closest to f, in this case B Bx, is applied first. In fxy, the derivative with respect to the variable closest to f, in this case x, is applied first. Example 2.3.1 Let f (x, y) = emy cos(nx). Then fx = ´nemy sin(nx) fy = memy cos(nx) fxx = ´n2emy cos(nx) fyx = ´mnemy sin(nx) fxy = ´mnemy sin(nx) fyy = m2emy cos(nx) Example 2.3.1 Example 2.3.2 Let f (x, y) = eαx+βy. Then fx = αeαx+βy fy = βeαx+βy fxx = α2eαx+βy fyx = βαeαx+βy fxy = αβeαx+βy fyy = β2eαx+βy 110 PARTIAL DERIVATIVES 2.3 HIGHER ORDER DERIVATIVES More generally, for any integers m, n ě 0, Bm+n f Bxm Byn = αmβneαx+βy Example 2.3.2 Example 2.3.3 If f (x1, x2, x3, x4) = x4 1 x3 2 x2 3 x4, then B4 f Bx1 Bx2 Bx3 Bx4 = B3 Bx1 Bx2 Bx3 x4 1 x3 2 x2 3 = B2 Bx1 Bx2 2 x4 1 x3 2 x3 = B Bx1 6 x4 1 x2 2 x3 = 24 x3 1 x2 2 x3 and B4 f Bx4 Bx3 Bx2 Bx1 = B3 Bx4 Bx3 Bx2 4x3 1 x3 2 x2 3 x4 = B2 Bx4 Bx3 12 x3 1 x2 2 x2 3 x4 = B Bx4 24 x3 1 x2 2 x3 x4 = 24 x3 1 x2 2 x3 Example 2.3.3 Notice that in Example 2.3.1, fxy = fyx = ´mnemy sin(nx) and in Example 2.3.2 fxy = fyx = αβeαx+βy and in Example 2.3.3 B4 f Bx1 Bx2 Bx3 Bx4 = B4 f Bx4 Bx3 Bx2 Bx1 = 24 x3 1 x2 2 x3 In all of these examples, it didn’t matter what order we took the derivatives in. The fol-lowing theorem13 shows that this was no accident. 13 The history of this important theorem is pretty convoluted. See “A note on the history of mixed partial derivatives” by Thomas James Higgins which was published in Scripta Mathematica 7 (1940), 59-62. 111 PARTIAL DERIVATIVES 2.3 HIGHER ORDER DERIVATIVES If the partial derivatives B2 f BxBy and B2 f ByBx exist and are continuous at (x0, y0), then B2 f BxBy(x0, y0) = B2 f ByBx(x0, y0) Theorem 2.3.4 (Clairaut’s Theorem14 or Schwarz’s Theorem15). 2.3.1 § § Optional — The Proof of Theorem 2.3.4 §§§ Outline Here is an outline of the proof of Theorem 2.3.4. The (numbered) details are in the subsec-tion below. Fix real numbers x0 and y0 and define F(h, k) = 1 hk f (x0 + h, y0 + k) ´ f (x0, y0 + k) ´ f (x0 + h, y0) + f (x0, y0) We define F(h, k) in this way because both partial derivatives B2 f BxBy(x0, y0) and B2 f ByBx(x0, y0) are limits of F(h, k) as h, k Ñ 0. Precisely, we show in item (1) in the details below that B By B f Bx(x0, y0) = lim kÑ0 lim hÑ0 F(h, k) B Bx B f By(x0, y0) = lim hÑ0 lim kÑ0 F(h, k) Note that the two right hand sides here are identical except for the order in which the limits are taken. Now, by the mean value theorem (four times), F(h, k) (2) = 1 h B f By(x0 + h, y0 + θ1k) ´ B f By(x0, y0 + θ1k) (3) = B Bx B f By(x0 + θ2h, y0 + θ1k) F(h, k) (4) = 1 k B f Bx(x0 + θ3h, y0 + k) ´ B f Bx(x0 + θ3h, y0) (5) = B By B f Bx(x0 + θ3h, y0 + θ4k) for some numbers 0 ă θ1, θ2, θ3, θ4 ă 1. All of the numbers θ1, θ2, θ3, θ4 depend on x0, y0, h, k. Hence B Bx B f By(x0 + θ2h, y0 + θ1k) = B By B f Bx(x0 + θ3h, y0 + θ4k) 14 Alexis Clairaut (1713–1765) was a French mathematician, astronomer, and geophysicist. 15 Hermann Schwarz (1843–1921) was a German mathematician. 112 PARTIAL DERIVATIVES 2.3 HIGHER ORDER DERIVATIVES for all h and k. Taking the limit (h, k) Ñ (0, 0) and using the assumed continuity of both partial derivatives at (x0, y0) gives lim (h,k)Ñ(0,0) F(h, k) = B Bx B f By(x0, y0) = B By B f Bx(x0, y0) as desired. To complete the proof we just have to justify the details (1), (2), (3), (4) and (5). §§§ The Details (1) By definition, B By B f Bx(x0, y0) = lim kÑ0 1 k B f Bx(x0, y0 + k) ´ B f Bx(x0, y0) = lim kÑ0 1 k lim hÑ0 f (x0 + h, y0 + k) ´ f (x0, y0 + k) h ´ lim hÑ0 f (x0 + h, y0) ´ f (x0, y0) h = lim kÑ0 lim hÑ0 f (x0 + h, y0 + k) ´ f (x0, y0 + k) ´ f (x0 + h, y0) + f (x0, y0) hk = lim kÑ0 lim hÑ0 F(h, k) Similarly, B Bx B f By(x0, y0) = lim hÑ0 1 h B f By(x0 + h, y0) ´ B f By(x0, y0) = lim hÑ0 1 h lim kÑ0 f (x0 + h, y0 + k) ´ f (x0 + h, y0) k ´ lim kÑ0 f (x0, y0 + k) ´ f (x0, y0) k = lim hÑ0 lim kÑ0 f (x0 + h, y0 + k) ´ f (x0 + h, y0) ´ f (x0, y0 + k) + f (x0, y0) hk = lim hÑ0 lim kÑ0 F(h, k) (2) The mean value theorem (Theorem 2.13.4 in the CLP-1 text) says that, for any differ-entiable function φ(x), • the slope of the line joining the points x0, φ(x0) and x0 + k, φ(x0 + k) on the graph of φ is the same as • the slope of the tangent to the graph at some point between x0 and x0 + k. That is, there is some 0 ă θ1 ă 1 such that φ(x0 + k) ´ φ(x0) k = dφ dx (x0 + θ1k) 113 PARTIAL DERIVATIVES 2.3 HIGHER ORDER DERIVATIVES x y y “ ϕpxq x0 x0k x0θ1k Applying this with x replaced by y and φ replaced by G(y) = f (x0 + h, y) ´ f (x0, y) gives G(y0 + k) ´ G(y0) k = dG dy (y0 + θ1k) for some 0 ă θ1 ă 1 = B f By(x0 + h, y0 + θ1k) ´ B f By(x0, y0 + θ1k) Hence, for some 0 ă θ1 ă 1, F(h, k) = 1 h G(y0 + k) ´ G(y0) k = 1 h B f By(x0 + h, y0 + θ1k) ´ B f By(x0, y0 + θ1k) (3) Define H(x) = B f By(x, y0 + θ1k). By the mean value theorem, F(h, k) = 1 h [H(x0 + h) ´ H(x0)] = dH dx (x0 + θ2h) for some 0 ă θ2 ă 1 = B Bx B f By(x0 + θ2h, y0 + θ1k) (4) Define A(x) = f (x, y0 + k) ´ f (x, y0). By the mean value theorem, F(h, k) = 1 k A(x0 + h) ´ A(x0) h = 1 k dA dx (x0 + θ3h) for some 0 ă θ3 ă 1 = 1 k B f Bx(x0 + θ3h, y0 + k) ´ B f Bx(x0 + θ3h, y0) (5) Define B(y) = B f Bx(x0 + θ3h, y). By the mean value theorem F(h, k) = 1 k [B(y0 + k) ´ B(y0)] = dB dy (y0 + θ4k) for some 0 ă θ4 ă 1 = B By B f Bx(x0 + θ3h, y0 + θ4k) This completes the proof of Theorem 2.3.4. 114 PARTIAL DERIVATIVES 2.3 HIGHER ORDER DERIVATIVES 2.3.2 § § Optional — An Example of B2 f BxBy(x0, y0) ‰ B2 f ByBx(x0, y0) In Theorem 2.3.4, we showed that B2 f BxBy(x0, y0) = B2 f ByBx(x0, y0) if the partial derivatives B2 f BxBy and B2 f ByBx exist and are continuous at (x0, y0). Here is an example which shows that if the partial derivatives B2 f BxBy and B2 f ByBx are not continuous at (x0, y0), then it is possible that B2 f BxBy(x0, y0) ‰ B2 f ByBx(x0, y0). Define f (x, y) = # xy x2´y2 x2+y2 if (x, y) ‰ (0, 0) 0 if (x, y) = (0, 0) This function is continuous everywhere. Note that f (x, 0) = 0 for all x and f (0, y) = 0 for all y. We now compute the first order partial derivatives. For (x, y) ‰ (0, 0), B f Bx(x, y) = y x2 ´ y2 x2 + y2 + xy 2x x2 + y2 ´ xy2x(x2 ´ y2) (x2 + y2)2 = y x2 ´ y2 x2 + y2 + xy 4xy2 (x2 + y2)2 B f By(x, y) = x x2 ´ y2 x2 + y2 ´ xy 2y x2 + y2 ´ xy2y(x2 ´ y2) (x2 + y2)2 = x x2 ´ y2 x2 + y2 ´ xy 4yx2 (x2 + y2)2 For (x, y) = (0, 0), B f Bx(0, 0) = d dx f (x, 0) x=0 = d dx0 x=0 = 0 B f By(0, 0) = d dy f (0, y) y=0 = d dy0 y=0 = 0 By way of summary, the two first order partial derivatives are fx(x, y) = $ & % y x2´y2 x2+y2 + 4x2y3 (x2+y2)2 if (x, y) ‰ (0, 0) 0 if (x, y) = (0, 0) fy(x, y) = $ & % x x2´y2 x2+y2 ´ 4x3y2 (x2+y2)2 if (x, y) ‰ (0, 0) 0 if (x, y) = (0, 0) Both B f Bx(x, y) and B f By(x, y) are continuous. Finally, we compute B2 f BxBy(0, 0) = d dx fy(x, 0) x=0 = lim hÑ0 1 h fy(h, 0) ´ fy(0, 0) = lim hÑ0 1 h h h2 ´ 02 h2 + 02 ´ 0 = 1 B2 f ByBx(0, 0) = d dy fx(0, y) y=0 = lim kÑ0 1 k [ fx(0, k) ´ fx(0, 0)] = lim kÑ0 1 k k 02 ´ k2 02 + k2 ´ 0 = ´1 115 PARTIAL DERIVATIVES 2.4 THE CHAIN RULE 2.4Ĳ The Chain Rule You already routinely use the one dimensional chain rule d dt f x(t) = df dx x(t) dx dt (t) in doing computations like d dt sin(t2) = cos(t2) 2t In this example, f (x) = sin(x) and x(t) = t2. We now generalize the chain rule to functions of more than one variable. For con-creteness, we concentrate on the case in which all functions are functions of two variables. That is, we find the partial derivatives BF Bs and BF Bt of a function F(s, t) that is defined as a composition F(s, t) = f x(s, t) , y(s, t) We are using the name F for the new function F(s, t) as a reminder that it is closely related to, though not the same as, the function f (x, y). The partial derivative BF Bs is the rate of change of F when s is varied with t held constant. When s is varied, both the x-argument, x(s, t), and the y-argument, y(s, t), in f x(s, t) , y(s, t) vary. Consequently, the chain rule for f x(s, t) , y(s, t) is a sum of two terms — one resulting from the variation of the x-argument and the other resulting from the variation of the y-argument. Assume that all first order partial derivatives of f (x, y), x(s, t) and y(s, t) exist and are continuous. Then the same is true for F(s, t) = f x(s, t) , y(s, t) and BF Bs (s, t) = B f Bx x(s, t) , y(s, t) Bx Bs (s, t) + B f By x(s, t) , y(s, t) By Bs (s, t) BF Bt (s, t) = B f Bx x(s, t) , y(s, t) Bx Bt (s, t) + B f By x(s, t) , y(s, t) By Bt (s, t) Theorem 2.4.1 (The Chain Rule). We will give the proof of this theorem in §2.4.4, below. It is common to state this chain rule as BF Bs = B f Bx Bx Bs + B f By By Bs BF Bt = B f Bx Bx Bt + B f By By Bt That is, it is common to suppress the function arguments. But you should make sure that you understand what the arguments are before doing so. Theorem 2.4.1 is given for the case that F is the composition of a function of two vari-ables, f (x, y), with two functions, x(s, t) and y(s, t), of two variables each. There is nothing magical about the number two. There are obvious variants for any numbers of variables. For example, 116 PARTIAL DERIVATIVES 2.4 THE CHAIN RULE if F(t) = f x(t), y(t), z(t) , then dF dt (t) = B f Bx x(t) , y(t) , z(t) dx dt (t) + B f By x(t) , y(t) , z(t) dy dt (t) + B f Bz x(t) , y(t) , z(t) dz dt (t) Equation 2.4.2. and if F(s, t) = f x(s, t) , then BF Bt (s, t) = df dx x(s, t) Bx Bt (s, t) Equation 2.4.3. There will be a large number of examples shortly. First, here is a memory aid. 2.4.1 § § Memory Aids for the Chain Rule We recommend strongly that you use the following procedure, without leaving out any steps, the first couple of dozen times that you use the chain rule. Step 1 List explicitly all the functions involved and specify the arguments of each func-tion. Ensure that all different functions have different names. Invent new names for some of the functions if necessary. In the case of the chain rule in Theorem 2.4.1, the list would be f (x, y) x(s, t) y(s, t) F(s, t) = f x(s, t), y(s, t) While the functions f and F are closely related, they are not the same. One is a function of x and y while the other is a function of s and t. Step 2 Write down the template BF Bs = B f Bs Note that • The function F appears once in the numerator on the left. The function f, from which F is constructed by a change of variables, appears once in the numerator on the right. • The variable in the denominator on the left appears once in the denominator on the right. Step 3 Fill in the blanks with every variable that makes sense. In particular, since f is a function of x and y, it may only be differentiated with respect to x and y. So we add together two copies of our template — one for x and one for y: BF Bs = B f Bx Bx Bs + B f By By Bs 117 PARTIAL DERIVATIVES 2.4 THE CHAIN RULE Note that x and y are functions of s so that the derivatives Bx Bs and By Bs make sense. The first term, B f Bx Bx Bs , arises from the variation of x with respect to s and the second term, B f By By Bs , arises from the variation of y with respect to s. Step 4 Put in the functional dependence explicitly. Fortunately, there is only one func-tional dependence that makes sense. The left hand side is a function of s and t. Hence the right hand side must also be a function of s and t. As f is a function of x and y, this is achieved by evaluating f at x = x(s, t) and y = y(s, t). BF Bs (s, t) = B f Bx x(s, t), y(s, t) Bx Bs (s, t) + B f By x(s, t), y(s, t) By Bs (s, t) If you fail to put in the arguments, or at least if you fail to remember what the argu-ments are, you may forget that B f Bx and B f By depend on s and t. Then, if you have to compute a second derivative of F, you will probably fail to differentiate the factors B f Bx x(s, t), y(s, t) and B f By x(s, t), y(s, t) . To help remember the formulae of Theorem 2.4.1, it is sometimes also useful to pretend that our variables are physical quantities with f, F having units of grams, x, y having units of meters and s, t having units of seconds. Note that • the left hand side, BF Bs , has units grams per second. • Each term on the right hand side contains the partial derivative of f with respect to a different independent variable. That independent variable appears once in the denominator and once in the numerator, so that its units (in this case meters) cancel out. Thus both of the terms B f Bx Bx Bs and B f By By Bs on the right hand side also have the units grams per second. • Hence both sides of the equation have the same units. Here is a pictorial procedure that uses a tree diagram to help remember the chain rule B Bs f x(s, t), y(s, t) = B f Bx Bx Bs + B f By By Bs . As in the figure on the left below, ˝ write, on the top row, “f”. ˝ Write, on the middle row, each of the variables that the function f (x, y) depends on, namely“x” and “y”. ˝ Write, on the bottom row, – below x, each of the variables that the function x(s, t) depends on, namely “s” and “t”, and – below y, each of the variables that the function y(s, t) depends on, namely “s” and “t”. ˝ Draw a line joining each function with each of the variables that it depends on. ˝ Then, as in the figure on the right below, write beside each line, the partial derivative of the function at the top of the line with respect to the variable at the bottom of the line. 118 PARTIAL DERIVATIVES 2.4 THE CHAIN RULE f x y s t s t f x y s t s t Bf Bx Bf By Bx Bt By Bt Bx Bs By Bs ˝ Finally – observe, from the figure below, that there are two paths from f, on the top, to s, on the bottom. One path goes from f at the top, through x in the middle to s at the bottom. The other path goes from f at the top, through y in the middle to s at the bottom. – For each such path, multiply together the partial derivatives beside the lines of the path. In this example, the two products are B f Bx Bx Bs , for the first path, and B f By By Bs , for the second path. – Then add together those products, giving, in this example, B f Bx Bx Bs + B f By By Bs . – Put in the arguments, as in Step 4, above. ˝ That’s it. We have B Bs f x(s, t), y(s, t) = B f Bx x(s, t), y(s, t) Bx Bs (s, t) + B f By x(s, t), y(s, t) By Bs (s, t) f x y s t s t Bf Bx Bf By Bx Bs By Bs Bx Bt By Bt Example 2.4.4 The right hand side of the chain rule d dt f x(t) , y(t) , z(t) = B f Bx x(t) , y(t) , z(t) dx dt (t) + B f By x(t) , y(t) , z(t) dy dt (t) + B f Bz x(t) , y(t) , z(t) dz dt (t) of Equation (2.4.2), without arguments, is B f Bx dx dt + B f By dy dt + B f Bz dz dt . The corresponding tree diagram is 119 PARTIAL DERIVATIVES 2.4 THE CHAIN RULE f x y z t t t Bf Bx Bf By Bf Bz dx dt dy dt dz dt Because x(t), y(t) and z(t) are each functions of just one variable, the derivatives beside the lower lines in the tree are ordinary, rather than partial, derivatives. Example 2.4.4 2.4.2 § § Chain Rule Examples Let’s do some routine examples first and work our way to some trickier ones. Example 2.4.5 B Bs f x(s, t), y(s, t) In this example we find B Bs f x(s, t), y(s, t) for f (x, y) = exy x(s, t) = s y(s, t) = cos t Define F(s, t) = f x(s, t) , y(s, t) . The appropriate chain rule for this example is the upper equation of Theorem 2.4.1. BF Bs (s, t) = B f Bx x(s, t) , y(s, t) Bx Bs (s, t) + B f By x(s, t) , y(s, t) By Bs (s, t) For the given functions f (x, y) = exy B f Bx(x, y) = yexy B f Bx x(s, t), y(s, t) = y(s, t)ex(s,t) y(s,t) = cos t es cos t B f By(x, y) = xexy B f By x(s, t), y(s, t) = x(s, t)ex(s,t) y(s,t) = s es cos t Bx Bs = 1 By Bs = 0 so that BF Bs (s, t) = B f Bx hkkkkkkikkkkkkj ␣ cos t es cos t( Bx Bs hk kik kj (1) + B f By hkkkkikkkkj ␣ s es cos t( By Bs hk kik kj (0) = cos t es cos t Example 2.4.5 120 PARTIAL DERIVATIVES 2.4 THE CHAIN RULE Example 2.4.6 d dt f x(t), y(t) In this example we find d dt f x(t), y(t) for f (x, y) = x2 ´ y2 x(t) = cos t y(t) = sin t Define F(t) = f x(t), y(t) . Since F(t) is a function of one variable its derivative is denoted dF dt rather than BF Bt . The appropriate chain rule for this example (see (2.4.2)) is dF dt (t) = B f Bx x(t), y(t) dx dt (t) + B f By x(t), y(t) dy dt (t) For the given functions f (x, y) = x2 ´ y2 B f Bx(x, y) = 2x B f Bx x(t), y(t) = 2x(t) = 2 cos t B f By(x, y) = ´2y B f By x(t), y(t) = ´2y(t) = ´2 sin t dx dt = ´ sin t dy dt = cos t so that dF dt (t) = (2 cos t)(´ sin t) + (´2 sin t)(cos t) = ´4 sin t cos t Of course, in this example we can compute F(t) explicitly F(t) = f x(t), y(t) = x(t)2 ´ y(t)2 = cos2 t ´ sin2 t and then differentiate F1(t) = 2(cos t)(´ sin t) ´ 2(sin t)(cos t) = ´4 sin t cos t Example 2.4.6 Example 2.4.7 B Bt f (x + ct) Define u(x, t) = x + ct and w(x, t) = f (x + ct) = f u(x, t) . Then B Bt f (x + ct) = Bw Bt (x, t) = df du u(x, t) Bu Bt (x, t) = c f 1(x + ct) Example 2.4.7 121 PARTIAL DERIVATIVES 2.4 THE CHAIN RULE Example 2.4.8 B2 Bt2 f (x + ct) Define w(x, t) = f (x + ct) and W(x, t) = Bw Bt (x, t) = c f 1(x + ct) = F u(x, t) where F(u) = c f 1(u) and u(x, t) = x + ct. Then B2 Bt2 f (x + ct) = BW Bt (x, t) = dF du u(x, t) Bu Bt (x, t) = c f 2(x + ct) c = c2 f 2(x + ct) Example 2.4.8 Example 2.4.9 (Equation of state) Suppose that we are told that F(P, V, T) = 0 and that we are to find BP BT. Before we can find BP BT, we first have to decide what it means. This happens regularly in applications. In fact, this particular problem comes from thermodynamics. The variables P, V, T are the pressure, volume and temperature, respectively, of some gas. These three variables are not independent. They are related by an equation of state, here denoted F(P, V, T) = 0. Given values for any two of P, V, T, the third can be found by solving F(P, V, T) = 0. We are being asked to find BP BT. This implicitly instructs us to treat P, in this problem, as the dependent variable. So a careful wording of this problem (which you will never encounter in the “real world”) would be the following. The function P(V, T) is defined by F P(V, T), V, T) = 0. Find BP BT V. That is, find the rate of change of pressure as the temperature is varied, while holding the volume fixed. Since we are not told explicitly what F is, we cannot solve explicitly for P(V, T). So, instead we differentiate both sides of F P(V, T), V, T = 0 with respect to T, while holding V fixed. Think of the left hand side, F P(V, T), V, T , as being F P(V, T), Q(V, T), R(V, T) with Q(V, T) = V and R(V, T) = T. By the chain rule, B BT F P(V, T), Q(V, T), R(V, T) = F1 BP BT + F2 BQ BT + F3 BR BT = 0 with Fj referring to the partial derivative of F with respect to its jth argument. Experienced chain rule users never introduce Q and R. Instead, they just write BF BP BP BT + BF BV BV BT + BF BT BT BT = 0 Recalling that V and T are the independent variables and that, in computing B BT, V is to be treated as a constant, BV BT = 0 BT BT = 1 Now putting in the functional dependence BF BP P(V, T), V, T BP BT(V, T) + BF BT P(V, T), V, T = 0 122 PARTIAL DERIVATIVES 2.4 THE CHAIN RULE and solving BP BT(V, T) = ´ BF BT P(V, T), V, T BF BP P(V, T), V, T Example 2.4.9 Example 2.4.10 Suppose that f (x, y) = 0 and that we are to find d2y dx2. Once again, x and y are not independent variables. Given a value for either x or y, the other is determined by solving f (x, y) = 0. Since we are asked to find d2y dx2, it is y that is to be viewed as a function of x, rather than the other way around. So f (x, y) = 0 really means that, in this problem, f x, y(x) = 0 for all x. Differentiating both sides of this equation with respect to x, f x, y(x) = 0 for all x ù ñ d dx f x, y(x) = 0 Note that d dx f x, y(x) is not the same as fx x, y(x) . The former is, by definition, the rate of change with respect to x of g(x) = f x, y(x) . Precisely, dg dx = lim ∆xÑ0 g(x + ∆x) ´ g(x) ∆x = lim ∆xÑ0 f x + ∆x , y(x + ∆x) ´ f x , y(x) ∆x (˚) On the other hand, by definition, fx(x, y) = lim ∆xÑ0 f (x + ∆x, y) ´ f (x, y) ∆x ù ñ fx x, y(x) = lim ∆xÑ0 f x + ∆x , y(x) ´ f x , y(x) ∆x (˚˚) The right hand sides of (˚) and (˚˚) are not the same. In dg dx, as ∆x varies the value of y that is substituted into the first f (¨ ¨ ¨ ) on the right hand side, namely y(x + ∆x), changes as ∆x changes. That is, we are computing the rate of change of f along the (curved) path y = y(x). In (˚˚), the corresponding value of y is y(x) and is independent of ∆x. That is, we are computing the rate of change of f along a horizontal straight line. As a concrete example, suppose that f (x, y) = x + y. Then, 0 = f x , y(x) = x + y(x) gives y(x) = ´x so that d dx f x, y(x) = d dx f (x, ´x) = d dx[x + (´x)] = d dx0 = 0 But f (x, y) = x + y implies that fx(x, y) = 1 for all x and y so that fx(x, y(x)) = fx(x, y) ˇ ˇ ˇ y=´x = 1 ˇ ˇ ˇ y=´x = 1 123 PARTIAL DERIVATIVES 2.4 THE CHAIN RULE Now back to f x, y(x) = 0 for all x ù ñ d dx f x, y(x) = 0 ù ñ fx x, y(x) dx dx + fy x, y(x) dy dx(x) = 0 by the chain rule ù ñ dy dx(x) = ´ fx x, y(x) fy x, y(x) ù ñ d2y dx2(x) = ´ d dx " fx x, y(x) fy x, y(x) # = ´ fy x, y(x) d dx[ fx x, y(x) ] ´ fx x, y(x) d dx[ fy x, y(x) ] fy x, y(x) 2 (:) by the quotient rule. Now it suffices to substitute in d dx fx x, y(x) and d dx fy x, y(x) . For the former apply the chain rule to h(x) = u x, y(x) with u(x, y) = fx x, y . d dx fx x, y(x) = dh dx(x) = ux x, y(x) dx dx + uy x, y(x) dy dx(x) = fxx x, y(x) dx dx + fxy x, y(x) dy dx(x) = fxx x, y(x) ´ fxy x, y(x) " fx x, y(x) fy x, y(x) # Substituting this and d dx fy x, y(x) = fyx x, y(x) dx dx + fyy x, y(x) dy dx(x) = fyx x, y(x) ´ fyy x, y(x) " fx x, y(x) fy x, y(x) # into the right hand side of (:) gives the final answer. d2y dx2(x) = ´ fy fxx ´ fy fxy fx fy ´ fx fyx + fx fyy fx fy f 2 y = ´ f 2 y fxx ´ 2fx fy fxy + f 2 x fyy f 3 y with all of fx, fy, fxx, fxy, fyy having arguments x , y(x) . Example 2.4.10 We now move on to the proof of Theorem 2.4.1. To give you an idea of how the proof will go, we first review the proof of the familiar one dimensional chain rule. 124 PARTIAL DERIVATIVES 2.4 THE CHAIN RULE 2.4.3 § § Review of the Proof of d dt f x(t) = df dx x(t) dx dt (t) As a warm up, let’s review the proof of the one dimensional chain rule d dt f x(t) = df dx x(t) dx dt (t) assuming that dx dt exists and that df dx is continuous. We wish to find the derivative of F(t) = f x(t) . By definition F1(t) = lim hÑ0 F(t + h) ´ F(t) h = lim hÑ0 f x(t + h) ´ f x(t) h Notice that the numerator is the difference of f (x) evaluated at two nearby values of x, namely x1 = x(t + h) and x0 = x(t). The mean value theorem is a good tool for studying the difference in the values of f (x) at two nearby points. Recall that the mean value theorem says that, for any given x0 and x1, there exists an (in general unknown) c between them so that f (x1) ´ f (x0) = f 1(c) (x1 ´ x0) For this proof, we choose x0 = x(t) and x1 = x(t + h). The mean value theorem tells us that there exists a ch so that f x(t + h) ´ f x(t) = f (x1) ´ f (x0) = f 1(ch) x(t + h) ´ x(t) We have put the subscript h on ch to emphasise that ch, which is between x0 = x(t) and x1 = x(t + h), may depend on h. Now since ch is trapped between x(t) and x(t + h) and since x(t + h) Ñ x(t) as h Ñ 0, we have that ch must also tend to x(t) as h Ñ 0. Plugging this into the definition of F1(t), F1(t) = lim hÑ0 f x(t + h) ´ f x(t) h = lim hÑ0 f 1(ch) x(t + h) ´ x(t) h = lim hÑ0 f 1(ch) lim hÑ0 x(t + h) ´ x(t) h = f 1x(t) x1(t) as desired. 2.4.4 § § Proof of Theorem 2.4.1 We’ll now prove the formula for B Bs f x(s, t) , y(s, t) that is given in Theorem 2.4.1. The proof uses the same ideas as the proof of the one variable chain rule, that we have just reviewed. 125 PARTIAL DERIVATIVES 2.4 THE CHAIN RULE We wish to find the partial derivative with respect to s of F(s, t) = f x(s, t) , y(s, t) . By definition BF Bs (s, t) = lim hÑ0 F(s + h, t) ´ F(s, t) h = lim hÑ0 f x(s + h, t) , y(s + h, t) ´ f x(s, t) , y(s, t) h The numerator is the difference of f (x, y) evaluated at two nearby values of (x, y), namely (x1, y1) = x(s + h, t) , y(s + h, t) and (x0, y0) = x(s, t) , y(s, t) . In going from (x0, y0) to (x1, y1), both the x and y-coordinates change. By adding and subtracting we can separate the change in the x-coordinate from the change in the y-coordinate. f (x1, y1) ´ f (x0, y0) = ␣ f (x1, y1) ´ f (x0, y1) ( + ␣ f (x0, y1) ´ f (x0, y0) ( The first half, ␣ f (x1, y1) ´ f (x0, y1) ( , has the same y argument in both terms and so is the difference of the function of one variable g(x) = f (x, y1) (viewing y1 just as a constant) evaluated at the two nearby values, x0, x1, of x. Consequently, we can make use of the mean value theorem as we did in §2.4.3 above. There is a cx,h between x0 = x(s, t) and x1 = x(s + h, t) such that f (x1, y1) ´ f (x0, y1) = g(x1) ´ g(x0) = g1(cx,h)[x1 ´ x0] = B f Bx(cx,h , y1) [x1 ´ x0] = B f Bx cx,h , y(s + h, t) x(s + h, t) ´ x(s, t) We have introduced the two subscripts in cx,h to remind ourselves that it may depend on h and that it lies between the two x-values x0 and x1. Similarly, the second half, ␣ f (x0, y1) ´ f (x0, y0) ( , is the difference of the function of one variable h(y) = f (x0, y) (viewing x0 just as a constant) evaluated at the two nearby values, y0, y1, of y. So, by the mean value theorem, f (x0, y1) ´ f (x0, y0) = h(y1) ´ h(y0) = h1(cy,h)[y1 ´ y0] = B f By(x0, cy,h) [y1 ´ y0] = B f By x(s, t) , cy,h y(s + h, t) ´ y(s, t) for some (unknown) cy,h between y0 = y(s, t) and y1 = y(s + h, t). Again, the two sub-scripts in cy,h remind ourselves that it may depend on h and that it lies between the two y-values y0 and y1. So, noting that, as h tends to zero, cx,h, which is trapped between x(s, t) and x(s + h, t), must tend to x(s, t), and cy,h, which is trapped between y(s, t) and 126 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES y(s + h, t), must tend to y(s, t), BF Bs (s, t)) = lim hÑ0 f x(s + h, t) , y(s + h, t) ´ f x(s, t) , y(s, t) h = lim hÑ0 B f Bx cx,h , y(s + h, t) x(s + h, t) ´ x(s, t) h + lim hÑ0 B f By x(s, t) , cy,h y(s + h, t) ´ y(s, t) h = lim hÑ0 B f Bx cx,h , y(s + h, t) lim hÑ0 x(s + h, t) ´ x(s, t) h + lim hÑ0 B f By x(s, t) , cy,h lim hÑ0 y(s + h, t) ´ y(s, t) h = B f Bx x(s, t) , y(s, t) Bx Bs (s, t) + B f By x(s, t) , y(s, t) By Bs (s, t) We can of course follow the same procedure to evaluate the partial derivative with respect to t. This concludes the proof of Theorem 2.4.1. 2.5Ĳ Tangent Planes and Normal Lines The tangent line to the curve y = f (x) at the point x0, f (x0) is the straight line that fits the curve best16 at that point. Finding tangent lines was probably one of the first applications of derivatives that you saw. See, for example, Theorem 2.3.2 in the CLP-1 text. The analog of the tangent line one dimension up is the tangent plane. The tangent plane to a surface S at a point (x0, y0, z0) is the plane that fits S best at (x0, y0, z0). For example, the tangent plane to the hemisphere S = ␣(x, y, z) ˇ ˇ x2 + y2 + (z ´ 1)2 = 1, 0 ď z ď 1 ( at the origin is the xy-plane, z = 0. z y x We are now going to determine, as our first application of partial derivatives, the tan-gent plane to a general surface S at a general point (x0, y0, z0) lying on the surface. We 16 It is possible, but beyond the scope of this text, to give a precise meaning to “fits best”. 127 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES will also determine the line which passes through (x0, y0, z0) and whose direction is per-pendicular to S at (x0, y0, z0). It is called the normal line to S at (x0, y0, z0). For example, the following figure shows the side view of the tangent plane (in black) and normal line (in blue) to the surface z = x2 + y2 (in red) at the point (0, 1, 1). y z z “ x2 y2 tangent plane normal line p0, 1, 1q side view Recall, from (1.4.1), that to specify any plane, we need ˝ one point on the plane and ˝ a vector perpendicular to the plane, i.e. a normal vector, and recall, from (1.5.1), that to specify any line, we need ˝ one point on the line and ˝ a direction vector for the line. We already have one point that is on both the tangent plane of interest and the normal line of interest — namely x0, y0, z0 . Furthermore we can use any (nonzero) vector that is perpendicular to S at (x0, y0, z0) as both the normal vector to the tangent plane and the direction vector of the normal line. So our main task is to determine a normal vector to the surface S at (x0, y0, z0). That’s what we do now, first for surfaces of the form z = f (x, y) and then, more generally, for surfaces of the form G(x, y, z) = 0. 2.5.1 § § Surfaces of the Form z = f (x, y). We construct a vector perpendicular to the surface z = f (x, y) at x0 , y0 , f (x0, y0) by, first, constructing two tangent vectors to the specified surface at the specified point, and, second, taking the cross product of those two tangent vectors. Consider the red curve in the figure below. It is the intersection of our surface z = f (x, y) with the plane y = y0. 128 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES z y x z “ fpx, yq y“y0 x“x0 x0, y0, fpx0, y0q ˘ px0, y0, 0q Here is a side view of the red curve. The vector from the point x0 , y0 , f (x0, y0) , on the z x z “ fpx, y0q, y “ y0 px0 , y0 , fpx0,y0qq px0h , y0 , fpx0h,y0qq px0,y0,0q px0h,y0,0q red curve, to the point x0 + h , y0 , f (x0 + h, y0) , also on the red curve, is almost tangent to the red curve, if h is very small. As h tends to 0, that vector, which is ⟨h , 0 , f (x0 + h, y0) ´ f (x0, y0)⟩ becomes exactly tangent to the curve. However its length also tends to 0. If we divide by h, and then take the limit h Ñ 0, we get lim hÑ0 1 h ⟨h , 0 , f (x0 + h, y0) ´ f (x0, y0)⟩= lim hÑ0 1 , 0 , f (x0 + h, y0) ´ f (x0, y0) h Since the limit limhÑ0 f (x0+h,y0)´f (x0,y0) h is the definition of the partial derivative fx(x0, y0), we get that lim hÑ0 1 h ⟨h , 0 , f (x0 + h, y0) ´ f (x0, y0)⟩= ⟨1 , 0 , fx(x0, y0)⟩ is a nonzero vector that is exactly tangent to the red curve and hence is also tangent to our surface z = f (x, y) at the point x0 , y0 , f (x0, y0) . 129 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES For the second tangent vector, we repeat the process with the blue curve in the figure at the beginning of this subsection. That blue curve is the intersection of our surface z = f (x, y) with the plane x = x0. Here is a front view of the blue curve. z y z “ fpx0, yq, x “ x0 px0 , y0 , fpx0,y0qq px0 , y0h , fpx0,y0hqq px0,y0,0q px0,y0h,0q When h is very small, the vector 1 h ⟨0 , h , f (x0, y0 + h) ´ f (x0, y0)⟩ from the point x0 , y0 , f (x0, y0) , on the blue curve, to x0 , y0 + h , f (x0, y0 + h) , also on the blue curve, (and lengthened by a factor 1 h) is almost tangent to the blue curve. Taking the limit h Ñ 0 gives the tangent vector lim hÑ0 1 h ⟨0 , h , f (x0, y0 + h) ´ f (x0, y0)⟩= lim hÑ0 0 , 1 , f (x0, y0 + h) ´ f (x0, y0) h = 0 , 1 , fy(x0, y0) to the blue curve at the point a , b , f (a, b) . Now that we have two vectors in the tangent plane to the surface z = f (x, y) at x0 , y0 , f (x0, y0) , we can find a normal vector to the tangent plane by taking their cross product. Their cross product is ⟨1 , 0 , fx(x0, y0)⟩ˆ 0 , 1 , fy(x0, y0) = det   ˆ ı ı ı ˆ ȷ ȷ ȷ ˆ k 1 0 fx(x0, y0) 0 1 fy(x0, y0)   = ´fx(x0, y0) ˆ ı ı ı ´ fy(x0, y0) ˆ ȷ ȷ ȷ + ˆ k and we have that the vector ´fx(x0, y0) ˆ ı ı ı ´ fy(x0, y0) ˆ ȷ ȷ ȷ + ˆ k is perpendicular to the surface z = f (x, y) at x0 , y0 , f (x0, y0) . The tangent plane to the surface z = f (x, y) at x0 , y0 , f (x0, y0) is the plane through x0 , y0 , f (x0, y0) with normal vector ´fx(x0, y0) ˆ ı ı ı ´ fy(x0, y0) ˆ ȷ ȷ ȷ + ˆ k. This plane has equa-tion ´fx(x0, y0) (x ´ x0) ´ fy(x0, y0) (y ´ y0) + z ´ f (x0, y0) = 0 130 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES or, after a little rearrangement, z = f (x0, y0) + fx(x0, y0) (x ´ x0) + fy(x0, y0) (y ´ y0) Now that we have the normal vector, finding the equation of the normal line to the surface z = f (x, y) at the point x0 , y0 , f (x0, y0) is straightforward. Writing it in parametric form, ⟨x, y, z⟩= ⟨x0, y0, f (x0, y0)⟩+ t ´fx(x0, y0) , ´fy(x0, y0) , 1 By way of summary (a) The vector ´fx(x0, y0) ˆ ı ı ı ´ fy(x0, y0) ˆ ȷ ȷ ȷ + ˆ k is normal to the surface z = f (x, y) at x0 , y0 , f (x0, y0) . (b) The equation of the tangent plane to the surface z = f (x, y) at the point x0 , y0 , f (x0, y0) may be written as z = f (x0, y0) + fx(x0, y0) (x ´ x0) + fy(x0, y0) (y ´ y0) (c) The parametric equation of the normal line to the surface z = f (x, y) at the point x0 , y0 , f (x0, y0) is ⟨x, y, z⟩= ⟨x0, y0, f (x0, y0)⟩+ t ´fx(x0, y0) , ´fy(x0, y0) , 1 or, writing it component by component, x = x0 ´ t fx(x0, y0) y = y0 ´ t fy(x0, y0) z = f (x0, y0) + t Theorem 2.5.1 (Tangent Plane and Normal Line). Example 2.5.2 As a warm-up example, we’ll find the tangent plane and normal line to the surface z = x2 + y2 at the point (1, 0, 1). To do so, we just apply Theorem 2.5.1 with x0 = 1, y0 = 0 and f (x, y) = x2 + y2 f (1, 0) = 1 fx(x, y) = 2x fx(1, 0) = 2 fy(x, y) = 2y fy(1, 0) = 0 So the tangent plane is z = f (x0, y0) + fx(x0, y0) (x ´ x0) + fy(x0, y0) (y ´ y0) = 1 + 2(x ´ 1) + 0(y ´ 0) = ´1 + 2x and the normal line is ⟨x, y, z⟩= ⟨x0, y0, f (x0, y0)⟩+ t ´fx(x0, y0) , ´fy(x0, y0) , 1 = ⟨1, 0, 1⟩+ t ⟨´2 , 0 , 1⟩ = ⟨1 ´ 2t , 0 , 1 + t⟩ 131 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES Example 2.5.2 That was pretty simple — find the partial derivatives and substitute in the coordinates. Let’s do something a bit more challenging. Example 2.5.3 (Optional) Find the distance from (0, 3, 0) to the surface z = x2 + y2. Solution. Write f (x, y) = x2 + y2. Let’s denote by a, b, f (a, b) the point on z = f (x, y) that is nearest (0, 3, 0). Before we really get into the problem, let’s make a simple sketch and think about what the lines from (0, 3, 0) to the surface look like and, in particular, the angles between these lines and the surface. The line from (0, 3, 0) to a, b, f (a, b) , the z y x z “ x2 y2 p0, 3, 0q z y x z “ x2 y2 p0, 3, 0q pa, b, a2 b2q point on z = f (x, y) nearest (0, 3, 0), is distinguished from the other lines from (0, 3, 0) to the surface, by being perpendicular to the surface. We will provide a detailed justification for this claim below. Let’s first exploit the fact that the vector from (0, 3, 0) to a, b, f (a, b) must be per-pendicular to the surface to determine a, b, f (a, b) , and consequently the distance from (0, 3, 0) to the surface. By Theorem 2.5.1.a, with x0 = a and y0 = b, the vector ´fx(a, b) ˆ ı ı ı ´ fy(a, b) ˆ ȷ ȷ ȷ + ˆ k = ´2a ˆ ı ı ı ´ 2b ˆ ȷ ȷ ȷ + ˆ k (˚) is normal to the surface z = f (x, y) at a, b, f (a, b) . So the vector from (0, 3, 0) to a, b, f (a, b) , namely a ˆ ı ı ı + (b ´ 3) ˆ ȷ ȷ ȷ + f (a, b) ˆ k = a ˆ ı ı ı + (b ´ 3) ˆ ȷ ȷ ȷ + (a2 + b2) ˆ k (˚˚) must be parallel to (˚). This does not force the vector (˚) to equal (˚˚), but it does force the existence of some number t obeying a ˆ ı ı ı + (b ´ 3) ˆ ȷ ȷ ȷ + (a2 + b2) ˆ k = t ´ 2a ˆ ı ı ı ´ 2b ˆ ȷ ȷ ȷ + ˆ k or equivalently $ ’ & ’ % a = ´2a t b ´ 3 = ´2b t a2 + b2 = t We now have a system of three equations in the three unknowns a, b and t. If we can solve them, we will have found the point on the surface that we want. 132 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES • The first equation is a(1 + 2t) = 0 so that either a = 0 or t = ´1 2. • The third equation forces t ě 0, so a = 0, and the last equation reduces to t = b2. • Substituting this into the middle equation gives b ´ 3 = ´2b3 or equivalently 2b3 + b ´ 3 = 0 In general, cubic equations are very hard to solve17. But, in this case, we can guess one solution18, namely b = 1. So (b ´ 1) must be a factor of 2b3 + b ´ 3 and a little division then gives us 0 = 2b3 + b ´ 3 = (b ´ 1)(2b2 + 2b + 3) We can now find the roots of the quadratic factor by using the high school formula ´2 ˘ a 22 ´ 4(2)(3) 4 Since 22 ´ 4(2)(3) ă 0, the factor 2b2 + 2b + 3 has no real roots. So the only real solution to the cubic equation 2b3 + b ´ 3 = 0 is b = 1. In summary, • a = 0, b = 1 and • the point on z = x2 + y2 nearest (0, 3, 0) is (0, 1, 1) and • the distance from (0, 3, 0) to z = x2 + y2 is the distance from (0, 3, 0) to (0, 1, 1), which is a (´2)2 + 12 = ? 5. Finally back to the claim that, because a, b, f (a, b) is the point on z = f (x, y) that is nearest19 (0, 3, 0), the vector from (0, 3, 0) to a, b, f (a, b) must be perpendicular to the surface z = f (x, y) at a, b, f (a, b) . Note that the square of the distance from (0, 3, 0) to a general point x, y, f (x, y) on z = f (x, y) is D(x, y) = x2 + (y ´ 3)2 + f (x, y)2 If x = a, y = b minimizes D(x, y) then, in particular, 17 The method for solving cubics was developed in the 15th century by del Ferro, Cardano and Ferrari (Cardano’s student). Ferrari then went on to discover a formula for the roots of a quartic. Both the cubic and quartic formulae are extremely cumbersome, and no such formula exists for polynomials of degree 5 and higher. This is the famous Abel-Ruffini theorem. 18 See Appendix A.16 in the CLP-2 text. There it is shown that any integer root of a polynomial with integer coefficients must divide the constant term exactly. So in this case only ˘1 and ˘3 could be integer roots. So it is good to check to see if any of these are solutions before moving on to more sophisticated techniques. 19 Note that we are assuming that a, b, f (a, b) is the point on the surface that is nearest (0, 3, 0). That there exists such a point is intuitively obvious from a sketch of the surface. The mathematical proof that there exists such a point is beyond the scope of this text. 133 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES • restricting our attention to the slice y = b of the surface, x = a minimizes g(x) = D(x, b) = x2 + (b ´ 3)2 + f (x, b)2 so that 0 = g1(a) = B Bx h x2 + (b ´ 3)2 + f (x, b)2iˇ ˇ ˇ ˇ x=a = 2a + 2f (a, b) fx(a, b) = 2 ⟨a , b ´ 3 , f (a, b)⟩¨ ⟨1 , 0 , fx(a, b)⟩ and • restricting our attention to the slice x = a of the surface, y = b minimizes h(y) = D(a, y) = a2 + (y ´ 3)2 + f (a, y)2 so that 0 = h1(b) = B By h a2 + (y ´ 3)2 + f (a, y)2iˇ ˇ ˇ ˇ y=b = 2(b ´ 3) + 2f (a, b) fy(a, b) = 2 ⟨a , b ´ 3 , f (a, b)⟩¨ 0 , 1 , fy(a, b) We have expressed the final right hand sides of both of the above bullets as the dot product of the vector ⟨a , b ´ 3 , f (a, b)⟩with something because ˝ ⟨a , b ´ 3 , f (a, b)⟩is the vector from (0, 3, 0) to the point (a , b , f (a, b) on the surface and ˝ the vanishing of the dot product of two vectors implies that the two vectors are perpendicular. Thus, that ⟨a , b ´ 3 , f (a, b)⟩¨ ⟨1 , 0 , fx(a, b)⟩= ⟨a , b ´ 3 , f (a, b)⟩¨ 0 , 1 , fy(a, b) = 0 tells us that the vector ⟨a , b ´ 3 , f (a, b)⟩from (0, 3, 0) to a, b, f (a, b) is perpendicular to both ⟨1 , 0 , fx(a, b)⟩and 0 , 1 , fy(a, b) and hence is parallel to their cross product ⟨1 , 0 , fx(a, b)⟩ˆ 0 , 1 , fy(a, b) , which we already know is a normal vector to the surface z = f (x, y) at a, b, f (a, b) . This shows that the point on the surface that minimises the distance to (0, 3, 0) is joined to (0, 3, 0) by a line that is parallel to the normal vector — just as we required. Example 2.5.3 2.5.2 § § Surfaces of the Form G(x, y, z) = 0. We now use a little trickery to construct a vector perpendicular to the surface G(x, y, z) = 0 at the point x0 , y0 , z0 . Imagine that you are walking on the surface and that at time 0 you are at the point x0 , y0 , z0 . Let r(t) = x(t) , y(t) , z(t) denote your position at time t. Because you are walking along the surface, we know that r(t) always lies on the 134 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES z y x Gpx, y, zq “ 0 rp0q “ x0, y0, z0 ˘ rptq surface and so G x(t) , y(t) , z(t) = 0 for all t. Differentiating this equation with respect to t gives, by the chain rule, BG Bx x(t) , y(t) , z(t) x1(t) + BG By x(t) , y(t) , z(t) y1(t) + BG Bz x(t) , y(t) , z(t) z1(t) = 0 Then setting t = 0 gives BG Bx x0 , y0 , z0 x1(0) + BG By x0 , y0 , z0 y1(0) + BG Bz x0 , y0 , z0 z1(0) = 0 Expressing this as a dot product allows us to turn this into a statement about vectors. BG Bx x0 , y0 , z0 , BG By x0 , y0 , z0 , BG Bz x0 , y0 , z0 ¨ r1(0) = 0 (˚) The first vector in this dot product is sufficiently important that it is given its own name. The gradient20 of the function G(x, y, z) at the point x0 , y0 , z0 is BG Bx x0 , y0 , z0 , BG By x0 , y0 , z0 , BG Bz x0 , y0 , z0 It is denoted ∇ ∇ ∇G(x0, y0, z0). Definition 2.5.4 (Gradient). So (˚) tells us that the gradient ∇ ∇ ∇G(x0, y0, z0), is perpendicular to the vector r1(0). 20 The gradient will also play a big role in Section 2.7. 135 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES Now if t is very close to zero, the vector r(t) ´ r(0), from r(0) to r(t), is almost tangent to the path that we are walking on. The limit r1(0) = lim tÑ0 r(t) ´ r(0) t is thus exactly tangent to our path, and consequently to the surface G(x, y, z) = 0 at (x0, y0, z0). This is true for all paths on the surface that pass through (x0, y0, z0) at time t = 0, which tells us that ∇ ∇ ∇G(x0, y0, z0) is perpendicular to the surface at (x0, y0, z0). We have just found a normal vector! The above argument goes through unchanged for surfaces of the form21 G(x, y, z) = K, for any constant K. So we have Let K be a constant and (x0, y0, z0) be a point on the surface G(x, y, z) = K. As-sume that the gradient ∇ ∇ ∇G(x0, y0, z0) = BG Bx x0 , y0 , z0 , BG By x0 , y0 , z0 , BG Bz x0 , y0 , z0 of G at (x0, y0, z0) is nonzero. (a) The vector ∇ ∇ ∇G(x0, y0, z0) is normal to the surface G(x, y, z) = K at (x0, y0, z0). (b) The equation of the tangent plane to the surface G(x, y, z) = K at (x0, y0, z0) is ∇ ∇ ∇G(x0, y0, z0) ¨ ⟨x ´ x0 , y ´ y0 , z ´ z0⟩= 0 (c) The parametric equation of the normal line to the surface G(x, y, z) = K at (x0, y0, z0) is ⟨x, y, z⟩= ⟨x0, y0, z0⟩+ t∇ ∇ ∇G(x0, y0, z0) Theorem 2.5.5 (Tangent Plane and Normal Line). Remark 2.5.6. Theorem 2.5.1 about the tangent planes and normal lines to the surface z = f (x, y) is actually a very simple consequence of Theorem 2.5.5 about the tangent planes and normal lines to the surface G(x, y, z) = 0. This is just because we can always rewrite the equation z = f (x, y) as z ´ f (x, y) = 0 and apply Theorem 2.5.5 with G(x, y, z) = z ´ f (x, y). Since ∇ ∇ ∇G(x0, y0, z0) = ´fx(x0, y0) ˆ ı ı ı ´ fy(x0, y0) ˆ ȷ ȷ ȷ + ˆ k Theorem 2.5.5 then gives22 Theorem 2.5.1. Here are a couple of routine examples. 21 Alternatively, one could rewrite G = K as G ´ K = 0 and replace G by G ´ K in the above argument. 22 Indeed we could write Theorem 2.5.1 as a corollary of Theorem 2.5.5. But in a textbook one tries to start with the concrete and move to the more general. 136 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES Example 2.5.7 Problem: Find the tangent plane and the normal line to the surface z = x2 + 5xy ´ 2y2 at the point (1, 2, 3). Solution. As a preliminary check, note that 12 + 5 ˆ 1 ˆ 2 ´ 2(2)2 = 3 which verifies that the point (1, 2, 3) is indeed on the surface. This is a good reality check and also increases our confidence that the question is asking what we think that it is ask-ing. Rewrite the equation of the surface as G(x, y, z) = x2 + 5xy ´ 2y2 ´ z = 0. Then the gradient ∇ ∇ ∇G(x, y, z) = (2x + 5y) ˆ ı ı ı + (5x ´ 4y) ˆ ȷ ȷ ȷ ´ ˆ k so that, by Theorem 2.5.5, n = ∇ ∇ ∇G(1, 2, 3) = 12ˆ ı ı ı ´ 3 ˆ ȷ ȷ ȷ ´ ˆ k is a normal vector to the surface at (1, 2, 3). Equipped23 with the normal, it is easy to work out an equation for the tangent plane. n ¨ ⟨x ´ 1 , y ´ 2 , z ´ 3⟩= ⟨12 , ´3 , ´1⟩¨ ⟨x ´ 1 , y ´ 2 , z ´ 3⟩= 0 or 12x ´ 3y ´ z = 3 We can quickly check that the point (1, 2, 3) does indeed lie on the plane: 12 ˆ 1 ´ 3 ˆ 2 ´ 3 = 3 The normal line is ⟨x ´ 1 , y ´ 2 , z ´ 3⟩= t n = t ⟨12 , ´3 , ´1⟩ or x ´ 1 12 = y ´ 2 ´3 = z ´ 3 ´1 = t Example 2.5.7 Another warm-up example. This time the surface is a hyperboloid of one sheet. Example 2.5.8 Problem: Find the tangent plane and the normal line to the surface x2 + y2 ´ z2 = 4 at the point (2, ´3, 3). Solution. As a preliminary check, note that the point (2, ´3, 3) is indeed on the surface: 22 + (´3)2 ´ (3)2 = 4 23 The spelling “equipt” is a bit archaic. There must be a joke here about quips. 137 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES The equation of the surface is G(x, y, z) = x2 + y2 ´ z2 = 4. Then the gradient of G is ∇ ∇ ∇G(x, y, z) = 2x ˆ ı ı ı + 2y ˆ ȷ ȷ ȷ ´ 2z ˆ k so that, at (2, ´3, 3), ∇ ∇ ∇G(2, ´3, 3) = 4ˆ ı ı ı ´ 6 ˆ ȷ ȷ ȷ ´ 6 ˆ k and so, by Theorem 2.5.5, n = 1 2 4ˆ ı ı ı ´ 6 ˆ ȷ ȷ ȷ ´ 6 ˆ k = 2ˆ ı ı ı ´ 3 ˆ ȷ ȷ ȷ ´ 3 ˆ k is a normal vector to the surface at (2, ´3, 3). The tangent plane is n ¨ ⟨x ´ 2 , y + 3 , z ´ 3⟩= ⟨2 , ´3 , ´3⟩¨ ⟨x ´ 2 , y + 3 , z ´ 3⟩= 0 or 2x ´ 3y ´ 3z = 4 Again, as a check, we can verify that our point (2, ´3, 3) is indeed on the plane: 2 ˆ 2 ´ 3 ˆ (´3) ´ 3 ˆ 3 = 4 The normal line is ⟨x ´ 2 , y + 3 , z ´ 3⟩= t n = t ⟨2 , ´3 , ´3⟩ or x ´ 2 2 = y + 3 ´3 = z ´ 3 ´3 = t Example 2.5.8 138 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES The vector ∇ ∇ ∇G(x, y, z) is not a normal vector to the surface G(x, y, z)=K at (x0, y0, z0). The vector ∇ ∇ ∇G(x0, y0, z0) is a normal vector to G(x, y, z) = K at (x0, y0, z0) (provided G(x0, y0, z0) = K). As an example of the consequences of failing to evaluate ∇ ∇ ∇G(x, y, z) at the point (x0, y0, z0), consider the problem Find the tangent plane to the surface x2 + y2 + z2 = 1 at the point (0, 0, 1). In this case, the surface is G(x, y, z) = x2 + y2 + z2 = 1. The gradient of G is ∇ ∇ ∇G(x, y, z) = 2x ˆ ı ı ı + 2y ˆ ȷ ȷ ȷ + 2z ˆ k. To correctly apply part (b) of Theorem 2.5.5, we evaluate ∇ ∇ ∇G(0, 0, 1) = 2 ˆ k and find that the tangent plane at (0, 0, 1) is ∇ ∇ ∇G(0, 0, 1) ¨ ⟨x ´ 0 , y ´ 0 , z ´ 1⟩= 0 or 2(z ´ 1) = 0 or z = 1 This is of course correct — the tangent plane to the unit sphere at the north pole is indeed horizontal. But if we were to incorrectly apply part (b) of Theorem 2.5.5 by failing to evaluate ∇ ∇ ∇G(x, y, z) at (0, 0, 1), we would find that the “tangent plane” is ∇ ∇ ∇G(x, y, z) ¨ ⟨x ´ 0 , y ´ 0 , z ´ 1⟩= 0 or 2x(x ´ 0) + 2y(y ´ 0) + 2z(z ´ 1) = 0 or x2 + y2 + z2 ´ z = 0 This is horribly wrong. It is not even a plane, as any plane has an equation of the form ax + by + cz = d, with a, b, c and d constants. Warning 2.5.9. Now we’ll move on to some more involved examples. Example 2.5.10 Suppose that we wish to find the highest and lowest points on the surface G(x, y, z) = x2 ´ 2x + y2 ´ 4y + z2 ´ 6z = 2. That is, we wish to find the points on the surface with the maximum value of z and with the minimum24 value of z. Completing three squares, G(x, y, z) = x2 ´ 2x + y2 ´ 4y + z2 ´ 6z = (x ´ 1)2 + (y ´ 2)2 + (z ´ 3)2 ´ 14. So the surface G(x, y, z) = 2 is a sphere, whose highest point is the north pole and whose lowest point is the south pole. But let’s pretend that G(x, y, z) = 2 is some complicated surface that we can’t easily picture. We’ll find its highest and lowest points by exploiting the fact that the tangent plane to G = 2 is horizontal at the highest and lowest points. Equivalently, the normal vector to G = 2 is vertical at the highest and lowest points. To see that this is the case, look 24 Recall that “minimum” means the most negative, not the closest to zero. 139 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES at the figure below. If the tangent plane at (x0, y0, z0) is not horizontal, then the tangent plane contains points near (x0, y0, z0) with z bigger than z0 and points near (x0, y0, z0) with z smaller than z0. Near (x0, y0, z0), the tangent plane is a good approximation to the surface. So the surface also contains25 such points. y z x2 ´ 2x y2 ´ 4y z2 ´ 6z “ 2 px0, y0, z0q side view The gradient is ∇ ∇ ∇G(x, y, z) = (2x ´ 2) ˆ ı ı ı + (2y ´ 4) ˆ ȷ ȷ ȷ + (2z ´ 6) ˆ k It is vertical when the ˆ ı ı ı and ˆ ȷ ȷ ȷ components are both zero. This happens when 2x ´ 2 = 0 and 2y ´ 4 = 0, i.e. when x = 1 and y = 2. So the normal vector to the surface G = 2 at the point (x, y, z) is vertical when x = 1, y = 2 and (don’t forget that (x, y, z) has to be on G = 2) G(1, 2, z) = 12 ´ 2 ˆ 1 + 22 ´ 4 ˆ 2 + z2 ´ 6z = 2 ð ñ z2 ´ 6z ´ 7 = 0 ð ñ (z ´ 7)(z + 1) = 0 ð ñ z = 7, ´1 The highest point is (1, 2, 7) and the lowest point is (1, 2, ´1), as expected. Example 2.5.10 We could have short-cut the last example by using that the surface was a sphere. Here is an example in the same spirit for which we don’t have an easy short-cut. Example 2.5.11 In the last example, we found the points on a specified surface having the largest and smallest values of z. We’ll now ramp up the level of difficulty a bit and find the points on the surface x2 + 2y2 + 3z2 = 72 that have the largest and smallest values of x + y + 3z. To develop a strategy for tackling this problem, consider the following sketch. The 25 While this is intuitively obvious, proving it is beyond the scope of this text. 140 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES x y 3z “ C x2 2y2 3z2 “ 72 red ellipse in the sketch is intended to represent (schematically) our surface x2 + 2y2 + 3z2 = 72 which is an ellipsoid. The middle diagonal (black) line is intended to represent (schemat-ically) the plane x + y + 3z = C for some more or less randomly chosen value of the constant C. At each point on that plane, the function, x + y + 3z, (that we are trying to maximize and minimize) takes the value C. In particular, for the C chosen in the figure, x + y + 3z = C does intersect our surface, indicating that x + y + 3z does indeed take the value C somewhere on our surface. To maximize x + y + 3z, imagine slowly increasing the value of C. As we do so, the plane x + y + 3z = C moves to the right. We want to stop increasing C at the biggest value of C for which the plane x + y + 3z = C intersects our surface x2 + 2y2 + 3z2 = 72. For that value of C the plane x + y + 3z = C, which is represented by the right hand blue line in the sketch, is tangent to our surface. Similarly, to minimize x + y + 3z, imagine slowly decreasing the value of C. As we do so, the plane x + y + 3z = C moves to the left. We want to stop decreasing C at the smallest value of C for which the plane x + y + 3z = C intersects our surface x2 + 2y2 + 3z2 = 72. For that value of C the plane x + y + 3z = C, which is represented by the left hand blue line in the sketch, is again tangent to our surface. The previous Example 2.5.10 was similar, except that the plane was z = C. We are now ready to compute. We need to find the points (a, b, c) (in the sketch, they are the black dot points of tangency) for which • (a, b, c) is on the surface and • the normal vector to the surface x2 + 2y2 + 3z2 = 72 at (a, b, c) is parallel to ⟨1, 1, 3⟩, which is a normal vector to the plane x + y + 3z = C Since the gradient of x2 + 2y2 + 3z2 is ⟨2x , 4y , 6z⟩= 2 ⟨x , 2y , 3z⟩, these two conditions are, in equations, a2 + 2b2 + 3c2 = 72 ⟨a , 2b , 3c⟩= t ⟨1, 1, 3⟩ for some number t The second equation says that a = t, b = t 2 and c = t. Substituting this into the first equation gives t2 + 1 2t2 + 3t2 = 72 ð ñ 9 2t2 = 72 ð ñ t2 = 16 ð ñ t = ˘4 141 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES So • the point on the surface x2 + 2y2 + 3z2 = 72 at which x + y + 3z takes its maximum value is (a, b, c) = t, t 2, t ˇ ˇ ˇ t=4 = (4, 2, 4) and • x + y + 3z takes the value 4 + 2 + 3 ˆ 4 = 18 there. • The point on the surface x2 + 2y2 + 3z2 = 72 at which x + y + 3z takes its minimum value is (a, b, c) = t, t 2, t ˇ ˇ ˇ t=´4 = (´4, ´2, ´4) and • x + y + 3z takes the value ´4 ´ 2 + 3 ˆ (´4) = ´18 there. Example 2.5.11 Example 2.5.12 Problem: Find the distance from the point (1, 1, 1) to the plane x + 2y + 3z = 20. Solution 1. First note that the point (1, 1, 1) is not itself on the plane x + 2y + 3z = 20 because 1 + 2 ˆ 1 + 3 ˆ 1 = 6 ‰ 20 Denote by (a, b, c) the point on the plane x + 2y + 3z = 20 that is nearest (1, 1, 1). Then the vector from (1, 1, 1) to (a, b, c), namely ⟨a ´ 1 , b ´ 1 , c ´ 1⟩, must be perpendicular26 to the plane. As the gradient of x + 2y + 3z, namely ⟨1 , 2 , 3⟩, is a normal vector to the plane, ⟨a ´ 1 , b ´ 1 , c ´ 1⟩must be parallel to ⟨1 , 2 , 3⟩. So there must be some number t so that ⟨a ´ 1 , b ´ 1 , c ´ 1⟩= t ⟨1 , 2 , 3⟩ or a = t + 1, b = 2t + 1, c = 3t + 1 As (a, b, c) must be on the plane, we know that a + 2b + 3c = 20 and so (t + 1) + 2(2t + 1) + 3(3t + 1) = 20 ù ñ 14t = 14 ù ñ t = 1 The distance from (1, 1, 1) to the plane x + 2y + 3z = 20 is the length of the vector ⟨a ´ 1 , b ´ 1 , c ´ 1⟩= t ⟨1 , 2 , 3⟩= ⟨1 , 2 , 3⟩which is ? 14. x 2y 3z “ 20 P “ pa, b, cq Q “ p1, 1, 1q R “ p20, 0, 0q ⟨1, 2, 3⟩ θ 26 We saw why this vector must be perpendicular to the plane in Example 2.5.3. 142 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES Solution 2. Denote by P = (a, b, c) the point on the plane x + 2y + 3z = 20 that is nearest the point Q = (1, 1, 1). Pick any other point on the plane and call it R. For example (x, y, z) = (20, 0, 0) obeys x + 2y + 3z = 20 and so R = (20, 0, 0) is a point on the plane. The triangle PQR is right angled. Denote by θ the angle between the hypotenuse QR and the side QP. The distance from Q = (1, 1, 1) to the plane is the length of the line segment QP, which is distance = \|QP\| = \|QR\| cos θ Now, the dot product between the vector from Q to R, which is ⟨19, ´1, ´1⟩, with the vector ⟨1, 2, 3⟩, which is normal to the plane and hence parallel to the side QP is ⟨19, ´1, ´1⟩¨ ⟨1, 2, 3⟩= 14 = \| ⟨19, ´1, ´1⟩\| \| ⟨1, 2, 3⟩\| cos θ = \|QR\| ? 14 cos θ so that, finally, distance = \|QR\| cos θ = 14 ? 14 = ? 14 Example 2.5.12 Example 2.5.13 Problem: Let F(x, y, z) = 0 and G(x, y, z) = 0 be two surfaces. These two surfaces intersect along a curve. Find a tangent vector to this curve at the point (x0, y0, z0). Solution. Call the tangent vector T. Then T has to be ˝ tangent to the surface F(x, y, z) = 0 at (x0, y0, z0) and ˝ tangent to the surface G(x, y, z) = 0 at (x0, y0, z0). Consequently T has to be ˝ perpendicular to the vector ∇ ∇ ∇F(x0, y0, z0), which is normal to F(x, y, z) = 0 at (x0, y0, z0), and at the same time has to be ˝ perpendicular to the vector ∇ ∇ ∇G(x0, y0, z0), which is normal to G(x, y, z) = 0 at (x0, y0, z0). Recall that an easy way to construct a vector that is perpendicular to two other vectors is to take their cross product. So we take T = ∇ ∇ ∇F(x0, y0, z0) ˆ ∇ ∇ ∇G(x0, y0, z0) = det   ˆ ı ı ı ˆ ȷ ȷ ȷ ˆ k Fx Fy Fz Gx Gy Gz   = FyGz ´ FzGy ˆ ı ı ı + FzGx ´ FxGz ˆ ȷ ȷ ȷ + FxGy ´ FyGx ˆ k where all partial derivatives are evaluated at (x, y, z) = (x0, y0, z0). 143 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES Example 2.5.13 Let’s put Example 2.5.13 into action. Example 2.5.14 Problem: Consider the curve that is the intersection of the surfaces x2 + y2 + z2 = 5 and x2 + y2 = 4z Find a tangent vector to this curve at the point ? 3 , 1 , 1 . Solution. As a preliminary check, we verify that the point ? 3 , 1 , 1 really is on the curve. To do so, we check that ? 3 , 1 , 1 satisfies both equations: ? 3 2 + 12 + 12 = 5 ? 3 2 + 12 = 4 ˆ 1 We’ll find the specified tangent vector by using the strategy of Example 2.5.13. Write F(x, y, z) = x2 + y2 + z2 and G(x, y, z) = x2 + y2 ´ 4z. Then ˝ the vector ∇ ∇ ∇F( ? 3, 1, 1) = ⟨2x , 2y , 2z⟩ ˇ ˇ ˇ (x,y,z)=( ? 3,1,1) = 2 D? 3 , 1 , 1 E is normal to the surface F(x, y, z) = 5 at ? 3 , 1 , 1 , and ˝ the vector ∇ ∇ ∇G( ? 3, 1, 1) = ⟨2x , 2y , ´4⟩ ˇ ˇ ˇ (x,y,z)=( ? 3,1,1) = 2 D? 3 , 1 , ´2 E is normal to the surface G(x, y, z) = 0 at ? 3 , 1 , 1 . So a tangent vector is D? 3 , 1 , 1 E ˆ D? 3 , 1 , ´2 E = det   ˆ ı ı ı ˆ ȷ ȷ ȷ ˆ k ? 3 1 1 ? 3 1 ´2   = ´ 2 ´ 1 ˆ ı ı ı + ? 3 + 2 ? 3 ˆ ȷ ȷ ȷ + ? 3 ´ ? 3 ˆ k = ´3ˆ ı ı ı + 3 ? 3 ˆ ȷ ȷ ȷ There is an easy common factor of 3 in both components. So we can create a slightly neater tangent vector by dividing the length of ´3ˆ ı ı ı + 3 ? 3 ˆ ȷ ȷ ȷ by 3, giving ´1 , ? 3 , 0 . Example 2.5.14 Example 2.5.15 ((Optional) computer graphics hidden-surface elimination ) When you look at a solid three dimensional object, you do not see all of the surface of the 144 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES object — parts of the surface are hidden from your view by other parts of the object. For example, the following sketch shows, schematically, a ray of light leaving your eye and hitting the surface of the object at the light dot. The object is solid, so the light cannot penetrate any further. But, if it could, it would follow the dotted line, hitting the surface of the object three more times. Your eye can see the light dot, but cannot see the other three dark dots. Recreating this effect in computer generated graphics is called “hidden-surface elimi-nation”. In general, implementing hidden-surface elimination can be quite complicated. Often a technique called “ray tracing” is used27. However, it is easy if you know about vectors and gradients, and you are only looking at a single convex body. By definition, a solid is convex if, whenever two points are in the solid, then the line segment joining the two points is also contained in the solid. not convex convex So suppose that we are looking at a convex solid, that the equation of the surface of the solid is G(x, y, z) = 0, and that our eye is at (xe, ye, ze). • First consider a light ray that leaves our eye and then just barely nicks the solid at the point (x, y, z), as in the figure on the left below. The light ray is a tangent line to the surface at (x, y, z). So the direction vector of the light ray, ⟨x ´ xe, y ´ ye, z ´ ze⟩, is tangent to the surface at (x, y, z) and consequently is perpendicular to the normal vector, n = ∇ ∇ ∇G(x, y, z), of the surface at (x, y, z). Thus ⟨x ´ xe, y ´ ye, z ´ ze⟩¨ ∇ ∇ ∇G(x, y, z) = 0 27 You can find out more about it by plugging “ray tracing” into the search engine of your choice. 145 PARTIAL DERIVATIVES 2.5 TANGENT PLANES AND NORMAL LINES Gpx,y,zq“0 pxe,ye,zeq ⟨x´xe,y´ye,z´ze⟩ ∇ ∇ ∇Gpx,y,zq px,y,zq Gpx,y,zq“0 v n px,y,zq pxe,ye,zeq px1,y1,z1q v n1 • Now consider a light ray that leaves our eye and then passes through the solid, as in the figure on the right above. Call the point at which the light ray first enters the solid (x, y, z) and the point at which the light ray leaves the solid (x1, y,1 z1). – Let v be a vector that has the same direction as, i.e. is a positive multiple of, the vector ⟨x ´ xe, y ´ ye, z ´ ze⟩. – Let n be an outward pointing normal to the solid at (x, y, z). It will be either ∇ ∇ ∇G(x, y, z) or ´∇ ∇ ∇G(x, y, z). – Let n1 be an outward pointing normal to the solid at (x1, y1, z1). It will be either ∇ ∇ ∇G(x1, y1, z1) or ´∇ ∇ ∇G(x1, y1, z1). Then – at the point (x, y, z) where the ray enters the solid, which is a visible point, the direction vector v points into the solid. The angle θ between v and the outward pointing normal n is greater than 90˝, so that the dot product v ¨ n = \|v\| \|n\| cos θ ă 0. But – at the point (x1, y1, z1) where the ray leaves the solid, which is a hidden point, the direction vector v points out of the solid. The angle θ between v and the outward pointing normal n1 is less than 90˝, so that the dot product v ¨ n1 = \|v\| \|n1\| cos θ ą 0. Our conclusion is that, if we are looking in the direction v, and if the outward pointing normal28 to the surface of the solid at (x, y, z) is ∇ ∇ ∇G(x, y, z) then the point (x, y, z) is hid-den if and only if v ¨ ∇ ∇ ∇G(x, y, z) ą 0. This method was used by the computer graphics program that created the shaded figures29 in Examples 1.7.1 and 1.7.2, which are reproduced here. 28 If ∇ ∇ ∇G(x, y, z) is the inward pointing normal, just replace G by ´G. 29 Those figures are not convex. But it was still possible to use the method discussed above because any light ray from our eye that passes through the figure intersects the figure at most twice. It first enters the figure at a visible point and then exits the figure at a hidden point. 146 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR Example 2.5.15 Tangent planes, in addition to being geometric objects, provide a simple but powerful tool for approximating functions of two variables near a specified point. We saw some-thing very similar in the CLP-1 text where we approximated functions of one variable by their tangent lines. This brings us to our next topic — approximating functions. 2.6Ĳ Linear Approximations and Error A frequently used, and effective, strategy for building an understanding of the behaviour of a complicated function near a point is to approximate it by a simple function. The following suite of such approximations is standard fare in Calculus I courses. See, for example, §3.4 in the CLP-1 text. g(t0 + ∆t) « g(t0) constant approximation g(t0 + ∆t) « g(t0) + g1(t0) ∆t linear, or tangent line, approximation g(t0 + ∆t) « g(t0) + g1(t0) ∆t + 1 2g2(t0) ∆t2 quadratic approximation More generally, for any natural number n, the approximation g(t0 + ∆t) « g(t0) + g1(t0) ∆t + 1 2g2(t0) ∆t2 + ¨ ¨ ¨ + 1 n!g(n)(t0) ∆tn is known as the Taylor polynomial of order n. You may have also found a formula for the error introduced in making this approximation. The error En(∆t) is defined by g(t0 + ∆t) = g(t0) + g1(t0)∆t + 1 2!g2(t0)∆t2 + ¨ ¨ ¨ + 1 n!g(n)(t0)∆tn + En(∆t) and obeys30 En(∆t) = 1 (n+1)!g(n+1)(t0 + c∆t)∆tn+1 for some (unknown) 0 ď c ď 1. It is a simple matter to use these one dimensional approximations to generate the anal-ogous multidimensional approximations. To introduce the ideas, we’ll generate the linear approximation to a function, f (x, y), of two variables, near the point (x0, y0). Define g(t) = f x0 + t ∆x , y0 + t ∆y 30 You may have seen it written as En(x) = 1 (n+1)! g(n+1)(c)(x ´ a)n+1 147 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR We have defined g(t) so that g(0) = f x0 , y0 and g(1) = f x0 + ∆x , y0 + ∆y Consequently, setting t0 = 0 and ∆t = 1, f x0 + ∆x , y0 + ∆y = g(1) = g(t0 + ∆t) « g(t0) + g1(t0) ∆t = g(0) + g1(0) We can now compute g1(0) using the multivariable chain rule of (2.4.2): g1(t) = B f Bx x0 + t ∆x , y0 + t ∆y ∆x + B f By x0 + t ∆x , y0 + t ∆y ∆y so that, f x0 + ∆x , y0 + ∆y « f x0 , y0 + B f Bx x0 , y0 ∆x + B f By x0 , y0 ∆y Equation 2.6.1. Of course exactly the same procedure works for functions of three or more variables. In particular f x0 + ∆x , y0 + ∆y , z0 + ∆z « f x0 , y0 , z0 + B f Bx x0 , y0 , z0 ∆x + B f By x0 , y0 , z0 ∆y + B f Bz x0 , y0 , z0 ∆z Equation 2.6.2. While these linear approximations are quite simple, they tend to be pretty decent provided ∆x and ∆y are small. See the optional §2.6.1 for a more precise statement. Remark 2.6.3. Applying (2.6.1), with ∆x = x ´ x0 and ∆y = y ´ y0. gives f x , y « f x0 , y0 + B f Bx x0 , y0 (x ´ x0) + B f By x0 , y0 (y ´ y0) Looking at part (b) of Theorem 2.5.1, we see that this just says that the tangent plane to the surface z = f (x, y) at the point x0 , y0 , f (x0, y0) remains close to the surface when (x, y) is close to (x0, y0). Example 2.6.4 Let f (x, y) = b x2 + y2 148 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR Then B f Bx(x, y) = 1 2 2x a x2 + y2 fx(x0, y0) = x0 b x2 0 + y2 0 B f By(x, y) = 1 2 2y a x2 + y2 fy(x0, y0) = y0 b x2 0 + y2 0 so that the linear approximation to f (x, y) at (x0, y0) is f x0 + ∆x , y0 + ∆y « f x0 , y0 + fx x0 , y0 ∆x + fy x0 , y0 ∆y = b x2 0 + y2 0 + x0 b x2 0 + y2 0 ∆x + y0 b x2 0 + y2 0 ∆y Example 2.6.4 We now formalise the above terminology and notation, and add some new terminol-ogy and notation (for functions of two variables — the variants for functions of three or more variables are obvious). (a) The linear approximation to the function f (x, y) at the point (x0, y0) is f (x0, y0) + B f Bx x0 , y0 (x ´ x0) + B f By x0 , y0 (y ´ y0) People often write ∆f for the change, f x0 + ∆x , y0 + ∆y ´ f x0 , y0 , in the value of f, ∆x = x ´ x0 for the change in the value of x, and ∆y = y ´ y0 for the change in the value of y. In this notation, the linear approximation gives ∆f « B f Bx x0 , y0 ∆x + B f By x0 , y0 ∆y (b) If they want to emphasize that ∆x and ∆y are really small (they may even say31 “infinitesimal”), they’ll write dx, dy instead and define df = B f Bx x0 , y0 dx + B f By x0 , y0 dy That is, df is the change of the value of the linear approximation to f when x is changed by dx and y is changed by dy. It approximates the corresponding change f x0 + dx , y0 + dy ´ f x0 , y0 in the value of f. People sometimes call dx, dy and df “differentials” and sometimes df is called the “total dif-ferential of f” to indicate that it includes the impact of small changes in both x and y. Definition 2.6.5. 149 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR Suppose that we wish to approximate a quantity Q and that the approximation turns out to be Q + ∆Q. Then • the absolute error in the approximation is \|∆Q\| and • the relative error in the approximation is ˇ ˇ ˇ∆Q Q ˇ ˇ ˇ and • the percentage error in the approximation is 100 ˇ ˇ ˇ∆Q Q ˇ ˇ ˇ Definition 2.6.6. In Example 3.4.5 of the CLP-1 text we found an approximate value for the number ? 4.1 by using a linear approximation to the single variable function f (x) = ?x. We can make similar use of linear approximations to multivariable functions. Example 2.6.7 Problem: Find an approximate value for (0.998)3 1.003 . Solution: Set f (x, y) = x3 y . We are to find (approximately) f (0.998 , 1.003). We can easily find f (1, 1) = 13 1 = 1 and since B f Bx = 3x2 y and B f By = ´x3 y2 we can also easily find B f Bx(1, 1) = 312 1 = 3 B f By(1, 1) = ´113 12 = ´1 So, setting ∆x = ´0.002 and ∆y = 0.003, we have 0.9983 1.003 = f (0.998 , 1.003) = f (1 + ∆x , 1 + ∆y) « f 1, 1 + B f Bx 1, 1 ∆x + B f By 1, 1 ∆y « 1 + 3(´0.002) ´ 1(0.003) = 0.991 By way of comparison, the exact answer is 0.9910389 to seven decimal places. Example 2.6.7 31 Don’t take the terminology “infinitesimal” too seriously. It is just intended to signal “very small”. 150 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR Example 2.6.8 Problem: Find an approximate value for (4.2)1/2 + (26.7)1/3 + (256.4)1/4. Solution: Set f (x, y, z) = x1/2 + y1/3 + z1/4. We are to find (approximately) f (4.2 , 26.7 , 256.4). We can easily find f (4, 27, 256) = (4)1/2 + (27)1/3 + (256)1/4 = 2 + 3 + 4 = 9 and since B f Bx = 1 2x1/2 B f By = 1 3y2/3 B f Bz = 1 4z3/4 we can also easily find B f Bx(4, 27, 256) = 1 2(4)1/2 = 1 2 ˆ 1 2 B f By(4, 27, 256) = 1 3(27)2/3 = 1 3 ˆ 1 9 B f Bz (4, 27, 256) = 1 4(256)3/4 = 1 4 ˆ 1 64 So, setting ∆x = 0.2, ∆y = ´0.3, and ∆z = 0.4, we have (4.2)1/2 + (26.7)1/3 + (256.4)1/4 = f (4.2 , 26.7 , 256.4) = f (4 + ∆x , 27 + ∆y , 256 + ∆z) « f 4, 27, 256 + B f Bx 4, 27, 256 ∆x + B f By 4, 27, 256 ∆y + B f Bz 4, 27, 256 ∆z « 9 + 0.2 2 ˆ 2 ´ 0.3 3 ˆ 9 + 0.4 4 ˆ 64 = 9 + 1 20 ´ 1 90 + 1 640 = 9.0405 to four decimal places. The exact answer is 9.03980 to five decimal places. That’s a differ-ence of about 1009.0405 ´ 9.0398 9 % = 0.008% Note that we could have used the single variable approximation techniques in the CLP-1 text to separately approximate (4.2)1/2, (26.7)1/3 and (256.4)1/4 and then added the results together. Indeed what we have done here is equivalent. Example 2.6.8 Example 2.6.9 Problem: A triangle has sides a = 10.1cm and b = 19.8cm which include an angle 35˝. Ap-proximate the area of the triangle. θ b a 151 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR Solution: The triangle has height h = a sin θ and hence has area A(a, b, θ) = 1 2bh = 1 2ab sin θ The sin θ in this formula hides a booby trap built into this problem. In preparing the linear approximation we will need to use the derivative of sin θ. But the standard derivative d dθ sin θ = cos θ only applies when θ is expressed in radians — not in degrees. See Warning 3.4.23 in the CLP-1 text. So we are obliged to convert 35˝ into 35˝ = (30 + 5) π 180 radians = π 6 + π 36 radians We need to compute (approximately) A(10.1 , 19.8 , π/6 + π/36 . We will, of course32, choose a0 = 10 b0 = 20 θ0 = π/6 ∆a = 0.1 ∆b = ´0.2 ∆θ = π/36 By way of preparation, we evaluate A a0, b0, θ0 = 1 2a0b0 sin θ0 = 1 2(10)(20)1 2 = 50 BA Ba a0, b0, θ0 = 1 2b0 sin θ0 = 1 2(20)1 2 = 5 BA Bb a0, b0, θ0 = 1 2a0 sin θ0 = 1 2(10)1 2 = 5 2 BA Bθ a0, b0, θ0 = 1 2a0b0 cos θ0 = 1 2(10)(20) ? 3 2 = 50 ? 3 So the linear approximation gives Area = A(10.1 , 19.8 , π/6 + π/36 = A(a0 + ∆a , b0 + ∆b , θ0 + ∆θ « A a0, b0, θ0 + BA Ba a0, b0, θ0 ∆a + BA Bb a0, b0, θ0 ∆b + BA Bθ a0, b0, θ0 ∆θ = 50 + 5 ˆ 0.1 + 5 2 ˆ (´0.2) + 50 ? 3 π 36 = 50 + 5 10 ´ 5 10 + 50 ? 3 π 36 = 50 1 + ? 3 π 36 « 57.56 to two decimal places. The exact answer is 57.35 to two decimal places. Our approxima-tion has an error of about 100 57.56 ´ 57.35 57.35 % = 0.37% 32 There are other choices possible. For example, we could write 35˝ = 45˝ ´ 10˝. To get a good approxi-mation we try to make ∆θ as small as possible, while keeping the arithmetic reasonably simple. 152 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR Example 2.6.9 Another practical use of these linear approximations is to quantify how errors made in measured quantities propagate in computations using those measured quantities. Let’s explore this idea a little by recycling the last example. Example 2.6.10 (Example 2.6.9, continued) Suppose, that, as in Example 2.6.9, we are attempting to determine the area of a triangle by measuring the lengths of two of its sides together with the angle between them and then using the formula A(a, b, θ) = 1 2ab sin θ Of course, in the real world33, we cannot measure lengths and angles exactly. So if we need to know the area to within 1%, the question becomes: “How accurately do we have to measure the side lengths and included angle if we want the area that we compute to have an error of no more than about 1%?” Let’s call the exact side lengths and included angle a0, b0 and θ0, respectively, and the measured side lengths and included angle a0 + ∆a, b0 + ∆b and θ0 + ∆θ. So ∆a, ∆b and ∆θ represent the errors in our measurements. Then, by (2.6.2), the error in our computed area will be approximately ∆A « BA Ba a0, b0, θ0 ∆a + BA Bb a0, b0, θ0 ∆b + BA Bθ a0, b0, θ0 ∆θ = ∆a 2 b0 sin θ0 + ∆b 2 a0 sin θ0 + ∆θ 2 a0b0 cos θ0 and the percentage error in our computed area will be 100 \|∆A\| A(a0, b0, θ0) « ˇ ˇ ˇ ˇ100∆a a0 + 100∆b b0 + 100∆θcos θ0 sin θ0 ˇ ˇ ˇ ˇ By the triangle inequality, \|u + v\| ď \|u\| + \|v\|, and the fact that \|uv\| = \|u\| \|v\|, ˇ ˇ ˇ ˇ100∆a a0 + 100∆b b0 + 100∆θcos θ0 sin θ0 ˇ ˇ ˇ ˇ ď 100 ˇ ˇ ˇ ˇ ∆a a0 ˇ ˇ ˇ ˇ + 100 ˇ ˇ ˇ ˇ ∆b b0 ˇ ˇ ˇ ˇ + 100\|∆θ\| ˇ ˇ ˇ ˇ cos θ0 sin θ0 ˇ ˇ ˇ ˇ We want this to be less than 1. Of course we do not know exactly what a0, b0 and θ0 are. But suppose that we are confident that a0 ě 10, b0 ě 10 and π 6 ď θ0 ď π 2 so that cot θ0 ď cot π 6 = ? 3 ď 2. Then 100 ˇ ˇ ˇ ˇ ∆a a0 ˇ ˇ ˇ ˇ ď 100 ˇ ˇ ˇ ˇ ∆a 10 ˇ ˇ ˇ ˇ = 10 \|∆a\| 100 ˇ ˇ ˇ ˇ ∆b b0 ˇ ˇ ˇ ˇ ď 100 ˇ ˇ ˇ ˇ ∆b 10 ˇ ˇ ˇ ˇ = 10 \|∆b\| 100\|∆θ\| ˇ ˇ ˇ ˇ cos θ0 sin θ0 ˇ ˇ ˇ ˇ ď 100\|∆θ\| 2 = 200 \|∆θ\| 33 Of course in our “real world” everyone uses calculus. 153 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR and 100 \|∆A\| A(a0, b0, θ0) À 10 \|∆a\| + 10 \|∆b\| + 200 \|∆θ\| So it will suffice to have measurement errors \|∆a\|, \|∆b\| and \|∆θ\| obey 10 \|∆a\| + 10 \|∆b\| + 200 \|∆θ\| ă 1 Example 2.6.10 Example 2.6.11 A Question: Suppose that three variables are measured with percentage error ε1, ε2 and ε3 respec-tively. In other words, if the exact value of variable number i is xi and measured value of variable number i is xi + ∆xi then 100 ˇ ˇ ˇ ˇ ∆xi xi ˇ ˇ ˇ ˇ = εi Suppose further that a quantity P is then computed by taking the product of the three variables. So the exact value of P is P(x1, x2, x3) = x1x2x3 and the measured value is P(x1 + ∆x1 , x2 + ∆x2 , x3 + ∆x3). What is the percentage error in this measured value of P? The Answer: The percentage error in the measured value P(x1 + ∆x1 , x2 + ∆x2 , x3 + ∆x3) is 100 ˇ ˇ ˇ ˇ P(x1 + ∆x1 , x2 + ∆x2 , x3 + ∆x3) ´ P(x1, x2, x3) P(x1, x2, x3) ˇ ˇ ˇ ˇ We can get a much simpler approximate expression for this percentage error, which is good enough for virtually all applications, by applying P(x1 + ∆x1 , x2 + ∆x2 , x3 + ∆x3) « P(x1, x2, x3) + Px1(x1, x2, x3) ∆x1 + Px2(x1, x2, x3) ∆x2 + Px3(x1, x2, x3) ∆x3 The three partial derivatives are Px1(x1, x2, x3) = B Bx1 x1x2x3 = x2x3 Px2(x1, x2, x3) = B Bx2 x1x2x3 = x1x3 Px3(x1, x2, x3) = B Bx3 x1x2x3 = x1x2 154 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR So P(x1 + ∆x1 , x2 + ∆x2 , x3 + ∆x3) « P(x1, x2, x3) + x2x3 ∆x1 + x1x3 ∆x2 + x1x2 ∆x3 and the (approximate) percentage error in P is 100 ˇ ˇ ˇ ˇ P(x1 + ∆x1, x2 + ∆x2, x3 + ∆x3) ´ P(x1, x2, x3) P(x1, x2, x3) ˇ ˇ ˇ ˇ « 100 ˇ ˇ ˇ ˇ x2x3∆x1 + x1x3∆x2 + x1x2∆x3 P(x1, x2, x3) ˇ ˇ ˇ ˇ = 100 ˇ ˇ ˇ ˇ x2x3∆x1 + x1x3∆x2 + x1x2∆x3 x1x2x3 ˇ ˇ ˇ ˇ = ˇ ˇ ˇ ˇ100∆x1 x1 + 100∆x2 x2 + 100∆x3 x3 ˇ ˇ ˇ ˇ ď ε1 + ε2 + ε3 More generally, if we take a product of n, rather than three, variables the percentage error in the product becomes at most (approximately) n ř i=1 εi. This is the basis of the experimen-talist’s rule of thumb that when you take products, percentage errors add. Still more generally, if we take a “product” śn i=1 xmi i , the percentage error in the “prod-uct” becomes at most (approximately) n ř i=1 \|mi\|εi. Example 2.6.11 2.6.1 § § Quadratic Approximation and Error Bounds Recall that, in the CLP-1 text, we started with the constant approximation, then improved it to the linear approximation by adding in degree one terms, then improved that to the quadratic approximation by adding in degree two terms, and so on. We can do the same thing here. Once again, set g(t) = f x0 + t ∆x , y0 + t ∆y and recall that g(0) = f x0 , y0 and g(1) = f x0 + ∆x , y0 + ∆y We’ll now see what the quadratic approximation g(t0 + ∆t) « g(t0) + g1(t0) ∆t + 1 2g2(t0) ∆t2 and the corresponding exact formula (see (3.4.32) in the CLP-1 text) g(t0 + ∆t) = g(t0) + g1(t0) ∆t + 1 2g2(t0 + c∆t) ∆t2 for some 0 ď c ď 1 155 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR tells us about f. We have already found, using the chain rule, that g1(t) = B f Bx x0 + t ∆x , y0 + t ∆y ∆x + B f By x0 + t ∆x , y0 + t ∆y ∆y We now need to evaluate g2(t). Temporarily write f1 = B f Bx and f2 = B f By so that g1(t) = f1 x0 + t ∆x , y0 + t ∆y ∆x + f2 x0 + t ∆x , y0 + t ∆y ∆y Then we have, again using the chain rule, d dt f1 x0 + t ∆x , y0 + t ∆y = B f1 Bx x0 + t ∆x , y0 + t ∆y ∆x + B f1 By x0 + t ∆x , y0 + t ∆y ∆y = B2 f Bx2 x0 + t ∆x , y0 + t ∆y ∆x + B2 f ByBx x0 + t ∆x , y0 + t ∆y ∆y (˚) and d dt f2 x0 + t ∆x , y0 + t ∆y = B f2 Bx x0 + t ∆x , y0 + t ∆y ∆x + B f2 By x0 + t ∆x , y0 + t ∆y ∆y = B2 f BxBy x0 + t ∆x , y0 + t ∆y ∆x + B2 f By2 x0 + t ∆x , y0 + t ∆y ∆y (˚˚) Adding ∆x times (˚) to ∆y times (˚˚) and recalling that B2 f ByBx = B2 f BxBy, gives g2(t) = B2 f Bx2 x0 + t ∆x , y0 + t ∆y ∆x2 + 2 B2 f BxBy x0 + t ∆x , y0 + t ∆y ∆x∆y + B2 f By2 x0 + t ∆x , y0 + t ∆y ∆y2 Now setting t0 = 0 and ∆t = 1, the quadratic approximation f x0 + ∆x , y0 + ∆y = g(1) « g(0) + g1(0) + 1 2g2(0) is f x0 + ∆x , y0 + ∆y « f x0 , y0 + B f Bx x0 , y0 ∆x + B f By x0 , y0 ∆y + 1 2 "B2 f Bx2 x0, y0 ∆x2 + 2 B2 f BxBy x0, y0 ∆x∆y + B2 f By2 x0, y0 ∆y2 Equation 2.6.12. and the corresponding exact formula f x0 + ∆x , y0 + ∆y = g(1) = g(0) + g1(0) + 1 2g2(c) is 156 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR f x0 + ∆x , y0 + ∆y = f x0 , y0 + B f Bx x0 , y0 ∆x + B f By x0 , y0 ∆y + 1 2 "B2 f Bx2 r(c) ∆x2 + 2 B2 f BxBy r(c) ∆x∆y + B2 f By2 r(c) ∆y2 where r(c) = x0 + c ∆x , y0 + c ∆y and c is some (unknown) number satisfying 0 ď c ď 1. Equation 2.6.13. If we can bound the second derivatives ˇ ˇ ˇ ˇ B2 f Bx2 r(c) ˇ ˇ ˇ ˇ , ˇ ˇ ˇ ˇ B2 f BxBy r(c) ˇ ˇ ˇ ˇ , ˇ ˇ ˇ ˇ B2 f By2 r(c) ˇ ˇ ˇ ˇ ď M we can massage (2.6.13) into the form ˇ ˇ ˇ ˇf x0 + ∆x , y0 + ∆y ´ ! f x0 , y0 + B f Bx x0 , y0 ∆x + B f By x0 , y0 ∆y )ˇ ˇ ˇ ˇ ď M 2 \|∆x\|2 + 2\|∆x\| \|∆y\| + \|∆y\|2 Equation 2.6.14. Why might we want to do this? The left hand side of (2.6.14) is exactly the error in the linear approximation (2.6.1). So the right hand side is a rigorous bound on the error in the linear approximation. Example 2.6.15 (Example 2.6.7, continued) Suppose that we approximate (0.998)3 1.003 as in Example 2.6.7 and we want a rigorous bound on the approximation. We can get such a rigorous bound by applying (2.6.13). Set f (x, y) = x3 y and x0 = 1 ∆x = ´0.002 y0 = 1 ∆y = 0.003 Then the exact answer is f x0 + ∆x , y0 + ∆y and the approximate answer is f x0 , y0 + B f Bx x0 , y0 ∆x + B f By x0 , y0 ∆y, so that, by (2.6.13), the error in the approximation is exactly 1 2 ˇ ˇ ˇ ˇ B2 f Bx2 r(c) ∆x2 + 2 B2 f BxBy r(c) ∆x∆y + B2 f By2 r(c) ∆y2 ˇ ˇ ˇ ˇ 157 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR with r(c) = 1 ´ 0.002c , 1 + 0.003c for some, unknown, 0 ď c ď 1. For our function f f (x, y) = x3 y B f Bx(x, y) = 3x2 y B f By(x, y) = ´x3 y2 B2 f Bx2 (x, y) = 6x y B2 f BxBy(x, y) = ´3x2 y2 B2 f By2 (x, y) = 2x3 y3 We don’t know what r(c) = 1 ´ 0.002c , 1 + 0.003c is. But we know that 0 ď c ď 1, so we definitely know that the x component of r(c) is smaller that 1 and the y component of r(c) is bigger than 1. So ˇ ˇ ˇ ˇ B2 f Bx2 r(c) ˇ ˇ ˇ ˇ ď 6 ˇ ˇ ˇ ˇ B2 f BxBy r(c) ˇ ˇ ˇ ˇ ď 3 ˇ ˇ ˇ ˇ B2 f By2 r(c) ˇ ˇ ˇ ˇ ď 2 and error ď 1 2 h 6∆x2 + 2 ˆ 3\|∆x ∆y\| + 2∆y2i ď 3(0.002)2 + 3(0.002)(0.003) + (0.003)2 = 0.000039 By way of comparison, the exact error is 0.0000389, to seven decimal places. Example 2.6.15 Example 2.6.16 In this example, we find the quadratic approximation of f (x, y) = a 1 + 4x2 + y2 at (x0, y0) = (1, 2) and use it to compute approximately f (1.1 , 2.05). We know that we will need all partial derivatives up to order 2, so we first compute them and evaluate them at (x0, y0) = (1, 2). f (x, y) = b 1 + 4x2 + y2 f (x0, y0) = 3 fx(x, y) = 4x a 1 + 4x2 + y2 fx(x0, y0) = 4 3 fy(x, y) = y a 1 + 4x2 + y2 fy(x0, y0) = 2 3 fxx(x, y) = 4 a 1 + 4x2 + y2 ´ 16x2 [1 + 4x2 + y2]3/2 fxx(x0, y0) = 4 3 ´ 16 27 = 20 27 fxy(x, y) = ´ 4xy [1 + 4x2 + y2]3/2 fxy(x0, y0) = ´ 8 27 fyy(x, y) = 1 a 1 + 4x2 + y2 ´ y2 [1 + 4x2 + y2]3/2 fyy(x0, y0) = 1 3 ´ 4 27 = 5 27 158 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR We now just substitute them into (2.6.12) to get that the quadratic approximation to f about (x0, y0) is f x0 + ∆x , y0 + ∆y « f (x0, y0) + fx(x0, y0)∆x + fy(x0, y0)∆y + 1 2 fxx(x0, y0)∆x2 + 2fxy(x0, y0)∆x∆y + fyy(x0, y0)∆y2 = 3 + 4 3∆x + 2 3∆y + 10 27∆x2 ´ 8 27∆x∆y + 5 54∆y2 In particular, with ∆x = 0.1 and ∆y = 0.05, f (1.1 , 2.05) « 3 + 4 3(0.1) + 2 3(0.05) + 10 27(0.01) ´ 8 27(0.005) + 5 54(0.0025) = 3.1691 The actual value, to four decimal places, is 3.1690. The percentage error is about 0.004%. Example 2.6.16 Example 2.6.17 In this example, we find the quadratic approximation of f (x, y) = e2x sin(3y) about (x0, y0) = (0, 0) in two different ways. The first way uses the canned formula (2.6.12). We compute all partial derivatives up to order 2 at (x0, y0). f (x, y) = e2x sin(3y) f (x0, y0) = 0 fx(x, y) = 2e2x sin(3y) fx(x0, y0) = 0 fy(x, y) = 3e2x cos(3y) fy(x0, y0) = 3 fxx(x, y) = 4e2x sin(3y) fxx(x0, y0) = 0 fxy(x, y) = 6e2x cos(3y) fxy(x0, y0) = 6 fyy(x, y) = ´9e2x sin(3y) fyy(x0, y0) = 0 So the quadratic approximation to f about (0, 0) is f x , y « f (x, y) + fx(x, y)x + fy(0, 0)y + 1 2 fxx(0, 0)x2 + 2fxy(0, 0)xy + fyy(0, 0)y2 = 3y + 6xy That’s pretty simple — just compute a bunch of partial derivatives and substitute into the formula (2.6.12). But there is also a sneakier, and often computationally more efficient, method to get the same result. It exploits the single variable Taylor expansions ex = 1 + x + 1 2!x2 + ¨ ¨ ¨ sin y = y ´ 1 3!y3 + ¨ ¨ ¨ 159 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR Replacing x by 2x in the first and y by 3y in the second and multiplying the two together, keeping track only of terms of degree at most two, gives f (x, y) = e2x sin(3y) = h 1 + (2x) + 1 2!(2x)2 + ¨ ¨ ¨ ih (3y) ´ 1 3!(3y)3 + ¨ ¨ ¨ i = h 1 + 2x + 2x2 + ¨ ¨ ¨ ih 3y ´ 9 2y3 + ¨ ¨ ¨ i = 3y + 6xy + 6x2y + ¨ ¨ ¨ ´ 9 2y3 ´ 9xy3 ´ 9x2y3 + ¨ ¨ ¨ = 3y + 6xy + ¨ ¨ ¨ just as in the first computation. Example 2.6.17 2.6.2 § § Optional — Taylor Polynomials We have just found linear and quadratic approximations to the function f (x, y), for (x, y) near the point (x0, y0). In CLP-1, we found not only linear and quadratic approximations, but in fact a whole hierarchy of approximations. For each integer n ě 0, the nth order Taylor polynomial for f (x) about x = a was defined, in Definition 3.4.11 of the CLP-1 text, to be n ÿ k=0 1 k! f (k)(a) ¨ (x ´ a)k We’ll now define, and find, the Taylor polynomial of order n for the function f (x, y) about (x, y) = (x0, y0). It is going to be a polynomial of degree at most n in ∆x and ∆y. The most general such polynomial is Tn(∆x, ∆y) = ÿ ℓ,mě0 ℓ+mďn aℓ,m (∆x)ℓ(∆y)m with all of the coefficients aℓ,m being constants. The specific coefficients for the Taylor polynomial are determined by the requirement that all partial derivatives of Tn(∆x, ∆y) at ∆x = ∆y = 0 are the same as the corresponding partial derivatives of f x0 + ∆x , y0 + ∆y at ∆x = ∆y = 0. By way of preparation for our computation of the derivatives of Tn(∆x, ∆y), consider d dtt4 = 4t3 d2 dt2 t4 = (4)(3)t2 d3 dt3 t4 = (4)(3)(2)t d4 dt4 t4 = (4)(3)(2)(1) = 4! d5 dt5 t4 = 0 d6 dt6 t4 = 0 160 PARTIAL DERIVATIVES 2.6 LINEAR APPROXIMATIONS AND ERROR and d dtt4 ˇ ˇ ˇ ˇ t=0 = 0 d2 dt2 t4 ˇ ˇ ˇ ˇ t=0 = 0 d3 dt3 t4 ˇ ˇ ˇ ˇ t=0 = 0 d4 dt4 t4 ˇ ˇ ˇ ˇ t=0 = 4! d5 dt5 t4 ˇ ˇ ˇ ˇ t=0 = 0 d6 dt6 t4 ˇ ˇ ˇ ˇ t=0 = 0 More generally, for any natural numbers p, m, dp dtp tm = # m(m ´ 1) ¨ ¨ ¨ (m ´ p + 1)tm´p if p ď m 0 if p ą m so that dp dtp tm ˇ ˇ ˇ ˇ t=0 = # m! if p = m 0 if p ‰ m Consequently Bp B(∆x)p Bq B(∆y)q (∆x)ℓ(∆y)m ˇ ˇ ˇ ˇ ∆x=∆y=0 = # ℓ! m! if p = ℓand q = m 0 if p ‰ ℓor q ‰ m and Bp+q Tn B(∆x)p B(∆y)q (0, 0) = ÿ ℓ,mě0 ℓ+mďn aℓ,m Bp B(∆x)p Bq B(∆y)q (∆x)ℓ(∆y)m ˇ ˇ ˇ ˇ ∆x=∆y=0 = # p! q! ap,q if p + q ď n 0 if p + q ą n Our requirement that the derivatives of f and Tn match is the requirement that, for all p + q ď n, Bp+q Tn B(∆x)p B(∆y)q (0, 0) = Bp+q B(∆x)p B(∆y)q f x0 + ∆x , y0 + ∆y ˇ ˇ ˇ ∆x=∆y=0 = Bp+q f Bxp Byq (x0, y0) This requirement gives p! q! ap,q = Bp+q f Bxp Byq (x0, y0) So the Taylor polynomial of order n for the function f (x, y) about (x, y) = (x0, y0) is the right hand side of f x0 + ∆x , y0 + ∆y « ÿ ℓ,mě0 ℓ+mďn 1 ℓ! m! Bℓ+m f BxℓBym (x0, y0) (∆x)ℓ(∆y)m Equation 2.6.18. 161 PARTIAL DERIVATIVES 2.7 DIRECTIONAL DERIVATIVES AND THE GRADIENT This is for functions, f (x, y), of two variables. There are natural extensions of this for functions of any (finite) number of variables. For example, the Taylor polynomial of order n for a function, f (x, y, z), of three variables is the right hand side of f x0 + ∆x , y0 + ∆y , z0 + ∆z « ÿ k,ℓ,mě0 k+ℓ+mďn 1 k! ℓ! m! Bk+ℓ+m f Bxk ByℓBzm (x0, y0, z0) (∆x)k(∆y)ℓ(∆z)m 2.7Ĳ Directional Derivatives and the Gradient The principal interpretation of df dx(a) is the rate of change of f (x), per unit change of x, at x = a. The natural analog of this interpretation for multivariable functions is the directional derivative, which we now introduce through a question. §§§ A Question Suppose that you are standing at (a, b) near a campfire. The temperature you feel at (x, y) is f (x, y). You start to move with velocity v = ⟨v1, v2⟩. What rate of change of temperature do you feel? §§§ The Answer Let’s set the beginning of time, t = 0, to the time at which you leave (a, b). Then ˝ at time 0 you are at (a, b) and feel the temperature f (a, b) and ˝ at time t you are at (a + v1t , b + v2t) and feel the temperature f (a + v1t , b + v2t). So ˝ the change in temperature between time 0 and time t is f (a + v1t , b + v2t) ´ f (a, b), ˝ the average rate of change of temperature, per unit time, between time 0 and time t is f (a+v1t , b+v2t)´f (a,b) t and the ˝ instantaneous rate of change of temperature per unit time as you leave (a, b) is lim tÑ0 f (a+v1t , b+v2t)´f (a,b) t . Concentrate on the t dependence in this limit by writing f (a + v1t , b + v2t) = g(t). Then lim tÑ0 f (a + v1t , b + v2t) ´ f (a, b) t = lim tÑ0 g(t) ´ g(0) t = dg dt (0) = d dt f (a + v1t , b + v2t) ˇ ˇ ˇ t=0 By the chain rule, we can write the right hand side in terms of partial derivatives of f. d dt f (a + v1t , b + v2t) = fx(a + v1t , b + v2t) v1 + fy(a + v1t , b + v2t) v2 162 PARTIAL DERIVATIVES 2.7 DIRECTIONAL DERIVATIVES AND THE GRADIENT So, the instantaneous rate of change per unit time as you leave (a, b) is lim tÑ0 f (a + v1t , b + v2t) ´ f (a, b) t = fx(a + v1t , b + v2t) v1 + fy(a + v1t , b + v2t) v2 ˇ ˇ ˇ t=0 = fx(a, b) v1 + fy(a, b) v2 = fx(a, b) , fy(a, b) ¨ ⟨v1, v2⟩ Notice that we have expressed the rate of change as the dot product of the velocity vector with a vector of partial derivatives of f. We have seen such a vector of partial derivatives of f before; in Definition 2.5.4, we defined the gradient of the three variable function G(x, y, z) at the point x0 , y0 , z0 to be Gx x0 , y0 , z0 , Gy x0 , y0 , z0 , Gz x0 , y0 , z0 . Here we see the natural two dimensional analog. The vector fx(a, b) , fy(a, b) is denoted ∇ ∇ ∇f (a, b) and is called “the gradient of the function f at the point (a, b)”. Definition 2.7.1. In general, the gradient of f is a vector with one component for each variable of f. The jth component is the partial derivative of f with respect to the jth variable. Now because the dot product ∇ ∇ ∇f (a, b) ¨ v appears frequently, we introduce some handy notation. Given any vector v = ⟨v1, v2⟩, the expression fx(a, b), fy(a, b) ¨ ⟨v1, v2⟩= ∇ ∇ ∇f (a, b) ¨ v is denoted Dv f (a, b). Notation 2.7.2. Armed with this useful notation we can answer our question very succinctly. The rate of change of f per unit time as you leave (a, b) moving with velocity v is Dv f (a, b) = ∇ ∇ ∇f (a, b) ¨ v Equation 2.7.3. We can compute the rate of change of temperature per unit distance (as opposed to per unit time) in a similar way. The change in temperature between time 0 and time t is f (a + v1t, b + v2t) ´ f (a, b). Between time 0 and time t, you have travelled a distance \|v\|t. So the instantaneous rate of change of temperature per unit distance as you leave (a, b) is lim tÑ0 f (a + v1t, b + v2t) ´ f (a, b) t\|v\| This is exactly 1 \|v\| times lim tÑ0 f (a+v1t,b+v2t)´f (a,b) t which we computed above to be Dv f (a, b). So 163 PARTIAL DERIVATIVES 2.7 DIRECTIONAL DERIVATIVES AND THE GRADIENT Given any nonzero vector v, the rate of change of f per unit distance as you leave (a, b) moving in direction v is ∇ ∇ ∇f (a, b) ¨ v \|v\| = Dv/\|v\| f (a, b) Equation 2.7.4. Dv/\|v\| f (a, b) is called the directional derivative of the function f (x, y) at the point (a, b) in the direction34 v. Definition 2.7.5. §§§ The Implications We have just seen that the instantaneous rate of change of f per unit distance as we leave (a, b) moving in direction v is a dot product, which we can write as ∇ ∇ ∇f (a, b) ¨ v \|v\| = \|∇ ∇ ∇f (a, b)\| cos θ where θ is the angle between the gradient vector ∇ ∇ ∇f (a, b) and the direction vector v. Writ-ing it in this way allows us to make some useful observations. Since cos θ is always be-tween ´1 and +1 • the direction of maximum rate of increase is that having θ = 0. So to get maximum rate of increase per unit distance, as you leave (a, b), you should move in the same direction as the gradient ∇ ∇ ∇f (a, b). Then the rate of increase per unit distance is \|∇ ∇ ∇f (a, b)\|. • The direction of minimum (i.e. most negative) rate of increase is that having θ = 180˝. To get minimum rate of increase per unit distance you should move in the direction opposite ∇ ∇ ∇f (a, b). Then the rate of increase per unit distance is ´\|∇ ∇ ∇f (a, b)\|. • The directions giving zero rate of increase are those perpendicular to ∇ ∇ ∇f (a, b). If you move in a direction perpendicular to ∇ ∇ ∇f (a, b), then f (x, y) remains constant as you leave (a, b). At that instant, you are moving so that f (x, y) remains constant and consequently you are moving along the level curve f (x, y) = f (a, b). So ∇ ∇ ∇f (a, b) is perpendicular to the level curve f (x, y) = f (a, b) at (a, b). The corresponding statement in three dimensions is that ∇ ∇ ∇F(a, b, c) is perpendicular to the level surface F(x, y, z) = F(a, b, c) at (a, b, c). Hence a good way to find a vector normal to the sur-face F(x, y, z) = F(a, b, c) at the point (a, b, c) is to compute the gradient ∇ ∇ ∇F(a, b, c). This is precisely what we saw back in Theorem 2.5.5. 34 Some people require direction vectors to have unit length. We don’t. 164 PARTIAL DERIVATIVES 2.7 DIRECTIONAL DERIVATIVES AND THE GRADIENT Now that we have defined the directional derivative, here are some examples. Example 2.7.6 Problem: Find the directional derivative of the function f (x, y) = ex+y2 at the point (0, 1) in the direction ´ˆ ı ı ı + ˆ ȷ ȷ ȷ. Solution. To compute the directional derivative, we need the gradient. To compute the gradient, we need some partial derivatives. So we start with the partial derivatives of f at (0, 1): fx(0, 1) = ex+y2ˇ ˇ ˇ x=0 y=1 = e fy(0, 1) = 2yex+y2ˇ ˇ ˇ x=0 y=1 = 2e So the gradient of f at (0, 1) is ∇ ∇ ∇f (0, 1) = fx(0, 1) ˆ ı ı ı + fy(0, 1) ˆ ȷ ȷ ȷ = e ˆ ı ı ı + 2e ˆ ȷ ȷ ȷ and the direction derivative in the direction ´ˆ ı ı ı + ˆ ȷ ȷ ȷ is D ´ˆ ı ı ı+ˆ ȷ ȷ ȷ \|´ˆ ı ı ı+ˆ ȷ ȷ ȷ\| f (0, 1) = ∇ ∇ ∇f (0, 1) ¨ ´ˆ ı ı ı + ˆ ȷ ȷ ȷ \| ´ ˆ ı ı ı + ˆ ȷ ȷ ȷ\| = e ˆ ı ı ı + 2e ˆ ȷ ȷ ȷ ¨ ´ˆ ı ı ı + ˆ ȷ ȷ ȷ ? 2 = e ? 2 Example 2.7.6 Example 2.7.7 Problem: Find the directional derivative of the function w(x, y, z) = xyz + ln(xz) at the point (1, 3, 1) in the direction ⟨1 , 0 , 1⟩. In what directions is the directional derivative zero? Solution. First, the partial derivatives of w at (1, 3, 1) are wx(1, 3, 1) = yz + 1 x ˇ ˇ ˇ ˇ (1,3,1) = 3 ˆ 1 + 1 1 = 4 wy(1, 3, 1) = xz ˇ ˇ ˇ ˇ (1,3,1) = 1 ˆ 1 = 1 wz(1, 3, 1) = xy + 1 z ˇ ˇ ˇ ˇ (1,3,1) = 1 ˆ 3 + 1 1 = 4 so the gradient of w at (1, 3, 1) is ∇ ∇ ∇w(1, 3, 1) = wx(1, 3, 1) , wy(1, 3, 1) , wz(1, 3, 1) = ⟨4 , 1 , 4⟩ and the direction derivative in the direction ⟨1 , 0 , 1⟩is D ⟨1 , 0 , 1⟩ \|⟨1 , 0 , 1⟩\| w(1, 3, 1) = ∇ ∇ ∇w(1, 3, 1) ¨ ⟨1 , 0 , 1⟩ \| ⟨1 , 0 , 1⟩\| = ⟨4 , 1 , 4⟩¨ ⟨1 , 0 , 1⟩ ? 2 = 8 ? 2 = 4 ? 2 165 PARTIAL DERIVATIVES 2.7 DIRECTIONAL DERIVATIVES AND THE GRADIENT The directional derivative of w at (1, 3, 1) in the direction t ‰ 0 is zero if and only if 0 = D t \|t\| w(1, 3, 1) = ∇ ∇ ∇w(1, 3, 1) ¨ t \|t\| = ⟨4 , 1 , 4⟩¨ t \|t\| which is the case if and only if t is perpendicular to ⟨4 , 1 , 4⟩. So if we walk in the direction of any vector in the plane, 4x + y + 4z = 0 (which has normal vector ⟨4 , 1 , 4⟩) then the directional derivative is zero. Example 2.7.7 Example 2.7.8 Let f (x, y) = 5 ´ x2 ´ 2y2 (a, b) = ´ 1, ´1 In this example, we’ll explore the behaviour of the function f (x, y) near the point (a, b). Note that for any fixed f0 ă 5, f (x, y) = f0 is the ellipse x2 + 2y2 = 5 ´ f0. So the graph z = f (x, y) consists of a bunch of horizontal ellipses stacked one on top of each other. • Since the ellipse x2 + 2y2 = 5 ´ f0 has x-semi-axis a 5 ´ f0 and y-semi-axis b 5´f0 2 , – the ellipses start with a point on the z axis when f0 = 5 and – increase in size as f0 decreases. • The part of the graph z = f (x, y) in the first octant is sketched in the top figure below. • Several level curves, f (x, y) = f0, are sketched in the bottom figure below. • The gradient vector ∇ ∇ ∇f (a, b) = ⟨´2x, ´4y⟩ ˇ ˇ (´1,´1) = ⟨2, 4⟩= 2 ⟨1, 2⟩ at (´1, ´1) is also illustrated in the bottom sketch. We have that, at (a, b) = (´1, ´1), ˝ the unit vector giving the direction of maximum rate of increase is the unit vector in the direction of the gradient vector 2 ⟨1, 2⟩, which is 1 ? 5 ⟨1, 2⟩. The maximum rate of increase is \| ⟨2, 4⟩\| = 2 ? 5. ˝ The unit vector giving the direction of minimum rate of increase is ´ 1 ? 5 ⟨1, 2⟩and that minimum rate is ´\| ⟨2, 4⟩\| = ´2 ? 5. ˝ The directions giving zero rate of increase are perpendicular to ∇ ∇ ∇f (a, b). One vector perpendicular35 to ⟨1, 2⟩is ⟨2, ´1⟩. So the unit vectors giving the direction of zero rate of increase are the ˘ 1 ? 5(2, ´1). These are the directions of the tangent vector at (a, b) to the level curve of f through (a, b), which is the curve f (x, y) = f (a, b). 35 Check this by taking the dot product of ⟨1, 2⟩and ⟨2, ´1⟩. 166 PARTIAL DERIVATIVES 2.7 DIRECTIONAL DERIVATIVES AND THE GRADIENT z y x z “ 4 z “ fpx, yq z “ 3 z “ 2 z “ 1 z “ 0 x y fpx,yq“4 fpx,yq“0 p´1, ´1q Example 2.7.8 Example 2.7.9 Problem: What is the rate of change of f (x, y, z) = x2 + y2 + z2 at (3, 5, 4) moving in the positive x-direction along the curve of intersection of the surfaces G(x, y, z) = 25 and H(x, y, z) = 0 where G(x, y, z) = 2x2 ´ y2 + 2z2 and H(x, y, z) = x2 ´ y2 + z2 Solution. As a first check note that (3, 5, 4) really does lie on both surfaces because G(3, 5, 4) = 2 32 ´ 52 + 2 42 = 18 ´ 25 + 32 = 25 H(3, 5, 4) = 32 ´ 52 + 42 = 9 ´ 25 + 16 = 0 We compute gradients to get the normal vectors to the surfaces G(x, y, z) = 25 and H(x, y, z) = 0 at (3, 5, 4). ∇ ∇ ∇G(3, 5, 4) = h 4x ˆ ı ı ı ´ 2y ˆ ȷ ȷ ȷ + 4z ˆ k i (3,5,4) = 12ˆ ı ı ı ´ 10 ˆ ȷ ȷ ȷ + 16 ˆ k = 2 6ˆ ı ı ı ´ 5 ˆ ȷ ȷ ȷ + 8 ˆ k ∇ ∇ ∇H(3, 5, 4) = h 2x ˆ ı ı ı ´ 2y ˆ ȷ ȷ ȷ + 2z ˆ k i (3,5,4) = 6ˆ ı ı ı ´ 10 ˆ ȷ ȷ ȷ + 8 ˆ k = 2 3ˆ ı ı ı ´ 5 ˆ ȷ ȷ ȷ + 4 ˆ k 167 PARTIAL DERIVATIVES 2.7 DIRECTIONAL DERIVATIVES AND THE GRADIENT The direction of interest is tangent to the curve of intersection. So the direction of interest is tangent to both surfaces and hence is perpendicular to both gradients. Consequently one tangent vector to the curve at (3, 5, 4) is ∇ ∇ ∇G(3, 5, 4) ˆ ∇ ∇ ∇H(3, 5, 4) = 4 6ˆ ı ı ı ´ 5 ˆ ȷ ȷ ȷ + 8 ˆ k ˆ 3ˆ ı ı ı ´ 5 ˆ ȷ ȷ ȷ + 4 ˆ k = 4 det   ˆ ı ı ı ˆ ȷ ȷ ȷ ˆ k 6 ´5 8 3 ´5 4   = 4 20ˆ ı ı ı ´ 15 ˆ k = 20 4ˆ ı ı ı ´ 3 ˆ k and the unit tangent vector to the curve at (3, 5, 4) that has positive x component is 4ˆ ı ı ı ´ 3 ˆ k \|4ˆ ı ı ı ´ 3 ˆ k\| = 4 5 ˆ ı ı ı ´ 3 5 ˆ k The desired rate of change is D 4 5 ˆ ı ı ı´ 3 5 ˆ k f (3, 5, 4) = ∇ ∇ ∇f (3, 5, 4) ¨ 4 5 ˆ ı ı ı ´ 3 5 ˆ k = 2x ˆ ı ı ı+2y ˆ ȷ ȷ ȷ+2z ˆ k=(3,5,4) hkkkkkkkkkikkkkkkkkkj 6ˆ ı ı ı + 10 ˆ ȷ ȷ ȷ + 8 ˆ k ¨ 4 5 ˆ ı ı ı ´ 3 5 ˆ k = 0 Actually, we could have known that the rate of change would be zero. • Any point (x, y, z) on the curve obeys both y2 = x2 + z2 and 2x2 ´ y2 + 2z2 = 25. • Substituting y2 = x2 + z2 into 2x2 ´ y2 + 2z2 = 25 gives x2 + z2 = 25. • So, at any point on the curve, x2 + z2 = 25 and y2 = x2 + z2 = 25 so that x2 + y2 + z2 = 50. • That is, f (x, y, z) = x2 + y2 + z2 takes the value 50 at every point of the curve. • So of course the rate of change of f along the curve is 0. Example 2.7.9 Let’s change things up a little. In the next example, we are told the rates of change in two different directions. From this we are to determine the rate of change in a third direction. Example 2.7.10 Problem: The rate of change of a given function f (x, y) at the point P0 = (1, 2) in the direc-tion towards P1 = (2, 3) is 2 ? 2 and in the direction towards P2 = (1, 0) is ´3. What is the rate of change of f at P0 towards the origin P3 = (0, 0)? Solution. We can easily determine the rate of change of f at the point P0 in any direction once we know the gradient ∇ ∇ ∇f (1, 2) = a ˆ ı ı ı + b ˆ ȷ ȷ ȷ. So we will first use the two given rates of change to determine a and b, and then we determine the rate of change towards (0, 0). 168 PARTIAL DERIVATIVES 2.7 DIRECTIONAL DERIVATIVES AND THE GRADIENT The two rates of change that we are given are those in the directions of the vectors Ý Ý Ñ P0P1 = ⟨1, 1⟩ Ý Ý Ñ P0P2 = ⟨0, ´2⟩ As you might guess, the notation Ý Ñ PQ means the vector whose tail is at P and whose head is at Q. So the given rates of change tell us that 2 ? 2 = D ⟨1,1⟩ \|⟨1,1⟩\| f (1, 2) = ∇ ∇ ∇f (1, 2) ¨ ⟨1, 1⟩ \| ⟨1, 1⟩\| = ⟨a, b⟩¨ ⟨1, 1⟩ ? 2 = a ? 2 + b ? 2 ´3 = D ⟨0,´2⟩ \|⟨0,´2⟩\| f (1, 2) = ∇ ∇ ∇f (1, 2) ¨ ⟨0, ´2⟩ \| ⟨0, ´2⟩\| = ⟨a, b⟩¨ ⟨0, ´2⟩ 2 = ´b These two lines give us two linear equations in the two unknowns a and b. The second equation directly gives us b = 3. Substituting b = 3 into the first equation gives a ? 2 + 3 ? 2 = 2 ? 2 ù ñ a + 3 = 4 ù ñ a = 1 A direction vector from P0 = (1, 2) towards P3 = (0, 0) is Ý Ý Ñ P0P3 = ⟨´1, ´2⟩ and the rate of change (per unit distance) in that direction is D ⟨´1,´2⟩ \|⟨´1,´2⟩\| f (1, 2) = ∇ ∇ ∇f (1, 2) ¨ ⟨´1, ´2⟩ \| ⟨´1, ´2⟩\| = ⟨a, b⟩¨ ⟨´1, ´2⟩ ? 5 = ⟨1, 3⟩¨ ⟨´1, ´2⟩ ? 5 = ´ 7 ? 5 Example 2.7.10 Example 2.7.11 (Optional) Problem: Find all points (a, b, c) for which the spheres (x ´ a)2 + (y ´ b)2 + (z ´ c)2 = 1 and x2 + y2 + z2 = 1 intersect orthogonally. That is, the tangent planes to the two spheres are to be perpendicular at each point of intersection. Solution. Let (x0, y0, z0) be a point of intersection. That is (x0 ´ a)2 + (y0 ´ b)2 + (z0 ´ c)2 = 1 x2 0 + y2 0 + z2 0 = 1 A normal vector to G(x, y, z) = (x ´ a)2 + (y ´ b)2 + (z ´ c)2 = 1 at (x0, y0, z0) is N = ∇ ∇ ∇G(x0, y0, z0) = ⟨2(x0 ´ a) , 2(y0 ´ b) , 2(z0 ´ c)⟩ A normal vector to g(x, y, z) = x2 + y2 + z2 = 1 at (x0, y0, z0) is n = ∇ ∇ ∇g(x0, y0, z0) = ⟨2x0 , 2y0 , 2z0⟩ 169 PARTIAL DERIVATIVES 2.7 DIRECTIONAL DERIVATIVES AND THE GRADIENT The two tangent planes are perpendicular if and only if ˆ N and ˆ n are perpendicular, which is the case if and only if 0 = ˆ N ¨ ˆ n = 4x0(x0 ´ a) + 4y0(y0 ´ b) + 4z0(z0 ´ c) or, dividing the equation by 4, x0(x0 ´ a) + y0(y0 ´ b) + z0(z0 ´ c) = 0 Let’s pause to take stock. We need to find all (a, b, c)’s such that the statement (x0, y0, z0) is a point of intersection of the two spheres (S1) implies the statement the normal vectors ˆ N and ˆ n are perpendicular (S2) In equations, we need to find all (a, b, c)’s such that the statement (x0, y0, z0) obeys x2 0 + y2 0 + z2 0 = 1 and (x0 ´ a)2 + (y0 ´ b)2 + (z0 ´ c)2 = 1 (S1) implies the statement (x0, y0, z0) obeys x0(x0 ´ a) + y0(y0 ´ b) + z0(z0 ´ c) = 0 (S2) Now if we expand (S2) then we can, with a little care, massage it into something that looks more like (S1). x0(x0 ´ a) + y0(y0 ´ b) + z0(z0 ´ c) = x2 0 + y2 0 + z2 0 ´ ax0 ´ by0 ´ cz0 = 1 2 ! x2 0 + y2 0 + z2 0 + (x0 ´ a)2 + (y0 ´ b)2 + (z0 ´ c)2 ´ a2 ´ b2 ´ c2) If (S1) is true, then x2 0 + y2 0 + z2 0 = 1 and (x0 ´ a)2 + (y0 ´ b)2 + (z0 ´ c)2 = 1 so that x0(x0 ´ a) + y0(y0 ´ b) + z0(z0 ´ c) = 1 2 ! 1 + 1 ´ a2 ´ b2 ´ c2) and statement (S2) is true if and only if a2 + b2 + c2 = 2 Our conclusion is that the set of allowed points (a, b, c) is the sphere of radius ? 2 centred on the origin. Example 2.7.11 Example 2.7.12 (Optional — The gradient in polar coordinates) Question: What is the gradient of a function in polar coordinates? Answer. As was the case in Examples 2.4.9 and 2.4.10, figuring out what the question is asking is half the battle. By Definition 2.5.4, the gradient of a function g(x, y) is the vector 170 PARTIAL DERIVATIVES 2.7 DIRECTIONAL DERIVATIVES AND THE GRADIENT gx(x, y), gy(x, y) . In this question we are told that we are given some function f (r, θ) of the polar coordinates36 r and θ. We are supposed to convert this function to Cartesian coordinates. This means that we are to consider the function g(x, y) = f r(x, y), θ(x, y) with r(x, y) = b x2 + y2 θ(x, y) = arctan y x Then we are to compute the gradient of g(x, y) and express the answer in terms of r and θ. By the chain rule, Bg Bx = B f Br Br Bx + B f Bθ Bθ Bx = B f Br 1 2 2x a x2 + y2 + B f Bθ ´y/x2 1 + (y/x)2 = B f Br x a x2 + y2 ´ B f Bθ y x2 + y2 = B f Br r cos θ r ´ B f Bθ r sin θ r2 since x = r cos θ and y = r sin θ = B f Br cos θ ´ B f Bθ sin θ r Similarly Bg By = B f Br Br By + B f Bθ Bθ By = B f Br 1 2 2y a x2 + y2 + B f Bθ 1/x 1 + (y/x)2 = B f Br y a x2 + y2 + B f Bθ x x2 + y2 = B f Br sin θ + B f Bθ cos θ r So gx, gy = fr ⟨cos θ, sin θ⟩+ fθ r ⟨´ sin θ, cos θ⟩ or, with all the arguments written explicitly, gx(x, y), gy(x, y) = fr r(x, y), θ(x, y) ⟨cos θ(x, y) , sin θ(x, y)⟩ + 1 r(x, y) fθ r(x, y), θ(x, y) ⟨´ sin θ(x, y) , cos θ(x, y)⟩ 36 Polar coordinates were defined in Example 2.1.8. 171 PARTIAL DERIVATIVES 2.8 A FIRST LOOK AT PARTIAL DIFFERENTIAL EQUATIONS Example 2.7.12 2.8Ĳ A First Look at Partial Differential Equations Many phenomena are modelled by equations that relate the rates of change of various quantities. As rates of change are given by derivatives, the resulting equations contain derivatives and so are called differential equations. We saw a number of such differential equations in §2.4 of the CLP-2 text. In particular, a partial differential equation is an equa-tion for an unknown function of two or more variables that involves the partial derivatives of the unknown function. The standard acronym for partial differential equation is PDE. PDEs37 play a central role in modelling a huge number of different phenomena. Here is a table giving a bunch of named PDEs and what they are used for. It is far from complete. Maxwell’s equations describes electromagnetic radiation Navier–Stokes equations describes fluid motion Heat equation describes heat flow Wave equation describes wave motion Schr¨ odinger equation describes atoms, molecules and crystals Black–Scholes equation used for pricing stock options Einstein’s equations connects gravity and geometry Laplace’s equation used in many applications, including electrostatics We are just going to scratch the surface of the study of partial differential equations. Many of you will take a separate course on the subject. Some very important PDEs are very hard. One of the million U.S. dollar prizes38 announced in 2000 by the Clay Institute concerns the Navier-Stokes equations. On the other hand, we already know enough to accomplish some PDE tasks. In particular, we can check if a given function really does satisfy a given PDE. Here are some examples. Example 2.8.1 x Bz Bx + y Bz By = 0 Show that the function z(x, y) = x+y x´y obeys xBz Bx + yBz By = 0 37 There is a divided community on what the plural of PDE should be. Most people use PDEs as the plural. But some people use PDE as its own plural. 38 See or wiki/Millennium_Prize_Problems 172 PARTIAL DERIVATIVES 2.8 A FIRST LOOK AT PARTIAL DIFFERENTIAL EQUATIONS Solution. We simply evaluate the two terms on the left hand side when z = z(x, y) = x+y x´y. xBz Bx = x B Bx x + y x ´ y = x(1)(x ´ y) ´ (x + y)(1) (x ´ y)2 = ´2xy (x ´ y)2 yBz By = y B By x + y x ´ y = y(1)(x ´ y) ´ (x + y)(´1) (x ´ y)2 = 2xy (x ´ y)2 So xBz Bx + yBz By = ´2xy (x ´ y)2 + 2xy (x ´ y)2 = 0 and z(x, y) = x+y x´y really does solve the PDE x Bz Bx + y Bz By = 0. Beware however, that while we have found one solution to the given PDE, we have not found all solutions. There are many others. Trivially, if z(x, y) = 7, or any other constant, then we certainly have x Bz Bx = 0 and y Bz By = 0 so that x Bz Bx + y Bz By = 0. Less trivially, in the next example, we’ll find a ton3940 of solutions. Example 2.8.1 Example 2.8.2 x Bz Bx + y Bz By = 0, again Let G(u) be any differentiable function. Show that the function z(x, y) = G y x obeys xBz Bx + yBz By = 0 for all x ‰ 0. Solution. We again simply evaluate the two terms on the left hand side when z = z(x, y) = G y x . By the chain rule xBz Bx = x B Bx G y x = xG1 y x B Bx y x = xG1 y x ´ y x2 = ´G1 y x y x yBz By = y B By G y x = yG1 y x B By y x = yG1 y x 1 x = G1 y x y x So xBz Bx + yBz By = ´G1 y x y x + G1 y x y x = 0 39 Or, if you prefer, we will find 1.01605 tonnes of solutions. Although the authors of this text believe strongly in the supremacy of the modern metric system over the archaic chaos of imperial units, they are less certain of the appropriateness of revising well established colloquialisms. It is not at all clear that rewriting “I have a ton of work to do” as “I have a tonne of work to do” achieves very much except to give the impression that the author is wasting time adding two letters when they are expressing the sheer quantity of tasks that require their attention. Speaking of which, the authors should end this footnote, and get on with the next example. 40 In the previous footnote, the authors, writing from Canada, are using imperial tons rather than U.S. tons. The interested reader is invited to proceed to their favourite search engine to discover just how much time they can waste investigating the history, similarities and differences of these systems. 173 PARTIAL DERIVATIVES 2.8 A FIRST LOOK AT PARTIAL DIFFERENTIAL EQUATIONS and z(x, y) = G y x really does solve the PDE x Bz Bx + y Bz By = 0. Note that we can rewrite the solution x+y x´y of Example 2.8.1 as 1+y/x 1´y/x, which is of the form G y x . Example 2.8.2 Example 2.8.3 (Harmonic) A function u(x, y) is said to be harmonic if it satisfies Laplace’s equation uxx + uyy = 0 We will now find all harmonic polynomials (in the variables x and y) of degree at most two. Any polynomial of degree at most two is of the form u(x, y) = a + bx + cy + αx2 + βxy + γy2 for some constants a, b, c, α, β, γ. We will need uxx and uyy, so we compute them now. u(x, y) = a + bx + cy + αx2 + βxy + γy2 ux(x, y) = b + 2αx + βy uxx(x, y) = 2α uy(x, y) = c + βx + 2γy uyy(x, y) = 2γ The polynomial u(x, y) is harmonic if and only if 0 = uxx(x, y) + uyy(x, y) = 2α + 2γ So the polynomial u(x, y) is harmonic if and only if α + γ = 0, i.e. if and only if the polynomial is of the form u(x, y) = a + degree 1 hkkikkj bx + cy + degree 2 hkkkkkkkkkkikkkkkkkkkkj α(x2 ´ y2) + βxy with a, b, c, α, and β all constants. Notice that since both terms in the equation involve a second derivative, we would not expect there to be any conditions on the constant and linear terms. There aren’t. Beware that, while we have found all harmonic degree-two polynomials, there are many other harmonic functions, like, for example ex cos y. Example 2.8.3 2.8.1 § § Optional — Solving the Advection and Wave Equations In this section we consider B2w Bx2 (x, t) ´ 1 c2 B2w Bt2 (x, t) = 0 174 PARTIAL DERIVATIVES 2.8 A FIRST LOOK AT PARTIAL DIFFERENTIAL EQUATIONS This is an extremely important41partial differential equation called the “wave equation” (in one spatial dimension) that is used in modelling water waves, sound waves, seismic waves, light waves and so on. The reason that we are looking at it here is that we can use what we have just learned to see that its solutions are waves travelling with speed c. To start, we’ll use gradients and the chain rule to find the solution of the slightly sim-pler equation Bw Bx (x, t) ´ 1 c Bw Bt (x, t) = 0 which is called an advection equation. By way of motivation for what will follow, note that • we can rewrite the above equation as 1 , ´1 c ¨ ∇ ∇ ∇w(x, t) = 0 • This equation tells that the gradient of any solution w(x, t) must always be perpen-dicular to the constant vector 1 , ´1 c . • A vector ⟨a, b⟩is perpendicular to 1 , ´1 c if and only if ⟨a, b⟩¨ 1 , ´1 c = 0 ð ñ a ´ b c = 0 ð ñ b = ac ð ñ ⟨a, b⟩= a ⟨1, c⟩ That is, a vector is perpendicular to 1 , ´1 c if and only if it is parallel to ⟨1, c⟩. • Thus the gradient of any solution w(x, t) must always be parallel to the constant vector ⟨1 , c⟩. • Recall that one of our implications following Definition 2.7.5 is that the gradient of w(x, t) must always be perpendicular to the level curves of w. • So the level curves of w(x, t) are always perpendicular to the constant vector 1 , c . They must be straight lines with equations of the form ⟨1 , c⟩¨ ⟨x ´ x0 , t ´ t0⟩= 0 or x + ct = u with u a constant t x x c t “ 0 x c t “ 1 x c t “ ´1 x c t “ ´2 x c t “ ´3 x c t “ 2 x c t “ 3 41 If you plug “wave equation” into your favourite search engine you will get more than a million hits. 175 PARTIAL DERIVATIVES 2.8 A FIRST LOOK AT PARTIAL DIFFERENTIAL EQUATIONS • That is, for each constant u, w(x, t) takes the same value at each point of the straight line x + ct = u. Call that value U(u). So w(x, t) = U(u) = U(x + ct) for some function U. This solution represents a wave packet moving to the left with speed c. You can see this by observing that all points (x, t) in space-time for which x + ct takes the same fixed value, say z, have the same value of U(x + ct), namely U(z). So if you move so that your position at time t is x = z ´ ct (i.e. move the left with speed c) you always see the same value of w. The figure below illustrates this. It contains the graphs of U(x), U(x + c) = U(x + ct) ˇ ˇ t=1 and U(x + 2c) = U(x + ct) ˇ ˇ t=2 for a bump shaped U(x). In the figure the location of the tick z on the x-axis was chosen so that so that U(z) = maxx U(x). x y z ´ 2c y “ Upx2cq t “ 2 z ´ c y “ Upxcq t “ 1 z y “ Upxq t “ 0 The above argument that lead to the solution w(x, t) = U(x + ct) was somewhat hand-wavy. But we can easily turn it into a much tighter argument by simply changing variables from (x, y) to (u, v) with u = x + ct. It doesn’t much matter what we choose (within rea-son) for the new variable v. Let’s take v = x ´ ct. Then x = u+v 2 and t = u´v 2c and it is easy to translate back and forth between x, t and u, v. Now define the function W(u, v) by w(x, t) = W(x + ct , x ´ ct) By the chain rule Bw Bx (x, t) = B Bx W(x + ct , x ´ ct) = BW Bu (x + ct , x ´ ct) B Bx(x + ct) + BW Bv (x + ct , x ´ ct) B Bx(x ´ ct) = BW Bu (x + ct , x ´ ct) + BW Bv (x + ct , x ´ ct) and Bw Bt (x, t) = B Bt W(x + ct , x ´ ct) = BW Bu (x + ct , x ´ ct) B Bt(x + ct) + BW Bv (x + ct , x ´ ct) B Bt(x ´ ct) = BW Bu (x + ct , x ´ ct) ˆ c + BW Bv (x + ct , x ´ ct) ˆ (´c) Subtracting 1 c times the second equation from the first equation gives Bw Bx (x, t) ´ 1 c Bw Bt (x, t) = 2BW Bv (x + ct , x ´ ct) 176 PARTIAL DERIVATIVES 2.8 A FIRST LOOK AT PARTIAL DIFFERENTIAL EQUATIONS So w(x, t) obeys the equation Bw Bx (x, t) ´ 1 c Bw Bt (x, t) = 0 for all x and t if and only if W(u, v) obeys the equation BW Bv (x + ct , x ´ ct) = 0 for all x and t, which, substituting in x = u+v 2 and t = u´v 2c , is the case if and only if W(u, v) obeys the equation BW Bv (u , v) = 0 for all u and v The equation BW Bv (u , v) = 0 means that W(u, v) is independent of v, so that W(u, v) is of the form W(u, v) = U(u), for some function U, and, so finally, w(x, t) = W(x + ct , x ´ ct) = U(x + ct) Now that we have solved our toy equation, let’s move on to the 1d wave equation. Example 2.8.4 (Wave Equation) We’ll now expand the above argument to find the general solution to B2w Bx2 (x, t) ´ 1 c2 B2w Bt2 (x, t) = 0 We’ll again make the change of variables from (x, y) to (u, v) with u = x + ct and v = x ´ ct and again define the function W(u, v) by w(x, t) = W(x + ct , x ´ ct) By the chain rule, we still have Bw Bx (x, t) = B Bx W(x + ct , x ´ ct) = BW Bu (x + ct , x ´ ct) + BW Bv (x + ct , x ´ ct) Bw Bt (x, t) = B Bt W(x + ct , x ´ ct) = BW Bu (x + ct , x ´ ct) ˆ c + BW Bv (x + ct , x ´ ct) ˆ (´c) We now need to differentiate a second time. Write W1(u, v) = BW Bu (u, v) and W2(u, v) = BW Bv (u, v) so that Bw Bx (x, t) = W1(x + ct , x ´ ct) + W2(x + ct , x ´ ct) Bw Bt (x, t) = c W1(x + ct , x ´ ct) ´ c W2(x + ct , x ´ ct) 177 PARTIAL DERIVATIVES 2.8 A FIRST LOOK AT PARTIAL DIFFERENTIAL EQUATIONS Using the chain rule again B2w Bx2 (x, t) = B Bx Bw Bx (x, t) = B Bx [W1(x + ct , x ´ ct)] + B Bx [W2(x + ct , x ´ ct)] = BW1 Bu + BW1 Bv + BW2 Bu + BW2 Bv = B2W Bu2 + B2 W Bv Bu + B2 W Bu Bv + B2W Bv2 B2w Bt2 (x, t) = B Bt Bw Bt (x, t) = c B Bt [W1(x + ct , x ´ ct)] ´ c B Bt [W2(x + ct , x ´ ct)] = c2BW1 Bu ´ c2BW1 Bv ´ c2BW2 Bu + c2BW2 Bv = c2B2W Bu2 ´ c2 B2W BvBu ´ c2 B2W BuBv + c2B2W Bv2 with all of the functions on the right hand sides having arguments (x + ct , x ´ ct). So, subtracting 1 c2 times the second from the first, we get B2w Bx2 (x, t) ´ 1 c2 B2w Bt2 (x, t) = 4 B2W BuBv(x + ct , x ´ ct) and w(x, t) obeys B2w Bx2 (x, t) ´ 1 c2 B2w Bt2 (x, t) = 0 for all x and t if and only if B2W BuBv(u , v) = 0 for all u and v. • This tells us that the u-derivative of BW Bv is zero, so that BW Bv is independent of u. That is BW Bv (u, v) = r V(v) for some function ˜ V. The reason that we have called it r V instead of V with become evident shortly. • Recall that to apply B Bv, you treat u as a constant and differentiate with respect to v. • So BW Bv (u, v) = r V(v) says that, when u is thought of as a constant, W is an antideriva-tive of r V. • That is, W(u, v) = ş ˜ V(v) dv + U, with U being an arbitrary constant. As u is being thought of as a constant, U is allowed to depend on u. So, denoting by V any antiderivative of ˜ V, we can write our solution in a very neat form. W(u, v) = U(u) + V(v) 178 PARTIAL DERIVATIVES 2.8 A FIRST LOOK AT PARTIAL DIFFERENTIAL EQUATIONS and the function we want is42 w(x, t) = W(x + ct , x ´ ct) = U(x + ct) + V(x ´ ct) As we saw above U(x + ct) represents a wave packet moving to the left with speed c. Similarly, V(x ´ ct) represents a wave packet moving to the right with speed c. Notice that w(x, t) = U(x + ct) + V(x ´ ct) is a solution regardless of what U and V are. The differential equation cannot tell us what U and V are. To determine them, we need more information about the system — usually in the form of initial conditions, like w(x, 0) = ¨ ¨ ¨ and Bw Bt (x, 0) = ¨ ¨ ¨ . General techniques for solving partial differential equa-tions lie beyond this text — but definitely require a good understanding of multivariable calculus. A good reason to keep on reading! Example 2.8.4 2.8.2 § § Really Optional — Derivation of the Wave Equation In this section we derive the wave equation B2w Bx2 (x, t) ´ 1 c2 B2w Bt2 (x, t) = 0 in one application. To be precise, we apply Newton’s law to an elastic string, and conclude that small amplitude transverse vibrations of the string obey the wave equation. Here is a sketch of a tiny element of the string. wpx, tq ∆x ∆w x Tpx ∆x, tq Tpx, tq θpx ∆x, tq θpx, tq The basic notation that we will use (most of which appears in the sketch) is w(x, t) = vertical displacement of the string from the x axis at position x and time t θ(x, t) = angle between the string and a horizontal line at position x and time t T(x, t) = tension in the string at position x and time t ρ(x) = mass density (per unit length) of the string at position x The forces acting on the tiny element of string at time t are 42 This is known as d’Alembert’s form of the solution. It is named after Jean le Rond d’Alembert, 1717– 1783, who was a French mathematician, physicist, philosopher and music theorist. 179 PARTIAL DERIVATIVES 2.8 A FIRST LOOK AT PARTIAL DIFFERENTIAL EQUATIONS (a) tension pulling to the right, which has magnitude T(x + ∆x, t) and acts at an angle θ(x + ∆x, t) above horizontal (b) tension pulling to the left, which has magnitude T(x, t) and acts at an angle θ(x, t) below horizontal and, possibly, (c) various external forces, like gravity. We shall assume that all of the external forces act vertically and we shall denote by F(x, t)∆x the net magnitude of the external force acting on the element of string. The length of the element of string is essentially ? ∆x2 + ∆w2 so that the mass of the ele-ment of string is essentially ρ(x) ? ∆x2 + ∆w2 and the vertical component of Newton’s law F = ma says that ρ(x) a ∆x2 + ∆w2 B2w B t2 (x, t) = T(x + ∆x, t) sin θ(x + ∆x, t) ´ T(x, t) sin θ(x, t) + F(x, t)∆x Dividing by ∆x and taking the limit as ∆x Ñ 0 gives ρ(x) d 1 + Bw Bx 2 B2w B t2 (x, t) = B Bx T(x, t) sin θ(x, t) + F(x, t) = BT Bx (x, t) sin θ(x, t) + T(x, t) cos θ(x, t) Bθ Bx(x, t) + F(x, t) (E1) We can dispose of all the θ’s by observing from the figure above that tan θ(x, t) = lim ∆xÑ0 ∆w ∆x = Bw Bx (x, t) which implies, using the figure on the right below, that sin θ(x, t) = Bw Bx (x, t) b 1 + Bw Bx (x, t) 2 cos θ(x, t) = 1 b 1 + Bw Bx (x, t) 2 θ tan θ 1 ? 1 tan2 θ θ(x, t) = arctan Bw Bx (x, t) Bθ Bx(x, t) = B2w Bx2 (x, t) 1 + Bw Bx (x, t) 2 Substituting these formulae into (E1) give a horrendous mess. However, we can get considerable simplification by looking only at small vibrations. By a small vibration, we mean that \|θ(x, t)\| ! 1 for all x and t. This implies that \| tan θ(x, t)\| ! 1, hence that ˇ ˇBw Bx (x, t) ˇ ˇ ! 1 and hence that d 1 + Bw Bx 2 « 1 sin θ(x, t) « Bw Bx (x, t) cos θ(x, t) « 1 Bθ Bx(x, t) « B2w Bx2 (x, t) (E2) Substituting these into equation (E1) give ρ(x)B2w B t2 (x, t) = BT Bx (x, t)Bw Bx (x, t) + T(x, t) B2w Bx2 (x, t) + F(x, t) (E3) 180 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES which is indeed relatively simple, but still exhibits a problem. This is one equation in the two unknowns w and T. Fortunately there is a second equation lurking in the background, that we haven’t used yet. Namely, the horizontal component of Newton’s law of motion. As a second simplification, we assume that there are only transverse vibrations. That is, our tiny string element moves only vertically. Then the net horizontal force on it must be zero. That is, T(x + ∆x, t) cos θ(x + ∆x, t) ´ T(x, t) cos θ(x, t) = 0 Dividing by ∆x and taking the limit as ∆x tends to zero gives B Bx T(x, t) cos θ(x, t) = 0 Thus T(x, t) cos θ(x, t) is independent of x. For small amplitude vibrations, cos θ is very close to one, for all x. So T is a function of t only, which is determined by how hard you are pulling on the ends of the string at time t. So for small, transverse vibrations, (E3) simplifies further to ρ(x)B2w B t2 (x, t) = T(t) B2w Bx2 (x, t) + F(x, t) (E4) In the event that the string density ρ is a constant, independent of x, the string tension T(t) is a constant independent of t (in other words you are not continually playing with the tuning pegs) and there are no external forces F we end up with the wave equation B2w B t2 (x, t) = c2 B2w Bx2 (x, t) where c = d T ρ as desired. The equation that is called the wave equation has built into it a lot of approximations. By going through the derivation, we have seen what those approximations are, and we can get some idea as to when they are applicable. 2.9Ĳ Maximum and Minimum Values One of the core topics in single variable calculus courses is finding the maxima and min-ima of functions of one variable. We’ll now extend that discussion to functions of more than one variable43. Rather than leaping into the deep end, we’ll not be too ambitious and concentrate on functions of two variables. That being said, many of the techniques work more generally. To start, we have the following natural extensions to some familiar definitions. 43 Life is not (always) one-dimensional and sometimes we have to embrace it. 181 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES Let the function f (x, y) be defined for all (x, y) in some subset R of R2. Let (a, b) be a point in R. • (a, b) is a local maximum of f (x, y) if f (x, y) ď f (a, b) for all (x, y) close to (a, b). More precisely, (a, b) is a local maximum of f (x, y) if there is an r ą 0 such that f (x, y) ď f (a, b) for all points (x, y) within a distance r of (a, b). • (a, b) is a local minimum of f (x, y) if f (x, y) ě f (a, b) for all (x, y) close to (a, b). • Local maximum and minimum values are also called extremal values. • (a, b) is an absolute maximum or global maximum of f (x, y) if f (x, y) ď f (a, b) for all (x, y) in R. • (a, b) is an absolute minimum or global minimum of f (x, y) if f (x, y) ě f (a, b) for all (x, y) in R. Definition 2.9.1. § § Local Maxima and Minima One of the first things you did when you were developing the techniques used to find the maximum and minimum values of f (x) was ask yourself44 Suppose that the largest value of f (x) is f (a). What does that tell us about a? After a little thought you answered If the largest value of f (x) is f (a) and f is differentiable at a, then f 1(a) = 0. x y y “ fpxq Let’s recall why that’s true. Suppose that the largest value of f (x) is f (a). Then for all h ą 0, f (a + h) ď f (a) ù ñ f (a + h) ´ f (a) ď 0 ù ñ f (a + h) ´ f (a) h ď 0 if h ą 0 44 Or perhaps your instructor asked you. 182 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES Taking the limit h Ñ 0 tells us that f 1(a) ď 0. Similarly45, for all h ă 0, f (a + h) ď f (a) ù ñ f (a + h) ´ f (a) ď 0 ù ñ f (a + h) ´ f (a) h ě 0 if h ă 0 Taking the limit h Ñ 0 now tells us that f 1(a) ě 0. So we have both f 1(a) ě 0 and f 1(a) ď 0 which forces f 1(a) = 0. You also observed at the time that for this argument to work, you only need f (x) ď f (a) for all x’s close to a, not necessarily for all x’s in the whole world. (In the above inequalities, we only used f (a + h) with h small.) Since we care only about f (x) for x near a, we can refine the above statement. If f (a) is a local maximum for f (x) and f is differentiable at a, then f 1(a) = 0. Precisely the same reasoning applies to minima. If f (a) is a local minimum for f (x) and f is differentiable at a, then f 1(a) = 0. Let’s use the ideas of the above discourse to extend the study of local maxima and local minima to functions of more than one variable. Suppose that the function f (x, y) is defined for all (x, y) in some subset R of R2, that (a, b) is a point of R that is not on the boundary of R, and that f has a local maximum at (a, b). See the figure below. z y x R pa,b , fpa,bqq pa,bq z “ fpx, yq Then the function f (x, y) must decrease in value as (x, y) moves away from (a, b) in any direction. No matter which direction d we choose, the directional derivative of f at (a, b) in direction d must be zero or smaller. Writing this in mathematical symbols, we get D d \|d\| f (a, b) = ∇ ∇ ∇f (a, b) ¨ d \|d\| ď 0 And the directional derivative of f at (a, b) in the direction ´d also must be zero or nega-tive. D´ d \|d\| f (a, b) = ∇ ∇ ∇f (a, b) ¨ ´d \|d\| ď 0 which implies that ∇ ∇ ∇f (a, b) ¨ d \|d\| ě 0 45 Recall that if h ă 0 and A ď B, then hA ě hB. This is because the product of any two negative numbers is positive, so that h ă 0, A ď B ù ñ A ´ B ď 0 ù ñ h(A ´ B) ě 0 ù ñ hA ě hB. 183 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES As ∇ ∇ ∇f (a, b) ¨ d \|d\| must be both positive (or zero) and negative (or zero) at the same time, it must be zero. In particular, choosing d = ˆ ı ı ı forces the x component of ∇ ∇ ∇f (a, b) to be zero, and choosing d = ˆ ȷ ȷ ȷ forces the y component of ∇ ∇ ∇f (a, b) to be zero. We have thus shown that ∇ ∇ ∇f (a, b) = 0. The same argument shows that ∇ ∇ ∇f (a, b) = 0 when (a, b) is a local minimum too. This is an important and useful result, so let’s theoremise it. Let the function f (x, y) be defined for all (x, y) in some subset R of R2. Assume that ˝ (a, b) is a point of R that is not on the boundary of R and ˝ (a, b) is a local maximum or local minimum of f and that ˝ the partial derivatives of f exist at (a, b). Then ∇ ∇ ∇f (a, b) = 0. Theorem 2.9.2. Let f (x, y) be a function and let (a, b) be a point in its domain. Then • if ∇ ∇ ∇f (a, b) exists and is zero we call (a, b) a critical point (or a stationary point) of the function, and • if ∇ ∇ ∇f (a, b) does not exist then we call (a, b) a singular point of the function. Definition 2.9.3. Note that some people (and texts) combine both of these cases and call (a, b) a critical point when either the gradient is zero or does not exist. Warning 2.9.4. Theorem 2.9.2 tells us that every local maximum or minimum (in the interior of the domain of a function whose partial derivatives exist) is a critical point. Beware that it does not46 tell us that every critical point is either a local maximum or a local minimum. Warning 2.9.5. 46 A very common error of logic that people make is “Affirming the consequent”. “If P then Q” is true, does not imply that “If Q then P” is true . The statement “If he is Shakespeare then he is dead” is true. But concluding from “That man is dead” that “He must be Shakespeare” is just silly. 184 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES In fact, we shall see later47, in Examples 2.9.13 and 2.9.15, critical points that are neither local maxima nor a local minima. None-the-less, Theorem 2.9.2 is very useful because often functions have only a small number of critical points. To find local maxima and minima of such functions, we only need to consider its critical and singular points. We’ll return later to the question of how to tell if a critical point is a local maximum, local minimum or neither. For now, we’ll just practice finding critical points. Example 2.9.6 f (x, y) = x2 ´ 2xy + 2y2 + 2x ´ 6y + 12 Find all critical points of f (x, y) = x2 ´ 2xy + 2y2 + 2x ´ 6y + 12. Solution. To find the critical points, we need to find the gradient. To find the gradient we need to find the first order partial derivatives. So, as a preliminary calculation, we find the two first order partial derivatives of f (x, y). fx(x, y) = 2x ´ 2y + 2 fy(x, y) = ´2x + 4y ´ 6 So the critical points are the solutions of the pair of equations 2x ´ 2y + 2 = 0 ´ 2x + 4y ´ 6 = 0 or equivalently (dividing by two and moving the constants to the right hand side) x ´ y = ´1 (E1) ´x + 2y = 3 (E2) This is a system of two equations in two unknowns (x and y). One strategy for solving a system like this is to • First use one of the equations to solve for one of the unknowns in terms of the other unknown. For example, (E1) tells us that y = x + 1. This expresses y in terms of x. We say that we have solved for y in terms of x. • Then substitute the result, y = x + 1 in our case, into the other equation, (E2). In our case, this gives ´x + 2(x + 1) = 3 ð ñ x + 2 = 3 ð ñ x = 1 • We have now found that x = 1, y = x + 1 = 2 is the only solution. So the only critical point is (1, 2). Of course it only takes a moment to verify that ∇ ∇ ∇f (1, 2) = ⟨0, 0⟩. It is a good idea to do this as a simple check of our work. An alternative strategy for solving a system of two equations in two unknowns, like (E1) and (E2), is to 47 And you also saw, for example in Example 3.6.4 of the CLP-1 text, that critical points that are also inflection points are neither local maxima nor local minima. 185 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES • add equations (E1) and (E2) together. This gives (E1) + (E2) : (1 ´ 1)x + (´1 + 2)y = ´1 + 3 ð ñ y = 2 The point here is that adding equations (E1) and (E2) together eliminates the un-known x, leaving us with one equation in the unknown y, which is easily solved. For other systems of equations you might have to multiply the equations by some numbers before adding them together. • We now know that y = 2. Substituting it into (E1) gives us x ´ 2 = ´1 ù ñ x = 1 • Once again (thankfully) we have found that the only critical point is (1, 2). Example 2.9.6 This was pretty easy because we only had to solve linear equations, which in turn was a consequence of the fact that f (x, y) was a polynomial of degree two. Here is an example with some slightly more challenging algebra. Example 2.9.7 f (x, y) = 2x3 ´ 6xy + y2 + 4y Find all critical points of f (x, y) = 2x3 ´ 6xy + y2 + 4y. Solution. As in the last example, we need to find where the gradient is zero, and to find the gradient we need the first order partial derivatives. fx = 6x2 ´ 6y fy = ´6x + 2y + 4 So the critical points are the solutions of 6x2 ´ 6y = 0 ´ 6x + 2y + 4 = 0 We can rewrite the first equation as y = x2, which expresses y as a function of x. We can then substitute y = x2 into the second equation, giving ´6x + 2y + 4 = 0 ð ñ ´6x + 2x2 + 4 = 0 ð ñ x2 ´ 3x + 2 = 0 ð ñ (x ´ 1)(x ´ 2) = 0 ð ñ x = 1 or 2 When x = 1, y = 12 = 1 and when x = 2, y = 22 = 4. So, there are two critical points: (1, 1), (2, 4). Alternatively, we could have also used the second equation to write y = 3x ´ 2, and then substituted that into the first equation to get 6x2 ´ 6(3x ´ 2) = 0 ð ñ x2 ´ 3x + 2 = 0 just as above. Example 2.9.7 186 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES And here is an example for which the algebra requires a bit more thought. Example 2.9.8 ( f (x, y) = xy(5x + y ´ 15)) Find all critical points of f (x, y) = xy(5x + y ´ 15). Solution. The first order partial derivatives of f (x, y) = xy(5x + y ´ 15) are fx(x, y) = y(5x + y ´ 15) + xy(5) = y(5x + y ´ 15) + y(5x) = y(10x + y ´ 15) fy(x, y) = x(5x + y ´ 15) + xy(1) = x(5x + y ´ 15) + x(y) = x(5x + 2y ´ 15) The critical points are the solutions of fx(x, y) = fy(x, y) = 0. That is, we need to find all x, y that satisfy the pair of equations y(10x + y ´ 15) = 0 (E1) x(5x + 2y ´ 15) = 0 (E2) The first equation, y(10x + y ´ 15) = 0, is satisfied if at least one of the two factors y, (10x + y ´ 15) is zero. So the first equation is satisfied if at least one of the two equations y = 0 (E1a) 10x + y = 15 (E1b) is satisfied. The second equation, x(5x + 2y ´ 15) = 0, is satisfied if at least one of the two factors x, (5x + 2y ´ 15) is zero. So the second equation is satisfied if at least one of the two equations x = 0 (E2a) 5x + 2y = 15 (E2b) is satisfied. So both critical point equations (E1) and (E2) are satisfied if and only if at least one of (E1a), (E1b) is satisfied and in addition at least one of (E2a), (E2b) is satisfied. So both critical point equations (E1) and (E2) are satisfied if and only if at least one of the following four possibilities hold. • (E1a) and (E2a) are satisfied if and only if x = y = 0 • (E1a) and (E2b) are satisfied if and only if y = 0, 5x + 2y = 15 ð ñ y = 0, 5x = 15 • (E1b) and (E2a) are satisfied if and only if 10x + y = 15, x = 0 ð ñ y = 15, x = 0 • (E1b) and (E2b) are satisfied if and only if 10x + y = 15, 5x + 2y = 15. We can use, for example, the second of these equations to solve for x in terms of y: x = 1 5(15 ´ 2y). When we substitute this into the first equation we get 2(15 ´ 2y) + y = 15, which we can solve for y. This gives ´3y = 15 ´ 30 or y = 5 and then x = 1 5(15 ´ 2 ˆ 5) = 1. In conclusion, the critical points are (0, 0), (3, 0), (0, 15) and (1, 5). 187 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES A more compact way to write what we have just done is fx(x, y) = 0 and fy(x, y) = 0 ð ñ y(10x + y ´ 15) = 0 and x(5x + 2y ´ 15) = 0 ð ñ ␣ y = 0 or 10x + y = 15 ( and ␣ x = 0 or 5x + 2y = 15 ( ð ñ ␣ y = 0, x = 0 ( or ␣ y = 0, 5x + 2y = 15 ( or ␣ 10x + y = 15, x = 0 ( or ␣ 10x + y = 15, 5x + 2y = 15 ( ð ñ ␣ x = y = 0 ( or ␣ y = 0, x = 3 ( or ␣ x = 0, y = 15 ( or ␣ x = 1, y = 5 ( Example 2.9.8 Let’s try a more practical example — something from the real world. Well, a mathe-matician’s “real world”. The interested reader should search-engine their way to a dis-cussion of “idealisation”, “game theory” “Cournot models” and “Bertrand models”. But don’t spend too long there. A discussion of breweries is about to take place. Example 2.9.9 In a certain community, there are two breweries in competition48, so that sales of each neg-atively affect the profits of the other. If brewery A produces x litres of beer per month and brewery B produces y litres per month, then the profits of the two breweries are given by P = 2x ´ 2x2 + y2 106 Q = 2y ´ 4y2 + x2 2 ˆ 106 respectively. Find the sum of the two profits if each brewery independently sets its own production level to maximize its own profit and assumes that its competitor does likewise. Then, assuming cartel behaviour, find the sum of the two profits if the two breweries cooperate so as to maximize that sum49. Solution. If A adjusts x to maximize P (for y held fixed) and B adjusts y to maximize Q (for x held fixed) then x and y are determined by the equations Px = 2 ´ 4x 106 = 0 (E1) Qy = 2 ´ 8y 2ˆ106 = 0 (E2) Equation (E1) yields x = 1 2106 and equation (E2) yields y = 1 2106. Knowing x and y we can determine P, Q and the total profit P + Q = 2(x + y) ´ 1 106 5 2x2 + 3y2 = 1061 + 1 ´ 5 8 ´ 3 4 = 5 8106 48 We have both types of music here — country and western. 49 This sort of thing is generally illegal. 188 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES On the other hand if (A, B) adjust (x, y) to maximize P + Q = 2(x + y) ´ 1 106 5 2x2 + 3y2 , then x and y are determined by (P + Q)x = 2 ´ 5x 106 = 0 (E1) (P + Q)y = 2 ´ 6y 106 = 0 (E2) Equation (E1) yields x = 2 5106 and equation (E2) yields y = 1 3106. Again knowing x and y we can determine the total profit P + Q = 2(x + y) ´ 1 106 5 2x2 + 3y2 = 106 4 5 + 2 3 ´ 2 5 ´ 1 3 = 11 15106 So cooperating really does help their profits. Unfortunately, like a very small tea-pot, consumers will be a little poorer50. Example 2.9.9 Moving swiftly away from the last pun, let’s do something a little more geometric. Example 2.9.10 Equal angle bends are made at equal distances from the two ends of a 100 metre long fence so the resulting three segment fence can be placed along an existing wall to make an en-closure of trapezoidal shape. What is the largest possible area for such an enclosure? Solution. This is a very geometric problem (fenced off from pun opportunities), and as such we should start by drawing a sketch and introducing some variable names. x sin θ x x 100 ´ 2x θ θ The area enclosed by the fence is the area inside the blue rectangle (in the figure on the right above) plus the area inside the two blue triangles. A(x, θ) = (100 ´ 2x)x sin θ + 2 ¨ 1 2 ¨ x sin θ ¨ x cos θ = (100x ´ 2x2) sin θ + x2 sin θ cos θ 50 Sorry about the pun. 189 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES To maximize the area, we need to solve 0 = BA Bx = (100 ´ 4x) sin θ + 2x sin θ cos θ 0 = BA Bθ = (100x ´ 2x2) cos θ + x2␣ cos2 θ ´ sin2 θ ( Note that both terms in the first equation contain the factor sin θ and all terms in the second equation contain the factor x. If either sin θ or x are zero the area A(x, θ) will also be zero, and so will certainly not be maximal. So we may divide the first equation by sin θ and the second equation by x, giving (100 ´ 4x) + 2x cos θ = 0 (E1) (100 ´ 2x) cos θ + x ␣ cos2 θ ´ sin2 θ ( = 0 (E2) These equations might look a little scary. But there is no need to panic. They are not as bad as they look because θ enters only through cos θ and sin2 θ, which we can easily write in terms of cos θ. Furthermore we can eliminate cos θ by observing that the first equation forces cos θ = ´100´4x 2x and hence sin2 θ = 1 ´ cos2 θ = 1 ´ (100´4x)2 4x2 . Substituting these into the second equation gives ´(100 ´ 2x)100 ´ 4x 2x + x (100 ´ 4x)2 2x2 ´ 1 = 0 ù ñ ´(100 ´ 2x)(100 ´ 4x) + (100 ´ 4x)2 ´ 2x2 = 0 ù ñ 6x2 ´ 200x = 0 ù ñ x = 100 3 cos θ = ´´100/3 200/3 = 1 2 θ = 60˝ and the maximum area enclosed is A = 100100 3 ´ 21002 32 ? 3 2 + 1 2 1002 32 ? 3 2 = 2500 ? 3 Example 2.9.10 Now here is a very useful (even practical!) statistical example — finding the line that best fits a given collection of points. Example 2.9.11 (Linear regression) An experiment yields n data points (xi, yi), i = 1, 2, ¨ ¨ ¨ , n. We wish to find the straight line y = mx + b which “best” fits the data. The definition of “best” is “minimizes the 190 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES x y px1,y1q px2,y2q px3,y3q pxn,ynq y “ mx b root mean square error”, i.e. minimizes E(m, b) = n ÿ i=1 (mxi + b ´ yi)2 Note that • term number i in E(m, b) is the square of the difference between yi, which is the ith measured value of y, and h mx + b i x=xi , which is the approximation to yi given by the line y = mx + b. • All terms in the sum are positive, regardless of whether the points (xi, yi) are above or below the line. Our problem is to find the m and b that minimizes E(m, b). This technique for drawing a line through a bunch of data points is called “linear regression”. It is used a lot51 52. Even in the real world — and not just the real world that you find in mathematics problems. The actual real world that involves jobs. Solution. We wish to choose m and b so as to minimize E(m, b). So we need to determine where the partial derivatives of E are zero. 0 = BE Bm = n ÿ i=1 2(mxi + b ´ yi)xi = m h n ř i=1 2x2 i i + b h n ř i=1 2xi i ´ h n ř i=1 2xiyi i 0 = BE Bb = n ÿ i=1 2(mxi + b ´ yi) = m h n ř i=1 2xi i + b h n ř i=1 2 i ´ h n ř i=1 2yi i There are a lot of symbols here. But remember that all of the xi’s and yi’s are given con-stants. They come from, for example, experimental data. The only unknowns are m and b. To emphasize this, and to save some writing, define the constants Sx = n ř i=1 xi Sy = n ř i=1 yi Sx2 = n ř i=1 x2 i Sxy = n ř i=1 xiyi 51 Proof by search engine. 52 And has been used for a long time. It was introduced by the French mathematician Adrien-Marie Legendre, 1752–1833, in 1805, and by the German mathematician and physicist Carl Friedrich Gauss, 1777–1855, in 1809. 191 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES The equations which determine the critical points are (after dividing by two) Sx2 m + Sx b = Sxy (E1) Sx m + n b = Sy (E2) These are two linear equations on the unknowns m and b. They may be solved in any of the usual ways. One is to use (E2) to solve for b in terms of m b = 1 n Sy ´ Sxm (E3) and then substitute this into (E1) to get the equation Sx2 m + 1 nSx Sy ´ Sxm = Sxy ù ñ nSx2 ´ S2 x m = nSxy ´ SxSy for m. We can then solve this equation for m and substitute back into (E3) to get b. This gives m = nSxy ´ SxSy nSx2 ´ S2 x b = Sy n nSx2 ´ S2 x nSx2 ´ S2 x ´ Sx n nSxy ´ SxSy nSx2 ´ S2 x = nSySx2 ´ nSxSxy n(nSx2 ´ S2 x) = ´SxSxy ´ SySx2 nSx2 ´ S2 x Another way to solve the system of equations is n(E1) ´ Sx(E2) : h nSx2 ´ S2 x i m = nSxy ´ SxSy ´Sx(E1) + Sx2(E2) : h nSx2 ´ S2 x i b = ´SxSxy + SySx2 which gives the same solution. So given a bunch of data points, it only takes a quick bit of arithmetic — no calculus required — to apply the above formulae and so to find the best fitting line. Of course while you don’t need any calculus to apply the formulae, you do need calculus to understand where they came from. The same technique can be extended to other types of curve fitting problems. For example, polynomial regression. Example 2.9.11 § § The Second Derivative Test Now let’s start thinking about how to tell if a critical point is a local minimum or max-imum. Remember what happens for functions of one variable. Suppose that x = a is a critical point of the function f (x). Any (sufficiently smooth) function is well approxi-mated, when x is close to a, by the first few terms of its Taylor expansion f (x) = f (a) + f 1(a) (x ´ a) + 1 2 f 2(a) (x ´ a)2 + 1 3! f (3)(a) (x ´ a)3 + ¨ ¨ ¨ 192 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES As a is a critical point, we know that f 1(a) = 0 and f (x) = f (a) + 1 2 f 2(a) (x ´ a)2 + 1 3! f (3)(a) (x ´ a)3 + ¨ ¨ ¨ If f 2(a) ‰ 0, f (x) is going to look a lot like f (a) + 1 2 f 2(a) (x ´ a)2 when x is really close to a. In particular • if f 2(a) ą 0, then we will have f (x) ą f (a) when x is close to (but not equal to) a, so that a will be a local minimum and • if f 2(a) ă 0, then we will have f (x) ă f (a) when x is close to (but not equal to) a, so that a will be a local maximum, but • if f 2(a) = 0, then we cannot draw any conclusions without more work. A similar, but messier, analysis is possible for functions of two variables. Here are some simple quadratic examples that provide a warmup for that messier analysis. Example 2.9.12 f (x, y) = x2 + 3xy + 3y2 ´ 6x ´ 3y ´ 6 Consider f (x, y) = x2 + 3xy + 3y2 ´ 6x ´ 3y ´ 6. The gradient of f is ∇ ∇ ∇f (x, y) = 2x + 3y ´ 6 ˆ ı ı ı + 3x + 6y ´ 3 ˆ ȷ ȷ ȷ So (x, y) is a critical point of f if and only if 2x + 3y = 6 (E1) 3x + 6y = 3 (E2) Multiplying the first equation by 2 and subtracting the second equation gives x = 9 (2(E1) - (E2)) Then substituting x = 9 back into the first equation gives 2 ˆ 9 + 3y = 6 ù ñ y = ´4 So f (x, y) has precisely one critical point, namely (9 , ´4). Now let’s try to determine if f (x, y) has a local minimum, or a local maximum, or neither, at (9, ´4). A good way to determine the behaviour of f (x, y) for (x, y) near (9, ´4) is to make the change of variables53 x = 9 + ∆x y = ´4 + ∆y and study the behaviour of f for ∆x and ∆y near zero. f 9 + ∆x , ´4 + ∆y = (9 + ∆x)2 + 3(9 + ∆x)(´4 + ∆y) + 3(´4 + ∆y)2 ´ 6(9 + ∆x) ´ 3(´4 + ∆y) ´ 6 = (∆x)2 + 3∆x ∆y + 3(∆y)2 ´ 27 53 This is equivalent to translating the graph so that the critical point lies at (0, 0). 193 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES And a good way to study the sign of quadratic expressions like (∆x)2 + 3∆x ∆y + 3(∆y)2 is to complete the square. So far you have probably just completed the square for quadratic expressions that involve only a single variable. For example x2 + 3x + 3 = x + 3 2 2 ´ 9 4 + 3 When there are two variables around, like ∆x and ∆y, you can just pretend that one of them is a constant and complete the square as before. For example, if you pretend that ∆y is a constant, (∆x)2 + 3∆x ∆y + 3(∆y)2 = ∆x + 3 2∆y 2 + 3 ´ 9 4 (∆y)2 = ∆x + 3 2∆y 2 + 3 4(∆y)2 To this point, we have expressed f 9 + ∆x , ´4 + ∆y = ∆x + 3 2∆y 2 + 3 4(∆y)2 ´ 27 As the smallest values of ∆x + 3 2∆y 2 and 3 4(∆y)2 are both zero, we have that f (x, y) = f 9 + ∆x , ´4 + ∆y ě ´27 = f (9, ´4) for all (x, y) so that (9, ´4) is both a local minimum and a global minimum for f. Example 2.9.12 You have already encountered single variable functions that have a critical point which is neither a local max nor a local min. See Example 3.5.9 in the CLP-1 text. Here are a couple of examples which show that this can also happen for functions of two variables. We’ll start with the simplest possible such example. Example 2.9.13 f (x, y) = x2 ´ y2 The first partial derivatives of f (x, y) = x2 ´ y2 are fx(x, y) = 2x and fy(x, y) = ´2y. So the only critical point of this function is (0, 0). Is this a local minimum or maximum? Well let’s start with (x, y) at (0, 0) and then move (x, y) away from (0, 0) and see if f (x, y) gets bigger or smaller. At the origin f (0, 0) = 0. Of course we can move (x, y) away from (0, 0) in many different directions. • First consider moving (x, y) along the x-axis. Then (x, y) = (x, 0) and f (x, y) = f (x, 0) = x2. So when we start with x = 0 and then increase x, the value of the function f increases — which means that (0, 0) cannot be a local maximum for f. • Next let’s move (x, y) away from (0, 0) along the y-axis. Then (x, y) = (0, y) and f (x, y) = f (0, y) = ´y2. So when we start with y = 0 and then increase y, the value of the function f decreases — which means that (0, 0) cannot be a local minimum for f. 194 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES So moving away from (0, 0) in one direction causes the value of f to increase, while mov-ing away from (0, 0) in a second direction causes the value of f to decrease. Consequently (0, 0) is neither a local minimum or maximum for f. It is called a saddle point, because the graph of f looks like a saddle. (The full definition of “saddle point” is given immediately after this example.) Here are some figures showing the graph of f. The figure below show some level curves of f. Observe from the level curves that • f increases as you leave (0, 0) walking along the x axis • f decreases as you leave (0, 0) walking along the y axis x y f=0 f=1 f=1 f=−1 f=−1 f=4 f=4 f=−4 f=−4 f=9 f=9 f=−9 f=−9 Example 2.9.13 Approximately speaking, if a critical point (a, b) is neither a local minimum nor a local maximum, then it is a saddle point. For (a, b) to not be a local minimum, f has to take val-ues bigger than f (a, b) at some points nearby (a, b). For (a, b) to not be a local maximum, f has to take values smaller than f (a, b) at some points nearby (a, b). Writing this more mathematically we get the following definition. 195 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES The critical point (a, b) is called a saddle point for the function f (x, y) if, for each r ą 0, • there is at least one point (x, y), within a distance r of (a, b), for which f (x, y) ą f (a, b) and • there is at least one point (x, y), within a distance r of (a, b), for which f (x, y) ă f (a, b). Definition 2.9.14. Here is another example of a saddle point. This time we have to work a bit to see it. Example 2.9.15 f (x, y) = x2 ´ 2xy ´ y2 + 4y ´ 2 Consider f (x, y) = x2 ´ 2xy ´ y2 + 4y ´ 2. The gradient of f is ∇ ∇ ∇f (x, y) = 2x ´ 2y ˆ ı ı ı + ´ 2x ´ 2y + 4 ˆ ȷ ȷ ȷ So (x, y) is a critical point of f if and only if 2x ´ 2y = 0 ´2x ´ 2y = ´4 The first equation gives that x = y. Substituting y = x into the second equation gives ´2y ´ 2y = ´4 ù ñ x = y = 1 So f (x, y) has precisely one critical point, namely (1, 1). To determine if f (x, y) has a local minimum, or a local maximum, or neither, at (1, 1), we proceed as in Example 2.9.12. We make the change of variables x = 1 + ∆x y = 1 + ∆y to give f 1 + ∆x , 1 + ∆y = (1 + ∆x)2 ´ 2(1 + ∆x)(1 + ∆y) ´ (1 + ∆y)2 + 4(1 + ∆y) ´ 2 = (∆x)2 ´ 2∆x ∆y ´ (∆y)2 Completing the square, f 1 + ∆x , 1 + ∆y = (∆x)2 ´ 2∆x ∆y ´ (∆y)2 = (∆x ´ ∆y)2 ´ 2(∆y)2 Notice that f has now been written as the difference of two squares, much like the f in the saddle point Example 2.9.13. • If ∆x and ∆y are such that the first square (∆x ´ ∆y)2 is nonzero, but the second square (∆y)2 is zero, then f 1 + ∆x , 1 + ∆y = (∆x ´ ∆y)2 ą 0 = f (1, 1). That is, whenever ∆y = 0 and ∆x ‰ ∆y, then f 1 + ∆x , 1 + ∆y = (∆x ´ ∆y)2 ą 0 = f (1, 1). 196 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES • On the other hand, if ∆x and ∆y are such that the first square (∆x ´ ∆y)2 is zero but the second square (∆y)2 is nonzero, then f 1 + ∆x , 1 + ∆y = ´2(∆y)2 ă 0 = f (1, 1). That is, whenever ∆x = ∆y ‰ 0, then f 1 + ∆x , 1 + ∆y = ´2(∆y)2 ă 0 = f (1, 1). f “ 0 f ă 0 f ă 0 f ą 0 f ą 0 x y p1, 1q So • f (x, y) ą f (1, 1) at all points on the blue line in the figure above, and • f (x, y) ă f (1, 1) at all point on the red line. We conclude that (1, 1) is the only critical point for f (x, y), and furthermore that it is a saddle point. Example 2.9.15 The above three examples show that we can find all critical points of quadratic func-tions of two variables. We can also classify each critical point as either a minimum, a maximum or a saddle point. Of course not every function is quadratic. But by using the quadratic approximation (2.6.12) we can apply the same ideas much more generally. Suppose that (a, b) is a critical point of some function f (x, y). For ∆x and ∆y small, the quadratic approximation (2.6.12) gives f a + ∆x , b + ∆y « f a , b + fx a , b ∆x + fy a , b ∆y + 1 2 ! fxx a, b ∆x2 + 2fxy a, b ∆x∆y + fyy a, b ∆y2) = f a , b + 1 2 ! fxx a, b ∆x2 + 2fxy a, b ∆x∆y + fyy a, b ∆y2) (˚) since (a, b) is a critical point so that fx(a, b) = fy(a, b) = 0. Then using the technique of Examples 2.9.12 and 2.9.15, we get54 (details below) 54 There are analogous results in higher dimensions that are accessible to people who have learned some linear algebra. They are derived by diagonalizing the matrix of second derivatives, which is called the Hessian matrix. 197 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES Let r ą 0 and assume that all second order derivatives of the function f (x, y) are continuous at all points (x, y) that are within a distance r of (a, b). Assume that fx(a, b) = fy(a, b) = 0. Define D(x, y) = fxx(x, y) fyy(x, y) ´ fxy(x, y)2 It is called the discriminant of f. Then • if D(a, b) ą 0 and fxx(a, b) ą 0, then f (x, y) has a local minimum at (a, b), • if D(a, b) ą 0 and fxx(a, b) ă 0, then f (x, y) has a local maximum at (a, b), • if D(a, b) ă 0, then f (x, y) has a saddle point at (a, b), but • if D(a, b) = 0, then we cannot draw any conclusions without more work. Theorem 2.9.16 (Second Derivative Test). “Proof”. We are putting quotation marks around the word “Proof”, because we are not go-ing to justify the fact that it suffices to analyse the quadratic approximation in equation (˚). Let’s temporarily suppress the arguments (a, b). If fxx(a, b) ‰ 0, then by completing the square we can write fxx ∆x2 + 2fxy ∆x∆y + fyy ∆y2 = fxx ∆x + fxy fxx ∆y 2 + fyy ´ f 2 xy fxx ! ∆y2 = 1 fxx !fxx ∆x + fxy ∆y 2 + fxx fyy ´ f 2 xy ∆y2) Similarly, if fyy(a, b) ‰ 0, fxx ∆x2 + 2fxy ∆x∆y + fyy ∆y2 = 1 fyy !fxy ∆x + fyy ∆y 2 + fxx fyy ´ f 2 xy ∆x2) Note that this algebra breaks down if fxx(a, b) = fyy(a, b) = 0. We’ll deal with that case shortly. More importantly, note that • if fxx fyy ´ f 2 xy ą 0 then both fxx and fyy must be nonzero and of the same sign and furthermore, whenever ∆x or ∆y are nonzero, !fxx ∆x + fxy ∆y 2 + fxx fyy ´ f 2 xy ∆y2) ą 0 and !fxy ∆x + fyy ∆y 2 + fxx fyy ´ f 2 xy ∆x2) ą 0 so that, recalling (˚), – if fxx(a, b) ą 0, then (a, b) is a local minimum and – if fxx(a, b) ă 0, then (a, b) is a local maximum. 198 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES • If fxx fyy ´ f 2 xy ă 0 and fxx is nonzero then !fxx ∆x + fxy ∆y 2 + fxx fyy ´ f 2 xy ∆y2) is strictly positive whenever ∆x ‰ 0, ∆y = 0 and is strictly negative whenever fxx ∆x + fxy ∆y = 0, ∆y ‰ 0, so that (a, b) is a saddle point. Similarly, (a, b) is also a saddle point if fxx fyy ´ f 2 xy ă 0 and fyy is nonzero. • Finally, if fxy ‰ 0 and fxx = fyy = 0, then fxx ∆x2 + 2fxy ∆x ∆y + fyy ∆y2 = 2fxy ∆x ∆y is strictly positive for one sign of ∆x ∆y and is strictly negative for the other sign of ∆x ∆y. So (a, b) is again a saddle point. You might wonder why, in the local maximum/local minimum cases of Theorem 2.9.16, fxx(a, b) appears rather than fyy(a, b). The answer is only that x is before y in the alphabet55. You can use fyy(a, b) just as well as fxx(a, b). The reason is that if D(a, b) ą 0 (as in the first two bullets of the theorem), then because D(a, b) = fxx(a, b) fyy(a, b) ´ fxy(a, b)2 ą 0, we necessarily have fxx(a, b) fyy(a, b) ą 0 so that fxx(a, b) and fyy(a, b) must have the same sign — either both are positive or both are negative. You might also wonder why we cannot draw any conclusions when D(a, b) = 0 and what happens then. The second derivative test for functions of two variables was derived in precisely the same way as the second derivative test for functions of one variable is derived — you approximate the function by a polynomial that is of degree two in (x ´ a), (y ´ b) and then you analyze the behaviour of the quadratic polynomial near (a, b). For this to work, the contributions to f (x, y) from terms that are of degree two in (x ´ a), (y ´ b) had better be bigger than the contributions to f (x, y) from terms that are of degree three and higher in (x ´ a), (y ´ b) when (x ´ a), (y ´ b) are really small. If this is not the case, for example when the terms in f (x, y) that are of degree two in (x ´ a), (y ´ b) all have coefficients that are exactly zero, the analysis will certainly break down. That’s exactly what happens when D(a, b) = 0. Here are some examples. The functions f1(x, y) = x4 + y4 f2(x, y) = ´x4 ´ y4 f3(x, y) = x3 + y3 f4(x, y) = x4 ´ y4 all have (0, 0) as the only critical point and all have D(0, 0) = 0. The first, f1 has its minimum there. The second, f2, has its maximum there. The third and fourth have a saddle point there. Here are sketches of some level curves for each of these four functions (with all re-named to simply f). 55 The shackles of convention are not limited to mathematics. Election ballots often have the candidates listed in alphabetic order. 199 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES x y f“0 f“0.1 f“1 f“4 f“9 level curves of fpx, yq “ x4 y4 x y f“0 f“´0.1 f“´1 f“´4 f“´9 level curves of fpx, yq “ ´x4 ´ y4 x y f“0 f“1 f“´1 f“4 f“´4 level curves of fpx, yq “ x3 y3 x y f“0 f“0 f“1 f“1 f“´1 f“´1 f“4 f“4 f“´4 f“´4 level curves of fpx, yq “ x4 ´ y4 Example 2.9.17 f (x, y) = 2x3 ´ 6xy + y2 + 4y Find and classify all critical points of f (x, y) = 2x3 ´ 6xy + y2 + 4y. Solution. Thinking a little way ahead, to find the critical points we will need the gradient and to apply the second derivative test of Theorem 2.9.16 we will need all second order partial derivatives. So we need all partial derivatives of order up to two. Here they are. f = 2x3 ´ 6xy + y2 + 4y fx = 6x2 ´ 6y fxx = 12x fxy = ´6 fy = ´6x + 2y + 4 fyy = 2 fyx = ´6 (Of course, fxy and fyx have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.) We have already found, in Example 2.9.7, that the critical points are (1, 1), (2, 4). The classification is 200 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES critical point fxx fyy ´ f 2 xy fxx type (1, 1) 12 ˆ 2 ´ (´6)2 ă 0 saddle point (2, 4) 24 ˆ 2 ´ (´6)2 ą 0 24 local min We were able to leave the fxx entry in the top row blank, because • we knew that fxx(1, 1) fyy(1, 1) ´ f 2 xy(1, 1) ă 0, and • we knew, from Theorem 2.9.16, that fxx(1, 1) fyy(1, 1) ´ f 2 xy(1, 1) ă 0, by itself, was enough to ensure that (1, 1) was a saddle point. Here is a sketch of some level curves of our f (x, y). They are not needed to answer this x y p1,1q, fp1,1q“1 fp2,4q“0, p2,4q f“0 f“0.25 f“0.5 f“0.5 f“1 f“1 f“2 f“2 f“3 f“3 question, but can give you some idea as to what the graph of f looks like. Example 2.9.17 Example 2.9.18 ( f (x, y) = xy(5x + y ´ 15)) Find and classify all critical points of f (x, y) = xy(5x + y ´ 15). Solution. We have already computed the first order partial derivatives fx(x, y) = y(10x + y ´ 15) fy(x, y) = x(5x + 2y ´ 15) of f (x, y) in Example 2.9.8. Again, to classify the critical points we need the second order partial derivatives. They are fxx(x, y) = 10y fyy(x, y) = 2x fxy(x, y) = (1)(10x + y ´ 15) + y(1)= 10x + 2y ´ 15 fyx(x, y) = (1)(5x + 2y ´ 15) + x(5)= 10x + 2y ´ 15 201 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES (Once again, we have computed both fxy and fyx to guard against mechanical errors.) We have already found, in Example 2.9.8, that the critical points are (0, 0), (0, 15), (3, 0) and (1, 5). The classification is critical point fxx fyy ´ f 2 xy fxx type (0, 0) 0 ˆ 0 ´ (´15)2 ă 0 saddle point (0, 15) 150 ˆ 0 ´ 152 ă 0 saddle point (3, 0) 0 ˆ 6 ´ 152 ă 0 saddle point (1, 5) 50 ˆ 2 ´ 52 ą 0 75 local min Here is a sketch of some level curves of our f (x, y). f is negative in the shaded re-gions and f is positive in the unshaded regions. Again this is not needed to answer this x y p3,0q, fp3,0q“0 fp0,0q“0, p0,0q p0,15q, fp0,15q“0 fp1,5q“´25, p1,5q f“0 f“´10 f“´20 f“´20 f“´20 f“20 f“20 f“20 question, but can give you some idea as to what the graph of f looks like. Example 2.9.18 Example 2.9.19 Find and classify all of the critical points of f (x, y) = x3 + xy2 ´ 3x2 ´ 4y2 + 4. Solution. We know the drill now. We start by computing all of the partial derivatives of f up to order 2. f = x3 + xy2 ´ 3x2 ´ 4y2 + 4 fx = 3x2 + y2 ´ 6x fxx = 6x ´ 6 fxy = 2y fy = 2xy ´ 8y fyy = 2x ´ 8 fyx = 2y 202 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES The critical points are then the solutions of fx = 0, fy = 0. That is fx = 3x2 + y2 ´ 6x = 0 (E1) fy = 2y(x ´ 4) = 0 (E2) The second equation, 2y(x ´ 4) = 0, is satisfied if and only if at least one of the two equations y = 0 and x = 4 is satisfied. • When y = 0, equation (E1) forces x to obey 0 = 3x2 + 02 ´ 6x = 3x(x ´ 2) so that x = 0 or x = 2. • When x = 4, equation (E1) forces y to obey 0 = 3 ˆ 42 + y2 ´ 6 ˆ 4 = 24 + y2 which is impossible. So, there are two critical points: (0, 0), (2, 0). Here is a table that classifies the critical points. critical point fxx fyy ´ f 2 xy fxx type (0, 0) (´6) ˆ (´8) ´ 02 ą 0 ´6 ă 0 local max (2, 0) 6 ˆ (´4) ´ 02 ă 0 saddle point Example 2.9.19 Example 2.9.20 A manufacturer wishes to make an open rectangular box of given volume V using the least possible material. Find the design specifications. Solution. Denote by x, y and z, the length, width and height, respectively, of the box. y x z The box has two sides of area xz, two sides of area yz and a bottom of area xy. So the total surface area of material used is S = 2xz + 2yz + xy 203 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES However the three dimensions x, y and z are not independent. The requirement that the box have volume V imposes the constraint xyz = V We can use this constraint to eliminate one variable. Since z is at the end of the alphabet (poor z), we eliminate z by substituting z = V xy. So we have find the values of x and y that minimize the function S(x, y) = 2V y + 2V x + xy Let’s start by finding the critical points of S. Since Sx(x, y) = ´2V x2 + y Sy(x, y) = ´2V y2 + x (x, y) is a critical point if and only if x2y = 2V (E1) xy2 = 2V (E2) Solving (E1) for y gives y = 2V x2 . Substituting this into (E2) gives x4V2 x4 = 2V ù ñ x3 = 2V ù ñ x = 3 ? 2V and y = 2V (2V)2/3 = 3 ? 2V As there is only one critical point, we would expect it to give the minimum56. But let’s use the second derivative test to verify that at least the critical point is a local minimum. The various second partial derivatives are Sxx(x, y) = 4V x3 Sxx 3 ? 2V , 3 ? 2V = 2 Sxy(x, y) = 1 Sxy 3 ? 2V , 3 ? 2V = 1 Syy(x, y) = 4V y3 Syy 3 ? 2V , 3 ? 2V = 2 So Sxx 3 ? 2V , 3 ? 2V Syy 3 ? 2V , 3 ? 2V ´ Sxy 3 ? 2V , 3 ? 2V 2 = 3 ą 0 Sxx 3 ? 2V , 3 ? 2V = 2 ą 0 and, by Theorem 2.9.16.b, 3 ? 2V , 3 ? 2V is a local minimum and the desired dimensions are x = y = 3 ? 2V z = 3 c V 4 Note that our solution has x = y. That’s a good thing — the function S(x, y) is symmetric in x and y. Because the box has no top, the symmetry does not extend to z. Example 2.9.20 56 Indeed one can use the facts that 0 ă x ă 8, that 0 ă y ă 8, and that S Ñ 8 as x Ñ 0 and as y Ñ 0 and as x Ñ 8 and as y Ñ 8 to prove that the single critical point gives the global minimum. 204 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES 2.9.1 § § Absolute Minima and Maxima Of course a local maximum or minimum of a function need not be the absolute maximum of minimum. We’ll now consider how to find the absolute maximum and minimum. Let’s start by reviewing how one finds the absolute maximum and minimum of a function of one variable on an interval. For concreteness, let’s suppose that we want to find the extremal57 values of a function f (x) on the interval 0 ď x ď 1. If an extremal value is attained at some x = a which is in the interior of the interval, i.e. if 0 ă a ă 1, then a is also a local maximum or minimum and so has to be a critical point of f. But if an extremal value is attained at a boundary point a of the interval, i.e. if a = 0 or a = 1, then a need not be a critical point of f. This happens, for example, when f (x) = x. The largest value of f (x) on the interval 0 ď x ď 1 is 1 and is attained at x = 1, but f 1(x) = 1 is never zero, so that f has no critical points. x y y = f(x) = x 1 1 So to find the maximum and minimum of the function f (x) on the interval [0, 1], you 1. build up a list of all candidate points 0 ď a ď 1 at which the maximum or minimum could be attained, by finding all a’s for which either (a) 0 ă a ă 1 and f 1(a) = 0 or (b) 0 ă a ă 1 and f 1(a) does not exist58 or (c) a is a boundary point, i.e. a = 0 or a = 1, 2. and then you evaluate f (a) at each a on the list of candidates. The biggest of these candidate values of f (a) is the absolute maximum and the smallest of these candi-date values is the absolute minimum. The procedure for finding the maximum and minimum of a function of two variables, f (x, y) in a set like, for example, the unit disk x2 + y2 ď 1, is similar. You again 1. build up a list of all candidate points (a, b) in the set at which the maximum or minimum could be attained, by finding all (a, b)’s for which either59 (a) (a, b) is in the interior of the set (for our example, a2 + b2 ă 1) and fx(a, b) = fy(a, b) = 0 or (b) (a, b) is in the interior of the set and fx(a, b) or fy(a, b) does not exist or 57 Recall that “extremal value” means “either maximum value or minimum value”. 58 Recall that if f 1(a) does not exist, then a is called a singular point of f. 59 This is probably a good time to review the statement of Theorem 2.9.2. 205 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES (c) (a, b) is a boundary60 point, (for our example, a2 + b2 = 1), and could give the maximum or minimum on the boundary — more about this shortly — 2. and then you evaluate f (a, b) at each (a, b) on the list of candidates. The biggest of these candidate values of f (a, b) is the absolute maximum and the smallest of these candidate values is the absolute minimum. The boundary of a set, like x2 + y2 ď 1, in R2 is a curve, like x2 + y2 = 1. This curve is a one dimensional set, meaning that it is like a deformed x-axis. We can find the maximum and minimum of f (x, y) on this curve by converting f (x, y) into a function of one variable (on the curve) and using the standard function of one variable techniques. This is best explained by some examples. Example 2.9.21 Find the maximum and minimum of T(x, y) = (x + y)e´x2´y2 on the region defined by x2 + y2 ď 1 (i.e. on the unit disk). Solution. Let’s follow our checklist. First critical points, then points where the partial derivatives don’t exist, and finally the boundary. Interior Critical Points: If T takes its maximum or minimum value at a point in the interior, x2 + y2 ă 1, then that point must be either a critical point of T or a singular point of T. To find the critical points we compute the first order derivatives. Tx(x, y) = (1 ´ 2x2 ´ 2xy)e´x2´y2 Ty(x, y) = (1 ´ 2xy ´ 2y2)e´x2´y2 Because the exponential e´x2´y2 is never zero, the critical points are the solutions of Tx = 0 ð ñ 2x(x + y) = 1 Ty = 0 ð ñ 2y(x + y) = 1 • As both 2x(x + y) and 2y(x + y) are nonzero, we may divide the two equations, which gives x y = 1, forcing x = y. • Substituting this into either equation gives 2x(2x) = 1 so that x = y = ˘1/2. So the only critical points are (1/2, 1/2) and (´1/2, ´1/2). Both are in x2 + y2 ă 1. Singular points: In this problem, there are no singular points. Boundary: Points on the boundary satisfy x2 + y2 = 1. That is they lie on a circle. We may use the figure below to express x = cos t and y = sin t, in terms of the angle t. This will make the formula for T on the boundary quite a bit easier to deal with. On the boundary, T = (cos t + sin t)e´ cos2 t´sin2 t = (cos t + sin t)e´1 As all t’s are allowed, this function takes its max and min at zeroes of 60 It should intuitively obvious from a sketch that the boundary of the disk x2 + y2 ď 1 is the circle x2 + y2 = 1. But if you really need a formal definition, here it is. A point (a, b) is on the boundary of a set S if there is a sequence of points in S that converges to (a, b) and there is also a sequence of points in the complement of S that converges to (a, b). 206 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES x y (cos t, sin t) t 1 dT dt = ´ sin t + cos t e´1 That is, (cos t + sin t)e´1 takes its max and min • when sin t = cos t, • that is, when x = y and x2 + y2 = 1, • which forces x2 + x2 = 1 and hence x = y = ˘ 1 ? 2. All together, we have the following candidates for max and min, with the max and min indicated. point (1 2, 1 2) (´1 2, ´1 2) ( 1 ? 2, 1 ? 2) (´ 1 ? 2, ´ 1 ? 2) value of T 1 ?e « 0.61 ´ 1 ?e ? 2 e « 0.52 ´ ? 2 e max min The following sketch shows all of the critical points. It is a good idea to make such a sketch so that you don’t accidentally include a critical point that is outside of the allowed region. x y ( 1 2, 1 2) (−1 2, −1 2) ( 1 √ 2, 1 √ 2) (−1 √ 2, −1 √ 2) Example 2.9.21 207 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES In the last example, we analyzed the behaviour of f on the boundary of the region of interest by using the parametrization x = cos t, y = sin t of the circle x2 + y2 = 1. Sometimes using this parametrization is not so clean. And worse, some curves don’t have such a simple parametrization. In the next problem we’ll look at the boundary a little differently. Example 2.9.22 Find the maximum and minimum values of f (x, y) = x3 + xy2 ´ 3x2 ´ 4y2 + 4 on the disk x2 + y2 ď 1. Solution. Again, we first find all critical points, then find all singular points and, finally, analyze the boundary. Interior Critical Points: If f takes its maximum or minimum value at a point in the interior, x2 + y2 ă 1, then that point must be either a critical point of f or a singular point of f. To find the critical points61 we compute the first order derivatives. fx = 3x2 + y2 ´ 6x fy = 2xy ´ 8y The critical points are the solutions of fx = 3x2 + y2 ´ 6x = 0 (E1) fy = 2y(x ´ 4) = 0 (E2) The second equation, 2y(x ´ 4) = 0, is satisfied if and only if at least one of the two equations y = 0 and x = 4 is satisfied. • When y = 0, equation (E1) forces x to obey 0 = 3x2 + 02 ´ 6x = 3x(x ´ 2) so that x = 0 or x = 2. • When x = 4, equation (E1) forces y to obey 0 = 3 ˆ 42 + y2 ´ 6 ˆ 4 = 24 + y2 which is impossible. So, there are only two critical points: (0, 0), (2, 0). Singular points: In this problem, there are no singular points. Boundary: On the boundary, x2 + y2 = 1, we could again take advantage of having a circle and write x = cos t and y = sin t. But, for practice, we’ll use another method62. We know that (x, y) satisfies x2 + y2 = 1, and hence y2 = 1 ´ x2. Examining the formula for f (x, y), 61 We actually found the critical points in Example 2.9.19. But, for the convenience of the reader, we’ll repeat that here. 62 Even if you don’t believe that “you can’t have too many tools”, it is pretty dangerous to have to rely on just one tool. 208 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES we see that it contains only even63 powers of y, so we can eliminate y by substituting y2 = 1 ´ x2 into the formula. f = x3 + x(1 ´ x2) ´ 3x2 ´ 4(1 ´ x2) + 4 = x + x2 The max and min of x + x2 for ´1 ď x ď 1 must occur either • when x = ´1 (ñ y = f = 0) or • when x = +1 (ñ y = 0, f = 2) or • when 0 = d dx(x + x2) = 1 + 2x (ñ x = ´1 2, y = ˘ b 3 4, f = ´1 4). Here is a sketch showing all of the points that we have identified. x y (0, 0) (2, 0) (1, 0) (−1, 0) (−1 2, √ 3 2 ) (−1 2, − √ 3 2 ) Note that the point (2, 0) is outside the allowed region64. So all together, we have the following candidates for max and min, with the max and min indicated. point (0, 0) (´1, 0) (1, 0) ´ 1 2, ˘ ? 3 2 value of f 4 0 2 ´1 4 max min Example 2.9.22 Example 2.9.23 Find the maximum and minimum values of f (x, y) = xy ´ x3y2 when (x, y) runs over the square 0 ď x ď 1, 0 ď y ď 1. 63 If it contained odd powers too, we could consider the cases y ě 0 and y ď 0 separately and substitute y = ? 1 ´ x2 in the former case and y = ´ ? 1 ´ x2 in the latter case. 64 We found (2, 0) as a solution to the critical point equations (E1), (E2). That’s because, in the course of solving those equations, we ignored the constraint that x2 + y2 ď 1. 209 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES Solution. As usual, let’s examine the critical points, singular points and boundary in turn. Interior Critical Points: If f takes its maximum or minimum value at a point in the interior, 0 ă x ă 1, 0 ă y ă 1, then that point must be either a critical point of f or a singular point of f. To find the critical points we compute the first order derivatives. fx(x, y) = y ´ 3x2y2 fy(x, y) = x ´ 2x3y The critical points are the solutions of fx = 0 ð ñ y(1 ´ 3x2y) = 0 ð ñ y = 0 or 3x2y = 1 fy = 0 ð ñ x(1 ´ 2x2y) = 0 ð ñ x = 0 or 2x2y = 1 • If y = 0, we cannot have 2x2y = 1, so we must have x = 0. • If 3x2y = 1, we cannot have x = 0, so we must have 2x2y = 1. Dividing gives 1 = 3x2y 2x2y = 3 2 which is impossible. So the only critical point in the square is (0, 0). There f = 0. Singular points: Yet again there are no singular points in this problem. Boundary: The region is a square, so its boundary consists of its four sides. • First, we look at the part of the boundary with x = 0. On that entire side f = 0. • Next, we look at the part of the boundary with y = 0. On that entire side f = 0. • Next, we look at the part of the boundary with y = 1. There f = f (x, 1) = x ´ x3. To find the maximum and minimum of f (x, y) on the part of the boundary with y = 1, we must find the maximum and minimum of x ´ x3 when 0 ď x ď 1. Recall that, in general, the maximum and minimum of a function h(x) on the interval a ď x ď b, must occur either at x = a or at x = b or at an x for which either h1(x) = 0 or h1(x) does not exist. In this case, d dx(x ´ x3) = 1 ´ 3x2, so the max and min of x ´ x3 for 0 ď x ď 1 must occur – either at x = 0, where f = 0, – or at x = 1 ? 3, where f = 2 3 ? 3, – or at x = 1, where f = 0. • Finally, we look at the part of the boundary with x = 1. There f = f (1, y) = y ´ y2. As d dy(y ´ y2) = 1 ´ 2y, the only critical point of y ´ y2 is at y = 1 2. So the max and min of y ´ y2 for 0 ď y ď 1 must occur – either at y = 0, where f = 0, – or at y = 1 2, where f = 1 4, – or at y = 1, where f = 0. All together, we have the following candidates for max and min, with the max and min indicated. 210 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES point (0, 0) (0,0ďyď1) (0ďxď1,0) (1, 0) (1, 1 2) (1, 1) (0, 1) ( 1 ? 3, 1) value of f 0 0 0 0 1 4 0 0 2 3 ? 3 « 0.385 min min min min min min max x y (0, 0) (1, 0) (1, 1) (1, 1 2) (0, 1) ( 1 √ 3, 1) Example 2.9.23 Example 2.9.24 Find the maximum and minimum values of f (x, y) = xy + 2x + y when (x, y) runs over the triangular region with vertices (0, 0), (1, 0) and (0, 2). The triangular region is sketched in x y p0, 0q p1, 0q p0, 2q Solution. As usual, let’s examine the critical points, singular points and boundary in turn. Interior Critical Points: If f takes its maximum or minimum value at a point in the interior, then that point must be either a critical point of f or a singular point of f. The critical points are the solutions of fx(x, y) = y + 2 = 0 fy(x, y) = x + 1 = 0 So there is exactly one critical point, namely (´1, ´2). This is well outside the triangle and so is not a candidate for the location of the max and min. Singular points: Yet again there are no singular points for this f. Boundary: The region is a triangle, so its boundary consists of its three sides. 211 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES • First, we look at the side that runs from (0, 0) to (0, 2). On that entire side x = 0, so that f (0, y) = y. The smallest value of f on that side is f = 0 at (0, 0) and the largest value of f on that side is f = 2 at (0, 2). • Next, we look at the side that runs from (0, 0) to (1, 0). On that entire side y = 0, so that f (x, 0) = 2x. The smallest value of f on that side is f = 0 at (0, 0) and the largest value of f on that side is f = 2 at (1, 0). • Finally, we look at the side that runs from (0, 2) to (1, 0). Or first job is to find the equation of the line that contains (0, 2) and (1, 0). By way of review, we’ll find the equation using three different methods. – Method 1: You (probably) learned in high school that any line in the xy-plane65 has equation y = mx + b where b is the y intercept and m is the slope. In this case, the line crosses the y axis at y = 2 and so has y intercept b = 2. The line passes through (0, 2) and (1, 0) and so, as we see in the figure below, has slope m = ∆y ∆x = 0´2 1´0 = ´2. Thus the side of the triangle that runs from (0, 2) to (1, 0) is y = 2 ´ 2x with 0 ď x ď 1. x y p1, 0q p0, 2q ⟨1, ´2⟩ ∆x “ 1 ∆y “ ´2 – Method 2: Every line in the xy-plane has an equation of the form ax + by = c. In this case (0, 0) is not on the line so that c ‰ 0 and we can divide the equation by c, giving a c x + b cy = 1. Rename a c = A and b c = B. Thus, because the line does not pass through the origin, it has an equation of the form Ax + By = 1, for some constants A and B. In order for (0, 2) to lie on the line, x = 0, y = 2 has to be a solution of Ax + By = 1. That is, Ax ˇ ˇ x=0 + By ˇ ˇ y=2 = 1, so that B = 1/2. In order for (1, 0) to lie on the line, x = 1, y = 0 has to be a solution of Ax + By = 1. That is Ax ˇ ˇ x=1 + By ˇ ˇ y=0 = 1, so that A = 1. Thus the line has equation x + 1 2y = 1, or equivalently, y = 2 ´ 2x. – Method 3: The vector from (0, 2) to (1, 0) is ⟨1 ´ 0 , 0 ´ 2⟩= ⟨1, ´2⟩. As we see from the figure above, it is a direction vector for the line. One point on the line is (0, 2). So a parametric equation for the line (see Equation 1.3.1) is ⟨x ´ 0 , y ´ 2⟩= t ⟨1, ´2⟩ or x = t, y = 2 ´ 2t 65 To be picky, any line the xy-plane that is not parallel to the y axis. 212 PARTIAL DERIVATIVES 2.9 MAXIMUM AND MINIMUM VALUES By any of these three methods66, we have that the side of the triangle that runs from (0, 2) to (1, 0) is y = 2 ´ 2x with 0 ď x ď 1. On that side of the triangle f (x, 2 ´ 2x) = x(2 ´ 2x) + 2x + (2 ´ 2x) = ´2x2 + 2x + 2 Write g(x) = ´2x2 + 2x + 2. The maximum and minimum of g(x) for 0 ď x ď 1, and hence the maximum and minimum values of f on the hypotenuse of the triangle, must be achieved either at – x = 0, where f (0, 2) = g(0) = 2, or at – x = 1, where f (1, 0) = g(1) = 2, or when – 0 = g1(x) = ´4x + 2 so that x = 1 2, y = 2 ´ 2 2 = 1 and f (1/2, 1) = g(1/2) = ´2 4 + 2 2 + 2 = 5 2 All together, we have the following candidates for max and min, with the max and min indicated. point (0, 0) (0, 2) (1, 0) (1/2, 1) value of f 0 2 2 5/2 min max x y p0, 0q p1, 0q p0, 2q p´1, ´2q p 1 2, 1q Example 2.9.24 Example 2.9.25 Find the high and low points of the surface z = a x2 + y2 with (x, y) varying over the square \|x\| ď 1, \|y\| ď 1 . Solution. The function f (x, y) = a x2 + y2 has a particularly simple geometric interpre-tation — it is the distance from the point (x, y) to the origin. So 66 In the third method, x has just be renamed to t. 213 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS • the minimum of f (x, y) is achieved at the point in the square that is nearest the origin — namely the origin itself. So (0, 0, 0) is the lowest point on the surface and is at height 0. • The maximum of f (x, y) is achieved at the points in the square that are farthest from the origin — namely the four corners of the square ˘ 1, ˘1 . At those four points z = ? 2. So the highest points on the surface are (˘1, ˘1, ? 2). Even though we have already answered this question, it will be instructive to see what we would have found if we had followed our usual protocol. The partial derivatives of f (x, y) = a x2 + y2 are defined for (x, y) ‰ (0, 0) and are fx(x, y) = x a x2 + y2 fy(x, y) = y a x2 + y2 • There are no critical points because – fx = 0 only for x = 0, and – fy = 0 only for y = 0, but – (0, 0) is not a critical point because fx and fy are not defined there. • There is one singular point — namely (0, 0). The minimum value of f is achieved at the singular point. • The boundary of the square consists of its four sides. One side is ␣(x, y) ˇ ˇ x = 1, ´1 ď y ď 1 ( On this side f = a 1 + y2. As a 1 + y2 increases with \|y\|, the smallest value of f on that side is 1 (when y = 0) and the largest value of f is ? 2 (when y = ˘1). The same thing happens on the other three sides. The maximum value of f is achieved at the four corners. Note that fx and fy are both nonzero at all four corners. Example 2.9.25 2.10Ĳ Lagrange Multipliers In the last section we had to solve a number of problems of the form “What is the maxi-mum value of the function f on the curve C?” In those examples, the curve C was simple enough that we could reduce the problem to finding the maximum of a function of one variable. For more complicated problems this reduction might not be possible. In this sec-tion, we introduce another method for solving such problems. First some nomenclature. 214 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS A problem of the form “Find the maximum and minimum values of the function f (x, y) for (x, y) on the curve g(x, y) = 0.” is one type of constrained optimization problem. The function being maximized or minimized, f (x, y), is called the objective function. The function, g(x, y), whose zero set is the curve of interest, is called the constraint function. Definition 2.10.1. Such problems are quite common. As we said above, we have already encountered them in the last section on absolute maxima and minima, when we were looking for the extreme values of a function on the boundary of a region. In economics “utility functions” are used to model the relative “usefulness” or “desirability” or “preference” of various economic choices. For example, a utility function U(w, κ) might specify the relative level of satisfaction a consumer would get from purchasing a quantity w of wine and κ of coffee. If the consumer wants to spend $100 and wine costs $20 per unit and coffee costs $5 per unit, then the consumer would like to maximize U(w, κ) subject to the constraint that 20w + 5κ = 100. To this point we have always solved such constrained optimization problems either by • solving g(x, y) = 0 for y as a function of x (or for x as a function of y) or by • parametrizing the curve g(x, y) = 0. This means writing all points of the curve in the form x(t), y(t) for some functions x(t) and y(t). For example we used x(t) = cos t, y(t) = sin t as a parametrization of the circle x2 + y2 = 1 in Example 2.9.21. However quite often the function g(x, y) is so complicated that one cannot explicitly solve g(x, y) = 0 for y as a function of x or for x as a function of y and one also cannot explicitly parametrize g(x, y) = 0. Or sometimes you can, for example, solve g(x, y) = 0 for y as a function of x, but the resulting solution is so complicated that it is really hard, or even virtually impossible, to work with. Direct attacks become even harder in higher dimen-sions when, for example, we wish to optimize a function f (x, y, z) subject to a constraint g(x, y, z) = 0. There is another procedure called the method of “Lagrange67 multipliers” that comes to our rescue in these scenarios. Here is the three dimensional version of the method. There are obvious analogs in other dimensions. 67 Joseph-Louis Lagrange was actually born Giuseppe Lodovico Lagrangia in Turin, Italy in 1736. He moved to Berlin in 1766 and then to Paris in 1786. He eventually acquired French citizenship and then the French claimed he was a French mathematician, while the Italians continued to claim that he was an Italian mathematician. 215 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS Let f (x, y, z) and g(x, y, z) have continuous first partial derivatives in a region of R3 that contains the surface S given by the equation g(x, y, z) = 0. Further assume that ∇ ∇ ∇g(x, y, z) ‰ 0 on S. If f, restricted to the surface S, has a local extreme value at the point (a, b, c) on S, then there is a real number λ such that ∇ ∇ ∇f (a, b, c) = λ∇ ∇ ∇g(a, b, c) that is fx(a, b, c) = λ gx(a, b, c) fy(a, b, c) = λ gy(a, b, c) fz(a, b, c) = λ gz(a, b, c) The number λ is called a Lagrange multiplier. Theorem 2.10.2 (Lagrange Multipliers). Proof. Suppose that (a, b, c) is a point of S and that f (x, y, z) ě f (a, b, c) for all points (x, y, z) on S that are close to (a, b, c). That is (a, b, c) is a local minimum for f on S. Of course the argument for a local maximum is virtually identical. Imagine that we go for a walk on S, with the time t running, say, from t = ´1 to t = +1 and that at time t = 0 we happen to be exactly at (a, b, c). Let’s say that our position is x(t), y(t), z(t) at time t. Write F(t) = f x(t), y(t), z(t) So F(t) is the value of f that we see on our walk at time t. Then for all t close to 0, x(t), y(t), z(t) is close to x(0), y(0), z(0) = (a, b, c) so that F(0) = f x(0), y(0), z(0) = f (a, b, c) ď f x(t), y(t), z(t) = F(t) for all t close to zero. So F(t) has a local minimum at t = 0 and consequently F1(0) = 0. By the chain rule, Theorem 2.4.1, F1(0) = d dt f x(t), y(t), z(t) ˇ ˇ ˇ t=0 = fx a, b, c x1(0) + fy a, b, c y1(0) + fz a, b, c z1(0) = 0 (˚) We may rewrite this as a dot product: 0 = F1(0) = ∇ ∇ ∇f (a, b, c) ¨ x1(0) , y1(0) , z1(0) ù ñ ∇ ∇ ∇f (a, b, c) K x1(0) , y1(0) , z1(0) This is true for all paths on S that pass through (a, b, c) at time 0. In particular it is true for all vectors ⟨x1(0) , y1(0) , z1(0)⟩that are tangent to S at (a, b, c). So ∇ ∇ ∇f (a, b, c) is perpendic-ular to S at (a, b, c). 216 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS But we already know, by Theorem 2.5.5.a, that ∇ ∇ ∇g(a, b, c) is also perpendicular to S at (a, b, c). So ∇ ∇ ∇f (a, b, c) and ∇ ∇ ∇g(a, b, c) have to be parallel vectors. That is, ∇ ∇ ∇f (a, b, c) = λ∇ ∇ ∇g(a, b, c) for some number λ. That’s the Lagrange multiplier rule of our theorem. So to find the maximum and minimum values of f (x, y, z) on a surface g(x, y, z) = 0, assuming that both the objective function f (x, y, z) and constraint function g(x, y, z) have continuous first partial derivatives and that ∇ ∇ ∇g(x, y, z) ‰ 0, you 1. build up a list of candidate points (x, y, z) by finding all solutions to the equations fx(x, y, z) = λ gx(x, y, z) fy(x, y, z) = λ gy(x, y, z) fz(x, y, z) = λ gz(x, y, z) g(x, y, z) = 0 Note that there are four equations and four unknowns, namely x, y, z and λ. 2. Then you evaluate f (x, y, z) at each (x, y, z) on the list of candidates. The biggest of these candidate values is the absolute maximum and the smallest of these candidate values is the absolute minimum. Another way to write the system of equations in the first step is Lx(a, b, c, λ) = Ly(a, b, c, λ) = Lz(a, b, c, λ) = Lλ(a, b, c, λ) = 0 where L(x, y, z, λ) is the auxiliary function68 69 L(x, y, z, λ) = f (x, y, z) ´ λ g(x, y, z) Now for a bunch of examples. Example 2.10.3 Find the maximum and minimum of the function x2 ´ 10x ´ y2 on the ellipse whose equa-tion is x2 + 4y2 = 16. Solution. For this problem the objective function is f (x, y) = x2 ´ 10x ´ y2 and the con-straint function is g(x, y) = x2 + 4y2 ´16. To apply the method of Lagrange multipliers we need ∇ ∇ ∇f and ∇ ∇ ∇g. So we start by computing the first order derivatives of these functions. fx = 2x ´ 10 fy = ´2y gx = 2x gy = 8y 68 We call L an auxiliary function because, while we use it to help solve the problem, it doesn’t actually appear in either the statement of the question or in the answer itself. 69 Some people use L(x, y, z, λ) = f (x, y, z) + λ g(x, y, z) instead. This amounts to renaming λ to ´λ. While we care that λ has a value, we don’t care what it is. 217 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS So, according to the method of Lagrange multipliers, we need to find all solutions to 2x ´ 10 = λ(2x) ´2y = λ(8y) x2 + 4y2 ´ 16 = 0 Rearranging these equations gives (λ ´ 1)x = ´5 (E1) (4λ + 1)y = 0 (E2) x2 + 4y2 ´ 16 = 0 (E3) From (E2), we see that we must have either λ = ´1/4 or y = 0. • If λ = ´1/4, (E1) gives ´5 4x = ´5, i.e. x = 4, and then (E3) gives y = 0. • If y = 0, then (E3) gives x = ˘4 (and while we could easily use (E1) to solve for λ, we don’t actually need λ). So the method of Lagrange multipliers, Theorem 2.10.2 (actually the dimension two ver-sion of Theorem 2.10.2), gives that the only possible locations of the maximum and min-imum of the function f are (4, 0) and (´4, 0). To complete the problem, we only have to compute f at those points. point (4, 0) (´4, 0) value of f ´24 56 min max Hence the maximum value of x2 ´ 10x ´ y2 on the ellipse is 56 and the minimum value is ´24. x y x2 4y2 “ 16 p´4,0q p4,0q Example 2.10.3 In the previous example, the objective function and the constraint were specified ex-plicitly. That will not always be the case. In the next example, we have to do a little geometry to extract them. Example 2.10.4 Find the rectangle of largest area (with sides parallel to the coordinates axes) that can be inscribed in the ellipse x2 + 2y2 = 1. Solution. Since this question is so geometric, it is best to start by drawing a picture. 218 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS x y px, yq px, ´yq p´x, ´yq x2 2y2 “ 1 Call the coordinates of the upper right corner of the rectangle (x, y), as in the figure above. The four corners of the rectangle are (˘x, ˘y) so the rectangle has width 2x and height 2y and the objective function is f (x, y) = 4xy. The constraint function for this problem is g(x, y) = x2 + 2y2 ´ 1. Again, to use Lagrange multipliers we need the first order partial derivatives. fx = 4y fy = 4x gx = 2x gy = 4y So, according to the method of Lagrange multipliers, we need to find all solutions to 4y = λ(2x) (E1) 4x = λ(4y) (E2) x2 + 2y2 ´ 1 = 0 (E3) Equation (E1) gives y = 1 2λx. Substituting this into equation (E2) gives 4x = 2λ2x or 2x 2 ´ λ2 = 0 So (E2) is satisfied if either x = 0 or λ = ? 2 or λ = ´ ? 2. • If x = 0, then (E1) gives y = 0 too. But (0, 0) violates the constraint equation (E3). Note that, to have a solution, all of the equations (E1), (E2) and (E3) must be satisfied. • If λ = ? 2, then – (E2) gives x = ? 2y and then – (E3) gives 2y2 + 2y2 = 1 or y2 = 1 4 so that – y = ˘1/2 and x = ? 2y = ˘1/ ? 2. • If λ = ´ ? 2, then – (E2) gives x = ´ ? 2y and then – (E3) gives 2y2 + 2y2 = 1 or y2 = 1 4 so that – y = ˘1/2 and x = ´ ? 2y = ¯1/ ? 2. We now have four possible values of (x, y), namely 1/ ? 2 , 1/2 , ´ 1/ ? 2 , ´1/2 , 1/ ? 2 , ´1/2 and ´ 1/ ? 2 , 1/2 . They are the four corners of a single rectangle. We said that we wanted (x, y) to be the upper right corner, i.e. the corner in the first quadrant. It is 1/ ? 2 , 1/2 . 219 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS Example 2.10.4 Example 2.10.5 Find the ends of the major and minor axes of the ellipse 3x2 ´ 2xy + 3y2 = 4. They are the points on the ellipse that are farthest from and nearest to the origin. Solution. Let (x, y) be a point on 3x2 ´ 2xy + 3y2 = 4. This point is at the end of a major axis when it maximizes its distance from the centre, (0, 0) of the ellipse. It is at the end of a minor axis when it minimizes its distance from (0, 0). So we wish to maximize and minimize the distance a x2 + y2 subject to the constraint g(x, y) = 3x2 ´ 2xy + 3y2 ´ 4 = 0 Now maximizing/minimizing a x2 + y2 is equivalent70 to maximizing/minimizing its square a x2 + y22 = x2 + y2. So we are free to choose the objective function f (x, y) = x2 + y2 which we will do, because it makes the derivatives cleaner. Again, we use Lagrange multipliers to solve this problem, so we start by finding the partial derivatives. fx(x, y) = 2x fy(x, y) = 2y gx(x, y) = 6x ´ 2y gy(x, y) = ´2x + 6y We need to find all solutions to 2x = λ(6x ´ 2y) 2y = λ(´2x + 6y) 3x2 ´ 2xy + 3y2 ´ 4 = 0 Dividing the first two equations by 2, and then collecting together the x’s and the y’s gives (1 ´ 3λ)x + λy = 0 (E1) λx + (1 ´ 3λ)y = 0 (E2) 3x2 ´ 2xy + 3y2 ´ 4 = 0 (E3) To start, let’s concentrate on the first two equations. Pretend, for a couple of minutes, that we already know the value of λ and are trying to find x and y. Note that λ cannot be zero because if it is, (E1) forces x = 0 and (E2) forces y = 0 and (0, 0) is not on the ellipse, i.e. violates (E3). So we may divide by λ and (E1) gives y = ´1 ´ 3λ λ x Subbing this into (E2) gives λx ´ (1 ´ 3λ)2 λ x = 0 70 The function S(z) = z2 is a strictly increasing function for z ě 0. So, for a, b ě 0, the statement “a ă b” is equivalent to the statement “S(a) ă S(b)”. 220 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS Again, x cannot be zero, since then y = ´1´3λ λ x would give y = 0 and (0, 0) is still not on the ellipse. So we may divide λx ´ (1´3λ)2 λ x = 0 by x, giving λ ´ (1 ´ 3λ)2 λ = 0 ð ñ (1 ´ 3λ)2 ´ λ2 = 0 ð ñ 8λ2 ´ 6λ + 1 = (2λ ´ 1)(4λ ´ 1) = 0 We now know that λ must be either 1 2 or 1 4. Subbing these into either (E1) or (E2) gives λ = 1 2 ù ñ ´1 2x + 1 2y = 0 ù ñ x = y (E3) ù ñ 3x2 ´ 2x2 + 3x2 = 4 ù ñ x = ˘1 λ = 1 4 ù ñ 1 4x + 1 4y = 0 ù ñ x = ´y (E3) ù ñ 3x2 + 2x2 + 3x2 = 4 ù ñ x = ˘ 1 ? 2 Here “ (E3) ù ñ ” indicates that we have just used (E3). We now have (x, y) = ˘(1, 1), from λ = 1 2, and (x, y) = ˘ 1 ? 2, ´ 1 ? 2 from λ = 1 4. The distance from (0, 0) to ˘(1, 1), namely ? 2, is larger than the distance from (0, 0) to ˘ 1 ? 2, ´ 1 ? 2 , namely 1. So the ends of the minor axes are ˘ 1 ? 2, ´ 1 ? 2 and the ends of the major axes are ˘(1, 1). Those ends are sketched in the figure on the left below. Once we have the ends, it is an easy matter71 to sketch the ellipse as in the figure on the right below. x y p1,1q p´1,´1q p1,´1q{ ? 2 p´1,1q{ ? 2 x y p1,1q p´1,´1q p1,´1q{ ? 2 p´1,1q{ ? 2 3x2 ´ 2xy 3y2 “ 4 Example 2.10.5 Example 2.10.6 Find the values of w ě 0 and κ ě 0 that maximize the utility function U(w, κ) = 6w 2/3κ 1/3 subject to the constraint 4w + 2κ = 12 71 if you tilt your head so that the line through (1, 1) and (´1, ´1) appears horizontal 221 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS Solution. The constraint 4w + 2κ = 12 is simple enough that we can easily use it to express κ in terms of w, then substitute κ = 6 ´ 2w into U(w, κ), and then maximize U(w, 6 ´ 2w) = 6w 2/3(6 ´ 2w) 1/3 using the techniques of §3.5 in the CLP-1 textbook. However, for practice purposes, we’ll use Lagrange multipliers with the objective func-tion U(w, κ) = 6w 2/3κ 1/3 and the constraint function g(w, κ) = 4w + 2κ ´12. The first order derivatives of these functions are Uw = 4w´1/3κ 1/3 Uκ = 2w 2/3κ´2/3 gw = 4 gκ = 2 The boundary values w = 0 and κ = 0 give utility 0, which is obviously not going to be the maximum utility. So it suffices to consider only local maxima. According to the method of Lagrange multipliers, we need to find all solutions to 4w´1/3κ 1/3 = 4λ (E1) 2w 2/3κ´2/3 = 2λ (E2) 4w + 2κ ´ 12 = 0 (E3) Then • equation (E1) gives λ = w´1/3κ 1/3. • Substituting this into (E2) gives w 2/3κ´2/3 = λ = w´1/3κ 1/3 and hence w = κ. • Then substituting w = κ into (E3) gives 6κ = 12. So w = κ = 2 and the maximum utility is U(2, 2) = 12. Example 2.10.6 Example 2.10.7 Find the point on the sphere x2 + y2 + z2 = 1 that is farthest from (1, 2, 3). Solution. As before, we simplify the algebra by maximizing the square of the distance rather than the distance itself. So we are to maximize f (x, y, z) = (x ´ 1)2 + (y ´ 2)2 + (z ´ 3)2 subject to the constraint g(x, y, z) = x2 + y2 + z2 ´ 1 = 0 Since fx(x, y, z) = 2(x ´ 1) fy(x, y, z) = 2(y ´ 2) fz(x, y, z) = 2(z ´ 3) gx(x, y, z) = 2x gy(x, y, z) = 2y gz(x, y, z) = 2z we need to find all solutions to 2(x ´ 1) = λ(2x) ð ñ x = 1 1 ´ λ (E1) 2(y ´ 2) = λ(2y) ð ñ y = 2 1 ´ λ (E2) 2(z ´ 3) = λ(2z) ð ñ z = 3 1 ´ λ (E3) 0 = x2 + y2 + z2 ´ 1 (E4) 222 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS Substituting (E1), (E2) and (E3) into (E4) gives 1 + 4 + 9 (1 ´ λ)2 ´ 1 = 0 ù ñ (1 ´ λ)2 = 14 ù ñ 1 ´ λ = ˘ ? 14 We can then substitute these two values of λ back into the expressions for x, y, z in terms of λ to get the two points 1 ? 14(1, 2, 3) and ´ 1 ? 14(1, 2, 3). The vector from 1 ? 14(1, 2, 3) to (1, 2, 3), namely ! 1 ´ 1 ? 14 ) (1, 2, 3), is obviously shorter than the vector from ´ 1 ? 14(1, 2, 3) to (1, 2, 3), which is ! 1 + 1 ? 14 ) (1, 2, 3). So the nearest point is 1 ? 14(1, 2, 3) and the farthest point is ´ 1 ? 14(1, 2, 3) . Example 2.10.7 2.10.1 § § (Optional) An Example with Two Lagrange Multipliers In this optional section, we consider an example of a problem of the form “maximize (or minimize) f (x, y, z) subject to the two constraints g(x, y, z) = 0 and h(x, y, z) = 0”. We use the following variant of Theorem 2.10.2. Let f (x, y, z), g(x, y, z) and h(x, y, z) have continuous first partial derivatives in a region of R3 that contains the curve C given by the equations g(x, y, z) = h(x, y, z) = 0 Assume72 that ∇ ∇ ∇g(x, y, z) ˆ ∇ ∇ ∇h(x, y, z) ‰ 0 on C. If f, restricted to the curve C, has a local extreme value at the point (a, b, c) on C, then there are real numbers λ and µ such that ∇ ∇ ∇f (a, b, c) = λ∇ ∇ ∇g(a, b, c) + µ∇ ∇ ∇h(a, b, c) that is fx(a, b, c) = λ gx(a, b, c) + µ hx(a, b, c) fy(a, b, c) = λ gy(a, b, c) + µ hy(a, b, c) fz(a, b, c) = λ gz(a, b, c) + µ hz(a, b, c) Theorem 2.10.8 (Two Lagrange Multipliers). We can reformulate this theorem in terms of the auxiliary function L(x, y, z, λ, µ) = f (x, y, z) ´ λ g(x, y, z) ´ µ h(x, y, z) It is a function of five variables — the original variables x, y and z, and two auxiliary variables λ and µ. If there is a local extreme value at (a, b, c) then (a, b, c) must obey 72 This condition says that the normal vectors to g = 0 and h = 0 at (x, y, z) are not parallel. This ensures that the surfaces g = 0 and h = 0 are not tangent to each other at (x, y, z). They intersect in a curve. 223 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS 0 = Lx(a, b, c, λ, µ) = fx(a, b, c) ´ λgx(a, b, c) ´ µhx(a, b, c) 0 = Ly(a, b, c, λ, µ) = fy(a, b, c) ´ λgy(a, b, c) ´ µhy(a, b, c) 0 = Lz(a, b, c, λ, µ) = fz(a, b, c) ´ λgz(a, b, c) ´ µhz(a, b, c) 0 = Lλ(a, b, c, λ, µ) = g(a, b, c) 0 = Lµ(a, b, c, λ, µ) = h(a, b, c) Equation 2.10.9. for some λ and µ. So solving this system of five equations in five unknowns gives all possible candidates for the locations of local maxima and minima. We’ll go through an example shortly. Proof of Theorem 2.10.8. Before we get to the example itself, here is why the above ap-proach works. Assume that a local minimum occurs at (a, b, c), which is the grey point in the schematic figure below. Imagine that you start walking away from (a, b, c) along the curve g = h = 0. Your path is the grey line in the schematic figure below. Call your gpx, y, zq “ 0 hpx, y, zq “ 0 ∇ ∇ ∇g ∇ ∇ ∇h λ∇ ∇ ∇g µ∇ ∇ ∇h v velocity vector v. It is tangent to the curve g(x, y, z) = h(x, y, z) = 0. Because f has a local minimum at (a, b, c), f must be increasing (or constant) as we leave (a, b, c). So the directional derivative Dv f (a, b, c) = ∇ ∇ ∇f (a, b, c) ¨ v ě 0 Now start over. Again walk away from (a, b, c) along the curve g = h = 0, but this time moving in the opposite direction, with velocity vector ´v. Again f must be increasing (or constant) as we leave (a, b, c), so the directional derivative D´v f (a, b, c) = ∇ ∇ ∇f (a, b, c) ¨ (´v) ě 0 As both ∇ ∇ ∇f (a, b, c) ¨ v and ´∇ ∇ ∇f (a, b, c) ¨ v are at least zero, we now have that ∇ ∇ ∇f (a, b, c) ¨ v = 0 (˚) for all vectors v that are tangent to the curve g = h = 0 at (a, b, c). Let’s denote by T the set of all vectors v that are tangent to the curve g = h = 0 at (a, b, c) and let’s denote by T K the set of all vectors that are perpendicular to all vectors in T . So (˚) says that ∇ ∇ ∇f (a, b, c) must in T K. 224 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS We now find all vectors in T K. We can easily guess two such vectors. Since the curve g = h = 0 lies inside the surface g = 0 and ∇ ∇ ∇g(a, b, c) is normal to g = 0 at (a, b, c), we have ∇ ∇ ∇g(a, b, c) ¨ v = 0 (E1) Similarly, since the curve g = h = 0 lies inside the surface h = 0 and ∇ ∇ ∇h(a, b, c) is normal to h = 0 at (a, b, c), we have ∇ ∇ ∇h(a, b, c) ¨ v = 0 (E2) Picking any two constants λ and µ, multiplying (E1) by λ, multiplying (E2) by µ and adding gives that λ∇ ∇ ∇g(a, b, c) + µ∇ ∇ ∇h(a, b, c) ¨ v = 0 for all vectors v in T . Thus, for all λ and µ, the vector λ∇ ∇ ∇g(a, b, c) + µ∇ ∇ ∇h(a, b, c) is in T K. Now the vectors in T form a line. (They are all tangent to the same curve at the same point.) So, T K, the set of all vectors perpendicular to T , forms a plane. As λ and µ run over all real numbers, the vectors λ∇ ∇ ∇g(a, b, c) + µ∇ ∇ ∇h(a, b, c) form a plane. Thus we have found all vector in T K and we conclude that ∇ ∇ ∇f (a, b, c) must be of the form λ∇ ∇ ∇g(a, b, c) + µ∇ ∇ ∇h(a, b, c) for some real numbers λ and µ. The three components of the equation ∇ ∇ ∇f (a, b, c) = λ∇ ∇ ∇g(a, b, c) + µ∇ ∇ ∇h(a, b, c) are exactly the first three equations of (2.10.9). This completes the explanation of why Lagrange multipliers work in this setting. Example 2.10.10 Find the distance from the origin to the curve that is the intersection of the two surfaces z2 = x2 + y2 x ´ 2z = 3 Solution. Yet again, we simplify the algebra by maximizing the square of the distance rather than the distance itself. So we are to maximize f (x, y, z) = x2 + y2 + z2 subject to the constraints 0 = g(x, y, z) = x2 + y2 ´ z2 0 = h(x, y, z) = x ´ 2z ´ 3 Since fx = 2x fy = 2y fz = 2z gx = 2x gy = 2y gz = ´2z hx = 1 hy = 0 hz = ´2 the method of Lagrange multipliers requires us to find all solutions to 2x = λ(2x) + µ(1) (E1) 2y = λ(2y) + µ(0) ð ñ (1 ´ λ)y = 0 (E2) 2z = λ(´2z) + µ(´2) (E3) z2 = x2 + y2 (E4) x ´ 2z = 3 (E5) 225 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS Since equation (E2) factors so nicely we start there. It tells us that either y = 0 or λ = 1. Case λ = 1: When λ = 1 the remaining equations reduce to 0 = µ (E1) 0 = 4z + 2µ (E3) z2 = x2 + y2 (E4) x ´ 2z = 3 (E5) So • equation (E1) gives µ = 0. • Then substituting µ = 0 into (E3) gives z = 0. • Then substituting z = 0 into (E5) gives x = 3. • Then substituting z = 0 and x = 3 into (E4) gives 0 = 9 + y2, which is impossible, since 9 + y2 ě 9 ą 0 for all y. So we can’t have λ = 1. Case y = 0: When y = 0 the remaining equations reduce to 2(1 ´ λ)x = µ (E1) (1 + λ)z = ´µ (E3) z2 = x2 (E4) x ´ 2z = 3 (E5) These don’t clean up quite so nicely as in the λ = 1 case. But at least equation (E4) tells us that z = ˘x. So we have to consider those two possibilities. Subcase y = 0, z = x: When y = 0 and z = x, the remaining equations reduce to 2(1 ´ λ)x = µ (E1) (1 + λ)x = ´µ (E3) ´x = 3 (E5) So equation (E5) now tells us that x = ´3 so that (x, y, z) = (´3, 0, ´3). (We don’t really care what λ and µ are. But as they obey ´6(1 ´ λ) = µ, ´3(1 + λ) = ´µ we have, adding the two equations together ´9 + 3λ = 0 ù ñ λ = 3 and then, subbing into either equation, µ = 12.) Subcase y = 0, z = ´x: When y = 0 and z = ´x, the remaining equations reduce to 2(1 ´ λ)x = µ (E1) (1 + λ)x = µ (E3) 3x = 3 (E5) 226 PARTIAL DERIVATIVES 2.10 LAGRANGE MULTIPLIERS So equation (E5) now tells us that x = 1 so that (x, y, z) = (1, 0, ´1). (Again, we don’t really care what λ and µ are. But as they obey 2(1 ´ λ) = µ, (1 + λ) = µ we have, subtracting the second equation from the first, 1 ´ 3λ = 0 ù ñ λ = 1 3 and then, subbing into either equation, µ = 4 3.) Conclusion: We have two candidates for the location of the max and min, namely (´3, 0, ´3) and (1, 0, ´1). The first is a distance 3 ? 2 from the origin, giving the maxi-mum, and the second is a distance ? 2 from the origin, giving the minimum. In particular, the distance is ? 2. Example 2.10.10 227 MULTIPLE INTEGRALS Chapter 3 In your previous calculus courses you defined and worked with single variable integrals, like şb a f (x) dx. In this chapter, we define and work with multivariable integrals, like ť R f (x, y) dx dy and ţ V f (x, y, z) dx dy dz. We start with two variable integrals. 3.1Ĳ Double Integrals 3.1.1 § § Vertical Slices Suppose that you want to compute the mass of a plate that fills the region R in the xy-plane. Suppose further that the density of the plate, say in kilograms per square meter, depends on position. Call the density f (x, y). For simplicity we’ll assume that R is the region between the bottom curve y = B(x) and the top curve y = T(x) with x running from a to b. That is, R = ␣(x, y) ˇ ˇ a ď x ď b, B(x) ď y ď T(x) ( x y y “ Tpxq y “ Bpxq a b R We’ll shortly express that mass as a two dimensional integral. As a warmup, recall the procedure that we used to set up a (one dimensional) integral representing the area of R in Example 1.5.1 of the CLP-2 text. 228 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS • Pick a natural number n (that we will later send to infinity), and then • subdivide R into n narrow vertical slices, each of width ∆x = b´a n . Denote by xi = a + i ∆x the x-coordinate of the right hand edge of slice number i. x y y “ Tpxq y “ Bpxq x0 x1 x2 ¨ ¨ ¨ xn • For each i = 1, 2, . . . , n, slice number i has x running from xi´1 to xi. We approximate its area by the area of a rectangle. We pick a number x˚ i between xi´1 and xi and approximate the slice by a rectangle whose top is at y = T(x˚ i ) and whose bottom is at y = B(x˚ i ). The rectangle is outlined in blue in the figure below. x y xi´1 xi x˚ i Bpx˚ i q Tpx˚ i q y“Tpxq y“Bpxq • Thus the area of slice i is approximately T(x˚ i ) ´ B(x˚ i ) ∆x. • So the Riemann sum approximation of the area of R is Area « n ÿ i=1 T(x˚ i ) ´ B(x˚ i ) ∆x • By taking the limit as n Ñ 8 (i.e. taking the limit as the width of the rectangles goes to zero), we convert the Riemann sum into a definite integral (see Definition 1.1.9 in the CLP-2 text) and at the same time our approximation of the area becomes the exact area: Area = lim nÑ8 n ÿ i=1 T(x˚ i ) ´ B(x˚ i ) ∆x = ż b a T(x) ´ B(x) dx 229 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS Now we can expand that procedure to yield the mass of R rather than the area of R. We just have to replace our approximation T(x˚ i ) ´ B(x˚ i ) ∆x of the area of slice i by an approximation to the mass of slice i. To do so, we • Pick a natural number m (that we will later send to infinity), and then • subdivide slice number i into m tiny rectangles, each of width ∆x and of height ∆y = 1 m T(x˚ i ) ´ B(x˚ i ) . Denote by yj = B(x˚ i ) + j ∆y the y-coordinate of the top of rectangle number j. x y xi´1 xi x˚ i yj´1 yj y˚ j • At this point we approximate the density inside each rectangle by a constant. For each j = 1, 2, . . . , m, rectangle number j has y running from yj´1 to yj. We pick a number y˚ j between yj´1 and yj and approximate the density on rectangle number j in slice number i by the constant f x˚ i , y˚ j . • Thus the mass of rectangle number j in slice number i is approximately f x˚ i , y˚ j ∆x ∆y. • So the Riemann sum approximation of the mass of slice number i is Mass of slice i « m ÿ j=1 f x˚ i , y˚ j ∆x ∆y Note that the y˚ j ’s depend on i and m. • By taking the limit as m Ñ 8 (i.e. taking the limit as the height of the rectangles goes to zero), we convert the Riemann sum into a definite integral: Mass of slice i « ∆x ż T(x˚ i ) B(x˚ i ) f x˚ i , y dy = F(x˚ i ) ∆x where F(x) = ż T(x) B(x) f x, y dy Notice that, while we started with the density f (x, y) being a function of both x and y, by taking the limit of this Riemann sum, we have “integrated out” the dependence on y. As a result, F(x) is a function of x only, not of x and y. 230 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS • Finally taking the limit as n Ñ 8 (i.e. taking the limit as the slice width goes to zero), we get Mass = lim nÑ8 n ÿ i=1 ∆x ż T(x˚ i ) B(x˚ i ) f x˚ i , y dy = lim nÑ8 n ÿ i=1 F(x˚ i ) ∆x Now we are back in familiar 1-variable territory. The sum n ř i=1 F(x˚ i ) ∆x is a Riemann sum approximation to the integral şb a F(x) dx. So Mass = ż b a F(x) dx = ż b a "ż T(x) B(x) f x, y dy # dx This is our first double integral. There are a couple of different standard notations for this integral. ĳ R f x, y dx dy = ż b a "ż T(x) B(x) f x, y dy # dx = ż b a ż T(x) B(x) f x, y dy dx = ż b a dx ż T(x) B(x) dy f x, y The last three integrals here are called iterated integrals, for obvious reasons. Notation 3.1.1. Note that • to evaluate the integral ż b a ż T(x) B(x) f x, y dy dx, – first evaluate the inside integral şT(x) B(x) f x, y dy using the inside limits of in-tegration, and by treating x as a constant and using standard single variable integration techniques, such as those in the CLP-2 text. The result of the inside integral is a function of x only. Call it F(x). – Then evaluate the outside integral şb a F(x) dx, whose integrand is the answer to the inside integral. Again, this integral is evaluated using standard single variable integration techniques. • To evaluate the integral ż b a dx ż T(x) B(x) dy f x, y , – first evaluate the inside integral şT(x) B(x) dy f x, y using the limits of integration that are directly beside the dy. Indeed the dy is written directly beside şT(x) B(x) to 231 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS make it clear that the limits of integration B(x) and T(x) are for the y-integral. In the past you probably wrote this integral as şT(x) B(x) f x, y dy. The result of the inside integral is again a function of x only. Call it F(x). – Then evaluate the outside integral şb a dx F(x), whose integrand is the answer to the inside integral and whose limits of integration are directly beside the dx. At this point you may be wondering “Do we always have to use vertical slices?” and “Do we always have to integrate with respect to y first?” The answer is “no”. This brings us to consider “horizontal slices”. 3.1.2 § § Horizontal Slices We found, when computing areas of regions in the xy-plane, that it is often advantageous to use horizontal slices, rather than vertical slices. See, for example, Example 1.5.4 in the CLP-2 text. The same is true when setting up multidimensional integrals. So we now repeat the setup procedure of the last section, but starting with horizontal slices, rather than vertical slices. This procedure will be useful when dealing with regions of the form R = ␣(x, y) ˇ ˇ c ď y ď d, L(y) ď x ď R(y) ( x y x “ Lpyq x “ Rpyq y “ d y “ c R Here L(y) (“L” stands for “left”) is the smallest1 allowed value of x, when the y-coordinate is y, and R(y) (“R” stands for “right”) is the largest allowed value of x, when the y-coordinate is y. Suppose that we wish to evaluate the mass of a plate that fills the region R, and that the density of the plate is f (x, y). We follow essentially the same the procedure as we used with vertical slices, but with the roles of x and y swapped. • Pick a natural number n (that we will later send to infinity). Then • subdivide the interval c ď y ď d into n narrow subintervals, each of width ∆y = d´c n . Each subinterval cuts a thin horizontal slice from the region (see the figure below). 1 By the “smallest ” x we mean the x farthest to the left along the number line, not the x closest to 0. 232 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS • We approximate slice number i by a thin horizontal rectangle (indicated by the long darker gray rectangle in the figure below). On this slice, the y-coordinate runs over a very narrow range. We pick a number y˚ i , somewhere in that range. We approximate slice i by a rectangle whose left side is at x = L(y˚ i ) and whose right side is at x = R(y˚ i ). • If we were computing the area of R, we would now approximate the area of slice i by R(y˚ i ) ´ L(y˚ i ) ∆y, which is the area of the rectangle with width R(y˚ i ) ´ L(y˚ i ) and height ∆y. • To get the mass, just as we did above with vertical slices, we – pick another natural number m (that we will later send to infinity), and then – subdivide slice number i into m tiny rectangles, each of height ∆y and of width ∆x = 1 m R(y˚ i ) ´ L(y˚ i ) . – For each j = 1, 2, . . . , m, rectangle number j has x running over a very narrow range. We pick a number x˚ j somewhere in that range. See the small black rectangle in the figure below. x y x “ Lpyq x “ Rpyq x “ x˚ j y “ y˚ i y “ d y “ c Lpy˚ i q Rpy˚ i q Here is a magnified sketch of slice number i x y x˚ j xj´1 xj y “ y˚ i y “ yi´1 y “ yi Lpy˚ i q Rpy˚ i q 233 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS – On rectangle number j in slice number i, we approximate the density by f x˚ j , y˚ i , giving us that the mass of rectangle number j in slice number i is approximately f x˚ j , y˚ i ∆x ∆y. – So the Riemann sum approximation of the mass of (horizontal) slice number i is Mass of slice i « m ÿ j=1 f x˚ j , y˚ i ∆x ∆y – By taking the limit as m Ñ 8 (i.e. taking the limit as the width of the rectangles goes to zero), we convert the Riemann sum into a definite integral: Mass of slice i « ∆y ż R(y˚ i ) L(y˚ i ) f x, y˚ i dx = F(y˚ i ) ∆y where F(y) = ż R(y) L(y) f x, y dx Observe that, as x has been integrated out, F(y) is a function of y only, not of x and y. • Finally taking the limit as n Ñ 8 (i.e. taking the limit as the slice width goes to zero), we get Mass = lim nÑ8 n ÿ i=1 ∆y ż R(y˚ i ) L(y˚ i ) f x, y˚ i dx = lim nÑ8 n ÿ i=1 F(y˚ i ) ∆y Now n ř i=1 F(y˚ i ) ∆y is a Riemann sum approximation to the integral şd c F(y) dy. So Mass = ż d c F(y) dy = ż d c "ż R(y) L(y) f x, y dx # dy The standard notations of Notation 3.1.1 also apply to this integral. ĳ R f x, y dx dy = ż d c "ż R(y) L(y) f x, y dx # dy = ż d c ż R(y) L(y) f x, y dx dy = ż d c dy ż R(y) L(y) dx f x, y Notation 3.1.2. Note that 234 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS • to evaluate the integral ż d c ż R(y) L(y) f x, y dx dy, – first evaluate the inside integral şR(y) L(y) f x, y dx using the inside limits of inte-gration. The result of the inside integral is a function of y only. Call it F(y). – Then evaluate the outside integral şd c F(y) dy, whose integrand is the answer to the inside integral. • To evaluate the integral ż d c dy ż R(y) L(y) dx f x, y , – first evaluate the inside integral şR(y) L(y) dx f x, y using the limits of integration that are directly beside the dx. Again, the dx is written directly beside şR(y) L(y) to make it clear that the limits of integration L(y) and R(y) are for the x-integral. In the past you probably wrote this integral as şR(y) L(y) f x, y dx. The result of the inside integral is again a function of y only. Call it F(y). – Then evaluate the outside integral şd c dy F(y), whose integrand is the answer to the inside integral and whose limits of integration are directly beside the dy. By way of summary, we now have two integral representations for the mass of regions in the xy-plane. Let R be a region in the xy-plane and let the function f (x, y) be defined and continuous on R. (a) If R = ␣(x, y) ˇ ˇ a ď x ď b, B(x) ď y ď T(x) ( with B(x) and T(x) being continuous, and if the mass density in R is f (x, y), then the mass of R is ż b a "ż T(x) B(x) f x, y dy # dx = ż b a ż T(x) B(x) f x, y dy dx = ż b a dx ż T(x) B(x) dy f x, y (b) If R = ␣(x, y) ˇ ˇ c ď y ď d, L(y) ď x ď R(y) ( with L(y) and R(y) being continuous, and if the mass density in R is f (x, y), then the mass of R is ż d c "ż R(y) L(y) f x, y dx # dy = ż d c ż R(y) L(y) f x, y dx dy = ż d c dy ż R(y) L(y) dx f x, y Theorem 3.1.3. 235 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS Implicit in Theorem 3.1.3 is the statement that, if ␣(x, y) ˇ ˇ a ď x ď b, B(x) ď y ď T(x) ( = ␣(x, y) ˇ ˇ c ď y ď d, L(y) ď x ď R(y) ( and if f (x, y) is continuous, then ż b a ż T(x) B(x) f x, y dy dx = ż d c ż R(y) L(y) f x, y dx dy This is called Fubini’s theorem2. It will be discussed more in the optional §3.1.5. The integrals of Theorem 3.1.3 are often denoted ĳ R f (x, y) dxdy or ĳ R f (x, y) dA The symbol dA represents the area of an “infinitesimal” piece of R. Notation 3.1.4. Here is a simple example. We’ll do some more complicated examples in §3.1.4. Example 3.1.5 Let R be the triangular region above the x-axis, to the right of the y-axis and to the left of the line x + y = 1. Find the mass of R if it has density f (x, y) = y. Solution. We’ll do this problem twice — once using vertical strips and once using hori-zontal strips. First, here is a sketch of R. x y x y “ 1 R p1,0q p0,1q Solution using vertical strips. We’ll now set up a double integral for the mass using vertical strips. Note, from the figure 2 This theorem is named after the Italian mathematician Guido Fubini (1879–1943). 236 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS x y y “ Tpxq “ 1 ´ x p1,0q p0,1q that • the leftmost points in R have x = 0 and the rightmost point in R has x = 1 and • for each fixed x between 0 and 1, the point (x, y) in R with the smallest y has y = 0 and the point (x, y) in R with the largest y has y = 1 ´ x. Thus R = ␣(x, y) ˇ ˇ 0 = a ď x ď b = 1, 0 = B(x) ď y ď T(x) = 1 ´ x ( and, by part (a) of Theorem 3.1.3 Mass = ż b a dx ż T(x) B(x) dy f x, y = ż 1 0 dx ż 1´x 0 dy y Now the inside integral is ż 1´x 0 y dy = y2 2 1´x 0 = 1 2(1 ´ x)2 so that the Mass = ż 1 0 dx (1 ´ x)2 2 = ´(1 ´ x)3 6 1 0 = 1 6 Solution using horizontal strips. This time we’ll set up a double integral for the mass using horizontal strips. Note, from the figure x y x “ Rpyq “ 1 ´ y p1,0q p0,1q that 237 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS • the lowest points in R have y = 0 and the topmost point in R has y = 1 and • for each fixed y between 0 and 1, the point (x, y) in R with the smallest x has x = 0 and the point (x, y) in R with the largest x has x = 1 ´ y. Thus R = ␣(x, y) ˇ ˇ 0 = c ď y ď d = 1, 0 = L(y) ď x ď R(y) = 1 ´ y ( and, by part (b) of Theorem 3.1.3 Mass = ż d c dy ż R(y) L(y) dx f x, y = ż 1 0 dy ż 1´y 0 dx y Now the inside integral is ż 1´y 0 y dx = [xy]1´y 0 = y ´ y2 since the y integral treats x as a constant. So the Mass = ż 1 0 dy y ´ y2 = y2 2 ´ y3 3 1 0 = 1 2 ´ 1 3 = 1 6 Example 3.1.5 Double integrals share the usual basic properties that we are used to from integrals of functions of one variable. See, for example, Theorem 1.2.1 and Theorem 1.2.12 in the CLP-2 text. Indeed the following theorems follow from them. Let A, B, C be real numbers. Under the hypotheses of Theorem 3.1.3, (a) ĳ R ( f (x, y) + g(x, y)) dxdy = ĳ R f (x, y) dxdy + ĳ R g(x, y) dxdy (b) ĳ R ( f (x, y) ´ g(x, y)) dxdy = ĳ R f (x, y) dxdy ´ ĳ R g(x, y) dxdy (c) ĳ R C f (x, y) dxdy = C ĳ R f (x, y) dxdy Theorem 3.1.6 (Arithmetic of Integration). 238 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS Combining these three rules we have (d) ĳ R (A f (x, y) + Bg(x, y)) dxdy = A ĳ R f (x, y) dxdy + B ĳ R g(x, y) dxdy That is, integrals depend linearly on the integrand. (e) ĳ R dxdy = Area(R) If the region R in the xy-plane is the union of regions R1 and R2 that do not overlap (except possibly on their boundaries), then (f) ĳ R f (x, y) dxdy = ĳ R1 f (x, y) dxdy + ĳ R2 f (x, y) dxdy R1 R2 R R1 R2 R R Theorem 3.1.6 (continued). In the very special (but not that uncommon) case that R is the rectangle R = ␣(x, y) ˇ ˇ a ď x ď b, c ď y ď d ( and the integrand is the product f (x, y) = g(x)h(y), ĳ R f (x, y) dxdy = ż b a dx ż d c dy g(x)h(y) = ż b a dx g(x) ż d c dy h(y) since g(x) is a constant as far as the y-integral is concerned = "ż b a dx g(x) # "ż d c dy h(y) # since ż d c dy h(y) is a constant as far as the x-integral is concerned This is worth stating as a theorem 239 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS If the domain of integration R = ␣(x, y) ˇ ˇ a ď x ď b, c ď y ď d ( is a rectangle and the integrand is the product f (x, y) = g(x)h(y), then ĳ R f (x, y) dxdy = "ż b a dx g(x) # "ż d c dy h(y) # Theorem 3.1.7. Just as was the case for single variable integrals, sometimes we don’t actually need to know the value of a double integral exactly. We are instead interested in bounds on its value. The following theorem provides some simple tools for generating such bounds. They are the multivariable analogs of the single variable tools in Theorem 1.2.12 of the CLP-2 text. Under the hypotheses of Theorem 3.1.3, (a) If f (x, y) ě 0 for all (x, y) in R, then ĳ R f (x, y) dxdy ě 0 (b) If there are constants m and M such that m ď f (x, y) ď M for all (x, y) in R, then m Area(R) ď ĳ R f (x, y) dxdy ď M Area(R) (c) If f (x, y) ď g(x, y) for all (x, y) in R, then ĳ R f (x, y) dxdy ď ĳ R g(x, y) dxdy (d) We have ˇ ˇ ˇ ˇ ˇ ˇ ĳ R f (x, y) dxdy ˇ ˇ ˇ ˇ ˇ ˇ ď ĳ R \|f (x, y)\| dxdy Theorem 3.1.8 (Inequalities for Integrals). 240 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS 3.1.3 § § Volumes Now that we have defined double integrals, we should start putting them to use. One of the most immediate applications arises from interpreting f (x, y), not as a density, but rather as the height of the part of a solid above the point (x, y) in the xy-plane. Then Theorem 3.1.3 gives the volume between the xy-plane and the surface z = f (x, y). We’ll now see how this goes in the case of part (b) of Theorem 3.1.3. The case of part (a) works in the same way. So we assume that the solid V lies above the base region R = ␣(x, y) ˇ ˇ c ď y ď d, L(y) ď x ď R(y) ( and that V = ␣(x, y, z) ˇ ˇ (x, y) P R, 0 ď z ď f (x, y) ( The base region R (which is also the top view of V) is sketched in the figure on the left below and the part of V in the first octant is sketched in the figure on the right below. y x top view x “ Lpyq x “ Rpyq dx dy y “ d y “ c R z y x z “ fpx, yq To find the volume of V we shall • Pick a natural number n and slice R into strips of width ∆y = d´c n . • Subdivide slice number i into m tiny rectangles, each of height ∆y and of width ∆x = 1 m ¨ ¨ ¨ . • Compute, approximately, the volume of the part of V that is above each rectangle. • Take the limit m Ñ 8 and then the limit n Ñ 8. We have just been through this type of argument twice. So we’ll abbreviate the argument and just say • slice the base region R into long “infinitesimally” thin strips of width dy. • Subdivide each strip into “infinitesimal” rectangles each of height dy and of width dx. See the figure on the left above. 241 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS • The volume of the part of V that is above the rectangle centred on (x, y) is essentially f (x, y) dx dy. See the figure on the right above. • So the volume of the part of V that is above the strip centred on y is essentially3 dy şR(y) L(y) dx f (x, y) and • we arrive at the following conclusion. If V = ␣(x, y, z) ˇ ˇ (x, y) P R, 0 ď z ď f (x, y) ( where R = ␣(x, y) ˇ ˇ c ď y ď d, L(y) ď x ď R(y) ( then Volume(V) = ż d c dy ż R(y) L(y) dx f (x, y) Equation 3.1.9. Similarly If V = ␣(x, y, z) ˇ ˇ (x, y) P R, 0 ď z ď f (x, y) ( where R = ␣(x, y) ˇ ˇ a ď x ď b, B(x) ď y ď T(x) ( then Volume(V) = ż b a dx ż T(x) B(x) dy f (x, y) Equation 3.1.10. 3.1.4 § § Examples Oof — we have had lots of equations and theory. It’s time to put all of this to work. Let’s start with a mass example and then move on to a volume example. You will notice that the mathematics is really very similar. Just the interpretation changes. Example 3.1.11 (Mass) Let ν ą 0 be a constant and let R be the region above the curve x2 = 4νy and to the right of the curve y2 = 1 2νx. Find the mass of R if it has density f (x, y) = xy. 3 Think of the part of V that is above the strip as being a thin slice of bread. Then the factor dy in dy şR(y) L(y) dx f (x, y) is the thickness of the slice of bread. The factor şR(y) L(y) dx f (x, y) is the surface area of the constant y cross-section ␣(x, z) ˇ ˇ L(y) ď x ď R(y), 0 ď z ď f (x, y) ( , i.e. the surface area of the slice of bread. 242 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS Solution. For practice, we’ll do this problem twice — once using vertical strips and once using horizontal strips. We’ll start by sketching R. First note that, since y ě x2 4ν and x ě 2y2 ν , both x and y are positive throughout R. The two curves intersect at points (x, y) that satisfy both x = 2y2 ν and y = x2 4ν ù ñ x = 2y2 ν = 2 ν x2 4ν 2 = x4 8ν3 ð ñ x3 8ν3 ´ 1 x = 0 This equation has only two real4 solutions — x = 0 and x = 2ν. So the upward opening parabola y = x2 4ν and the rightward opening parabola x = 2y2 ν intersect at (0, 0) and (2ν, ν). x y y “ x2 4ν x “ 2y2 ν p2ν, νq R Solution using vertical strips. We’ll now set up a double integral for the mass using vertical strips and using the abbreviated argument of the end of the last section (on volumes). Note, from the figure above, that R = " (x, y) ˇ ˇ ˇ ˇ 0 = a ď x ď b = 2ν, x2 4ν = B(x) ď y ď T(x) = c νx 2 • Slice R into long “infinitesimally” thin vertical strips of width dx. • Subdivide each strip into “infinitesimal” rectangles each of height dy and of width dx. See the figure below. dx dy x y y “ Bpxq “ x2 4ν y “ Tpxq “ aνx 2 x “ b “ 2ν 4 It also has two complex solutions that play no role here. 243 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS • The mass of the rectangle centred on (x, y) is essentially f (x, y) dx dy = xy dx dy. • So the mass of the strip centred on x is essentially dx şT(x) B(x) dy f (x, y) (the integral over y adds up the masses of all of the different rectangles on the single vertical strip in question) and • we conclude that the Mass(R) = ż b a dx ż T(x) B(x) dy f (x, y) = ż 2ν 0 dx ż ? νx/2 x2/(4ν) dy xy Here the integral over x adds up the masses of all of the different strips. Recall that, when integrating y, x is held constant, so we may factor the constant x out of the inner y integral. ż ? νx/2 x2/(4ν) dy xy = x ż ? νx/2 x2/(4ν) dy y = x y2 2 ? νx/2 x2/(4ν) = νx2 4 ´ x5 32ν2 and the Mass(R) = ż 2ν 0 dx νx2 4 ´ x5 32ν2 = ν(2ν)3 3 ˆ 4 ´ (2ν)6 6 ˆ 32ν2 = ν4 3 Solution using horizontal strips. We’ll now set up a double integral for the mass using horizontal strips, again using the abbreviated argument of the end of the last section (on volumes). Note, from the figure at the beginning of this example, that R = " (x, y) ˇ ˇ ˇ ˇ 0 = c ď y ď d = ν, 2y2 ν = L(y) ď x ď R(y) = a 4νy • Slice R into long “infinitesimally” thin horizontal strips of width dy. • Subdivide each strip into “infinitesimal” rectangles each of height dy and of width dx. See the figure below. 244 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS dx dy x y x “ Rpyq “ ?4νy x “ Lpyq “ 2y2 ν y “ d “ ν • The mass of the rectangle centred on (x, y) is essentially f (x, y) dx dy = xy dx dy. • So the mass of the strip centred on y is essentially dy şR(y) L(y) dx f (x, y) (the integral over x adds up the masses of all of the different rectangles on the single horizontal strip in question) and • we conclude that the Mass(R) = ż d c dy ż R(y) L(y) dx f (x, y) = ż ν 0 dy ż ? 4νy 2y2/ν dx xy Here the integral over y adds up the masses of all of the different strips. Recalling that, when integrating x, y is held constant Mass(R) = ż ν 0 dy y "ż ? 4νy 2y2/ν dx x # = ż ν 0 dy y x2 2 ? 4νy 2y2/ν = ż ν 0 dy 2νy2 ´ 2y5 ν2 = 2ν(ν)3 3 ´ 2ν6 6ν2 = ν4 3 Example 3.1.11 Example 3.1.12 (Volume) Let R be the part of the xy-plane above the x-axis and below the parabola y = 1 ´ x2. Find the volume between R and the surface z = x2a 1 ´ y. Solution. Yet again, for practice, we’ll do this problem twice — once using vertical strips and once using horizontal strips. First, here is a sketch of R. 245 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS x y y “ 1 ´ x2 R p´1,0q p1,0q p0,1q Solution using vertical strips. We’ll now set up a double integral for the volume using vertical strips. Note, from the figure x y y “ Tpxq “ 1 ´ x2 p´1,0q p1,0q p0,1q that • the leftmost point in R has x = ´1 and the rightmost point in R has x = 1 and • for each fixed x between ´1 and 1, the point (x, y) in R with the smallest y has y = 0 and the point (x, y) in R with the largest y has y = 1 ´ x2. Thus R = ␣(x, y) ˇ ˇ ´ 1 = a ď x ď b = 1, 0 = B(x) ď y ď T(x) = 1 ´ x2 ( and, by (3.1.10) Volume = ż b a dx ż T(x) B(x) dy f x, y = ż 1 ´1 dx ż 1´x2 0 dy x2a 1 ´ y = 2 ż 1 0 dx ż 1´x2 0 dy x2a 1 ´ y since the inside integral F(x) = ş1´x2 0 dy x2a 1 ´ y is an even function of x. Now, for x ě 0, the inside integral is ż 1´x2 0 x2a 1 ´ y dy = x2 ż 1´x2 0 a 1 ´ y dy = x2 ´2 3(1 ´ y)3/2 1´x2 0 = 2 3x21 ´ x3 246 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS so that the Volume = 2 ż 1 0 dx 2 3x21 ´ x3 = 4 3 x3 3 ´ x6 6 1 0 = 2 9 Solution using horizontal strips. This time we’ll set up a double integral for the volume using horizontal strips. Note, from the figure x y x “ Lpyq “ ´?1 ´ y x “ Rpyq “ ?1 ´ y p´1,0q p1,0q p0,1q that • the lowest points in R have y = 0 and the topmost point in R has y = 1 and • for each fixed y between 0 and 1, the point (x, y) in R with the leftmost x has x = ´ a 1 ´ y and the point (x, y) in R with the rightmost x has x = a 1 ´ y. Thus R = ␣(x, y) ˇ ˇ 0 = c ď y ď d = 1, ´ a 1 ´ y = L(y) ď x ď R(y) = a 1 ´ y ( and, by (3.1.9) Volume = ż d c dy ż R(y) L(y) dx f x, y = ż 1 0 dy ż ? 1´y ´? 1´y dx x2a 1 ´ y Now the inside integral has an even integrand (in x) and so is ż ? 1´y ´? 1´y dx x2a 1 ´ y = 2 a 1 ´ y ż ? 1´y 0 x2 dx = 2 a 1 ´ y x3 3 ? 1´y 0 = 2 3(1 ´ y)2 So the Volume = 2 3 ż 1 0 dy (1 ´ y)2 = 2 3 ´(1 ´ y)3 3 1 0 = 2 9 Example 3.1.12 Example 3.1.13 (Volume) Find the volume common to the two cylinders x2 + y2 = a2 and x2 + z2 = a2. Solution. Our first job is figure out what the specified solid looks like. Note that 247 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS • The variable z does not appear in the equation x2 + y2 = a2. So, for every value of the constant z0, the part of the cylinder x2 + y2 = a2 in the plane z = z0, is the circle x2 + y2 = a2, z = z0. So the cylinder x2 + y2 = a2 consists of many circles stacked vertically, one on top of the other. The part of the cylinder x2 + y2 = a2 that lies above the xy-plane is sketched in the figure on the left below. • The variable y does not appear in the equation x2 + z2 = a2. So, for every value of the constant y0, the part of the cylinder x2 + z2 = a2 in the plane y = y0, is the circle x2 + z2 = a2, y = y0. So the cylinder x2 + z2 = a2 consists of many circles stacked horizontally, one beside the other. The part of the cylinder x2 + z2 = a2 that lies to the right of the xz-plane is sketched in the figure on the right below. y z x y z x We have to compute the volume common to these two intersecting cylinders. • The equations x2 + y2 = a2 and x2 + z2 = a2 do not change at all if x is replaced by ´x. Consequently both cylinders, and hence our solid, is symmetric about the yz-plane. In particular the volume of the part of the solid in the octant x ď 0, y ě 0, z ě 0 is the same as the volume in the first octant x ě 0, y ě 0, z ě 0. Similarly, the equations do not change at all if y is replaced by ´y or if z is replaced by ´z. Our solid is also symmetric about both the xz-plane and the xy-plane. Hence the volume of the part of our solid in each of the eight octants is the same. • So we will compute the volume of the part of the solid in the first octant, i.e. with x ě 0, y ě 0, z ě 0. The total volume of the solid is eight times that. The part of the solid in the first octant is sketched in the figure on the left below. A point (x, y, z) lies in the first cylinder if and only if x2 + y2 ď a2. It lies in the second cylinder if y z x y x top view x “ a R x2 y2 “ a2 or y “ ? a2 ´ x2 248 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS and only if x2 + z2 ď a2. So the part of the solid in the first octant is V1 = ␣(x, y, z) ˇ ˇ x ě 0, y ě 0, z ě 0, x2 + y2 ď a2, x2 + z2 ď a2 ( Notice that, in V1, z2 ď a2 ´ x2 so that z ď ? a2 ´ x2 and V1 = ␣(x, y, z) ˇ ˇ x ě 0, y ě 0, x2 + y2 ď a2, 0 ď z ď a a2 ´ x2 ( The top view of the part of the solid in the first octant is sketched in the figure on the right above. In that top view, x runs from 0 to a. For each fixed x, y runs from 0 to ? a2 ´ x2. So we may rewrite V1 = ␣(x, y, z) ˇ ˇ (x, y) P R, 0 ď z ď f (x, y) ( where R = ! (x, y) ˇ ˇ ˇ 0 ď x ď a, 0 ď y ď a a2 ´ x2 ) and f (x, y) = a a2 ´ x2 and “(x, y) P R” is read “(x, y) is an element of R.”. Note that f (x, y) is actually indepen-dent of y. This will make things a bit easier below. We can now compute the volume of V1 using our usual abbreviated protocol. • Slice R into long “infinitesimally” thin horizontal strips of height dx. • Subdivide each strip into “infinitesimal” rectangles each of width dy and of height dx. See the figure below. y x top view dx dy x “ a y “ ? a2 ´ x2 • The volume of the part of V1 above rectangle centred on (x, y) is essentially f (x, y) dx dy = a a2 ´ x2 dx dy • So the volume of the part of V1 above the strip centred on x is essentially dx ż ? a2´x2 0 a a2 ´ x2 dy (the integral over y adds up the volumes over all of the different rectangles on the single horizontal strip in question) and 249 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS • we conclude that the Volume(V1) = ż a 0 dx ż ? a2´x2 0 dy a a2 ´ x2 Here the integral over x adds up the volumes over all of the different strips. Recall-ing that, when integrating y, x is held constant Volume(V1) = ż a 0 dx a a2 ´ x2 "ż ? a2´x2 0 dy # = ż a 0 dx a2 ´ x2 = a2x ´ x3 3 a 0 = 2a3 3 and the total volume of the solid in question is Volume(V) = 8 Volume(V1) = 16a3 3 Example 3.1.13 Example 3.1.14 (Geometric Interpretation) Evaluate ż 2 0 ż a 0 a a2 ´ x2 dx dy. Solution. This integral represents the volume of a simple geometric figure and so can be evaluated without using any calculus at all. The domain of integration is R = ␣(x, y) ˇ ˇ 0 ď y ď 2, 0 ď x ď a ( and the integrand is f (x, y) = ? a2 ´ x2, so the integral represents the volume between the xy-plane and the surface z = ? a2 ´ x2, with (x, y) running over R. We can rewrite the equation of the surface as x2 + z2 = a2, which, as in Example 3.1.13, we recognize as the equation of a cylinder of radius a centred on the y-axis. We want the volume of the part of this cylinder that lies above R. It is sketched in the figure below. y z x x “ 0, z “ a x “ a, z “ 0 y “ 2 250 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS The constant y cross-sections of this volume are quarter circles of radius a and hence of area 1 4πa2. The inside integral, şa 0 ? a2 ´ x2 dx, is exactly this area. So, as y runs from 0 to 2, ż 2 0 ż a 0 a a2 ´ x2 dx dy = 1 4πa2 ˆ 2 = πa2 2 Example 3.1.14 Example 3.1.15 (Example 3.1.14, the hard way) It is possible, but very tedious, to evaluate the integral ş2 0 şa 0 ? a2 ´ x2 dx dy of Example 3.1.14, using single variable calculus techniques. We do so now as a review of a couple of those techniques. The inside integral is şa 0 ? a2 ´ x2 dx. The standard procedure for eliminating square roots like ? a2 ´ x2 from integrands is the method of trigonometric substitution, that was covered in §1.9 of the CLP-2 text. In this case, the appropriate substitution is x = a sin θ dx = a cos θ dθ The lower limit of integration x = 0, i.e. a sin θ = 0, corresponds to θ = 0, and the upper limit x = a, i.e. a sin θ = a, corresponds to θ = π 2 , so that ż a 0 a a2 ´ x2 dx = ż π/2 0 d a2 ´ a2 sin2 θ loooooomoooooon a2 cos2 θ a cos θ dθ = a2 ż π/2 0 cos2 θ dθ The orthodox procedure for evaluating the resulting trigonometric integral şπ/2 0 cos2 θ dθ, covered in §1.8 of the CLP-2 text, uses the trigonometric double angle formula cos(2θ) = 2 cos2 θ ´ 1 to write cos2 θ = 1 + cos(2θ) 2 and then ż a 0 a a2 ´ x2 dx = a2 ż π/2 0 cos2 θ dθ = a2 2 ż π/2 0 1 + cos(2θ) dθ = a2 2 θ + sin(2θ) 2 π/2 0 = πa2 4 However we remark that there is also an efficient, sneaky, way to evaluate definite integrals like şπ/2 0 cos2 θ dθ. Looking at the figures 251 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS y x ´π ´π{2 π π{2 1 y “ cos2 x y θ ´π ´π{2 π π{2 1 y “ sin2 θ we see that ż π/2 0 cos2 θ dθ = ż π/2 0 sin2 θ dθ Thus ż π/2 0 cos2 θ dθ = ż π/2 0 sin2 θ dθ = ż π/2 0 1 2 sin2 θ + cos2 θ dθ = 1 2 ż π/2 0 dθ = π 4 In any event, the inside integral ż a 0 a a2 ´ x2 dx = πa2 4 and the full integral ż 2 0 ż a 0 a a2 ´ x2 dx dy = πa2 4 ż 2 0 dy = πa2 2 just as we saw in Example 3.1.14. Example 3.1.15 Example 3.1.16 ( Order of Integration) The integral ż 2 ´1 ż x+2 x2 dy dx represents the area of a region in the xy-plane. Express the same area as a double integral with the order of integration reversed. Solution. The critical step in reversing the order of integration is to sketch the region in the xy-plane. Rewrite the given integral as ż 2 ´1 ż x+2 x2 dy dx = ż 2 ´1 "ż x+2 x2 dy # dx From this we see that, on the domain of integration, • x runs from ´1 to 2 and • for each fixed x, y runs from the parabola y = x2 to the straight line y = x + 2. 252 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS The given iterated integral corresponds to the (vertical) slicing in the figure on the left below. p2, 4q p´1, 1q dx dy x y y “ x2 y “ x 2 p2, 4q p´1, 1q dx dy dy x y x “ ?y x “ y ´ 2 x “ ´?y To reverse the order of integration we have to switch to horizontal slices as in the figure on the right above. There we see a new wrinkle: the formula giving the value of x at the left hand end of a slice depends on whether the y coordinate of the slice is bigger than, or smaller than y = 1. Looking at the figure on the right, we see that, on the domain of integration, • y runs from 0 to 4 and • for each fixed 0 ď y ď 1, x runs from x = ´?y to x = +?y. • for each fixed 1 ď y ď 4, x runs from x = y ´ 2 to x = +?y. So ż 2 ´1 dx ż x+2 x2 dy = ż 1 0 dy ż ?y ´?y dx + ż 4 1 dy ż ?y y´2 dx Example 3.1.16 There was a moral to the last example. Just because both orders of integration have to give the same answer doesn’t mean that they are equally easy to evaluate. Here is an extreme example illustrating that moral. Example 3.1.17 Evaluate the integral of sin x x over the region in the xy-plane that is above the x-axis, to the right of the line y = x and to the left of the line x = 1. Solution. Here is a sketch of the specified domain. 253 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS p1, 1q x y y “ x x “ 1 We’ll try to evaluate the specified integral twice — once using horizontal strips (the im-possibly hard way) and once using vertical strips (the easy way). Solution using horizontal strips. To set up the integral using horizontal strips, as in the figure on the left below, we observe that, on the domain of integration, • y runs from 0 to 1 and • for each fixed y, x runs from x = y to 1. So the iterated integral is ż 1 0 dy ż 1 y dx sin x x And we have a problem. The integrand sin x x does not have an antiderivative that can be expressed in terms of elementary functions5. It is impossible to evaluate ş1 y dx sin x x without resorting to, for example, numerical methods or infinite series6. p1, 1q dx dy x y x “ y x “ 1 p1, 1q dx dy x y y “ x x “ 1 Solution using vertical strips. To set up the integral using vertical strips, as in the figure on the right above, we observe that, on the domain of integration, • x runs from 0 to 1 and • for each fixed x, y runs from 0 to y = x. 5 Perhaps the best known function whose antiderivative cannot be expressed in terms of elementary functions is e´x2. It is the integrand of the error function erf(x) = 2 ?π şx 0 e´t2 dt that is used in computing “bell curve” probabilities. See Example 3.6.14 in the CLP-2 text. 6 See, for example, Example 3.6.14 in the CLP-2 text. 254 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS So the iterated integral is ż 1 0 dx ż x 0 dy sin x x This time, because x is treated as a constant in the inner integral, it is trivial to evaluate the iterated integral. ż 1 0 dx ż x 0 dy sin x x = ż 1 0 dx sin x x ż x 0 dy = ż 1 0 dx sin x = 1 ´ cos 1 Example 3.1.17 Here is an example which is included as an excuse to review some integration tech-nique from CLP-2. Example 3.1.18 Find the volume under the surface z = 1 ´ 3x2 ´ 2y2 and above the xy-plane. Solution. Before leaping into integration, we should try to understand what the surface and volume look like. For each constant z0, the part of the surface z = 1 ´ 3x2 ´ 2y2 that lies in the horizontal plane z = z0 is the ellipse 3x2 + 2y2 = 1 ´ z0. The biggest of these ellipses is that in the xy-plane, where z0 = 0. It is the ellipse 3x2 + 2y2 = 1. As z0 increases the ellipse shrinks, degenerating to a single point, namely (0, 0, 1), when z0 = 1. So the surface consists of a stack of ellipses and our solid is V = ␣(x, y, z) ˇ ˇ 3x2 + 2y2 ď 1, 0 ď z ď 1 ´ 3x2 ´ 2y2 ( This is sketched in the figure below z y x z “ 1 ´ 3x2 ´ 2y2 3x2 2y2 “ 1 R The top view of the base region R = ␣(x, y) ˇ ˇ 3x2 + 2y2 ď 1 ( is sketched in the figure below. 255 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS y x top view x “ ´ b 1´2y2 3 x “ b 1´2y2 3 dx dy y “ 1{ ? 2 y “ ´1{ ? 2 R Considering that the x-dependence in z = 1 ´ 3x2 ´ 2y2 is almost identical to the y-dependence in z = 1 ´ 3x2 ´ 2y2 (only the coefficients 2 and 3 are interchanged), using vertical slices is likely to lead to exactly the same level of difficulty as using horizontal slices. So we’ll just pick one — say vertical slices. The fattest part of R is on the y-axis. The intersection points of the ellipse with the y-axis have x = 0 and y obeying 3(0)2 + 2y2 = 1 or y = ˘1/ ? 2. So in R, ´1/ ? 2 ď y ď 1/ ? 2 and, for each such y, 3x2 ď 1 ´ 2y2 or ´ b 1´2y2 3 ď x ď b 1´2y2 3 . So using vertical strips as in the figure above Volume(V) = ĳ R 1 ´ 3x2 ´ 2y2 dx dy = ż 1/ ? 2 ´1/ ? 2 dy ż b 1´2y2 3 ´ b 1´2y2 3 dx 1 ´ 3x2 ´ 2y2 = 4 ż 1/ ? 2 0 dy ż b 1´2y2 3 0 dx 1 ´ 3x2 ´ 2y2 = 4 ż 1/ ? 2 0 dy h (1 ´ 2y2)x ´ x3i b 1´2y2 3 0 = 4 ż 1/ ? 2 0 dy c 1 ´ 2y2 3 h (1 ´ 2y2) ´ 1 ´ 2y2 3 i = 8 ż 1/ ? 2 0 dy 1 ´ 2y2 3 3/2 To evaluate this integral, we use the trig substitution7 2y2 = sin2 θ, or y = sin θ ? 2 dy = cos θ ? 2 dθ 7 See §1.9 in the CLP-2 text for a general discussion of trigonometric substitution. 256 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS to give Volume(V) = 8 ż π/2 0 dy hkkkikkkj dθ cos θ ? 2 cos2 θ 3 3/2 = 8 ? 54 ż π/2 0 dθ cos4 θ Then to integrate cos4 θ, we use the double angle formula8 cos2 θ = cos(2θ) + 1 2 ù ñ cos4 θ = cos(2θ) + 1 2 4 = cos2(2θ) + 2 cos(2θ) + 1 4 = cos(4θ)+1 2 + 2 cos(2θ) + 1 4 = 3 8 + 1 2 cos(2θ) + 1 8 cos(4θ) Finally, since şπ/2 0 cos(4θ) dθ = şπ/2 0 cos(2θ) dθ = 0, Volume(V) = 8 ? 54 3 8 π 2 = π 2 ? 6 Example 3.1.18 3.1.5 § § Optional — More about the Definition of ť R f (x, y) dxdy Technically, the integral ť R f (x, y) dx dy, where R is a bounded region in R2, is defined as follows. • Subdivide R by drawing lines parallel to the x and y axes. x y px˚ 13, y˚ 13q • Number the resulting rectangles contained in R, 1 through n. Notice that we are numbering all of the rectangles in R, not just those in one particular row or column. 8 We weren’t joking about his being a good review of single variable integration techniques. See Example 1.8.8 in the CLP-2 text. 257 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS • Denote by ∆Ai the area of rectangle #i. • Select an arbitrary point (x˚ i , y˚ i ) in rectangle #i. • Form the sum n ř i=1 f (x˚ i , y˚ i )∆Ai. Again note that the sum runs over all of the rectan-gles in R, not just those in one particular row or column. Now repeat this construction over and over again, using finer and finer grids. If, as the size9 of the rectangles approaches zero, this sum approaches a unique limit (independent of the choice of parallel lines and of points (x˚ i , y˚ i )), then we define ĳ R f (x, y) dx dy = lim n ÿ i=1 f (x˚ i , y˚ i ) ∆Ai If f (x, y) is continuous in a region R described by a ď x ď b B(x) ď y ď T(x) for continuous functions B(x), T(x), then ĳ R f (x, y) dx dy and ż b a dx ż T(x) B(x) dy f (x, y) both exist and are equal. Similarly, if R is described by c ď y ď d L(y) ď x ď R(y) for continuous functions L(y), R(y), then ĳ R f (x, y) dx dy and ż d c dy ż R(y) L(y) dx f (x, y) both exist and are equal. Theorem 3.1.19. The proof of this theorem is not particularly difficult, but is still beyond the scope of this text. The main ideas in the proof can already be seen in section 1.1.6 of the CLP-2 text. An important consequence of this theorem is 9 For example, let pi be the perimeter of rectangle number i and require that max1ďiďn pi tends to zero. This way both the heights and widths of all rectangles also tend to zero. 258 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS If f (x, y) is continuous in a region R described by both # a ď x ď b B(x) ď y ď T(x) + and # c ď y ď d L(y) ď x ď R(y) + for continuous functions B(x), T(x), L(y), R(y), then both ż b a dx ż T(x) B(x) dy f (x, y) and ż d c dy ż R(y) L(y) dx f (x, y) exist and are equal. Theorem 3.1.20 (Fubini). The hypotheses of both of these theorems can be relaxed a bit, but not too much. For example, if R = ␣(x, y) ˇ ˇ 0 ď x ď 1, 0 ď y ď 1 ( f (x, y) = # 1 if x, y are both rational numbers 0 otherwise then the integral ť R f (x, y) dx dy does not exist. This is easy to see. If all of the x˚ i ’s and y˚ i ’s are chosen to be rational numbers, then n ÿ i=1 f (x˚ i , y˚ i ) ∆Ai = n ÿ i=1 ∆Ai = Area(R) But if we choose all the x˚ i ’s and y˚ i ’s to be irrational numbers, then n ÿ i=1 f (x˚ i , y˚ i ) ∆Ai = n ÿ i=1 0 ∆Ai = 0 So the limit of n ř i=1 f (x˚ i , y˚ i ) ∆Ai, as the maximum diagonal of the rectangles approaches zero, depends on the choice of points (x˚ i , y˚ i ). So the integral ť R f (x, y) dx dy does not exist. Here is an even more pathological10 example. Example 3.1.21 In this example, we relax exactly one of the hypotheses of Fubini’s Theorem, namely the continuity of f, and construct an example in which both of the integrals in Fubini’s The-orem exist, but are not equal. In fact, we choose R = ␣(x, y) ˇ ˇ 0 ď x ď 1, 0 ď y ď 1 ( and we use a function f (x, y) that is continuous on R, except at exactly one point — the origin. 10 For mathematicians, “pathological” is a synonym for “cool”. 259 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS First, let δ1, δ2, δ3, ¨ ¨ ¨ be any sequence of real numbers obeying 1 = δ1 ą δ2 ą δ3 ą ¨ ¨ ¨ ą δn Ñ 0 For example δn = 1 n or δn = 1 2n´1 are both acceptable. For each positive integer n, let In = (δn+1, δn] = ␣ t ˇ ˇ δn+1 ă t ď δn ( and let gn(t) be any nonnegative continuous function obeying • gn(t) = 0 if t is not in In and • ş In g(t) dt = 1 There are many such functions. For example gn(t) = 2 δn ´ δn+1 2 $ ’ ’ & ’ ’ % δn ´ t if 1 2(δn+1 + δn) ď t ď δn t ´ δn+1 if δn+1 ď t ď 1 2(δn+1 + δn) 0 otherwise t y δn1 δn 2 δn´δn1 y “ gnptq Here is a summary of what we have done so far. • We subdivided the interval 0 ă x ď 1 into infinitely many subintervals In. As n increases, the subinterval In gets smaller and smaller and also gets closer and closer to zero. • We defined, for each n, a nonnegative continuous function gn that is zero everywhere outside of In and whose integral over In is one. Now we define the integrand f (x, y) in terms of these subintervals In and functions gn. f (x, y) = $ ’ ’ ’ ’ ’ ’ ’ ’ & ’ ’ ’ ’ ’ ’ ’ ’ % 0 if x = 0 0 if y = 0 gm(x)gn(y) if x P Im, y P In with m = n ´gm(x)gn(y) if x P Im, y P In with m = n + 1 0 otherwise You should think of (0, 1] ˆ (0, 1] as a union of a bunch of small rectangles Im ˆ In, as in the figure below. On most of these rectangles, f (x, y) is just zero. The exceptions are the darkly shaded rectangles In ˆ In on the “diagonal” of the figure and the lightly shaded rectangles In+1 ˆ In just to the left of the “diagonal”. On each darkly shaded rectangle, f (x, y) ě 0 and the graph of f (x, y) is the graph of gn(x)gn(y) which looks like a pyramid. On each lightly shaded rectangle, f (x, y) ď 0 and the graph of f (x, y) is the graph of ´gn+1(x)gn(y) which looks like a pyramidal hole in the ground. Now fix any 0 ď y ď 1 and let’s compute ş1 0 f (x, y) dx. That is, we are integrat-ing f along a line that is parallel to the x-axis. If y = 0, then f (x, y) = 0 for all x, so 260 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS δ1 δ1 δ2 δ2 δ3 δ3 δ4 δ4 δ5 δ5 I1 I1 I2 I2 I3 I3 I4 I4 ´ ´ ´ ´ ´ x y ş1 0 f (x, y) dx = 0. If 0 ă y ď 1, then there is exactly one positive integer n with y P In and f (x, y) is zero, except for x in In or In+1. So for y P In ż 1 0 f (x, y) dx = ÿ m=n,n+1 ż Im f (x, y) dx = ż In gn(x)gn(y) dx ´ ż In+1 gn+1(x)gn(y) dx = gn(y) ż In gn(x) dx ´ gn(y) ż In+1 gn+1(x) dx = gn(y) ´ gn(y) = 0 Here we have twice used that ş Im g(t) dt = 1 for all m. Thus ş1 0 f (x, y) dx = 0 for all y and hence ş1 0 dy h ş1 0 dx f (x, y) i = 0. Finally, fix any 0 ď x ď 1 and let’s compute ş1 0 f (x, y) dy. That is, we are integrat-ing f along a line that is parallel to the y-axis. If x = 0, then f (x, y) = 0 for all y, so ş1 0 f (x, y) dy = 0. If 0 ă x ď 1, then there is exactly one positive integer m with x P Im. If m ě 2, then f (x, y) is zero, except for y in Im and Im´1. But, if m = 1, then f (x, y) is zero, except for y in I1. (Take another look at the figure above.) So for x P Im, with m ě 2, ż 1 0 f (x, y) dy = ÿ n=m,m´1 ż In f (x, y) dy = ż Im gm(x)gm(y) dy ´ ż Im´1 gm(x)gm´1(y) dy = gm(x) ż Im gm(y) dy ´ gm(x) ż Im´1 gm´1(y) dy = gm(x) ´ gm(x) = 0 261 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS But for x P I1, ż 1 0 f (x, y) dy = ż I1 f (x, y) dy = ż I1 g1(x)g1(y) dy = g1(x) ż I1 g1(y) dy = g1(x) Thus ż 1 0 f (x, y) dy = # 0 if x ď δ2 g1(x) if x P I1 and hence ż 1 0 dx ż 1 0 dy f (x, y) = ż I1 g1(x) dx = 1 The conclusion is that for the f (x, y) above, which is defined for all 0 ď x ď 1, 0 ď y ď 1 and is continuous except at (0, 0), ż 1 0 dy ż 1 0 dx f (x, y) = 0 ż 1 0 dx ż 1 0 dy f (x, y) = 1 Example 3.1.21 3.1.6 § § Even and Odd Functions During the course of our study of integrals of functions of one variable, we found that the evaluation of certain integrals could be substantially simplified by exploiting symmetry properties of the integrand. Concretely, in §1.2.1 of the CLP-2 text, we gave the Let f (x) be a function of one variable. Then, • we say that f (x) is even when f (´x) = f (x) for all x, and • we say that f (x) is odd when f (´x) = ´f (x) for all x Definition 3.1.22 (Definition 1.2.8 in the CLP-2 text). and we saw that • f (x) = \|x\|, f (x) = cos x and f (x) = x2 are even functions and • f (x) = sin x, f (x) = tan x and f (x) = x3 are odd functions. • In fact, if f (x) is any even power of x, then f (x) is an even function and if f (x) is any odd power of x, then f (x) is an odd function. We also learned how to exploit evenness and oddness to simplify integration. 262 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS Let a ą 0. (a) If f (x) is an even function, then ż a ´a f (x)dx = 2 ż a 0 f (x)dx (b) If f (x) is an odd function, then ż a ´a f (x)dx = 0 Theorem 3.1.23 (Theorem 1.2.11 in the CLP-2 text). We will now see that we can similarly exploit evenness and oddness of functions of more than one variable. But for functions of more than one variable there is also more than one kind of oddness and evenness. In the Definition 3.1.22 (Definition 1.2.8 in the CLP-2 text) of evenness and oddness of the function f (x), we compared the value of f at x with the value of f at ´x. The points x and ´x are the same distance from the origin, 0, and are on opposite sides of 0. The point ´x is called the reflection of x across the origin. To prepare for our definitions of evenness and oddness of functions of two variables, we now define three different reflections in the two dimensional world of the xy-plane. Let x and y be two real numbers. • The reflection of (x, y) across the y-axis is (´x, y). • The reflection of (x, y) across the x-axis is (x, ´y). • The reflection of (x, y) across the origin is (´x, ´y). x y px, yq p´x, yq px, ´yq p´x, ´yq Definition 3.1.24. • To get from the point (x, y) to its image reflected across the y-axis, you – start from (x, y), and 263 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS – walk horizontally straight to the y-axis, and – cross the y-axis, and – continue horizontally the same distance as you have already travelled to (´x, y). Here are four examples. p5, 6q p´5, 6q p5, 2q p´5, 2q p5, ´2q p´5, ´2q p5, ´6q p´5, ´6q x y reﬂection across the y-axis • To get from the point (x, y) to its image reflected across the x-axis, you – start from (x, y), and – walk vertically straight to the x-axis, and – cross the x-axis, and – continue vertically the same distance as you have already travelled to the re-flected image (x, ´y). Here are four examples. p6, 4q p6, ´4q p2, 4q p2, ´4q p´2, 4q p´2, ´4q p´6, 4q p´6, ´4q x y reﬂection across the x-axis • To get from the point (x, y) to its image reflected across the origin, you – start from (x, y), and – walk radially straight to the origin, and – cross the origin, and – continue radially in the same direction the same distance as you have already travelled to the reflected image (´x, ´y). 264 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS Here are three examples. p3,6q p´3,´6q p6,6q p´6,´6q p7,2q p´7,´2q x y reﬂection across the origin For each of these three types of reflection, there is a corresponding kind of oddness and evenness. Let f (x, y) be a function of two variables. Then, • we say that f (x, y) is even (under reflection across the origin) when f (´x, ´y) = f (x, y) for all x and y, and • we say that f (x, y) is odd (under reflection across the origin) when f (´x, ´y) = ´f (x, y) for all x and y and • we say that f (x, y) is even under x Ñ ´x (i.e. under reflection across the y-axis) when f (´x, y) = f (x, y) for all x and y, and • we say that f (x, y) is odd under x Ñ ´x (i.e. under reflection across the y-axis) when f (´x, y) = ´f (x, y) for all x and y and • we say that f (x, y) is even under y Ñ ´y (i.e. under reflection across the x-axis) when f (x, ´y) = f (x, y) for all x and y, and • we say that f (x, y) is odd under y Ñ ´y (i.e. under reflection across the x-axis) when f (x, ´y) = ´f (x, y) for all x and y. Definition 3.1.25. Example 3.1.26 Let m and n be two integers and set f (x, y) = xmyn. Then f (´x, y) = (´x)myn = (´1)mxmyn = (´1)m f (x, y) f (x, ´y) = xm(´y)n = (´1)nxmyn = (´1)n f (x, y) f (´x, ´y) = (´x)m(´y)n = (´1)m+nxmyn = (´1)m+n f (x, y) 265 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS Consequently • if m is even, then f (x, y) is even under x Ñ ´x and • if m is odd, then f (x, y) is odd under x Ñ ´x and • if n is even, then f (x, y) is even under y Ñ ´y and • if n is odd, then f (x, y) is odd under y Ñ ´y and • if m + n is even, then f (x, y) is even (under reflection across the origin) and • if m + n is odd, then f (x, y) is odd (under reflection across the origin). Example 3.1.26 Recall from Theorem 3.1.23 (or Theorem 1.2.11 in the CLP-2 text) that we can exploit the evenness or oddness of the integrand, f (x), of the integral şa b f (x) dx to simplify the evaluation of the integral when b = ´a, i.e. when the domain of integration is invariant under reflection across the origin. Similarly, we will be able to simplify the evaluation of the double integral ť R f x, y dx dy when the integrand is even or odd and the domain of integration R is invariant under the corresponding reflection — meaning that the reflected R is identical to the original R. Here are some details for “reflection across the y-axis”. The details for the other reflections are similar. • If R is any subset of the xy-plane, the reflection of R across the y-axis = ␣(´x, y) ˇ ˇ (x, y) P R ( The set notation on the right hand side means “the set of all points (´x, y) with (x, y) a point of R”. • In the special case11 that R = ␣(x, y) ˇ ˇ c ď y ď d, L(y) ď x ď R(y) ( (see §3.1.2 on horizontal slices) then the reflection of R across the y-axis = ␣(x, y) ˇ ˇ c ď y ď d, ´R(y) ď x ď ´L(y) ( In the sketch below Ry is the reflection of R across the y-axis. 11 Here L(y) (“L” stands for “left”) is the leftmost allowed value of x when the y-coordinate is y, and R(y) (“R” stands for “right”) is the rightmost allowed value of x, when the y-coordinate is y. 266 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS x y x “ Lpyq x “ Rpyq y “ d y “ c R x y x “ ´Lpyq x “ ´Rpyq y “ d y “ c Ry • A subset R of the xy-plane is invariant under reflection across the y-axis (or is also known as “symmetric about the y-axis”) when (´x, y) is in R ð ñ (x, y) is in R Recall that the symbol ð ñ is read “if and only if”. In the special case that R = ␣(x, y) ˇ ˇ c ď y ď d, L(y) ď x ď R(y) ( R is invariant under reflection across the y-axis when L(y) = ´R(y). Here are some more sketches. The first sketch is of a rectangle that is invariant under reflection across the y-axis, but is not invariant under reflection across the x-axis. The remaining three sketches show a triangle and its reflections across the y-axis, across the x-axis and across the origin. p5, 5q p5, ´1q p´5, 5q p´5, ´1q x y symmetric about the y-axis R Ry p1, 6q p´1, 6q p6, 5q p´6, 5q x y Ry “reﬂection of R across the y-axis 267 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS R Rx p1, 6q p1, ´6q p6, 5q p6, ´5q x y Rx “reﬂection of R across the x-axis R Ro p1, 6q p´1, ´6q p6, 5q p´6, ´5q x y Ro “reﬂection of R across the origin We are finally ready for the analog of Theorem 3.1.23 (Theorem 1.2.11 in the CLP-2 text) for functions of two variables. By way of motivation for that theorem, consider the integral ť R f (x, y) dxdy, with the integrand, f (x, y), odd under x Ñ ´x, and the domain of integration, R, symmetric about the y-axis. Slice up R into tiny (think “infinitesmal”) squares, either by subdividing vertical slices into tiny squares, as in §3.1.1, or by subdi-viding horizontal slices into tiny squares, as in §3.1.2. Concentrate on any point (x0, y0) in R. The contribution to the integral coming from the square that contains (x0, y0) is x y px0, y0q p´x0, y0q (essentially12) f (x0, y0) ∆x ∆y. That contribution is cancelled by the contribution coming from the square containing (the reflected point) (´x0, y0), which is f (´x0, y0) ∆x ∆y = ´f (x0, y0) ∆x ∆y This is the case for all points (x0, y0) in R. Consequently ĳ R f (x, y) dxdy = 0 Here is the analog of Theorem 3.1.23 for functions of two variables. 12 In this motivation, we suppress the ∆x Ñ 0 and ∆y Ñ 0 limits. 268 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS (a) Let R be a subset of the xy-plane that is symmetric about the y-axis. If f (x, y) is odd under x Ñ ´x, then ĳ R f (x, y) dxdy = 0 Denote by R+ the set of all points in R that have x ě 0. If f (x, y) is even under x Ñ ´x, then ĳ R f (x, y) dxdy = 2 ĳ R+ f (x, y) dxdy (b) Let R be a subset of the xy-plane that is symmetric about the x-axis. If f (x, y) is odd under y Ñ ´y, then ĳ R f (x, y) dxdy = 0 Denote by R+ the set of all points in R that have y ě 0. If f (x, y) is even under y Ñ ´y, then ĳ R f (x, y) dxdy = 2 ĳ R+ f (x, y) dxdy (c) Let R be a subset of the xy-plane that is invariant under reflection across the origin. If f (x, y) is odd (under reflection across the origin), then ĳ R f (x, y) dxdy = 0 Denote by R+ either the set of all points in R that have x ě 0 or the set of all points in R that have y ě 0. If f (x, y) is even (under reflection across the origin), then ĳ R f (x, y) dxdy = 2 ĳ R+ f (x, y) dxdy Theorem 3.1.27 (2d Even and Odd). Proof. We will give only the proof for part (a) in the special case that R = ␣(x, y) ˇ ˇ c ď y ď d, L(y) ď x ď R(y) ( In part (a), we are assuming that R is symmetric about the y-axis, so that L(y) = ´R(y). 269 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS So, using horizontal strips, as described in §3.1.2, ĳ R f (x, y) dxdy = ż d c dy ż R(y) ´R(y) dx f (x, y) Fix any c ď y ď d. • If f (x, y) is odd under x Ñ ´x, then f (´x, y) = ´f (x, y) for all ´R(y) ď x ď R(y) and ż R(y) ´R(y) dx f (x, y) = 0 by part (b) of Theorem 3.1.23 (Theorem 1.2.11 in the CLP-2 text). • If f (x, y) is even under x Ñ ´x, then f (´x, y) = f (x, y) for all ´R(y) ď x ď R(y) and ż R(y) ´R(y) dx f (x, y) = 2 ż R(y) 0 dx f (x, y) by part (a) of Theorem 3.1.23. As the statements of the two bullets are true for each fixed c ď y ď d, we have that • if f (x, y) is odd under x Ñ ´x, then ĳ R f (x, y) dxdy = ż d c dy ż R(y) ´R(y) dx f (x, y) = ż d c dy 0 = 0 • and if f (x, y) is even under x Ñ ´x, then ĳ R f (x, y) dxdy = ż d c dy ż R(y) ´R(y) dx f (x, y) = ż d c dy 2 ż R(y) 0 dx f (x, y) = 2 ĳ R+ f (x, y) dxdy The proof of part (a) when R is not of the form R = ␣(x, y) ˇ ˇ c ď y ď d, L(y) ď x ď R(y) ( (for example if R has holes in it) is most easily done using the change of variables x = ´u, y = v in Theorem 3.8.3, which is part of the optional §3.8. The proof of part (b) is similar to the proof of part (a). The proof of part (c) is most easily done using the change of variables x = ´u, y = ´v in Theorem 3.8.3, which is part of the optional §3.8. Example 3.1.28 Evaluate the integral ť R ex sin(y + y3) dxdy over the triangular region R in the sketch 270 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS x y R p5, 3q p5, ´3q p´5, 0q Solution. Start by checking the evenness and oddness properties of the integrand f (x, y) = ex sin(y + y3). Since f (´x, y) = e´x sin(y + y3) f (x, ´y) = ex sin ´ y + (´y)3 = ex sin(´y ´ y3) = ´ex sin(y + y3) = ´f (x, y) f (´x, ´y) = ´e´x sin(y + y3) the integrand is odd under y Ñ ´y but is neither even nor odd under x Ñ ´x and (x, y) Ñ ´(x, y). Fortunately (or by rigging), the domain of integration R is invariant under y Ñ ´y (i.e. is symmetric about the x-axis) and so ĳ R ex sin(y + y3) dxdy = 0 by part (b) of Theorem 3.1.27 (Theorem 1.2.11 in the CLP-2 text). Example 3.1.28 Example 3.1.29 Evaluate the integral ť R(xey + yex + xexy + 7) dxdy over the region R whose outer bound-ary is the ellipse x2 + 4y2 = 1. Solution. First, let’s sketch the ellipse x2 + 4y2 = 1. Notice that its x intercepts are the points (x, 0) that obey x2 + 4(0)2 = 1. So the x-intercepts are (˘1, 0). Similarly its y intercepts are the points (0, y) that obey 02 + 4y2 = 1. So the y-intercepts are (0, ˘1/2). Here is a sketch of R. x y R p1, 0q p0, 1{2q From the sketch, it looks like R is invariant under x Ñ ´x (i.e. is symmetric about the y-axis) and is also invariant under y Ñ ´y (i.e. is symmetric about the x-axis) and is also 271 MULTIPLE INTEGRALS 3.1 DOUBLE INTEGRALS invariant under (x, y) Ñ ´(x, y). It is easy to check analytically that this is indeed the case. The point (x, y) is in R if and only if it is inside x2 + 4y2 = 1. That is the case if and only if x2 + 4y2 ď 1. Since (´x)2 + 4y2 = x2 + (´4y)2 = (´x)2 + 4(´y)2 = x2 + 4y2 we have (x, y) is in R ð ñ (´x, y) is in R ð ñ (x, ´y) is in R ð ñ (´x, ´y) is in R Now let’s check the evenness and oddness properties of the integrand. f (x, y) = xey + yex + xexy + 7 f (´x, y) = ´xey + ye´x ´ xe´xy + 7 f (x, ´y) = xe´y ´ yex + xe´xy + 7 f (´x, ´y) = ´xe´y ´ ye´x ´ xexy + 7 So f (x, y) is neither even nor odd under any of x Ñ ´x, y Ñ ´y, and (x, y) Ñ ´(x, y). BUT, look at the four terms of f (x, y) separately. • The first term of f (x, y), namely xey, is odd under x Ñ ´x. • The second term of f (x, y), namely yex, is odd under y Ñ ´y. • The third term of f (x, y), namely xexy, is odd under (x, y) Ñ ´(x, y). • The fourth term of f (x, y), namely 7, is even under all of x Ñ ´x, y Ñ ´y, and (x, y) Ñ ´(x, y). So, by parts (a), (b) and (c) of Theorem 3.1.27, in order, ĳ R (xey + yex + xexy + 7) dxdy = ĳ R xey dxdy + ĳ R yex dxdy + ĳ R xexy dxdy + 7 ĳ R dxdy = 0 + 0 + 0 + 7 Area(R) Since R is an ellipse with semi-major axis a = 1 and semi-minor axis b = 1 2, it has area πab = 1 2π and ĳ R (xey + yex + xexy + 7) dxdy = 7 2π Example 3.1.29 272 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES 3.2Ĳ Double Integrals in Polar Coordinates So far, in setting up integrals, we have always cut up the domain of integration into tiny rectangles by drawing in many lines of constant x and many lines of constant y. x y There is no law that says that we must cut up our domains of integration into tiny pieces in that way. Indeed, when the objects of interest are sort of round and centered on the origin, it is often advantageous13 to use polar coordinates, rather than Cartesian coordinates. 3.2.1 § § Polar Coordinates It may have been a while since you did anything in polar coordinates. So let’s review before we resume integrating. The polar coordinates14 of any point (x, y) in the xy-plane are r = the distance from (0, 0) to (x, y) θ = the (counter-clockwise) angle between the x axis and the line joining (x, y) to (0, 0) r px, yq x y θ Definition 3.2.1. 13 The “golden hammer” (also known as Maslow’s hammer and as the law of the instrument) refers to a tendency to always use the same tool, even when it isn’t the best tool for the job. It is just as bad in mathematics as it is in carpentry. 14 In the mathematical literature, the angular coordinate is usually denoted θ, as we do here. The symbol ϕ is also often used for the angular coordinate. In fact there is an ISO standard (#80000-2) which specifies that ϕ should be used in the natural sciences and in technology. See Appendix G. 273 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES Cartesian and polar coordinates are related, via a quick bit of trigonometry, by x = r cos θ y = r sin θ r = b x2 + y2 θ = arctan y x Equation 3.2.2. The following two figures show a number of lines of constant θ, on the left, and curves of constant r, on the right. x y lines of constant θ x y curves of constant r Note that the polar angle θ is only defined up to integer multiples of 2π. For example, the point (1, 0) on the x-axis could have θ = 0, but could also have θ = 2π or θ = 4π. It is sometimes convenient to assign θ negative values. When θ ă 0, the counter-clockwise15 angle θ refers to the clockwise angle \|θ\|. For example, the point (0, ´1) on the negative y-axis can have θ = ´π 2 and can also have θ = 3π 2 . x y r “ 1, θ “ ´π{2, θ “ 3π{2 3π{2 π{2 It is also sometimes convenient to extend the above definitions by saying that x = r cos θ and y = r sin θ even when r is negative. For example, the following figure shows (x, y) for r = 1, θ = π/4 and for r = ´1, θ = π/4. Both points lie on the line through 15 or anti-clockwise or widdershins. Yes, widdershins is a real word, though the Oxford English Dic-tionary lists its frequency of usage as between 0.01 and 0.1 times per million words. Of course both “counter-clockwise” and “anti-clockwise” assume that your clock is not a sundial in the southern hemi-sphere. 274 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES r “ 1, θ “ π{4 r “ ´1, θ “ π{4 x y π{4 the origin that makes an angle of 45˝ with the x-axis and both are a distance one from the origin. But they are on opposite sides of the origin. 3.2.2 § § Polar Curves Here are a couple of examples in which we sketch curves specified by equations in terms of polar coordinates. Example 3.2.3 (The Cardioid) Let’s sketch the curve r = 1 + cos θ Our starting point will be to understand how 1 + cos θ varies with θ. So it will be helpful to remember what the graph of cos θ looks like for 0 ď θ ď 2π. y θ π 2 3π 2 π 2π 1 ´1 y “ cos θ From this we see that the graph of y = 1 + cos θ is y θ π 2 3π 2 π 2π 1 2 y “ 1 cos θ Now let’s pick some easy θ values, find the corresponding r’s and sketch them. 275 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES • When θ = 0, we have r = 1 + cos 0 = 1 + 1 = 2. To sketch the point with θ = 0 and r = 2, we first draw in the half-line consisting of all points with θ = 0, r ą 0. That’s the positive x-axis, sketched in gray in the leftmost figure below. Then we put in a dot on that line a distance 2 from the origin. That’s the red dot in the leftmost figure below. • Now increase θ a bit (to another easy place to evaluate), say to θ = π 6 . As we do so r = 1 + cos θ decreases to r = 1 + cos π 6 = 1 + ? 3 2 « 1.87. To sketch the point with θ = π 6 and r « 1.87, we first draw in the half-line consisting of all points with θ = π 6 , r ą 0. That’s the upper gray line in the second figure below. Then we put in a dot on that line a distance 1.87 from the origin. That’s the upper red dot in the second figure below. x y θ “ 0 x y θ “ 0, π{6 x y θ “ 0, π{6, π{3, ¨ ¨ ¨ • Now increase θ still more, say to – θ = 2π 6 = π 3 , – followed by θ = 3π 6 = π 2 , – followed by θ = 4π 6 = 2π 3 , – followed by θ = 5π 6 , – followed by θ = 6π 6 = π. As θ increases, r = 1 + cos θ decreases, hitting r = 1 when θ = π 2 and ending at r = 0 when θ = π. For each of these θ’s, we first draw in the half-line consisting of all points with that θ and r ě 0. Those are the five gray lines in the figure on the right above. Then we put in a dot on each θ-line a distance r = 1 + cos θ from the origin. Those are the red dots on the gray lines in the figure on the right above. • We could continue the above procedure for π ď θ ď 2π. Or we can look at the graph of cos θ above and notice that the graph of cos θ for π ď θ ď 2π is exactly the mirror image, about θ = π, of the graph of cos θ for 0 ď θ ď π. That is, y θ π 2 3π 2 π 2π 1 ´1 θ “ π y “ cos θ 276 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES cos(π + θ) = cos(π ´ θ) so that r(π + θ) = r(π ´ θ). So we get the figure. x y • Finally, we fill in a smooth curve through the dots and we get the graph below. This curve is called a cardioid because it looks like a heart16. x y r “ 1 cospθq Example 3.2.3 Example 3.2.4 (The Three Petal Rose) Now we’ll use the same procedure as in the last example to sketch the graph of r = sin(3θ) Again it will be useful to remember what the graph of sin(3θ) looks like for 0 ď θ ď 2π. y θ π 6 π 3 π 2 2π 3 π 4π 3 5π 3 2π 1 ´1 y “ sinp3θq • We’ll first consider 0 ď θ ď π 3 , so that 0 ď 3θ ď π. On this interval r(θ) = sin(3θ) 16 Well, a mathematician’s heart. The name “cardioid” comes from the Greek word καρδια (which angli-cizes to kardia) for heart. 277 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES – starts with r(0) = 0, and then – increases as θ increases until – 3θ = π 2 , i.e. θ = π 6 , where r π/6 = 1, and then – decreases as θ increases until – 3θ = π, i.e. θ = π 3 , where r π/3 = 0, again. Here is a table giving a few values of r(θ) for 0 ď θ ď π 3 . Notice that we have chosen values of θ for which sin(3θ) is easy to compute. θ 3θ r(θ) 0 0 0 π/12 π/4 1/ ? 2 « 0.71 2π/12 π/2 1 3π/12 3π/4 1/ ? 2 « 0.71 4π/12 π 0 and here is a sketch exhibiting those values and another sketch of the part of the curve with 0 ď θ ď π 3 . x y θ“π{3 θ“π{6 x y θ“π{3 • Next consider π 3 ď θ ď 2π 3 , so that π ď 3θ ď 2π. On this interval r(θ) = sin(3θ) – starts with r π/3 = 0, and then – decreases as θ increases until – 3θ = 3π 2 , i.e. θ = π 2 , where r π/2 = ´1, and then – increases as θ increases until – 3θ = 2π, i.e. θ = 2π 3 , where r 2π/3 = 0, again. We are now encountering, for the first time, r(θ)’s that are negative. The figure on the left below contains, for each of θ = 4π 12 = π 3 , 5π 12 , 6π 12 = π 2 , 7π 12 and 8π 12 = 2π 3 – the (dashed) half-line consisting of all points with that θ and r ă 0 and – the dot with that θ and r(θ) = sin(3θ). The figure on the right below provides a sketch of the part of the curve r = sin(3θ) with π 3 ď θ ď 2π 3 . 278 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES x y θ“π{3 θ“2π{3 x y θ“π{3 θ“2π{3 • Finally consider 2π 3 ď θ ď π (because r(θ + π) = sin(3θ + 3π) = ´ sin(3θ) = ´r(θ), the part of the curve with π ď θ ď 2π just retraces the part with 0 ď θ ď π), so that 2π ď 3θ ď 3π. On this interval r(θ) = sin(3θ) – starts with r 2π/3 = 0, and then – increases as θ increases until – 3θ = 5π 2 , i.e. θ = 10π 12 , where r 5π/2 = 1, and then – decreases as θ increases until – 3θ = 3π, i.e. θ = 12π 12 = π, where r π = 0, again. The figure on the left below contains, for each of θ = 8π 12 = 2π 3 , 9π 12 , 10π 12 , 11π 12 and 12π 12 = π – the (solid) half-line consisting of all points with that θ and r ě 0 and – the dot with that θ and r(θ) = sin(3θ). The figure on the right below provides a sketch of the part of the curve r = sin(3θ) with 2π 3 ď θ ď π. x y θ“2π{3 x y θ“2π{3 Putting the three lobes together gives the full curve, which is called the “three petal rose”. 279 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES x y r “ sinp3θq θ“2π{3 θ“π{3 There is an infinite family of similar rose curves (also called rhodonea17 curves). Example 3.2.4 3.2.3 § § Integrals in Polar Coordinates We now return to the problem of using polar coordinates to set up double integrals. So far, we have used Cartesian coordinates, in the sense that we have cut up our domains of integration into tiny rectangles (on which the integrand is essentially constant) by drawing in many lines of constant x and many lines of constant y. To use polar coordinates, we instead draw in both lines of constant θ and curves of constant r. This cuts the xy-plane up into approximate rectangles. x y Here is an enlarged sketch of one such approximate rectangle. 17 The name rhodenea first appeared in the 1728 publication Flores geometrici of the Italian monk, theolo-gian, mathematician and engineer, Guido Grandi (1671–1742). 280 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES dθ dr r r dθ One side has length dr, the spacing between the curves of constant r. The other side is a portion of a circle of radius r that subtends, at the origin, an angle dθ, the angle between the lines of constant θ. As the circumference of the full circle is 2πr and as dθ is the fraction dθ 2π of a full circle18, the other side of the approximate rectangle has length dθ 2π2πr = rdθ. So the shaded region has area approximately dA = r dr dθ Equation 3.2.5. By way of comparison, using Cartesian coordinates we had dA = dx dy. This intuitive computation has been somewhat handwavy19. But using it in the usual integral setup procedure, in which we choose dr and dθ to be constants times 1 n and then take the limit n Ñ 0, gives, in the limit, error exactly zero. A sample argument, in which we see the error going to zero in the limit n Ñ 8, is provided in the (optional) section §3.2.4. Example 3.2.6 (Mass) Let 0 ď a ă b ď 2π be constants and let R be the region R = ␣(r cos θ, r sin θ) ˇ ˇ a ď θ ď b, B(θ) ď r ď T(θ) ( where the functions T(θ) and B(θ) are continuous and obey B(θ) ď T(θ) for all a ď θ ď b. Find the mass of R if it has density f (x, y). Solution. The figure on the left below is a sketch of R. Notice that r = T(θ) is the outer curve while r = B(θ) is the inner curve. Divide R into wedges (as in wedges of pie20 18 Recall that θ has to be measured in radians for this to be true. 19 “Handwaving” is sometimes used as a pejorative to refer to an argument that lacks substance. Here we are just using it to indicate that we have left out a bunch of technical details. In mathematics, “nose-following” is sometimes used as the polar opposite of handwaving. It refers to a very narrow, mechanical, line of reasoning. 281 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES x y θ “ a θ “ b r “ Tpθq r “ Bpθq R x y θ “ a θ “ b θ θdθ r “ Tpθq r “ Bpθq or wedges of cheese) by drawing in many lines of constant θ, with the various values of θ differing by a tiny amount dθ. The figure on the right above shows one such wedge, outlined in blue. Concentrate on any one wedge. Subdivide the wedge further into approximate rect-angles by drawing in many circles of constant r, with the various values of r differing by a tiny amount dr. The figure below shows one such approximate rectangle, in black. x y θ “ a θ “ b θ θdθ r “ Tpθq r “ Bpθq dr Now concentrate on one such rectangle. Let’s say that it contains the point with polar coordinates r and θ. As we saw in (3.2.5) above, • the area of that rectangle is essentially dA = r dr dθ. • As the mass density on the rectangle is essentially f r cos θ , r sin θ , the mass of the rectangle is essentially f r cos θ , r sin θ r dr dθ. • To get the mass of any one wedge, say the wedge whose polar angle runs from θ to θ + dθ, we just add up the masses of the approximate rectangles in that wedge, by integrating r from its smallest value on the wedge, namely B(θ), to its largest value on the wedge, namely T(θ). The mass of the wedge is thus dθ ż T(θ) B(θ) dr r f r cos θ , r sin θ 20 There is a pie/pi/pye pun in there somewhere. 282 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES • Finally, to get the mass of R, we just add up the masses of all of the different wedges, by integrating θ from its smallest value on R, namely a, to its largest value on R, namely b. In conclusion, Mass(R) = ż b a dθ ż T(θ) B(θ) dr r f r cos θ , r sin θ We have repeatedly used the word “essentially” above to avoid getting into the nitty-gritty details required to prove things rigorously. The mathematically correct proof of (3.2.7) follows the same intuition, but requires some more careful error bounds, as in the optional §3.2.4 below. Example 3.2.6 In the last example, we derived the important formula that the mass of the region R = ␣(r cos θ, r sin θ) ˇ ˇ a ď θ ď b, B(θ) ď r ď T(θ) ( with mass density f (x, y) is Mass(R) = ż b a dθ ż T(θ) B(θ) dr r f r cos θ , r sin θ Equation 3.2.7. We can immediately adapt that example to calculate areas and derive the formula that the area of the region R = ␣(r cos θ, r sin θ) ˇ ˇ a ď θ ď b, 0 ď r ď R(θ) ( is Area(R) = 1 2 ż b a R(θ)2 dθ Equation 3.2.8. We just have to set the density to 1. We do so in the next example. Example 3.2.9 (Polar Area) Let 0 ď a ă b ď 2π be constants. Find the area of the region R = ␣(r cos θ, r sin θ) ˇ ˇ a ď θ ď b, 0 ď r ď R(θ) ( where the function R(θ) ě 0 is continuous. Solution. To get the area of R we just need to assign it a density one and find the resulting mass. So, by (3.2.7), with f (x, y) = 1, B(θ) = 0 and T(θ) = R(θ), Area(R) = ż b a dθ ż R(θ) 0 dr r 283 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES In this case we can easily do the inner r integral, giving Area(R) = 1 2 ż b a R(θ)2 dθ The expression 1 2R(θ)2 dθ in (3.2.8) has a geometric interpretation. It is just the area of a wedge of a circular disk of radius R(θ) (with R(θ) treated as a constant) that subtends the angle dθ. To see this, note that area of the wedge is the fraction dθ 2π of the area of the entire x y Rpθq Rpθq dθ disk, which is πR(θ)2. So (3.2.8) just says that the area of R can be computed by cutting R up into tiny wedges and adding up the areas of all of the tiny wedges. Example 3.2.9 Example 3.2.10 (Polar Area) Find the area of one petal of the three petal rose r = sin(3θ). Solution. Looking at the last figure in Example 3.2.4, we see that we want the area of R = ␣(r cos θ, r sin θ) ˇ ˇ 0 ď θ ď π/3, 0 ď r ď sin(3θ) ( So, by (3.2.8) with a = 0, b = π/3, and R(θ) = sin(3θ), area(R) = 1 2 ż π/3 0 sin2(3θ) dθ = 1 4 ż π/3 0 1 ´ cos(6θ) dθ = 1 4 θ ´ 1 6 sin(6θ) π/3 0 = π 12 In the first step we used the double angle formula cos(2ϕ) = 1´2 sin2(ϕ). Unsurprisingly, trig identities show up a lot when polar coordinates are used. 284 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES Example 3.2.10 Example 3.2.11 (Volumes Using Polar Coordinates) A cylindrical hole of radius b is drilled symmetrically (i.e. along a diameter) through a metal sphere of radius a ě b. Find the volume of metal removed. Solution. Let’s use a coordinate system with the sphere centred on (0, 0, 0) and with the centre of the drill hole following the z-axis. In particular, the sphere is x2 + y2 + z2 ď a2. Here is a sketch of the part of the sphere in the first octant. The hole in the sphere made by the drill is outlined in red. By symmetry the total amount of metal removed will be eight times the amount from the first octant. That is, the volume of metal removed will z y x b R1 x2 y2 z2 “ a2 be eight times the volume of the solid V1 = ␣(x, y, z) ˇ ˇ (x, y) P R1, 0 ď z ď b a2 ´ x2 ´ y2 ( where the base region R1 = ␣(x, y) ˇ ˇ x2 + y2 ď b2, x ě 0, y ě 0 ( In polar coordinates V1 = ␣(r cos θ, r sin θ, z) ˇ ˇ (r cos θ, r sin θ) P R1, 0 ď z ď a a2 ´ r2 ( R1 = ␣(r cos θ, r sin θ) ˇ ˇ 0 ď r ď b, 0 ď θ ď π/2 ( We follow our standard divide and sum up strategy. We will cut the base region R1 into small pieces and sum up the volumes that lie above each small piece. • Divide R1 into wedges by drawing in many lines of constant θ, with the various values of θ differing by a tiny amount dθ. The figure on the left below shows one such wedge, outlined in blue. 285 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES z y x z y x • Concentrate on any one wedge. Subdivide the wedge further into approximate rect-angles by drawing in many circles of constant r, with the various values of r differ-ing by a tiny amount dr. The figure on the right above shows one such approximate rectangle, in black. • Concentrate on one such rectangle. Let’s say that it contains the point with polar coordinates r and θ. As we saw in (3.2.5) above, – the area of that rectangle is essentially dA = r dr dθ. – The part of V1 that is above that rectangle is like an office tower whose height is essentially ? a2 ´ r2, and whose base has area dA = r dr dθ. It is outlined in black in the figure below. So the volume of the part of V1 that is above the rectangle is essentially ? a2 ´ r2 r dr dθ. z y x r2 z2 “ a2 • To get the volume of the part of V1 above any one wedge (outlined in blue in the figure below), say the wedge whose polar angle runs from θ to θ + dθ, we just add up the volumes above the approximate rectangles in that wedge, by integrating r 286 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES from its smallest value on the wedge, namely 0, to its largest value on the wedge, namely b. The volume above the wedge is thus dθ ż b 0 dr r a a2 ´ r2 = dθ ż a2´b2 a2 du ´2 ?u where u = a2 ´ r2, du = ´2r dr = dθ " u3/2 ´3 #a2´b2 a2 = 1 3dθ h a3 ´ a2 ´ b23/2i Notice that this quantity is independent of θ. If you think about this for a moment, you can see that this is a consequence of the fact that our solid is invariant under rotations about the z-axis. z y x • Finally, to get the volume of V1, we just add up the volumes over all of the different wedges, by integrating θ from its smallest value on R1, namely 0, to its largest value on R1, namely π/2. Volume(V1) = 1 3 ż π/2 0 dθ h a3 ´ a2 ´ b23/2i = π 6 h a3 ´ a2 ´ b23/2i • In conclusion, the total volume of metal removed is Volume(V) = 8 Volume(V1) = 4π 3 h a3 ´ a2 ´ b23/2i Note that we can easily apply a couple of sanity checks to our answer. 287 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES • If the radius of the drill bit b = 0, no metal is removed at all. So the total volume removed should be zero. Our answer does indeed give 0 in this case. • If the radius of the drill bit b = a, the radius of the sphere, then the entire sphere disappears. So the total volume removed should be the volume of a sphere of radius a. Our answer does indeed give 4 3πa3 in this case. • If the radius, a, of the sphere and the radius, b, of the drill bit are measured in units of meters, then the remaining volume 4π 3 h a3 ´ a2 ´ b23/2i , has units meters3, as it should. Example 3.2.11 The previous two problems were given to us (or nearly given to us) in polar coor-dinates. We’ll now get a little practice converting integrals into polar coordinates, and recognising when it is helpful to do so. Example 3.2.12 (Changing to Polar Coordinates) Convert the integral ş1 0 şx 0 y a x2 + y2 dy dx to polar coordinates and evaluate the result. Solution. First recall that in polar coordinates x = r cos θ, y = r sin θ and dx dy = dA = r dr dθ so that the integrand (and dA) y b x2 + y2 dy dx = (r sin θ) r r dr dθ = r3 sin θ dr dθ is very simple. So whether or not this integral will be easy to evaluate using polar coordi-nates will be largely determined by the domain of integration. So our main task is to sketch the domain of integration. To prepare for the sketch, note that in the integral ż 1 0 ż x 0 y b x2 + y2 dy dx = ż 1 0 dx ż x 0 dy y b x2 + y2 • the variable x runs from 0 to 1 and • for each fixed 0 ď x ď 1, y runs from 0 to x. So the domain of integration is D = ␣(x, y) ˇ ˇ 0 ď x ď 1, 0 ď y ď x ( which is sketched in the figure on the left below. It is a right angled triangle. x y y “ x D x “ 1 x y θ “ π{4 r “ 1 cos θ 288 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES Next we express the domain of integration in terms of polar coordinates, by expressing the equations of each of the boundary lines in terms of polar coordinates. • The x-axis, i.e. y = r sin θ = 0, is θ = 0. • The line x = 1 is r cos θ = 1 or r = 1 cos θ. • Finally, (in the first quadrant) the line y = x ð ñ r sin θ = r cos θ ð ñ tan θ = sin θ cos θ = 1 ð ñ θ = π 4 So, in polar coordinates, we can write the domain of integration as R = ! (r, θ) ˇ ˇ ˇ 0 ď θ ď π 4 , 0 ď r ď 1 cos θ ) We can now slice up R using polar coordinates. • Divide R into wedges by drawing in many lines of constant θ, with the various values of θ differing by a tiny amount dθ. The figure on the right above shows one such wedge. – The first wedge has θ = 0. – The last wedge has θ = π 4 . • Concentrate on any one wedge. Subdivide the wedge further into approximate rect-angles by drawing in many circles of constant r, with the various values of r differ-ing by a tiny amount dr. The figure on the right above shows one such approximate rectangle, in black. – The rectangle that contains the point with polar coordinates r and θ has area (essentially) r dr dθ. – The first rectangle has r = 0. – The last rectangle has r = 1 cos θ. So our integral is ż 1 0 ż x 0 y b x2 + y2 dy dx = ż π/4 0 dθ ż 1 cos θ 0 dr r y? x2+y2 hkkkikkkj (r2 sin θ) Because the r-integral treats θ as a constant, we can pull the sin θ out of the inner r-integral. ż 1 0 ż x 0 y b x2 + y2 dy dx = ż π/4 0 dθ sin θ ż 1 cos θ 0 dr r3 = 1 4 ż π/4 0 dθ sin θ 1 cos4 θ Make the substitution u = cos θ, du = ´ sin θ dθ 289 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES When θ = 0, u = cos θ = 1 and when θ = π 4 , u = cos θ = 1 ? 2. So ż 1 0 ż x 0 y b x2 + y2 dy dx = 1 4 ż 1/ ? 2 1 (´du) 1 u4 = ´1 4 u´3 ´3 1/ ? 2 1 = 1 12 h 2 ? 2 ´ 1 i Example 3.2.12 Example 3.2.13 (Changing to Polar Coordinates) Evaluate ż 8 0 e´x2 dx. Solution. This is actually a trick question. In fact it is a famous trick question21. The integrand e´x2 does not have an antiderivative that can be expressed in terms of elemen-tary functions22. So we cannot evaluate this integral using the usual Calculus II methods. However we can evaluate it’s square ż 8 0 e´x2 dx 2 = ż 8 0 e´x2 dx ż 8 0 e´y2 dy = ż 8 0 dx ż 8 0 dy e´x2´y2 precisely because this double integral can be easily evaluated just by changing to polar coordinates! The domain of integration is the first quadrant ␣(x, y) ˇ ˇ x ě 0, y ě 0 ( . In polar coordinates, dxdy = r drdθ and the first quadrant is ␣(r cos θ , r sin θ) ˇ ˇ r ě 0, 0 ď θ ď π/2 ( So ż 8 0 e´x2 dx 2 = ż 8 0 dx ż 8 0 dy e´x2´y2 = ż π/2 0 dθ ż 8 0 dr r e´r2 As r runs all the way to +8, this is an improper integral, so we should be a little bit 21 The solution is attributed to the French Mathematician Sim´ eon Denis Poisson (1781–840) and was pub-lished in the textbook Cours d’Analyse de l’´ ecole polytechnique by Jacob Karl Franz Sturm (1803–1855). 22 On the other hand it is the core of the function erf(z) = 2 ?π şz 0 e´t2 dt, which gives Gaussian (i.e. bell curve) probabilities. “erf” stands for “error function”. 290 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES careful. ż 8 0 e´x2 dx 2 = lim RÑ8 ż π/2 0 dθ ż R 0 dr r e´r2 = lim RÑ8 ż π/2 0 dθ ż R2 0 du 2 e´u where u = r2, du = 2r dr = lim RÑ8 ż π/2 0 dθ ´e´u 2 R2 0 = lim RÑ8 π 2 " 1 2 ´ e´R2 2 # = π 4 and so we get the famous result ż 8 0 e´x2 dx = ?π 2 Example 3.2.13 Example 3.2.14 Find the area of the region that is inside the circle r = 4 cos θ and to the left of the line x = 1. Solution. First, let’s check that r = 4 cos θ really is a circle and figure out what circle it is. To do so, we’ll convert the equation r = 4 cos θ into Cartesian coordinates. Multiplying both sides by r gives r2 = 4r cos θ ð ñ x2 + y2 = 4x ð ñ (x ´ 2)2 + y2 = 4 So r = 4 cos θ is the circle of radius 2 centred on (2, 0). We’ll also need the intersection point(s) of x = r cos θ = 1 and r = 4 cos θ. At such an intersection point r cos θ = 1, r = 4 cos θ ù ñ 1 cos θ = 4 cos θ ù ñ cos2 θ = 1 4 ù ñ cos θ = 1 2 since r cos θ = 1 ą 0 ù ñ θ = ˘π 3 Here is a sketch of the region of interest, which we’ll call R. 291 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES x y r “ 4 cos θ R x “ 1 θ “ π{3 We could figure out the area of R by using some high school geometry, because R is a circular wedge with a triangle removed. (See Example 3.2.15, below.) Instead, we’ll treat x y its computation as an exercise in integration using polar coordinates. As R is symmetric about the x-axis, the area of R is twice the area of the part that is above the x-axis. We’ll denote by R1 the upper half of R. Note that we can write the equation x = 1 in polar coordinates as r = 1 cos θ. Here is a sketch of R1. x y θ ą π{3 θ ă π{3 r “ 4 cos θ R1 r “ 1 cos θ θ “ π{3 Observe that, on R1, for any fixed θ between 0 and π/2, • if θ ă π/3, then r runs from 0 to 1 cos θ, while 292 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES • if θ ą π/3, then r runs from 0 to 4 cos θ. This naturally leads us to split the domain of integration at θ = π 3 : Area(R1) = ż π/3 0 dθ ż 1/ cos θ 0 dr r + ż π/2 π/3 dθ ż 4 cos θ 0 dr r As ş r dr = r2 2 + C, Area(R1) = ż π/3 0 dθ sec2 θ 2 + ż π/2 π/3 dθ 8 cos2 θ = 1 2 tan θ ˇ ˇ ˇ π/3 0 + 4 ż π/2 π/3 dθ 1 + cos(2θ) = ? 3 2 + 4 θ + sin(2θ) 2 π/2 π/3 = ? 3 2 + 4 π 6 ´ ? 3 4 = 2π 3 ´ ? 3 2 and Area(R) = 2Area(R1) = 4π 3 ´ ? 3 Example 3.2.14 Example 3.2.15 (Optional — Example 3.2.14 by high school geometry) We’ll now again compute the area of the region R that is inside the circle r = 4 cos θ and to the left of the line x = 1. That was the region of interest in Example 3.2.14. This time we’ll just use some geometry. Think of R as being the wedge W, of the figure on the left below, with the triangle T , of the figure on the right below, removed. W x y π{3 x “ 1 A C 1 1 2 T x y x “ 1 B D C 1 1 2 ? 3 293 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES • First we’ll get the area of W. The cosine of the angle between the x axis and the radius vector from C to A is 1 2. So that angle is π 3 and W subtends an angle of 2π 3 . The entire circle has area π22, so that W, which is the fraction 2π/3 2π = 1 3 of the full circle, has area 4π 3 . • Now we’ll get the area of the triangle T . Think of T as having base BD. Then the length of the base of T is 2 ? 3 and the height of T is 1. So T has area 1 2(2 ? 3)(1) = ? 3. All together Area(R) = Area(W) ´ Area(T ) = 4π 3 ´ ? 3 Example 3.2.15 We used some hand waving in deriving the area formula (3.2.8): the word “essentially” appeared quite a few times. Here is how do that derivation more rigorously. 3.2.4 § § Optional— Error Control for the Polar Area Formula Let 0 ď a ă b ď 2π. In Examples 3.2.6 and 3.2.9 we derived the formula A = 1 2 ż b a R(θ)2 dθ for the area of the region R = ␣r cos θ, r sin θ ˇ ˇ a ď θ ď b, 0 ď r ď R(θ) ( In the course of that derivation we approximated the area of the shaded region in dθ dr r r dθ by dA = r dr dθ. We will now justify that approximation, under the assumption that 0 ď R(θ) ď M \|R1(θ)\| ď L for all a ď θ ď b. That is, R(θ) is bounded and its derivative exists and is bounded too. Divide the interval a ď θ ď b into n equal subintervals, each of length ∆θ = b´a n . Let θ˚ i be the midpoint of the ith interval. On the ith interval, θ runs from θ˚ i ´ 1 2∆θ to θ˚ i + 1 2∆θ. 294 MULTIPLE INTEGRALS 3.2 DOUBLE INTEGRALS IN POLAR COORDINATES By the mean value theorem R(θ) ´ R(θ˚ i ) = R1(c)(θ ´ θ˚ i ) for some c between θ and θ˚ i . Because \|R1(θ)\| ď L ˇ ˇR(θ) ´ R(θ˚ i ) ˇ ˇ ď L ˇ ˇθ ´ θ˚ i ˇ ˇ (˚) This tells us that the difference between R(θ) and R(θ˚ i ) can’t be too big compared to ˇ ˇθ ´ θ˚ i ˇ ˇ. On the ith interval, the radius r = R(θ) runs over all values of R(θ) with θ satisfying ˇ ˇθ ´ θ˚ i ˇ ˇ ď 1 2∆θ. By (˚), all of these values of R(θ) lie between ri = R(θ˚ i ) ´ 1 2L∆θ and Ri = R(θ˚ i ) + 1 2L∆θ. Consequently the part of R having θ in the ith subinterval, namely, Ri = ␣r cos θ, r sin θ ˇ ˇ θ˚ i ´ 1 2∆θ ď θ ď θ˚ i + 1 2∆θ, 0 ď r ď R(θ) ( must contain all of the circular sector ␣r cos θ, r sin θ ˇ ˇ θ˚ i ´ 1 2∆θ ď θ ď θ˚ i + 1 2∆θ, 0 ď r ď ri ( and must be completely contained inside the circular sector ␣r cos θ, r sin θ ˇ ˇ θ˚ i ´ 1 2∆θ ď θ ď θ˚ i + 1 2∆θ, 0 ď r ď Ri ( x y θ “ a θ “ b r “ fpθq x y ri Ri That is, we have found one circular sector that is bigger than the one we are approximat-ing, and one circular sector that is smaller. The area of a circular disk of radius ρ is πρ2. A circular sector of radius ρ that subtends an angle ∆θ is the fraction ∆θ 2π of the full disk and so has the area ∆θ 2ππρ2 = ∆θ 2 ρ2. x y ρ ∆θ So the area of Ri must lie between 1 2∆θ r2 i = 1 2∆θ R(θ˚ i ) ´ 1 2L∆θ 2 and 1 2∆θ R2 i = 1 2∆θ R(θ˚ i ) + 1 2L∆θ 2 295 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS Observe that R(θ˚ i ) ˘ 1 2L∆θ 2 = R(θ˚ i )2 ˘ LR(θ˚ i )∆θ + 1 4L2∆θ2 implies that, since 0 ď R(θ) ď M, R(θ˚ i )2 ´ LM∆θ + 1 4L2∆θ2 ď R(θ˚ i ) ˘ 1 2L∆θ 2 ď R(θ˚ i )2 + LM∆θ + 1 4L2∆θ2 Hence (multiplying by ∆θ 2 to turn them into areas) 1 2R(θ˚ i )2∆θ ´ 1 2LM∆θ2 + 1 8L2∆θ3 ď Area(Ri) ď 1 2R(θ˚ i )2∆θ + 1 2LM∆θ2 + 1 8L2∆θ3 and the total area A obeys the bounds n ÿ i=1 1 2R(θ˚ i )2∆θ ´ 1 2LM∆θ2 + 1 8L2∆θ3 ď Aď n ÿ i=1 1 2R(θ˚ i )2∆θ + 1 2LM∆θ2 + 1 8L2∆θ3 1 2 n ÿ i=1 R(θ˚ i )2∆θ ´ 1 2nLM∆θ2 + 1 8nL2∆θ3 ď Aď n ÿ i=1 1 2R(θ˚ i )2∆θ + 1 2nLM∆θ2 + 1 8nL2∆θ3 Since ∆θ = b´a n , 1 2 n ÿ i=1 R(θ˚ i )2∆θ´ LM 2 (b´a)2 n + L2 8 (b´a)3 n2 ď A ď 1 2 n ÿ i=1 R(θ˚ i )2∆θ+ LM 2 (b´a)2 n + L2 8 (b´a)3 n2 Now take the limit as n Ñ 8. Since lim nÑ8 " 1 2 n ÿ i=1 R(θ˚ i )2∆θ ˘ LM 2 (b ´ a)2 n + L2 8 (b ´ a)3 n2 # = 1 2 ż b a R(θ)2 dθ ˘ lim nÑ8 LM 2 (b ´ a)2 n + lim nÑ8 L2 8 (b ´ a)3 n2 = 1 2 ż b a R(θ)2 dθ (since L, M, a and b are all constants) we have that A = 1 2 ż b a R(θ)2 dθ exactly, as desired. 3.3Ĳ Applications of Double Integrals Double integrals are useful for more than just computing areas and volumes. Here are a few other applications that lead to double integrals. 296 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS §§§ Averages In Section 2.2 of the CLP-2 text, we defined the average value of a function of one vari-able. We’ll now extend that discussion to functions of two variables. First, we recall the definition of the average of a finite set of numbers. The average (mean) of a set of n numbers f1, f2, ¨ ¨ ¨ , fn is ¯ f = ⟨f ⟩= f1 + f2 + ¨ ¨ ¨ + fn n The notations ¯ f and ⟨f ⟩are both commonly used to represent the average. Definition 3.3.1. Now suppose that we want to take the average of a function f (x, y) with (x, y) running continuously over some region R in the xy-plane. A natural approach to defining what we mean by the average value of f over R is to • First fix any natural number n. • Subdivide the region R into tiny (approximate) squares each of width ∆x = 1 n and height ∆y = 1 n. This can be done by, for example, subdividing vertical strips into tiny squares, like in Example 3.1.11. • Name the squares (in any fixed order) R1, R2, ¨ ¨ ¨ , RN, where N is the total number of squares. • Select, for each 1 ď i ď N, one point in square number i and call it (x˚ i , y˚ i ). So (x˚ i , y˚ i ) P Ri. • The average value of f at the selected points is 1 N N ÿ i=1 f (x˚ i , y˚ i ) = řN i=1 f (x˚ i , y˚ i ) řN i=1 1 = řN i=1 f (x˚ i , y˚ i ) ∆x ∆y řN i=1 ∆x ∆y We have transformed the average into a ratio of Riemann sums. Once we have the Riemann sums it is clear what to do next. Taking the limit n Ñ 8, we get exactly ť R f (x,y) dx dy ť R dx dy . That’s why we define 297 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS Let f (x, y) be an integrable function defined on region R in the xy-plane. The average value of f on R is ¯ f = ⟨f ⟩= ĳ R f (x, y) dx dy ĳ R dx dy Definition 3.3.2. Example 3.3.3 (Average) Let a ą 0. A mountain, call it Half Dome23, has height z(x, y) = a a2 ´ x2 ´ y2 above each point (x, y) in the base region R = ␣(x, y) ˇ ˇ x2 + y2 ď a2, x ď 0 ( . Find its average height. z y x x2 y2 z2 “ a2 R Solution. By Definition 3.3.2 the average height is ¯ z = ť R z(x, y) dx dy ť R dx dy = ť R a a2 ´ x2 ´ y2 dx dy ť R dx dy The integrals in both the numerator and denominator are easily evaluated by interpreting them geometrically. • The numerator ť R z(x, y) dx dy = ť R a a2 ´ x2 ´ y2 dx dy can be interpreted as the volume of ! (x, y, z) ˇ ˇ ˇ x2 + y2 ď a2, x ď 0, 0 ď z ď b a2 ´ x2 ´ y2 ) = ␣(x, y, z) ˇ ˇ x2 + y2 + z2 ď a2, x ď 0, z ě 0 ( which is one quarter of the interior of a sphere of radius a. So the numerator is 1 3πa3. • The denominator ť R dx dy is the area of one half of a circular disk of radius a. So the denominator is 1 2πa2. 23 There is a real Half-Dome mountain in Yosemite National Park. It has a = 1445 m. 298 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS All together, the average height is ¯ z = 1 3πa3 1 2πa2 = 2 3 a Notice this this number is bigger than zero and less than the maximum height, which is a. That makes sense. Example 3.3.3 Example 3.3.4 (Example 3.3.3, the hard way) This last example was relatively easy because we could reinterpret the integrals as geomet-ric quantities. For practice, let’s go back and evaluate the numerator ť R a a2 ´ x2 ´ y2 dx dy of Example 3.3.3 as an iterated integral. Here is a sketch of the top view of the base region R. y x top view x “ ´ a a2 ´ y2 dx dy y “ a y “ ´a R Using the slicing in the figure ĳ R b a2 ´ x2 ´ y2 dx dy = ż a ´a dy ż 0 ´? a2´y2 dx b a2 ´ x2 ´ y2 Note that, in the inside integral ż 0 ´? a2´y2 dx b a2 ´ x2 ´ y2, the variable y is treated as a constant, so that the integrand a a2 ´ y2 ´ x2 = ? C2 ´ x2 with C being the constant a a2 ´ y2. The standard protocol for evaluating this integral uses the trigonometric sub-stitution x = C sin θ with ´ π 2 ď θ ď π 2 dx = C cos θ dθ Trigonometric substitution was discussed in detail in Section 1.9 in the CLP-2 text. Since x = 0 ù ñ C sin θ = 0 ù ñ θ = 0 x = ´ b a2 ´ y2 = ´C ù ñ C sin θ = ´C ù ñ θ = ´π 2 299 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS and b a2 ´ x2 ´ y2 = a C2 ´ C2 sin2 θ = C cos θ the inner integral ż 0 ´? a2´y2 dx b a2 ´ x2 ´ y2 = ż 0 ´π/2 C2 cos2 θ dθ = C2 ż 0 ´π/2 1 + cos(2θ) 2 dθ = C2 " θ + sin(2θ) 2 2 #0 ´π/2 = πC2 4 = π 4 a2 ´ y2 and the full integral ĳ R b a2 ´ x2 ´ y2 dx dy = π 4 ż a ´a a2 ´ y2 dy = π 2 ż a 0 a2 ´ y2 dy = π 2 a3 ´ a3 3 = 1 3πa3 just as we saw in Example 3.3.3. Example 3.3.4 Remark 3.3.5. We remark that there is an efficient, sneaky, way to evaluate definite inte-grals like ş0 ´π/2 cos2 θ dθ. Looking at the figures y x ´π ´π{2 π π{2 1 y “ cos2 x y x ´π ´π{2 π π{2 1 y “ sin2 x we see that ż 0 ´π/2 cos2 θ dθ = ż 0 ´π/2 sin2 θ dθ Thus ż 0 ´π/2 cos2 θ dθ = ż 0 ´π/2 sin2 θ dθ = ż 0 ´π/2 1 2 sin2 θ + cos2 θ dθ = 1 2 ż 0 ´π/2 dθ = π 4 300 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS It is not at all unusual to want to find the average value of some function f (x, y) with (x, y) running over some region R, but to also want some (x, y)’s to play a greater role in determining the average than other (x, y)’s. One common way to do so is to create a “weight function” w(x, y) ą 0 with w(x1,y1) w(x2,y2) giving the relative importance of (x1, y1) and (x2, y2). That is, (x1, y1) is w(x1,y1) w(x2,y2) times as important as (x2, y2). This leads to the definition ť R f (x, y) w(x, y) dx dy ť R w(x, y) dx dy is called the weighted average of f over R with weight w(x, y). Definition 3.3.6. Note that if f (x, y) = F, a constant, then the weighted average of f is just F, just as you would want. §§§ Centre of Mass One important example of a weighted average is the centre of mass. If you support a body at its centre of mass (in a uniform gravitational field) it balances perfectly. That’s the definition of the centre of mass of the body. In Section 2.3 of the CLP-2 text, we found that the centre of mass of a body that consists of mass distributed continuously along a straight line, with mass density ρ(x)kg/m and with x running from a to b, is at ¯ x = şb a x ρ(x) dx şb a ρ(x) dx That is, the centre of mass is at the average of the x-coordinate weighted by the mass density. In two dimensions, the centre of mass of a plate that covers the region R in the xy-plane and that has mass density ρ(x, y) is the point ( ¯ x, ¯ y) where ¯ x = the weighted average of x over R = ť R x ρ(x, y) dx dy ť R ρ(x, y) dx dy = ť R x ρ(x, y) dx dy Mass(R) ¯ y = the weighted average of y over R = ť R y ρ(x, y) dx dy ť R ρ(x, y) dx dy = ť R y ρ(x, y) dx dy Mass(R) Equation 3.3.7 (Centre of Mass). 301 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS If the mass density is a constant, the centre of mass is also called the centroid, and is the geometric centre of R. In this case ¯ x = ť R x dx dy ť R dx dy = ť R x dx dy Area(R) ¯ y = ť R y dx dy ť R dx dy = ť R y dx dy Area(R) Equation 3.3.8 (Centroid). Example 3.3.9 (Centre of Mass) In Section 2.3 of the CLP-2 text, we did not have access to multivariable integrals, so we used some physical intuition to derive that the centroid of a body that fills the region R = ␣(x, y) ˇ ˇ a ď x ď b, B(x) ď y ď T(x) ( in the xy-plane is ( ¯ x, ¯ y) where ¯ x = şb a x[T(x) ´ B(x)] dx A ¯ y = şb a [T(x)2 ´ B(x)2] dx 2A and A = şb a[T(x) ´ B(x)] dx is the area of R. Now that we do have access to multivariable integrals, we can derive these formulae directly from (3.3.8). Using vertical slices, as in this figure, x b Tpxq Bpxq x y y “ Tpxq y “ Bpxq a R we see that the area of R is A = ĳ R dx dy = ż b a dx ż T(x) B(x) dy = ż b a dx T(x) ´ B(x) 302 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS and that (3.3.8) gives ¯ x = 1 A ĳ R x dx dy = 1 A ż b a dx ż T(x) B(x) dy x = 1 A ż b a dx x T(x) ´ B(x) ¯ y = 1 A ĳ R y dx dy = 1 A ż b a dx ż T(x) B(x) dy y = 1 A ż b a dx T(x)2 2 ´ B(x)2 2 just as desired. Example 3.3.9 We’ll start with a simple mechanical example. Example 3.3.10 (Quarter Circle) In Example 2.3.4 of the CLP-2 text, we found the centroid of the quarter circular disk D = ␣(x, y) ˇ ˇ x ě 0, y ě 0, x2 + y2 ď r2 ( by using the formulae of the last example. We’ll now find it again using (3.3.8). Since the area of D is 1 4πr2, we have ¯ x = ť D x dx dy 1 4πr2 ¯ y = ť D y dx dy 1 4πr2 We’ll evaluate ť D x dx dy by using horizontal slices, as in the figure on the left below. x y x “ a r2 ´ y2 p0, rq x y y “ ? r2 ´ x2 pr, 0q Looking at that figure, we see that • y runs from 0 to r and • for each y in that range, x runs from 0 to a r2 ´ y2. So ĳ D x dx dy = ż r 0 dy ż ? r2´y2 0 dx x = ż r 0 dy x2 2 ? r2´y2 0 = 1 2 ż r 0 dy r2 ´ y2 = 1 2 r3 ´ r3 3 = r3 3 303 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS and ¯ x = 4 πr2 r3 3 = 4r 3π This is the same answer as we got in Example 2.3.4 of the CLP-2 text. But because we were able to use horizontal slices, the integral in this example was a little easier to evaluate than the integral in CLP-2. Had we used vertical slices, we would have ended up with exactly the integral of CLP-2. By symmetry, we should have ¯ y = ¯ x. We’ll check that by evaluating ť D y dx dy by using vertical slices slices, as in the figure on the right above. From that figure, we see that • x runs from 0 to r and • for each x in that range, y runs from 0 to ? r2 ´ x2. So ĳ D y dx dy = ż r 0 dx ż ? r2´x2 0 dy y = 1 2 ż r 0 dx r2 ´ x2 This is exactly the integral 1 2 şr 0 dy r2 ´ y2 that we evaluated above, with y renamed to x. So ť D y dx dy = r3 3 too and ¯ y = 4 πr2 r3 3 = 4r 3π = ¯ x as expected. Example 3.3.10 Example 3.3.11 (Example 3.2.14, continued) Find the centroid of the region that is inside the circle r = 4 cos θ and to the left of the line x = 1. Solution. Recall that we saw in Example 3.2.14 that r = 4 cos θ was indeed a circle, and in fact is the circle (x ´ 2)2 + y2 = 4. Here is a sketch of that circle and of the region of interest, R. x y r “ 4 cos θ R x “ 1 θ “ π{3 304 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS From the sketch, we see that R is symmetric about the x-axis. So we expect that its cen-troid, ( ¯ x, ¯ y), has ¯ y = 0. To see this from the integral definition, note that the integral ť R y dx dy • has domain of integration, namely R, invariant under y Ñ ´y (i.e. under reflection across the x-axis), and • has integrand, namely y, that is odd under y Ñ ´y. So ť R y dx dy = 0 and consequently ¯ y = 0. We now just have to find ¯ x: ¯ x = ť R x dx dy ť R dx dy We have already found, in Example 3.2.14, that ĳ R dx dy = 4π 3 ´ ? 3 So we just have to compute ť R x dx dy. Using R1 to denote the top half of R, and using polar coordinates, like we did in Example 3.2.14, ĳ R1 x dx dy = ż π/3 0 dθ ż 1/ cos θ 0 dr r x hkkkikkkj (r cos θ) + ż π/2 π/3 dθ ż 4 cos θ 0 dr r x hkkkikkkj (r cos θ) = ż π/3 0 dθ cos θ ż 1/ cos θ 0 dr r2 + ż π/2 π/3 dθ cos θ ż 4 cos θ 0 dr r2 = ż π/3 0 dθ sec2 θ 3 + ż π/2 π/3 dθ 64 3 cos4 θ The first integral is easy, provided we remember that tan θ is an antiderivative for sec2 θ. For the second integral, we’ll need the double angle formula cos2 θ = 1+cos(2θ) 2 : cos4 θ = cos2 θ 2 = 1 + cos(2θ) 2 2 = 1 4 h 1 + 2 cos(2θ) + cos2(2θ) i = 1 4 1 + 2 cos(2θ) + 1 + cos(4θ) 2 = 3 8 + cos(2θ) 2 + cos(4θ) 8 so ĳ R1 x dx dy = 1 3 tan θ ˇ ˇ ˇ π/3 0 + 64 3 3θ 8 + sin(2θ) 4 + sin(4θ) 32 π/2 π/3 = 1 3 ˆ ? 3 + 64 3 3 8 ˆ π 6 ´ ? 3 4 ˆ 2 + ? 3 32 ˆ 2 = 4π 3 ´ 2 ? 3 The integral we want, namely ť R x dx dy, 305 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS • has domain of integration, namely R, invariant under y Ñ ´y (i.e. under reflection across the x-axis), and • has integrand, namely x, that is even under y Ñ ´y. So ť R x dx dy = 2 ť R1 x dx dy and, all together, ¯ x = 2 4π 3 ´ 2 ? 3 4π 3 ´ ? 3 = 8π 3 ´ 4 ? 3 4π 3 ´ ? 3 « 0.59 As a check, note that 0 ď x ď 1 on R and more of R is closer to x = 1 than to x = 0. So it makes sense that ¯ x is between 1 2 and 1. Example 3.3.11 Example 3.3.12 (Reverse Centre of Mass) Evaluate ż 2 0 ż ? 2x´x2 ´? 2x´x2 2x + 3y dy dx. Solution. This is another integral that can be evaluated without using any calculus at all. This time by relating it to a centre of mass. By (3.3.8), ĳ R x dx dy = ¯ x Area(R) ĳ R y dx dy = ¯ y Area(R) so that we can easily evaluate ť R x dx dy and ť R y dx dy provided R is sufficiently simple and symmetric that we can easily determine its area and its centroid. That is the case for the integral in this example. Rewrite ż 2 0 ż ? 2x´x2 ´? 2x´x2 2x + 3y dy dx = 2 ż 2 0 dx "ż ? 2x´x2 ´? 2x´x2 dy x # + 3 ż 2 0 dx "ż ? 2x´x2 ´? 2x´x2 dy y # On the domain of integration • x runs from 0 to 2 and • for each fixed 0 ď x ď 2, y runs from ´ ? 2x ´ x2 to + ? 2x ´ x2 Observe that y = ˘ ? 2x ´ x2 is equivalent to y2 = 2x ´ x2 = 1 ´ (x ´ 1)2 ð ñ (x ´ 1)2 + y2 = 1 Our domain of integration is exactly the disk R = ␣(x, y) ˇ ˇ (x ´ 1)2 + y2 ď 1 ( of radius 1 centred on (1, 0). So R has area π and centre of mass ( ¯ x, ¯ y) = (1, 0) and 306 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS p1, 0q p2, 0q dx dy x “ 1 x y px ´ 1q2 y2 “ 1 ż 2 0 ż ? 2x´x2 ´? 2x´x2 2x + 3y dy dx = 2 ĳ R x dx dy + 3 ĳ R y dx dy = 2 ¯ x Area(R) + 3 ¯ y Area(R) = 2π Example 3.3.12 Example 3.3.13 (Compound bodies) In §2.3 of the CLP-2 text, we saw that it was possible to think of a body as being made up of a number of component parts, and then it was possible to compute the centre of mass of the body as a whole by using the centres of mass of the component parts. In that dis-cussion, we could only consider bodies consisting of a finite number of point particles arrayed along a straight line. We will now see that the same procedure can be used with continuous mass distributions. Suppose that we have a dumbbell which consists of • one end contained in a region R1 and having mass density ρ1(x, y) and • a second end contained in a region R2 and having mass density ρ2(x, y) and • an infinitely strong, weightless (idealized) rod joining the two ends. Suppose that both R1 and R2 are contained in the larger region R, that ρ1(x, y) = 0 for all (x, y) not in R1 and that ρ2(x, y) = 0 for all (x, y) not in R2. Then the mass and centre R R1 R2 307 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS of mass of the R1 end are M1 = ĳ R1 ρ1(x, y) dx dy and ( ¯ X1, ¯ Y1) with ¯ X1 = ť R1 x ρ1(x, y) dx dy M1 , ¯ Y1 = ť R1 y ρ1(x, y) dx dy M1 and the mass and centre of mass of the R2 end are M2 = ĳ R2 ρ2(x, y) dx dy and ( ¯ X2, ¯ Y2) with ¯ X2 = ť R2 x ρ2(x, y) dx dy M2 , ¯ Y2 = ť R2 y ρ2(x, y) dx dy M2 The mass and centre of mass of the entire dumbbell are M = ĳ R tρ1(x, y) + ρ2(x, y)u dx dy = ĳ R ρ1(x, y) dx dy + ĳ R ρ2(x, y) dx dy = ĳ R1 ρ1(x, y) dx dy + ĳ R2 ρ2(x, y) dx dy = M1 + M2 and ( ¯ X, ¯ Y) with ¯ X = ť R xtρ1(x, y) + ρ2(x, y)u dx dy M = ť R1 x ρ1(x, y) dx dy + ť R2 x ρ2(x, y) dx dy M = M1 ¯ X1 + M2 ¯ X2 M1 + M2 ¯ Y = ť R ytρ1(x, y) + ρ2(x, y)u dx dy M = ť R1 y ρ1(x, y) dx dy + ť R2 y ρ2(x, y) dx dy M = M1 ¯ Y1 + M2 ¯ Y2 M1 + M2 So we can compute the centre of mass of the entire dumbbell by treating it as being made up of two point particles, one of mass M1 located at the centre of mass ( ¯ X1, ¯ Y1) of the R1 end, and one of mass M2 located at the centre of mass (X2, Y2) of the R2 end. Example 3.3.13 308 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS Example 3.3.14 (Example 2.3.6 of CLP-2 revisited) In Example 2.3.6 of the CLP-2 text, we found the centroid, i.e. the centre of mass with den-sity 1, of the region R1 in the sketch p2, 2q 2 1 2 1 2 R1 quarter circle We now again compute the centroid of R1, but this time we use the ideas of Example 3.3.13 above. We think of R1 as one end of the “dumbbell” whose other end, R2, is the quarter circular disk which fills out R1 to the 2 ˆ 2 square in the sketch below. That is, the full dumbbell is R = R1 Y R2 = ␣(x, y) ˇ ˇ 0 ď x ď 2, 0 ď y ď 2 ( with R1 = ␣(x, y) ˇ ˇ 0 ď x ď 2, 0 ď y ď 2, x2 + y2 ě 1 ( R2 = ␣(x, y) ˇ ˇ x ě 0, y ě 0, x2 + y2 ď 1 ( p2, 2q p0, 0q 2 1 2 1 2 R1 R2 • The full dumbbell R is a 2 ˆ 2 square and so has area, and hence mass, M = 2 ˆ 2 = 4. Just by symmetry, the centre of mass of R is ( ¯ X, ¯ Y) = (1, 1). • The end R2 is one quarter of a circular disk of radius 1 and so has area, and hence mass, M2 = π 4 . In Example 3.3.10 above with r = 1, we found that the centre of mass of R2 is ( ¯ X2, ¯ Y2) = 4 3π, 4 3π). So M1 = M ´ M2 = 4 ´ π 4 and just filling all of this data into the formulae at the end of Example 3.3.13, above, gives 1 = ¯ X= M1 ¯ X1 + M2 ¯ X2 M1 + M2 = (4 ´ π 4 ) ¯ X1 + ( π 4 )( 4 3π) 4 = 1 ´ π 16 ¯ X1 + 1 12 1 = ¯ Y = M1 ¯ Y1 + M2 ¯ Y2 M1 + M2 = (4 ´ π 4 ) ¯ Y1 + ( π 4 )( 4 3π) 4 = 1 ´ π 16 ¯ Y1 + 1 12 309 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS Solving gives ¯ X1 = ¯ Y1 = 11 12 1 ´ π 16 = 44 48 ´ 3π which is exactly the answer that we found in Example 2.3.6 of the CLP-2 text. Example 3.3.14 §§§ Moment of Inertia Consider a plate that fills the region R in the xy-plane, that has mass density ρ(x, y) kg/m2, and that is rotating at ω rad/s about some axis. Let’s call the axis of rotation A. We are now going to determine the kinetic energy of that plate. Recall24 that, by definition, the kinetic energy of a point particle of mass m that is moving with speed v is 1 2mv2. To get the kinetic energy of the entire plate, cut it up into tiny rectangles25, say of size dx ˆ dy. Think of each rectangle as being (essentially) a point particle. If the point (x, y) on the plate is a distance D(x, y) from the axis of rotation A, then as the plate rotates, the point (x, y) sweeps out a circle of radius D(x, y). The figure on the right below shows that circle as seen from high up on the axis of rotation. The circular arc that the point dx dy D A ω D (x, y) sweeps out in one second subtends the angle ω radians, which is the fraction ω 2π of a full circle and so has length ω 2π 2πD(x, y) = ω D(x, y). Consequently the rectangle that contains the point (x, y) • has speed ω D(x, y), and • has area dx dy, and so • has mass ρ(x, y) dx dy, and 24 If you don’t recall, don’t worry. We wouldn’t lie to you. Or check it on Wikipedia. They wouldn’t lie to you either. 25 The relatively small number of “rectangles” around the boundary of R won’t actually be rectangles. But, as we have seen in the optional §3.2.4, one can still make things rigorous despite the rectangles being a bit squishy around the edges. 310 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS • has kinetic energy 1 2 m hkkkkkkkkikkkkkkkkj ρ(x, y) dx dy v2 hkkkkkkikkkkkkj (ω D(x, y))2 = 1 2ω2 D(x, y) 2ρ(x, y) dx dy So (via our usual Riemann sum limit procedure) the kinetic energy of R is ĳ R 1 2ω2 D(x, y)2 ρ(x, y) dx dy = 1 2ω2 ĳ R D(x, y)2 ρ(x, y) dx dy = 1 2 IA ω2 where IA = ĳ R D(x, y)2ρ(x, y) dx dy is called the moment of inertial of R about the axis A. In particular the moment of inertia of R about the y-axis is Iy = ĳ R x2 ρ(x, y) dx dy and the moment of inertia of R about the x-axis is Ix = ĳ R y2 ρ(x, y) dx dy Definition 3.3.15 (Moment of Inertia). Notice that the expression 1 2 IA ω2 for the kinetic energy has a very similar form to 1 2mv2, just with the velocity v replaced by the angular velocity ω, and with the mass m replaced by IA, which can be thought of as being a bit like a mass. So far, we have been assuming that the rotation was taking place in the xy-plane — a two dimensional world. Our analysis extends naturally to three dimensions, though the resulting integral formulae for the moment of inertia will then be triple integrals, which we have not yet dealt with. We shall soon do so, but let’s first do an example in two dimensions. Example 3.3.16 (Disk) Find the moment of inertia of the interior, R, of the circle x2 + y2 = a2 about the x-axis. Assume that it has density one. Solution. The distance from any point (x, y) inside the disk to the axis of rotation (i.e. the 311 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS R y \|y\| x x-axis) is \|y\|. So the moment of inertia of the interior of the disk about the x-axis is Ix = ĳ R y2 dxdy Switching to polar coordinates26, Ix = ż 2π 0 dθ ż a 0 dr r y2 hkkkikkkj (r sin θ)2 = ż 2π 0 dθ sin2 θ ż a 0 dr r3 = a4 4 ż 2π 0 dθ sin2 θ = a4 4 ż 2π 0 dθ 1 ´ cos(2θ) 2 = a4 8 θ ´ sin(2θ) 2 2π 0 = 1 4πa4 For an efficient, sneaky, way to evaluate ş2π 0 sin2 θ dθ, see Remark 3.3.5. Example 3.3.16 Example 3.3.17 (Cardioid) Find the moment of inertia of the interior, R, of the cardiod r = a(1 + cos θ) about the z-axis. Assume that the cardiod lies in the xy-plane and has density one. Solution. We sketched the cardiod (with a = 1) in Example 3.2.3. As we said above, the 26 See how handy they are! 312 MULTIPLE INTEGRALS 3.3 APPLICATIONS OF DOUBLE INTEGRALS R x y ? x2y2 r “ ap1 cos θq formula for IA in Definition 3.3.15 is valid even when the axis of rotation is not contained in the xy-plane. We just have to be sure that our D(x, y) really is the distance from (x, y) to the axis of rotation. In this example the axis of rotation is the z-axis so that D(x, y) = a x2 + y2 and that the moment of inertia is IA = ĳ R (x2 + y2) dxdy Switching to polar coordinates using dxdy = r drdθ and x2 + y2 = r2, IA = ż 2π 0 dθ ż a(1+cos θ) 0 dr r ˆ r2 = ż 2π 0 dθ ż a(1+cos θ) 0 dr r3 = a4 4 ż 2π 0 dθ 1 + cos θ 4 = a4 4 ż 2π 0 dθ 1 + 4 cos θ + 6 cos2 θ + 4 cos3 θ + cos4 θ Now ż 2π 0 dθ cos θ = sin θ ˇ ˇ ˇ 2π 0 = 0 ż 2π 0 dθ cos2 θ = ż 2π 0 dθ 1 + cos(2θ) 2 = 1 2 θ + sin(2θ) 2 2π 0 = π ż 2π 0 dθ cos3 θ = ż 2π 0 dθ cos θ 1 ´ sin2 θ u=sin θ = ż 0 0 du (1 ´ u2) = 0 To integrate cos4 θ, we use the double angle formula cos2 θ = cos(2θ) + 1 2 ù ñ cos4 θ = cos(2θ) + 1 2 4 = cos2(2θ) + 2 cos(2θ) + 1 4 = cos(4θ)+1 2 + 2 cos(2θ) + 1 4 = 3 8 + 1 2 cos(2θ) + 1 8 cos(4θ) 313 MULTIPLE INTEGRALS 3.4 SURFACE AREA to give ż 2π 0 dθ cos4 θ = ż 2π 0 dθ 3 8 + 1 2 cos(2θ) + 1 8 cos(4θ) = 3 8 ˆ 2π + 1 2 ˆ 0 + 1 8 ˆ 0 = 3 4π All together IA = a4 4 2π + 4 ˆ 0 + 6 ˆ π + 4 ˆ 0 + 3 4π = 35 16πa4 Example 3.3.17 3.4Ĳ Surface Area Suppose that we wish to find the area of part, S, of the surface z = f (x, y). We start by cutting S up into tiny pieces. To do so, • we draw a bunch of curves of constant x (the blue curves in the figure below). Each such curve is the intersection of S with the plane x = x0 for some constant x0. And we also • draw a bunch of curves of constant y (the red curves in the figure below). Each such curve is the intersection of S with the plane y = y0 for some constant y0. z y x z “ fpx, yq Concentrate on any one the tiny pieces. Here is a greatly magnified sketch of it, looking at it from above. 314 MULTIPLE INTEGRALS 3.4 SURFACE AREA P2 P3 P1 P0 x“x0dx y varying x“x0 y varying x varying y“y0dy x varying y“y0 We wish to compute its area, which we’ll call dS. Now this little piece of surface need not be parallel to the xy-plane, and indeed need not even be flat. But if the piece is really tiny, it’s almost flat. We’ll now approximate it by something that is flat, and whose area we know. To start, we’ll determine the corners of the piece. To do so, we first determine the bounding curves of the piece. Look at the figure above, and recall that, on the surface z = f (x, y). • The upper blue curve was constructed by holding x fixed at the value x0, and sketch-ing the curve swept out by x0 ˆ ı ı ı + y ˆ ȷ ȷ ȷ + f (x0, y) ˆ k as y varied, and • the lower blue curve was constructed by holding x fixed at the slightly larger value x0 + dx, and sketching the curve swept out by (x0 + dx) ˆ ı ı ı + y ˆ ȷ ȷ ȷ + f (x0 + dx, y) ˆ k as y varied. • The red curves were constructed similarly, by holding y fixed and varying x. So the four intersection points in the figure are P0 = x0 ˆ ı ı ı + y0 ˆ ȷ ȷ ȷ + f (x0, y0) ˆ k P1 = x0 ˆ ı ı ı + (y0 + dy) ˆ ȷ ȷ ȷ + f (x0, y0 + dy) ˆ k P2 = (x0 + dx) ˆ ı ı ı + y0 ˆ ȷ ȷ ȷ + f (x0 + dx, y0) ˆ k P3 = (x0 + dx) ˆ ı ı ı + (y0 + dy) ˆ ȷ ȷ ȷ + f (x0 + dx, y0 + dy) ˆ k Now, for any small constants dX and dY, we have the linear approximation27 f (x0 + dX, y0 + dY) « f (x0 , y0) + B f Bx(x0 , y0) dX + B f By(x0 , y0) dY Applying this three times, once with dX = 0, dY = dy (to approximate P1), once with dX = dx, dY = 0 (to approximate P2), and once with dX = dx, dY = dy (to approximate P3), P1 « P0 + dy ˆ ȷ ȷ ȷ + B f By(x0 , y0) dy ˆ k P2 « P0 + dx ˆ ı ı ı + B f Bx(x0 , y0) dx ˆ k P3 « P0 + dx ˆ ı ı ı + dy ˆ ȷ ȷ ȷ + hB f Bx(x0 , y0) dx + B f By(x0 , y0) dy i ˆ k 27 Recall (2.6.1). 315 MULTIPLE INTEGRALS 3.4 SURFACE AREA Of course we have only approximated the positions of the corners and so have introduced errors. However, with more work, one can bound those errors (like we in the optional §3.2.4) and show that in the limit dx, dy Ñ 0, all of the error terms that we dropped contribute exactly 0 to the integral. The small piece of our surface with corners P0, P1, P2, P3 is approximately a parallelo-gram with sides Ý Ý Ñ P0P1 « Ý Ý Ñ P2P3 « dy ˆ ȷ ȷ ȷ + B f By(x0 , y0) dy ˆ k Ý Ý Ñ P0P2 « Ý Ý Ñ P1P3 « dx ˆ ı ı ı + B f Bx(x0 , y0) dx ˆ k P2 P3 P1 P0 θ Ý Ý Ý Ñ P0P1 Ý Ý Ý Ñ P0P2 Denote by θ the angle between the vectors Ý Ý Ñ P0P1 and Ý Ý Ñ P0P2. The base of the parallelogram, Ý Ý Ñ P0P1, has length ˇ ˇÝ Ý Ñ P0P1 ˇ ˇ, and the height of the parallelogram is ˇ ˇÝ Ý Ñ P0P2 ˇ ˇ sin θ. So the area of the parallelogram is28, by Theorem 1.2.23, dS = \|Ý Ý Ñ P0P1\| \|Ý Ý Ñ P0P2\| sin θ = ˇ ˇÝ Ý Ñ P0P1 ˆ Ý Ý Ñ P0P2 ˇ ˇ « ˇ ˇ ˇ ˇ ˆ ȷ ȷ ȷ + B f By(x0 , y0) ˆ k ˆ ˆ ı ı ı + B f Bx(x0 , y0) ˆ k ˇ ˇ ˇ ˇdxdy The cross product is easily evaluated: ˆ ȷ ȷ ȷ + B f By(x0 , y0) ˆ k ˆ ˆ ı ı ı + B f Bx(x0 , y0) ˆ k = det    ˆ ı ı ı ˆ ȷ ȷ ȷ ˆ k 0 1 B f By(x0, y0) 1 0 B f Bx(x0, y0)    = fx(x0, y0) ˆ ı ı ı + fy(x0, y0) ˆ ȷ ȷ ȷ ´ ˆ k as is its length: ˇ ˇ ˇ ˇ ˆ ȷ ȷ ȷ + B f By(x0 , y0) ˆ k ˆ ˆ ı ı ı + B f Bx(x0 , y0) ˆ k ˇ ˇ ˇ ˇ = b 1 + fx(x0, y0)2 + fy(x0, y0)2 Throughout this computation, x0 and y0 were arbitrary. So we have found the area of each tiny piece of the surface S. 28 As we mentioned above, the approximation below becomes exact when the limit dx, dy Ñ 0 is taken in the definition of the integral. See §3.3.5 in the CLP-4 text. 316 MULTIPLE INTEGRALS 3.4 SURFACE AREA For the surface z = f (x, y), dS = b 1 + fx(x, y)2 + fy(x, y)2 dxdy Similarly, for the surface x = g(y, z), dS = b 1 + gy(y, z)2 + gz(y, z)2 dydz and for the surface y = h(x, z), dS = b 1 + hx(x, z)2 + hz(x, z)2 dxdz Equation 3.4.1. Consequently, we have (a) The area of the part of the surface z = f (x, y) with (x, y) running over the region D in the xy-plane is ĳ D b 1 + fx(x, y)2 + fy(x, y)2 dxdy (b) The area of the part of the surface x = g(y, z) with (y, z) running over the region D in the yz-plane is ĳ D b 1 + gy(y, z)2 + gz(y, z)2 dydz (c) The area of the part of the surface y = h(x, z) with (x, z) running over the region D in the xz-plane is ĳ D b 1 + hx(x, z)2 + hz(x, z)2 dxdz Theorem 3.4.2. Example 3.4.3 (Area of a cone) As a first example, we compute the area of the part of the cone z = b x2 + y2 with 0 ď z ď a or, equivalently, with x2 + y2 ď a2. 317 MULTIPLE INTEGRALS 3.4 SURFACE AREA h a Note that z = a x2 + y2 is the side of the cone. It does not include the top. To find its area, we will apply (3.4.1) to z = f (x, y) = b x2 + y2 with (x, y) running over x2 + y2 ď a2 That forces us to compute the first order partial derivatives fx(x, y) = x a x2 + y2 fy(x, y) = y a x2 + y2 Substituting them into the first formula in (3.4.1) yields dS = b 1 + fx(x, y)2 + fy(x, y)2 dx dy = d 1 + x a x2 + y2 2 + y a x2 + y2 2 dx dy = d 1 + x2 + y2 x2 + y2 dx dy = ? 2 dx dy So Area = ĳ x2+y2ďa2 ? 2 dx dy = ? 2 ĳ x2+y2ďa2 dx dy = ? 2πa2 because ť x2+y2ďa2 dx dy is exactly the area of a circular disk of radius a. Example 3.4.3 Example 3.4.4 (Area of a cylinder) Let a, b ą 0. Find the surface area of S = ␣(x, y, z) ˇ ˇ x2 + z2 = a2, 0 ď y ď b ( 318 MULTIPLE INTEGRALS 3.4 SURFACE AREA Solution. The intersection of x2 + z2 = a2 with any plane of constant y is the circle of radius a centred on x = z = 0. So S is a bunch of circles stacked sideways. It is a cylinder on its side (with both ends open). By symmetry, the area of S is four times the area of the part of S that is in the first octanct, which is S1 = ! (x, y, z) ˇ ˇ ˇ z = f (x, y) = a a2 ´ x2, 0 ď x ď a, 0 ď y ď b ) y z x S y z x S1 pa, b, 0q Since fx(x, y) = ´ x ? a2 ´ x2 fy(x, y) = 0 the first formula in (3.4.1) yields dS = b 1 + fx(x, y)2 + fy(x, y)2 dx dy = d 1 + ´ x ? a2 ´ x2 2 dx dy = d 1 + x2 a2 ´ x2 dxdy = a ? a2 ´ x2 dxdy So Area(S1) = ż a 0 dx ż b 0 dy a ? a2 ´ x2 = ab ż a 0 dx 1 ? a2 ´ x2 The indefinite integral of 1 ? a2´x2 is arcsin x a + C. (See the table of integrals in Appendix D. Alternatively, use the trig substitution x = a sin θ.) So Area(S1) = ab h arcsin x a ia 0 = ab arcsin 1 ´ arcsin 0 = π 2 ab and Area(S) = 4Area(S1) = 2πab We could have also come to this conclusion by using a little geometry, rather than using calculus. Cut open the cylinder by cutting along a line parallel to the y-axis, and 319 MULTIPLE INTEGRALS 3.4 SURFACE AREA then flatten out the cylinder. This gives a rectangle. One side of the rectangle is just a circle of radius a, straightened out. So the rectangle has sides of lengths 2πa and b and has area 2πab. Example 3.4.4 Example 3.4.5 (Area of a hemisphere) This time we compute the surface area of the hemisphere x2 + y2 + z2 = a2 z ě 0 (with a ą 0). You probably know, from high school, that the answer is 1 2 ˆ 4πa2 = 2πa2. But you have probably not seen a derivation29 of this answer. Note that, since x2 + y2 = a2 ´ z2 on the hemisphere, the set of (x, y)’s for which there is a z with (x, y, z) on the hemisphere is exactly ␣(x, y) P R2 ˇ ˇ x2 + y2 ď a2 ( . So the hemisphere is S = ! (x, y, z) ˇ ˇ ˇ z = b a2 ´ x2 ´ y2, x2 + y2 ď a2) We will compute the area of S by applying (3.4.1) to z = f (x, y) = b a2 ´ x2 ´ y2 with (x, y) running over x2 + y2 ď a2 The first formula in (3.4.1) yields dS = b 1 + fx(x, y)2 + fy(x, y)2 dxdy = d 1 + ´x a a2 ´ x2 ´ y2 2 + ´y a a2 ´ x2 ´ y2 2 dxdy = d 1 + x2 + y2 a2 ´ x2 ´ y2 dxdy = d a2 a2 ´ x2 ´ y2 dxdy So the area is ť x2+y2ďa2 a ? a2´x2´y2 dxdy. To evaluate this integral, we switch to polar 29 There is a pun hidden here, because you can (with a little thought) also get the surface area by differen-tiating the volume with respect to the radius. 320 MULTIPLE INTEGRALS 3.4 SURFACE AREA coordinates, substituting x = r cos θ, y = r sin θ. This gives area = ĳ x2+y2ďa2 a a a2 ´ x2 ´ y2 dxdy = ż a 0 dr r ż 2π 0 dθ a ? a2 ´ r2 = 2πa ż a 0 dr r ? a2 ´ r2 = 2πa ż 0 a2 ´du/2 ?u with u = a2 ´ r2, du = ´2r dr = 2πa h ´ ?u i0 a2 = 2πa2 as it should be. Example 3.4.5 Example 3.4.6 Find the surface area of the part of the paraboloid z = 2 ´ x2 ´ y2 lying above the xy-plane. Solution. The equation of the surface is of the form z = f (x, y) with f (x, y) = 2 ´ x2 ´ y2. So fx(x, y) = ´2x fy(x, y) = ´2y and, by the first part of (3.4.1), dS = b 1 + fx(x, y)2 + fy(x, y)2 dxdy = b 1 + 4x2 + 4y2 dxdy The point (x, y, z), with z = 2 ´ x2 ´ y2, lies above the xy-plane if and only if z ě 0, or, equivalently, 2 ´ x2 ´ y2 ě 0. So the domain of integration is ␣(x, y) ˇ ˇ x2 + y2 ď 2 ( and Surface Area = ĳ x2+y2ď2 b 1 + 4x2 + 4y2 dxdy Switching to polar coordinates, Surface Area = ż 2π 0 ż ? 2 0 a 1 + 4r2 r dr dθ = 2π 1 12 1 + 4r23/2? 2 0 = π 6 [27 ´ 1] = 13 3 π Example 3.4.6 321 MULTIPLE INTEGRALS 3.5 TRIPLE INTEGRALS 3.5Ĳ Triple Integrals Triple integrals, that is integrals over three dimensional regions, are just like double inte-grals, only more so. We decompose the domain of integration into tiny cubes, for example, compute the contribution from each cube and then use integrals to add up all of the dif-ferent pieces. We’ll go through the details now by means of a number of examples. Example 3.5.1 Find the mass inside the sphere x2 + y2 + z2 = 1 if the density is ρ(x, y, z) = \|xyz\|. Solution. The absolute values can complicate the computations. We can avoid those complications by exploiting the fact that, by symmetry, the total mass of the sphere will be eight times the mass in the first octant. We shall cut the first octant part of the sphere into tiny pieces using Cartesian coordinates. That is, we shall cut it up using planes of constant z, planes of constant y, and planes of constant x, which we recall look like y z x p0, 0, zq surface of constant z (a plane) y z x p0, y, 0q surface of constant y (a plane) y z x px, 0, 0q surface of constant x (a plane) • First slice the (the first octant part of the) sphere into horizontal plates by inserting many planes of constant z, with the various values of z differing by dz. The figure on the left below shows the part of one plate in the first octant outlined in red. Each plate – has thickness dz, – has z almost constant throughout the plate (it only varies by dz), and – has (x, y) running over x ě 0, y ě 0, x2 + y2 ď 1 ´ z2. – The bottom plate starts at z = 0 and the top plate ends at z = 1. See the figure on the right below. z y x x2 y2 z2 “ 1 z y x p0, 0, 1q 322 MULTIPLE INTEGRALS 3.5 TRIPLE INTEGRALS • Concentrate on any one plate. Subdivide it into long thin “square” beams by insert-ing many planes of constant y, with the various values of y differing by dy. The figure on the left below shows the part of one beam in the first octant outlined in blue. Each beam – has cross-sectional area dy dz, – has z and y essentially constant throughout the beam, and – has x running over 0 ď x ď a 1 ´ y2 ´ z2. – The leftmost beam has, essentially, y = 0 and the rightmost beam has, essen-tially, y = ? 1 ´ z2. See the figure on the right below. z y x x “ a 1 ´ y2 ´ z2 z y x p0, ? 1´z2,zq p ? 1´z2,0,zq • Concentrate on any one beam. Subdivide it into tiny approximate cubes by inserting many planes of constant x, with the various values of x differing by dx. The figure on the left below shows the top of one approximate cube in black. Each cube – has volume dx dy dz, and – has x, y and z all essentially constant throughout the cube. – The first cube has, essentially, x = 0 and the last cube has, essentially, x = a 1 ´ y2 ´ z2. See the figure on the right below. z y x x “ a 1 ´ y2 ´ z2 z y x p0, y, zq p? 1´y2´z2 , y , zq Now we can build up the mass. • Concentrate on one approximate cube. Let’s say that it contains the point (x, y, z). – The cube has volume essentially dV = dx dy dz and – essentially has density ρ(x, y, z) = xyz and so – essentially has mass xyz dx dy dz. 323 MULTIPLE INTEGRALS 3.5 TRIPLE INTEGRALS • To get the mass of any one beam, say the beam whose y coordinate runs from y to y + dy, we just add up the masses of the approximate cubes in that beam, by integrating x from its smallest value on the beam, namely 0, to its largest value on the beam, namely a 1 ´ y2 ´ z2. The mass of the beam is thus dy dz ż ? 1´y2´z2 0 dx xyz • To get the mass of any one plate, say the plate whose z coordinate runs from z to z + dz, we just add up the masses of the beams in that plate, by integrating y from its smallest value on the plate, namely 0, to its largest value on the plate, namely ? 1 ´ z2. The mass of the plate is thus dz ż ? 1´z2 0 dy ż ? 1´y2´z2 0 dx xyz • To get the mass of the part of the sphere in the first octant, we just add up the masses of the plates that it contains, by integrating z from its smallest value in the octant, namely 0, to its largest value on the sphere, namely 1. The mass in the first octant is thus ż 1 0 dz ż ? 1´z2 0 dy ż ? 1´y2´z2 0 dx xyz = ż 1 0 dz ż ? 1´z2 0 dy yz "ż ? 1´y2´z2 0 dx x # = ż 1 0 dz ż ? 1´z2 0 dy 1 2yz 1 ´ y2 ´ z2 = ż 1 0 dz ż ? 1´z2 0 dy z(1 ´ z2) 2 y ´ z 2y3 = ż 1 0 dz " z(1 ´ z2)2 4 ´ z(1 ´ z2)2 8 # = ż 1 0 dz z (1 ´ z2)2 8 = ż 0 1 du ´2 u2 8 with u = 1 ´ z2, du = ´2z dz = 1 48 • So the mass of the total (eight octant) sphere is 8 ˆ 1 48 = 1 6. Example 3.5.1 Consider, for example, the limits of integration for the integral ż 1 0 dz ż ? 1´z2 0 dy ż ? 1´y2´z2 0 dx xyz = ż 1 0 ż ? 1´z2 0 ż ? 1´y2´z2 0 xyz dx ! dy ! dz that we have just evaluated. 324 MULTIPLE INTEGRALS 3.5 TRIPLE INTEGRALS • When we are integrating over the innermost integral, with respect to x, the quantities y and z are treated as constants. In particular, y and z may appear in the limits of integration for the x-integral, but x may not appear in those limits. • When we are integrating over y, we have already integrated out x; x no longer exists. The quantity z is treated as a constant. In particular, z, but neither x nor y, may appear in the limits of integration for the y-integral. • Finally, when we are integrating over z, we have already integrated out x and y; they no longer exist. None of x, y or z, may appear in the limits of integration for the z-integral. Example 3.5.2 In practice, often the hardest part of dealing with a triple integral is setting up the limits of integration. In this example, we’ll concentrate on exactly that. Let V be the solid region in R3 bounded by the planes x = 0, y = 0, z = 0, y = 4 ´ x, and the surface z = 4 ´ x2. We are now going to write ţ V f (x, y, z) dV as an iterated integral (i.e. find the limits of integration) in two different ways. Here f is just some general, unspecified, function. First, we’ll figure out what V looks like. The following three figures show • the part of the first octant with y ď 4 ´ x (except that it continues vertically upward) • the part of the first octant with z ď 4 ´ x2 (except that it continues to the right) • the part of the first octant with both y ď 4 ´ x and z ď 4 ´ x2. That’s V = ␣(x, y, z) ˇ ˇ x ě 0, y ě 0, z ě 0, x + y ď 4, z ď 4 ´ x2 ( z y x z y x z y x The iterated integral ţ V f (x, y, z) dz dy dx = ş ( ş ( ş f (x, y, z) dz) dy) dx: For this iterated integral, the outside integral is with respect to x, so we first slice up V using planes of constant x, as in the figure below. 325 MULTIPLE INTEGRALS 3.5 TRIPLE INTEGRALS z y x p2, 2, 0q px, 4 ´ x, 4 ´ x2q y “ 4 ´ x z “ 4 ´ x2 Observe from that figure that, on V, • x runs from 0 to 2, and • for each fixed x in that range, y runs from 0 to 4 ´ x and • for each fixed (x, y) as above, z runs from 0 to 4 ´ x2. So ¡ V f (x, y, z) dz dy dx = ż 2 0 dx ż 4´x 0 dy ż 4´x2 0 dz f (x, y, z) = ż 2 0 ż 4´x 0 ż 4´x2 0 f (x, y, z) dz dy dx The iterated integral ţ V f (x, y, z) dy dx dz = ş ( ş ( ş f (x, y, z) dy) dx) dz: For this iterated integral, the outside integral is with respect to z, so we first slice up V using planes of constant z, as in the figure below. z y x p0, 4, 4q p?4 ´ z, 0, zq y “ 4 ´ x x “ ?4 ´ z 326 MULTIPLE INTEGRALS 3.5 TRIPLE INTEGRALS Observe from that figure that, on V, • z runs from 0 to 4, and • for each fixed z in that range, x runs from 0 to ? 4 ´ z and • for each fixed (x, z) as above, y runs from 0 to 4 ´ x. So ¡ V f (x, y, z) dy dx dz = ż 4 0 dz ż ? 4´z 0 dx ż 4´x 0 dy f (x, y, z) = ż 4 0 ż ? 4´z 0 ż 4´x 0 f (x, y, z) dy dx dz Example 3.5.2 Example 3.5.3 As was said in the last example, in practice, often the hardest parts of dealing with a triple integral concern the limits of integration. In this example, we’ll again concentrate on ex-actly that. This time, we will consider the integral I = ż 2 0 dy ż 2´y 0 dz ż 2´y 2 0 dx f (x, y, z) and we will reexpress I with the outside integral being over z. We will figure out the limits of integration for both the order ş dz ş dx ş dy f (x, y, z) and for the order ş dz ş dy ş dx f (x, y, z). Our first task is to get a good idea as to what the domain of integration looks like. We start by reading off of the given integral that • the outside integral says that y runs from 0 to 2, and • the middle integral says that, for each fixed y in that range, z runs from 0 to 2 ´ y and • the inside integral says that, for each fixed (y, z) as above, x runs from 0 to 2´y 2 . So the domain of integration is V = ␣(x, y, z) ˇ ˇ 0 ď y ď 2, 0 ď z ď 2 ´ y, 0 ď x ď 2´y 2 ( (˚) We’ll sketch V shortly. Because it is generally easier to make 2d sketches than it is to make 3d sketches, we’ll first make a 2d sketch of the part of V that lies in the vertical plane y = Y. Here Y is any constant between 0 and 2. Looking at the definition of V, we see that the point (x, Y, z) lies in V if and only if 0 ď z ď 2 ´ Y 0 ď x ď 2 ´ Y 2 Here, on the left, is a (2d) sketch of all (x, z)’s that obey those inequalities, and, on the right, is a (3d) sketch of all (x, Y, z)’s that obey those inequalities. 327 MULTIPLE INTEGRALS 3.5 TRIPLE INTEGRALS x z z “ 2 ´ Y x “ p2 ´ Y q{2 2´Y 2 , Y, 0 ˘ 0, Y, 2 ´ Y ˘ z y x So our solid V consists of a bunch of vertical rectangles stacked sideways along the y-axis. The rectangle in the plane y = Y has side lengths 2´Y 2 and 2 ´ Y. As we move from the plane y = Y = 0, i.e. the xz-plane, to the plane y = Y = 2, the rectangle decreases in size linearly from a one by two rectangle, when Y = 0, to a zero by zero rectangle, i.e. a point, when Y = 2. Here is a sketch of V together with a typical y = Y rectangle. p1, 0, 0q p0, 0, 2q p0, 2, 0q z y x To reexpress the given integral with the outside integral being with respect to z, we have to slice up V into horizontal plates by inserting planes of constant z. So we have to figure out what the part of V that lies in the horizontal plane z = Z looks like. From the figure above, we see that, in V, the smallest value of z is 0 and the biggest value of z is 2. So Z is any constant between 0 and 2. Again looking at the definition of V in (˚) above, we see that the point (x, y, Z) lies in V if and only if y ě 0 y ď 2 y ď 2 ´ Z x ě 0 2x + y ď 2 Here, on the left, is a (2d) sketch showing the top view of all (x, y)’s that obey those inequalities, and, on the right, is a (3d) sketch of all (x, y, Z)’s that obey those inequalities. 328 MULTIPLE INTEGRALS 3.5 TRIPLE INTEGRALS y x y “ 2 2x y “ 2 y “ 2 ´ Z p1, 0q pZ{2, 2 ´ Zq p0, 2 ´ Zq p1, 0, Zq p0, 2 ´ Z, Zq pZ{2, 2 ´ Z, Zq z y x To express I as an integral with the order of integration ş dz ş dy ş dx f (x, y, z), we subdi-vide the plate at height z into vertical strips as in the figure y x y “ 2 x “ 2´y 2 y “ 2 ´ z p1, 0q pz{2, 2 ´ zq p0, 2 ´ zq Since • y is essentially constant on each strip with the leftmost strip having y = 0 and the rightmost strip having y = 2 ´ z and • for each fixed y in that range, x runs from 0 to 2´y 2 we have I = ż 2 0 dz ż 2´z 0 dy ż 2´y 2 0 dx f (x, y, z) Alternatively, to express I as an integral with the order of integration ş dz ş dx ş dy f (x, y, z), we subdivide the plate at height z into horizontal strips as in the figure 329 MULTIPLE INTEGRALS 3.6 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES y x y “ 2 y “ 2 ´ 2x y “ 2 ´ z p1, 0q pz{2, 2 ´ zq p0, 2 ´ zq Since • x is essentially constant on each strip with the first strip having x = 0 and the last strip having x = 1 and • for each fixed x between 0 and z/2, y runs from 0 to 2 ´ z and • for each fixed x between z/2 and 1, y runs from 0 to 2 ´ 2x we have I = ż 2 0 dz ż z/2 0 dx ż 2´z 0 dy f (x, y, z) + ż 2 0 dz ż 1 z/2 dx ż 2´2x 0 dy f (x, y, z) Example 3.5.3 3.6Ĳ Triple Integrals in Cylindrical Coordinates Many problems possess natural symmetries. We can make our work easier by using coor-dinate systems, like polar coordinates, that are tailored to those symmetries. We will look at two more such coordinate systems — cylindrical and spherical coordinates. 3.6.1 § § Cylindrical Coordinates In the event that we wish to compute, for example, the mass of an object that is invariant under rotations about the z-axis30, it is advantageous to use a natural generalization of po-lar coordinates to three dimensions. The coordinate system is called cylindrical coordinates. 30 like a pipe or a can of tuna fish 330 MULTIPLE INTEGRALS 3.6 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES Cylindrical coordinates are denoted31 r, θ and z and are defined by r = the distance from (x, y, 0) to (0, 0, 0) = the distance from (x, y, z) to the z-axis θ = the angle between the positive x axis and the line joining (x, y, 0) to (0, 0, 0) z = the signed distance from (x, y, z) to the xy-plane px, y, zq px, y, 0q y z x z r θ That is, r and θ are the usual polar coordinates and z is the usual z. Definition 3.6.1. The Cartesian and cylindrical coordinates are related by32 x = r cos θ y = r sin θ z = z r = b x2 + y2 θ = arctan y x z = z Equation 3.6.2. Here are sketches of surfaces of constant r, constant θ, and constant z. y z x r surface of constant r (a cylindrical shell) y z x θ surface of constant θ (a plane) y z x p0, 0, zq surface of constant z (a plane) 31 We are using the standard mathematics conventions for the cylindrical coordinates. Under the ISO conventions they are (ρ, ϕ, z). See Appendix G. 32 As was the case for polar coordinates, it is sometimes convenient to extend these definitions by saying that x = r cos θ and y = r sin θ even when r is negative. See the end of Section 3.2.1. 331 MULTIPLE INTEGRALS 3.6 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES 3.6.2 § § The Volume Element in Cylindrical Coordinates Before we can start integrating using these coordinates we need to determine the volume element. Recall that before integrating in polar coordinates, we had to establish that dA = r dr dθ. In the arguments that follow we establish that dV = r dr dθ dz. If we cut up a solid by • first slicing it into horizontal plates of thickness dz by using planes of constant z, z y x dz • and then subdividing the plates into wedges using surfaces of constant θ, say with the difference between successive θ’s being dθ, z y x dθ • and then subdividing the wedges into approximate cubes using surfaces of constant r, say with the difference between successive r’s being dr, z y x dr we end up with approximate cubes that look like 332 MULTIPLE INTEGRALS 3.6 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES r dr dz r dθ • When we introduced slices using surfaces of constant r, the difference between the successive r’s was dr, so the indicated edge of the cube has length dr. • When we introduced slices using surfaces of constant z, the difference between the successive z’s was dz, so the vertical edges of the cube have length dz. • When we introduced slices using surfaces of constant θ, the difference between the successive θ’s was dθ, so the remaining edges of the cube are circular arcs of radius essentially33 r that subtend an angle θ, and so have length r dθ. See the derivation of equation (3.2.5). So the volume of the approximate cube in cylindrical coordinates is (essentially34) dV = r dr dθ dz Equation 3.6.3. 3.6.3 § § Sample Integrals in Cylindrical Coordinates Now we can use (3.6.3) to handle a variant of Example 3.5.1 in which the density is invari-ant under rotations around the z-axis. Cylindrical coordinates are tuned to provide easier integrals to evaluate when the integrand is invariant under rotations about the z-axis, or when the domain of integration is cylindrical. Example 3.6.4 Find the mass of the solid body consisting of the inside of the sphere x2 + y2 + z2 = 1 if the density is ρ(x, y, z) = x2 + y2. Solution. Before we get started, note that x2 + y2 is the square of the distance from (x, y, z) to the z-axis. Consequently both the integrand, x2 + y2, and the domain of integration, x2 + y2 + z2 ď 1, and hence our solid, are invariant under rotations about the z-axis35. That makes this integral a good candidate for cylindrical coordinates. Again, by symmetry the total mass of the sphere will be eight times the mass in the first octant. We shall cut the first octant part of the sphere into tiny pieces using cylindrical coordinates. That is, we shall cut it up using planes of constant z, planes of constant θ, and surfaces of constant r. 33 The inner edge has radius r, but the outer edge has radius r + dr. However the error that this generates goes to zero in the limit dr, dθ, dz Ñ 0. 34 By “essentially”, we mean that the formula for dV works perfectly when we take the limit dr, dθ, dz Ñ 0 of Riemann sums. 35 Imagine that you are looking that the solid from, for example, far out on the x-axis. You close your eyes for a minute. Your evil twin then sneaks in, rotates the solid about the z-axis, and sneaks out. You open your eyes. You will not be able to tell that the solid has been rotated. 333 MULTIPLE INTEGRALS 3.6 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES • First slice the (the first octant part of the) sphere into horizontal plates by inserting many planes of constant z, with the various values of z differing by dz. The figure on the left below shows the part of one plate in the first octant outlined in red. Each plate – has thickness dz, – has z essentially constant on the plate, and – has (x, y) running over x ě 0, y ě 0, x2 + y2 ď 1 ´ z2. In cylindrical coordinates, r runs from 0 to ? 1 ´ z2 and θ runs from 0 to π/2. – The bottom plate has, essentially, z = 0 and the top plate has, essentially, z = 1. See the figure on the right below. z y x r2 z2 “ 1 θ “ π{2 θ “ 0 z y x p0, 0, 1q So far, this looks just like what we did in Example 3.5.1. • Concentrate on any one plate. Subdivide it into wedges by inserting many planes of constant θ, with the various values of θ differing by dθ. The figure on the left below shows one such wedge outlined in blue. Each wedge – has z and θ essentially constant on the wedge, and – has r running over 0 ď r ď ? 1 ´ z2. – The leftmost wedge has, essentially, θ = 0 and the rightmost wedge has, essen-tially, θ = π/2. See the figure on the right below. z y x r2 z2 “ 1 r “ ? 1 ´ z2 z y x θ “ π{2 θ “ 0 • Concentrate on any one wedge. Subdivide it into tiny approximate cubes by insert-ing many surfaces of constant r, with the various values of r differing by dr. The figure on the left below shows the top of one approximate cube in black. Each cube – has volume r dr dθ dz, by (3.6.3), and 334 MULTIPLE INTEGRALS 3.6 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES – has r, θ and z all essentially constant on the cube. – The first cube has, essentially, r = 0 and the last cube has, essentially, r = ? 1 ´ z2. See the figure on the right below. z y x z y x r “ ? 1 ´ z2 r “ 0 Now we can build up the mass. • Concentrate on one approximate cube. Let’s say that it contains the point with cylin-drical coordinates r, θ and z. – The cube has volume essentially dV = r dr dθ dz and – essentially has density ρ(x, y, z) = ρ(r cos θ, r sin θ, z) = r2 and so – essentially has mass r3 dr dθ dz. (See how nice the right coordinate system can be!) • To get the mass any one wedge, say the wedge whose θ coordinate runs from θ to θ + dθ, we just add up the masses of the approximate cubes in that wedge, by integrating r from its smallest value on the wedge, namely 0, to its largest value on the wedge, namely ? 1 ´ z2. The mass of the wedge is thus dθ dz ż ? 1´z2 0 dr r3 • To get the mass of any one plate, say the plate whose z coordinate runs from z to z + dz, we just add up the masses of the wedges in that plate, by integrating θ from its smallest value on the plate, namely 0, to its largest value on the plate, namely π/2. The mass of the plate is thus dz ż π/2 0 dθ ż ? 1´z2 0 dr r3 • To get the mass of the part of the sphere in the first octant, we just add up the masses of the plates that it contains, by integrating z from its smallest value in the octant, namely 0, to its largest value on the sphere, namely 1. The mass in the first octant is 335 MULTIPLE INTEGRALS 3.6 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES thus ż 1 0 dz ż π/2 0 dθ ż ? 1´z2 0 dr r3 = 1 4 ż 1 0 dz ż π/2 0 dθ (1 ´ z2) 2 = π 8 ż 1 0 dz (1 ´ z2) 2 = π 8 ż 1 0 dz (1 ´ 2z2 + z4) = π 8 8/15 hkkkkkkikkkkkkj 1 ´ 2 3 + 1 5 = 1 15π • So the mass of the total (eight octant) sphere is 8 ˆ 1 15π = 8 15π. Just by way of comparison, here is the integral in Cartesian coordinates that gives the mass in the first octant. (We found the limits of integration in Example 3.5.1.) ż 1 0 dz ż ? 1´z2 0 dy ż ? 1´y2´z2 0 dx x2 + y2 Example 3.6.4 In the next example, we compute the moment of inertia of a right circular cone. The Definition 3.3.15 of the moment of inertia was restricted to two dimensions. However, as was pointed out at the time, the same analysis extends naturally to the definition IA = ¡ V D(x, y, z)2 ρ(x, y, z) dx dy dz Equation 3.6.5. of the moment of inertia of a solid V in three dimensions. Here • ρ(x, y, z) is the mass density of the solid at the point (x, y, z) and • D(x, y, z) is the distance from (x, y, z) to the axis of rotation. Example 3.6.6 Find the moment of inertia of a right circular cone • of radius a, • of height h, and • of constant density with mass M about an axis through the vertex (i.e. the tip of the cone) and parallel to the base. Solution. Here is a sketch of the cone. 336 MULTIPLE INTEGRALS 3.6 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES h a Let’s pick a coordinate system with • the vertex at the origin, • the cone symmetric about the z-axis and • the axis of rotation being the y-axis. and call the cone V. h a z y x We shall use (3.6.5) to find the moment of inertia. In the current problem, the axis of rotation is the y-axis. The point on the y-axis that is closest to (x, y, z) is (0, y, 0) so that the distance from (x, y, z) to the axis is just D(x, y, z) = a x2 + z2 y z x px, y, zq p0, y, 0q x y z Our solid has constant density and mass M, so ρ(x, y, z) = M Volume(V) 337 MULTIPLE INTEGRALS 3.6 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES The formula Volume(V) = 1 3πa2h for the volume of a cone was derived in Example 1.6.1 of the CLP-2 text and in Ap-pendix B.5.2 of the CLP-1 text. However because of the similarity between the integral Volume(V) = ţ V dx dy dz and the integral ţ V(x2 + z2) dx dy dz, that we need for our computation of IA, it is easy to rederive the volume formula and we shall do so. We’ll evaluate both of the integrals above using cylindrical coordinates. • Start by slicing the cone into horizontal plates by inserting many planes of constant z, with the various values of z differing by dz. z y x Each plate – is a circular disk of thickness dz. – By similar triangles, as in the figure on the right below, the disk at height z has radius R obeying R z = a h ù ñ R = a hz dz z h a z ha z h a R – So the disk at height z has the cylindrical coordinates r running from 0 to a hz and θ running from 0 to 2π. – The bottom plate has, essentially, z = 0 and the top plate has, essentially, z = h. • Now concentrate on any one plate. Subdivide it into wedges by inserting many planes of constant θ, with the various values of θ differing by dθ. – The first wedge has, essentially θ = 0 and the last wedge has, essentially, θ = 2π. 338 MULTIPLE INTEGRALS 3.6 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES • Concentrate on any one wedge. Subdivide it into tiny approximate cubes36 by in-serting many surfaces of constant r, with the various values of r differing by dr. Each cube – has volume r dr dθ dz, by (3.6.3). – The first cube has, essentially, r = 0 and the last cube has, essentially, r = a hz. So the two integrals of interest are ¡ V dx dy dz = ż h 0 dz ż 2π 0 dθ ż a h z 0 dr r = ż h 0 dz ż 2π 0 dθ 1 2 a hz 2 = a2π h2 ż h 0 dz z2 = 1 3πa2h as expected, and ¡ V (x2 + z2) dx dy dz = ż h 0 dz ż 2π 0 dθ ż a h z 0 dr r x2+z2 hkkkkkkkkikkkkkkkkj r2 cos2 θ + z2 = ż h 0 dz ż 2π 0 dθ 1 4 a hz 4 cos2 θ + 1 2 a hz 2 z2 = ż h 0 dz 1 4 a4 h4 + a2 h2 πz4 since ż 2π 0 cos2 θ dθ = π by Remark 3.3.5 = 1 5 1 4 a4 h4 + a2 h2 πh5 Putting everything together, the moment of inertia is IA = ¡ V D(x,y,z)2 hkkkkikkkkj (x2 + z2) ρ(x,y,z) hkkikkj M 1 3πa2h dx dy dz = 3 M πa2h 1 5 1 4 a4 h4 + a2 h2 πh5 = 3 20 M a2 + 4h2 Example 3.6.6 36 Again they are wonky cubes, but we can bound the error and show that it goes to zero in the limit dr, dθ, dz Ñ 0. 339 MULTIPLE INTEGRALS 3.7 TRIPLE INTEGRALS IN SPHERICAL COORDINATES 3.7Ĳ Triple Integrals in Spherical Coordinates 3.7.1 § § Spherical Coordinates In the event that we wish to compute, for example, the mass of an object that is invariant under rotations about the origin, it is advantageous to use another generalization of polar coordinates to three dimensions. The coordinate system is called spherical coordinates. Spherical coordinates are denoted37 ρ, θ and φ and are defined by ρ = the distance from (0, 0, 0) to (x, y, z) φ = the angle between the z axis and the line joining (x, y, z) to (0, 0, 0) θ = the angle between the x axis and the line joining (x, y, 0) to (0, 0, 0) z y x p0, 0, zq px, y, 0q px, 0, 0q θ ϕ px, y, zq ρ ρ sin ϕ ρ cos ϕ ρ sin ϕ sin θ Definition 3.7.1. Here are two more figures giving the side and top views of the previous figure. z p0, 0, zq ρ cos ϕ px, y, 0q side view ρ sin ϕ ϕ ρ px, y, zq y x px, 0, 0q top view ρ sin ϕ sin θ ρ sin ϕ cos θ ρ sin ϕ θ px, y, 0q 37 We are using the standard mathematics conventions for the spherical coordinates. Under the ISO con-ventions they are (r, ϕ, θ). See Appendix G. 340 MULTIPLE INTEGRALS 3.7 TRIPLE INTEGRALS IN SPHERICAL COORDINATES The spherical coordinate θ is the same as the cylindrical coordinate θ. The spherical coor-dinate φ is new. It runs from 0 (on the positive z-axis) to π (on the negative z-axis). The Cartesian and spherical coordinates are related by x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ ρ = b x2 + y2 + z2 θ = arctan y x φ = arctan a x2 + y2 z Equation 3.7.2. Here are three figures showing • a surface of constant ρ, i.e. a surface x2 + y2 + z2 = ρ2 with ρ a constant (which looks like an onion skin), • a surface of constant θ, i.e. a surface y = x tan θ with38 θ a constant (which looks like the page of a book), and • a surface of constant φ, i.e. a surface z = a x2 + y2 tan φ with φ a constant (which looks a conical funnel). z y x ρ surface of constant ρ (a sphere) z y x θ surface of constant θ (a plane) z y x ϕ surface of constant ϕ (a cone) 3.7.2 § § The Volume Element in Spherical Coordinates If we cut up a solid39 by • first slicing it into segments (like segments of an orange) by using planes of constant θ, say with the difference between successive θ’s being dθ, 38 and with the sign of x being the same as the sign of cos θ 39 You know the drill. 341 MULTIPLE INTEGRALS 3.7 TRIPLE INTEGRALS IN SPHERICAL COORDINATES z y x dθ • and then subdividing the segments into “searchlights” (like the searchlight outlined in blue in the figure below) using surfaces of constant φ, say with the difference between successive φ’s being dφ, z y x dϕ • and then subdividing the searchlights into approximate cubes using surfaces of con-stant ρ, say with the difference between successive ρ’s being dρ, z y x dρ we end up with approximate cubes that look like 342 MULTIPLE INTEGRALS 3.7 TRIPLE INTEGRALS IN SPHERICAL COORDINATES z y x dρ ρ dϕ ρ sin ϕ dθ The dimensions of the approximate “cube” in spherical coordinates are (essentially) dρ by ρdφ by ρ sin φ dθ. (These dimensions are derived in more detail in the next section.) So the approximate cube has volume (essentially) dV = ρ2 sin φ dρ dθ dφ Equation 3.7.3. §§§ The Details Here is an explanation of the edge lengths given in the above figure. Each of the 12 edges of the cube is formed by holding two of the three coordinates ρ, θ, φ fixed and varying the third. • Four of the cube edges are formed by holding θ and φ fixed and varying ρ. The inter-section of a plane of fixed θ with a cone of fixed φ is a straight line emanating from the origin. When we introduced slices using spheres of constant ρ, the difference between the successive ρ’s was dρ, so those edges of the cube each have length dρ. z y x θ plane of ﬁxed θ z y x ϕ cone of ﬁxed ϕ z y x θ ϕ line of intersection 343 MULTIPLE INTEGRALS 3.7 TRIPLE INTEGRALS IN SPHERICAL COORDINATES • Four of the cube edges are formed by holding θ and ρ fixed and varying φ. The intersection of a plane of fixed θ (which contains the origin) with a sphere of fixed ρ (which is centred on the origin) is a circle of radius ρ centred on the origin. It is a line of longitude40. z y x ρ sphere of ﬁxed ρ z y x θ plane of ﬁxed θ z y x ρ θ circle of intersection When we introduced searchlights using surfaces of constant φ, the difference be-tween the successive φ’s was dφ. Thus those four edges of the cube are circular arcs of radius essentially ρ that subtend an angle dφ, and so have length ρ dφ. z y x ρ ρ dϕ ϕ dϕ • Four of the cube edges are formed by holding φ and ρ fixed and varying θ. The intersection of a cone of fixed φ with a sphere of fixed ρ is a circle. As both ρ and φ are fixed, the circle of intersection lies in the plane z = ρ cos φ. It is a line of latitude. The circle has radius ρ sin φ and is centred on 0, 0, ρ cos φ . 40 The problem of finding a practical, reliable method for determining the longitude of a ship at sea was a very big deal for a period of several centuries. Among the scientists who worked in this were Galileo, Edmund Halley (of Halley’s comet) and Robert Hooke (of Hooke’s law). 344 MULTIPLE INTEGRALS 3.7 TRIPLE INTEGRALS IN SPHERICAL COORDINATES z y x ρ sphere of ﬁxed ρ z y x ϕ cone of ﬁxed ϕ z y x ϕ ρ circle of intersection When we introduced segments using surfaces of constant θ, the difference between the successive θ’s was dθ. Thus these four edge of the cube are circular arcs of radius essentially ρ sin φ that subtend an angle dθ, and so have length ρ sin φ dθ. z y x ρ sin ϕ ϕ ρ dθ ρ sin ϕ dθ 3.7.3 § § Sample Integrals in Spherical Coordinates Example 3.7.4 (Ice Cream Cone) Find the volume of the ice cream41 cone that consists of the part of the interior of the sphere x2 + y2 + z2 = a2 that is above the xy-plane and that is inside the cone x2 + y2 = b2z2. Here a and b are any two strictly positive constants. Solution. Note that, in spherical coordinates x2 + y2 = ρ2 sin2 φ z2 = ρ2 cos2 φ x2 + y2 + z2 = ρ2 Consequently, in spherical coordinates, the equation of the sphere is ρ = a, and the equa-tion of the cone is tan2 φ = b2. Let’s write β = arctan b, with 0 ă β ă π/2. Here is a sketch of the part of the ice cream cone in the first octant. The volume of the full ice cream cone will be four times the volume of the part in the first octant. 41 A very mathematical ice cream. Rocky-rho’d? Choculus? 345 MULTIPLE INTEGRALS 3.7 TRIPLE INTEGRALS IN SPHERICAL COORDINATES z y x β ϕ “ β ρ “ a We shall cut the first octant part of the ice cream cone into tiny pieces using spherical coordinates. That is, we shall cut it up using planes of constant θ, cones of constant φ, and spheres of constant ρ. • First slice the (the first octant part of the) ice cream cone into segments by inserting many planes of constant θ, with the various values of θ differing by dθ. The figure on the left below shows one segment outlined in red. Each segment – has θ essentially constant on the segment, and – has φ running from 0 to β and ρ running from 0 to a. – The leftmost segment has, essentially, θ = 0 and the rightmost segment has, essentially, θ = π/2. See the figure on the right below. z y x z y x • Concentrate on any one segment. A side view of the segment is sketched in the figure on the left below. Subdivide it into long thin searchlights by inserting many cones of constant φ, with the various values of φ differing by dφ. The figure on the left below shows one searchlight outlined in blue. Each searchlight – has θ and φ essentially constant on the searchlight, and – has ρ running over 0 ď ρ ď a. – The leftmost searchlight has, essentially, φ = 0 and the rightmost searchlight has, essentially, φ = β. See the figure on the right below. 346 MULTIPLE INTEGRALS 3.7 TRIPLE INTEGRALS IN SPHERICAL COORDINATES z ϕ “ β ϕ ϕdϕ ρ “ a z ϕ “ β ρ “ a • Concentrate on any one searchlight. Subdivide it into tiny approximate cubes by inserting many spheres of constant ρ, with the various values of ρ differing by dρ. The figure on the left below shows the side view of one approximate cube in black. Each cube – has ρ, θ and φ all essentially constant on the cube and – has volume ρ2 sin φ dρ dθ dφ, by (3.7.3). – The first cube has, essentially, ρ = 0 and the last cube has, essentially, ρ = a. See the figure on the right below. z ϕ “ β ρ “ a z ϕ “ β ρ “ a Now we can build up the volume. • Concentrate on one approximate cube. Let’s say that it contains the point with spher-ical coordinates ρ, θ, φ. The cube has volume essentially dV = ρ2 sin φ dρ dθ dφ, by (3.7.3). • To get the volume any one searchlight, say the searchlight whose φ coordinate runs from φ to φ + dφ, we just add up the volumes of the approximate cubes in that searchlight, by integrating ρ from its smallest value on the searchlight, namely 0, to its largest value on the searchlight, namely a. The volume of the searchlight is thus dθ dφ ż a 0 dρ ρ2 sin φ • To get the volume of any one segment, say the segment whose θ coordinate runs from θ to θ + dθ, we just add up the volumes of the searchlights in that segment, by integrating φ from its smallest value on the segment, namely 0, to its largest value on the segment, namely β. The volume of the segment is thus dθ ż β 0 dφ sin φ ż a 0 dρ ρ2 347 MULTIPLE INTEGRALS 3.7 TRIPLE INTEGRALS IN SPHERICAL COORDINATES • To get the volume of V1, the part of the ice cream cone in the first octant, we just add up the volumes of the segments that it contains, by integrating θ from its smallest value in the octant, namely 0, to its largest value on the octant, namely π/2. • The volume in the first octant is thus Volume(V1) = ż π/2 0 dθ ż β 0 dφ sin φ ż a 0 dρ ρ2 = a3 3 ż π/2 0 dθ ż β 0 dφ sin φ = a3 3 1 ´ cos β ż π/2 0 dθ = πa3 6 1 ´ cos β • So the volume of V, the total (four octant) ice cream cone, is Volume(V) = 4 Volume(V1) = 4πa3 6 1 ´ cos β We can express β (which was not given in the statement of the original problem) in terms of b (which was in the statement of the original problem), just by looking at the triangle β 1 b ? 1 b2 The right hand and bottom sides of the triangle have been chosen so that tan β = b, which was the definition of β. So cos β = 1 ? 1+b2 and the volume of the ice cream cone is Volume(V) = 2πa3 3 1 ´ 1 ? 1 + b2 Note that, as in Example 3.2.11, we can easily apply a couple of sanity checks to our answer. • If b = 0, so that the cone is just x2 + y2 = 0, which is the line x = y = 0, the total volume should be zero. Our answer does indeed give 0 in this case. • In the limit b Ñ 8, the angle β Ñ π/2 and the ice cream cone opens up into a hemisphere of radius a. Our answer does indeed give the volume of the hemisphere, which is 1 2 ˆ 4 3πa3. Example 3.7.4 348 MULTIPLE INTEGRALS 3.7 TRIPLE INTEGRALS IN SPHERICAL COORDINATES Example 3.7.5 (Cored Apple) A cylindrical hole of radius b is drilled symmetrically through a perfectly spherical apple of radius a ě b. Find the volume of apple that remains. Solution. In Example 3.2.11 we computed the volume removed, basically using cylindrical coordinates. So we could get the answer to this question just by subtracting the answer of Example 3.2.11 from 4 3πa3. Instead, we will evaluate the volume remaining as an exercise in setting up limits of integration when using spherical coordinates. As in Example 3.2.11, let’s use a coordinate system with the sphere centred on (0, 0, 0) and with the centre of the drill hole following the z-axis. Here is a sketch of the apple that remains in the first octant. It is outlined in red. By symmetry the total amount of apple remaining will be eight times the amount from the first octant. z y x b x2 y2 z2 “ a2 • First slice the first octant part of the remaining apple into segments by inserting many planes of constant θ, with the various values of θ differing by dθ. The leftmost segment has, essentially, θ = 0 and the rightmost segment has, essentially, θ = π/2. • Each segment, viewed from the side, looks like z ϕ “ arcsin b a a b ϕ Subdivide it into long thin searchlights by inserting many cones of constant φ, with the various values of φ differing by dφ. The figure on below shows one searchlight outlined in blue. Each searchlight 349 MULTIPLE INTEGRALS 3.7 TRIPLE INTEGRALS IN SPHERICAL COORDINATES – has θ and φ essentially constant on the searchlight. – The top searchlight has, essentially, φ = arcsin b a and the bottom searchlight has, essentially, φ = π/2. z ϕ “ arcsin b a ϕ ϕdϕ a b • Concentrate on any one searchlight. Subdivide it into tiny approximate cubes by inserting many spheres of constant ρ, with the various values of ρ differing by dρ. The figure on the left below shows the side view of one approximate cube in black. Each cube – has ρ, θ and φ all essentially constant on the cube and – has volume dV = ρ2 sin φ dρ dθ dφ, by (3.7.3). – The figure on the right below gives an expanded view of the searchlight. From it, we see (after a little trig) that the first cube has, essentially, ρ = b sin φ and the last cube has, essentially, ρ = a (the radius of the apple). z ϕ “ arcsin b a ϕ ϕdϕ a b z ϕ b ρ b Now we can build up the volume. • Concentrate on one approximate cube. Let’s say that it contains the point with spher-ical coordinates ρ, θ, φ. The cube has volume essentially dV = ρ2 sin φ dρ dθ dφ, by (3.7.3). • To get the volume any one searchlight, say the searchlight whose φ coordinate runs from φ to φ + dφ, we just add up the volumes of the approximate cubes in that searchlight, by integrating ρ from its smallest value on the searchlight, namely b sin φ, 350 MULTIPLE INTEGRALS 3.7 TRIPLE INTEGRALS IN SPHERICAL COORDINATES to its largest value on the searchlight, namely a. The volume of the searchlight is thus dθ dφ ż a b sin φ dρ ρ2 sin φ • To get the volume of any one segment, say the segment whose θ coordinate runs from θ to θ + dθ, we just add up the volumes of the searchlights in that segment, by integrating φ from its smallest value on the segment, namely arcsin b a, to its largest value on the segment, namely π 2 . The volume of the searchlight is thus dθ ż π 2 arcsin b a ż a b sin φ dρ ρ2 sin φ • To get the volume of the remaining part of the apple in the first octant, we just add up the volumes of the segments that it contains, by integrating θ from its smallest value in the octant, namely 0, to its largest value on the octant, namely π/2. The volume in the first octant is thus Volume(V1) = ż π/2 0 dθ ż π 2 arcsin b a dφ ż a b sin φ dρ ρ2 sin φ • Now we just have to integrate Volume(V1) = 1 3 ż π/2 0 dθ ż π 2 arcsin b a dφ sin φ a3 ´ b3 sin3 φ = 1 3 ż π/2 0 dθ ż π 2 arcsin b a dφ h a3 sin φ ´ b3 csc2 φ i = 1 3 ż π/2 0 dθ h ´a3 cos φ + b3 cot φ i π 2 arcsin b a since ż csc2 φ dφ = ´ cot φ + C = π 6 h ´a3 cos φ + b3 cot φ i π 2 arcsin b a Now cos π 2 = cot π 2 = 0 and, if we write α = arcsin b a, Volume(V1) = π 6 h a3 cos α ´ b3 cot α i From the triangle below, we have cos α = ? a2´b2 a and cot α = ? a2´b2 b . α ? a2 ´ b2 b a 351 MULTIPLE INTEGRALS 3.8 OPTIONAL— INTEGRALS IN GENERAL COORDINATES So Volume(V1) = π 6 h a2a a2 ´ b2 ´ b2a a2 ´ b2 i = π 6 a2 ´ b23/2 The full (eight octant) volume of the remaining apple is thus Volume(V) = 8Volume(V1) = 4 3π a2 ´ b23/2 We can, yet again, apply the sanity checks of Example 3.2.11 to our answer. • If the radius of the drill bit b = 0, no apple is removed at all. So the total volume remaining should be 4 3πa3. Our answer does indeed give this. • If the radius of the drill bit b = a, the radius of the apple, then the entire apple disappears. So the remaining apple should have volume 0. Again, our answer gives this. As a final check note that the sum of the answer to Example 3.2.11 and the answer to this Example is 4 3πa3, as it should be. Example 3.7.5 3.8Ĳ Optional— Integrals in General Coordinates One of the most important tools used in dealing with single variable integrals is the change of variable (substitution) rule x = f (u) dx = f 1(u) du Equation 3.8.1. See Theorems 1.4.2 and 1.4.6 in the CLP-2 text. Expressing multivariable integrals using polar or cylindrical or spherical coordinates are really multivariable substitutions. For example, switching to spherical coordinates amounts replacing the coordinates x, y, z with the coordinates ρ, θ, φ by using the substitution X = r(ρ, θ, φ) dx dy dz = ρ2 sin φ dρ dθ dφ where X = ⟨x , y , z⟩ and r(ρ, θ, φ) = ⟨ρ cos θ sin φ , ρ sin θ sin φ , ρ cos φ⟩ We’ll now derive a generalization of the substitution rule (3.8.1) to two dimensions. It will include polar coordinates as a special case. Later, we’ll state (without proof) its general-ization to three dimensions. It will include cylindrical and spherical coordinates as special cases. 352 MULTIPLE INTEGRALS 3.8 OPTIONAL— INTEGRALS IN GENERAL COORDINATES Suppose that we wish to integrate over a region, R, in R2 and that we also wish42 to use two new coordinates, that we’ll call u and v, in place of x and y. The new coordinates u, v are related to the old coordinates x, y, by the functions43 x = x(u, v) y = y(u, v) To make formulae more compact, we’ll define the vector-valued function r(u, v) by r(u, v) = ⟨x(u, v) , y(u, v)⟩ As an example, if the new coordinates are polar coordinates, with r renamed to u and θ renamed to v, then x(u, v) = u cos v and y = u sin v. Note that if we hold v fixed and vary u, then r(u, v) sweeps out a curve. For example, if x(u, v) = u cos v and y = u sin v, then, if we hold v fixed and vary u, r(u, v) sweeps out a straight line (that makes the angle v with the x-axis), while, if we hold u ą 0 fixed and vary v, r(u, v) sweeps out a circle (of radius u centred on the origin). rpu, vq v ﬁxed u varying x y v rpu, vq u ﬁxed v varying x y u We start by cutting R (the shaded region in the figure below) up into small pieces by drawing a bunch of curves of constant u (the blue curves in the figure below) and a bunch of curves of constant v (the red curves in the figure below). Concentrate on any one of the small pieces. Here is a greatly magnified sketch. 42 We’ll keep our third wish in reserve. 43 We are abusing notation a little here by using x and y both as coordinates and as functions. We could write x = f (u, v) and y = g(u, v), but it is easier to remember x = x(u, v) and y = y(u, v). 353 MULTIPLE INTEGRALS 3.8 OPTIONAL— INTEGRALS IN GENERAL COORDINATES P0 P1 P3 P2 u varying v“v0 u varying v“v0dv u“u0du v varying u“u0 v varying For example, the lower red curve was constructed by holding v fixed at the value v0, varying u and sketching r(u, v0), and the upper red curve was constructed by holding v fixed at the slightly larger value v0 + dv, varying u and sketching r(u, v0 + dv). So the four intersection points in the figure are P2 = r(u0, v0 + dv) P3 = r(u0 + du, v0 + dv) P0 = r(u0, v0) P1 = r(u0 + du, v0) Now, for any small constants dU and dV, we have the linear approximation44 r(u0 + dU, v0 + dV) « r(u0 , v0) + Br Bu(u0 , v0) dU + Br Bv(u0 , v0) dV Applying this three times, once with dU = du, dV = 0 (to approximate P1), once with dU = 0, dV = dv (to approximate P2), and once with dU = du, dV = dv (to approximate P3), P0 = r(u0 , v0) P1 = r(u0 + du, v0) « r(u0 , v0) + Br Bu(u0 , v0) du P2 = r(u0, v0 + dv) « r(u0 , v0) + Br Bv(u0 , v0) dv P3 = r(u0 + du, v0 + dv) « r(u0 , v0) + Br Bu(u0 , v0) du + Br Bv(u0 , v0) dv We have dropped all Taylor expansion terms that are of degree two or higher in du, dv. The reason is that, in defining the integral, we take the limit du, dv Ñ 0. Because of that limit, all of the dropped terms contribute exactly 0 to the integral. We shall not prove this. But we shall show, in the optional §3.8.1, why this is the case. The small piece of R surface with corners P0, P1, P2, P3 is approximately a parallelo-gram with sides Ý Ý Ñ P0P1 « Ý Ý Ñ P2P3 « Br Bu(u0 , v0) du = Bx Bu(u0, v0) , By Bu(u0, v0) du Ý Ý Ñ P0P2 « Ý Ý Ñ P1P3 « Br Bv(u0 , v0) dv = Bx Bv(u0, v0) , By Bv(u0, v0) dv P0 P1 P3 P2 θ Ý Ý Ý Ñ P0P1 Ý Ý Ý Ñ P0P2 44 Recall (2.6.1). 354 MULTIPLE INTEGRALS 3.8 OPTIONAL— INTEGRALS IN GENERAL COORDINATES Here the notation, for example, Ý Ý Ñ P0P1 refers to the vector whose tail is at the point P0 and whose head is at the point P1. Recall, from (1.2.17) that area of parallelogram with sides ⟨a, b⟩and ⟨c, d⟩= ˇ ˇ ˇ ˇdet a b c d ˇ ˇ ˇ ˇ = ˇ ˇad ´ bc ˇ ˇ So the area of our small piece of R is essentially dA = ˇ ˇ ˇ ˇ ˇdet " Bx Bu By Bu Bx Bv By Bv #ˇ ˇ ˇ ˇ ˇ du dv Equation 3.8.2. Recall that det M denotes the determinant of the matrix M. Also recall that we don’t really need determinants for this text, though it does make for nice compact notation. The formula (3.8.2) is the heart of the following theorem, which tells us how to translate an integral in one coordinate system into an integral in another coordinate system. Let the functions x(u, v) and y(u, v) have continuous first partial derivatives and let the function f (x, y) be continuous. Assume that x = x(u, v), y = y(u, v) provides a one-to-one correspondence between the points (u, v) of the region U in the uv-plane and the points (x, y) of the region R in the xy-plane. Then ĳ R f (x, y) dx dy = ĳ U f x(u, v) , y(u, v) ˇ ˇ ˇ ˇ ˇdet " Bx Bu(u, v) By Bu(u, v) Bx Bv(u, v) By Bv(u, v) #ˇ ˇ ˇ ˇ ˇ du dv Theorem 3.8.3. The determinant det " Bx Bu(u, v) By Bu(u, v) Bx Bv(u, v) By Bv(u, v) # that appears in (3.8.2) and Theorem 3.8.3 is known as the Jacobian45. Example 3.8.4 (dA for x Ø y) We’ll start with a pretty trivial example in which we simply rename x to Y and y to X. That is x(X, Y) = Y y(X, Y) = X 45 It is not named after the Jacobin Club, a political movement of the French revolution. It is not named after the Jacobite rebellions that took place in Great Britain and Ireland between 1688 and 1746. It is not named after the Jacobean era of English and Scottish history. It is named after the German mathematician Carl Gustav Jacob Jacobi (1804–1851). He died from smallpox. 355 MULTIPLE INTEGRALS 3.8 OPTIONAL— INTEGRALS IN GENERAL COORDINATES Since Bx BX = 0 By BX = 1 Bx BY = 1 By BY = 0 (3.8.2), but with u renamed to X and v renamed to Y, gives dA = ˇ ˇ ˇ ˇ ˇdet " 0 1 1 0 #ˇ ˇ ˇ ˇ ˇ dX dY = dX dY which should really not be a shock. Example 3.8.4 Example 3.8.5 (dA for Polar Coordinates) Polar coordinates have x(r, θ) = r cos θ y(r, θ) = r sin θ Since Bx Br = cos θ By Br = sin θ Bx Bθ = ´r sin θ By Bθ = r cos θ (3.8.2), but with u renamed to r and v renamed to θ, gives dA = ˇ ˇ ˇ ˇ ˇdet " cos θ sin θ ´r sin θ r cos θ #ˇ ˇ ˇ ˇ ˇ drdθ = r cos2 θ + r sin2 θ drdθ = r dr dθ which is exactly what we found in (3.2.5). Example 3.8.5 Example 3.8.6 (dA for Parabolic Coordinates) Parabolic46 coordinates are defined by x(u, v) = u2 ´ v2 2 y(u, v) = uv 46 The name comes from the fact that both the curves of constant u and the curves of constant v are parabolas. 356 MULTIPLE INTEGRALS 3.8 OPTIONAL— INTEGRALS IN GENERAL COORDINATES Since Bx Bu = u By Bu = v Bx Bv = ´v By Bv = u (3.8.2) gives dA = ˇ ˇ ˇ ˇ ˇdet " u v ´v u #ˇ ˇ ˇ ˇ ˇ dudv = (u2 + v2) du dv Example 3.8.6 In practice applying the change of variables Theorem 3.8.3 can be quite tricky. Here is just one simple (and rigged) example. Example 3.8.7 Evaluate ĳ R y 1 + x dx dy where R = ␣(x, y) ˇ ˇ 0 ď x ď 1, 1 + x ď y ď 2 + 2x ( Solution. We can simplify the integrand considerably by making the change of variables s = x x = s t = y 1 + x y = t(1 + x) = t(1 + s) Of course to evaluate the given integral by applying Theorem 3.8.3 we also need to know ˝ the domain of integration in terms of s and t and ˝ dx dy in terms of ds dt. By (3.8.2), recalling that x(s, t) = s and y(s, t) = t(1 + s), dx dy = ˇ ˇ ˇ ˇ ˇdet "Bx Bs By Bs Bx Bt By Bt #ˇ ˇ ˇ ˇ ˇ ds dt = ˇ ˇ ˇ ˇ ˇdet " 1 t 0 1 + s #ˇ ˇ ˇ ˇ ˇ ds dt = (1 + s) ds dt To determine what the change of variables does to the domain of integration, we’ll sketch R and then reexpress the boundary of R in terms of the new coordinates s and t. Here is the sketch of R in the original coordinates (x, y). 357 MULTIPLE INTEGRALS 3.8 OPTIONAL— INTEGRALS IN GENERAL COORDINATES x y y“2p1xq y“1x x “ 1 x “ 0 R The region R is a quadrilateral. It has four sides. • The left side is part of the line x = 0. Recall that x = s. So, in terms of s and t, this line is s = 0. • The right side is part of the line x = 1. In terms of s and t, this line is s = 1. • The bottom side is part of the line y = 1 + x, or y 1+x = 1. Recall that t = y 1+x. So, in terms of s and t, this line is t = 1. • The top side is part of the line y = 2(1 + x), or y 1+x = 2. In terms of s and t, this line is t = 2. Here is another copy of the sketch of R. But this time the equations of its four sides are expressed in terms of s and t. x y t “ 2 t “ 1 s “ 1 s “ 0 R So, expressed in terms of s and t, the domain of integration R is much simpler: ␣(s, t) ˇ ˇ 0 ď s ď 1, 1 ď t ď 2 ( As dx dy = (1 + s) ds dt and the integrand y 1+x = t, the integral is, by Theorem 3.8.3, ĳ R y 1 + x dx dy = ż 1 0 ds ż 2 1 dt (1 + s)t = ż 1 0 ds (1 + s) t2 2 2 1 = 3 2 s + s2 2 1 0 = 3 2 ˆ 3 2 = 9 4 358 MULTIPLE INTEGRALS 3.8 OPTIONAL— INTEGRALS IN GENERAL COORDINATES Example 3.8.7 There are natural generalizations of (3.8.2) and Theorem 3.8.3 to three (and also to higher) dimensions, that are derived in precisely the same way as (3.8.2) was derived. The derivation is based on the fact, discussed in the optional Section 1.2.4, that the volume of the parallelepiped (three dimensional parallelogram) determined by the three vectors a b c a = ⟨a1, a2, a3⟩, b = ⟨b1, b2, b3⟩and c = ⟨c1, c2, c3⟩is given by the formula volume of parallelepiped with edges a, b, c = ˇ ˇ ˇ ˇ ˇ ˇ det   a1 a2 a3 b1 b2 b3 c1 c2 c3   ˇ ˇ ˇ ˇ ˇ ˇ where the determinant of a 3ˆ3 matrix can be defined in terms of some 2ˆ2 determinants by det   a1 a2 a3 b1 b2 b3 c1 c2 c3  = a1 det   a1 a2 a3 b1 b2 b3 c1 c2 c3  −a2 det   a1 a2 a3 b1 b2 b3 c1 c2 c3  + a3 det   a1 a2 a3 b1 b2 b3 c1 c2 c3   = a1 (b2c3 −b3c2) −a2 (b1c3 −b3c1) + a3 (b1c2 −b2c1) If we use x = x(u, v, w) y = y(u, v, w) z = z(u, v, w) to change from old coordinates x, y, z to new coordinates u, v, w, then dV = ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ det     Bx Bu By Bu Bz Bu Bx Bv By Bv Bz Bv Bx Bw By Bw Bz Bw     ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ du dv dw Equation 3.8.8. 359 MULTIPLE INTEGRALS 3.8 OPTIONAL— INTEGRALS IN GENERAL COORDINATES Example 3.8.9 (dV for Cylindrical Coordinates) Cylindrical coordinates have x(r, θ, z) = r cos θ y(r, θ, z) = r sin θ z(r, θ, z) = z Since Bx Br = cos θ By Br = sin θ Bz Br = 0 Bx Bθ = ´r sin θ By Bθ = r cos θ Bz Bθ = 0 Bx Bz = 0 By Bz = 0 Bz Bz = 1 (3.8.8), but with u renamed to r and v renamed to θ, gives dV = ˇ ˇ ˇ ˇ ˇ ˇ ˇ det    cos θ sin θ 0 ´r sin θ r cos θ 0 0 0 1    ˇ ˇ ˇ ˇ ˇ ˇ ˇ dr dθ dz = ˇ ˇ ˇ ˇ ˇcos θ det " r cos θ 0 0 1 # ´ sin θ det " ´r sin θ 0 0 1 # + 0 det " ´r sin θ r cos θ 0 0 #ˇ ˇ ˇ ˇ ˇ dr dθ dz = r cos2 θ + r sin2 θ dr dθ dz = r dr dθ dz which is exactly what we found in (3.6.3). Example 3.8.9 Example 3.8.10 (dV for Spherical Coordinates) Spherical coordinates have x(ρ, θ, φ) = ρ cos θ sin φ y(ρ, θ, φ) = ρ sin θ sin φ z(ρ, θ, φ) = ρ cos φ Since Bx Bρ = cos θ sin φ By Bρ = sin θ sin φ Bz Bρ = cos φ Bx Bθ = ´ρ sin θ sin φ By Bθ = ρ cos θ sin φ Bz Bθ = 0 Bx Bφ = ρ cos θ cos φ By Bφ = ρ sin θ cos φ Bz Bφ = ´ρ sin φ 360 MULTIPLE INTEGRALS 3.8 OPTIONAL— INTEGRALS IN GENERAL COORDINATES (3.8.8), but with u renamed to ρ, v renamed to θ and w renamed to φ, gives dV = ˇ ˇ ˇ ˇ ˇ ˇ ˇ det    cos θ sin φ sin θ sin φ cos φ ´ρ sin θ sin φ ρ cos θ sin φ 0 ρ cos θ cos φ ρ sin θ cos φ ´ρ sin φ    ˇ ˇ ˇ ˇ ˇ ˇ ˇ dρ dθ dφ = ˇ ˇ ˇ ˇ ˇcos θ sin φ det " ρ cos θ sin φ 0 ρ sin θ cos φ ´ρ sin φ # ´ sin θ sin φ det " ´ρ sin θ sin φ 0 ρ cos θ cos φ ´ρ sin φ # + cos φ det " ´ρ sin θ sin φ ρ cos θ sin φ ρ cos θ cos φ ρ sin θ cos φ #ˇ ˇ ˇ ˇ ˇ dρ dθ dφ = ρ2ˇ ˇ ´ cos2 θ sin3 φ ´ sin2 θ sin3 φ ´ sin φ cos2 φ ˇ ˇ dρ dθ dφ = ρ2ˇ ˇ ´ sin φ sin2 φ ´ sin φ cos2 φ ˇ ˇ dρ dθ dφ = ρ2 sin φ dρ dθ dφ which is exactly what we found in (3.7.3). Example 3.8.10 3.8.1 § § Optional — Dropping Higher Order Terms in du, dv In the course of deriving (3.8.2), that is, the dA formula for P0 P1 P3 P2 u varying v“v0 u varying v“v0dv u“u0du v varying u“u0 v varying we approximated, for example, the vectors Ý Ý Ñ P0P1 = r(u0 + du, v0) ´ r(u0 , v0)= Br Bu(u0 , v0) du + E1 « Br Bu(u0 , v0) du Ý Ý Ñ P0P2 = r(u0, v0 + dv) ´ r(u0 , v0) = Br Bv(u0 , v0) dv + E2 « Br Bv(u0 , v0) dv where E1 is bounded47 by a constant times (du)2 and E2 is bounded by a constant times (dv)2. That is, we assumed that we could just ignore the errors and drop E1 and E2 by setting them to zero. 47 Remember the error in the Taylor polynomial approximations. See (2.6.13) and (2.6.14). 361 MULTIPLE INTEGRALS 3.8 OPTIONAL— INTEGRALS IN GENERAL COORDINATES So we approximated ˇ ˇÝ Ý Ñ P0P1 ˆ Ý Ý Ñ P0P2 ˇ ˇ = ˇ ˇ ˇ h Br Bu(u0 , v0) du + E1 i ˆ hBr Bv(u0 , v0) dv + E2 iˇ ˇ ˇ = ˇ ˇ ˇBr Bu(u0 , v0) du ˆ Br Bv(u0 , v0) dv + E3 ˇ ˇ ˇ « ˇ ˇ ˇBr Bu(u0 , v0) du ˆ Br Bv(u0 , v0) dv ˇ ˇ ˇ where the length of the vector E3 is bounded by a constant times (du)2 dv + du (dv)2. We’ll now see why dropping terms like E3 does not change the value of the integral at all48. Suppose that our domain of integration consists of all (u, v)’s in a rectangle of width W and height H, as in the figure below. u v du W dv H Subdivide the rectangle into a grid of n ˆ n small subrectangles by drawing lines of con-stant v (the red lines in the figure) and lines of constant u (the blue lines in the figure). Each subrectangle has width du = W n and height dv = H n . Now suppose that in setting up the integral we make, for each subrectangle, an error that is bounded by some constant times (du)2 dv + du (dv)2 = W n 2 H n + W n H n 2 = WH(W + H) n3 Because there are a total of n2 subrectangles, the total error that we have introduced, for all of these subrectangles, is no larger than a constant times n2 ˆ WH(W + H) n3 = WH(W + H) n When we define our integral by taking the limit n Ñ 0 of the Riemann sums, this error converges to exactly 0. As a consequence, it was safe for us to ignore the error terms when we established the change of variables formulae. 48 See the optional §1.1.6 of the CLP-2 text for an analogous argument concerning Riemann sums. 362 TRIGONOMETRY Appendix A A.1Ĳ Trigonometry — Graphs sin θ cos θ tan θ ´π ´ π 2 π 2 π 3π 2 2π ´1 1 ´π ´ π 2 π 2 π 3π 2 2π ´1 1 ´π ´ π 2 π 2 π 3π 2 2π A.2Ĳ Trigonometry — Special Triangles 363 TRIGONOMETRY A.3 TRIGONOMETRY — SIMPLE IDENTITIES From the above pair of special triangles we have sin π 4 = 1 ? 2 sin π 6 = 1 2 sin π 3 = ? 3 2 cos π 4 = 1 ? 2 cos π 6 = ? 3 2 cos π 3 = 1 2 tan π 4 = 1 tan π 6 = 1 ? 3 tan π 3 = ? 3 A.3Ĳ Trigonometry — Simple Identities • Periodicity sin(θ + 2π) = sin(θ) cos(θ + 2π) = cos(θ) • Reflection sin(´θ) = ´ sin(θ) cos(´θ) = cos(θ) • Reflection around π/4 sin π 2 ´ θ = cos θ cos π 2 ´ θ = sin θ • Reflection around π/2 sin (π ´ θ) = sin θ cos (π ´ θ) = ´ cos θ • Rotation by π sin (θ + π) = ´ sin θ cos (θ + π) = ´ cos θ • Pythagoras sin2 θ + cos2 θ = 1 tan2 θ + 1 = sec2 θ 1 + cot2 θ = csc2 θ • sin and cos building blocks tan θ = sin θ cos θ csc θ = 1 sin θ sec θ = 1 cos θ cot θ = cos θ sin θ = 1 tan θ 364 TRIGONOMETRY A.4 TRIGONOMETRY — ADD AND SUBTRACT ANGLES A.4Ĳ Trigonometry — Add and Subtract Angles • Sine sin(α ˘ β) = sin(α) cos(β) ˘ cos(α) sin(β) • Cosine cos(α ˘ β) = cos(α) cos(β) ¯ sin(α) sin(β) • Tangent tan(α + β) = tan α + tan β 1 ´ tan α tan β tan(α ´ β) = tan α ´ tan β 1 + tan α tan β • Double angle sin(2θ) = 2 sin(θ) cos(θ) cos(2θ) = cos2(θ) ´ sin2(θ) = 2 cos2(θ) ´ 1 = 1 ´ 2 sin2(θ) tan(2θ) = 2 tan(θ) 1 ´ tan2 θ cos2 θ = 1 + cos(2θ) 2 sin2 θ = 1 ´ cos(2θ) 2 tan2 θ = 1 ´ cos(2θ) 1 + cos(2θ) • Products to sums sin(α) cos(β) = sin(α + β) + sin(α ´ β) 2 sin(α) sin(β) = cos(α ´ β) ´ cos(α + β) 2 cos(α) cos(β) = cos(α ´ β) + cos(α + β) 2 • Sums to products sin α + sin β = 2 sin α + β 2 cos α ´ β 2 sin α ´ sin β = 2 cos α + β 2 sin α ´ β 2 cos α + cos β = 2 cos α + β 2 cos α ´ β 2 cos α ´ cos β = ´2 sin α + β 2 sin α ´ β 2 365 TRIGONOMETRY A.5 INVERSE TRIGONOMETRIC FUNCTIONS A.5Ĳ Inverse Trigonometric Functions arcsin x arccos x arctan x Domain: ´1 ď x ď 1 Domain: ´1 ď x ď 1 Domain: all real numbers Range: ´π 2 ď arcsin x ď π 2 Range: 0 ď arccos x ď π Range: ´π 2 ă arctan x ă π 2 ´1 1 ´π/2 π/2 ´1 1 π/2 π ´ π 2 π 2 Since these functions are inverses of each other we have arcsin(sin θ) = θ ´π 2 ď θ ď π 2 arccos(cos θ) = θ 0 ď θ ď π arctan(tan θ) = θ ´π 2 ă θ ă π 2 and also sin(arcsin x) = x ´1 ď x ď 1 cos(arccos x) = x ´1 ď x ď 1 tan(arctan x) = x any real x arccsc x arcsec x arccot x Domain: \|x\| ě 1 Domain: \|x\| ě 1 Domain: all real numbers Range: ´π 2 ď arccsc x ď π 2 Range: 0 ď arcsec x ď π Range: 0 ă arccot x ă π arccsc x ‰ 0 arcsec x ‰ π 2 ´1 1 ´ π 2 π 2 ´1 1 π 2 π π 2 π 366 TRIGONOMETRY A.5 INVERSE TRIGONOMETRIC FUNCTIONS Again arccsc(csc θ) = θ ´π 2 ď θ ď π 2 , θ ‰ 0 arcsec(sec θ) = θ 0 ď θ ď π, θ ‰ π 2 arccot(cot θ) = θ 0 ă θ ă π and csc(arccsc x) = x \|x\| ě 1 sec(arcsec x) = x \|x\| ě 1 cot(arccot x) = x any real x 367 POWERS AND LOGARITHMS Appendix B B.1Ĳ Powers In the following, x and y are arbitrary real numbers, q is an arbitrary constant that is strictly bigger than zero and e is 2.7182818284, to ten decimal places. • e0 = 1, q0 = 1 • ex+y = exey, ex´y = ex ey , qx+y = qxqy, qx´y = qx qy • e´x = 1 ex , q´x = 1 qx • exy = exy, qxy = qxy • d dxex = ex, d dxeg(x) = g1(x)eg(x), d dxqx = (ln q) qx • ş ex dx = ex + C, ş eax dx = 1 aeax + C if a ‰ 0 • ex = 8 ř n=0 xn n! • lim xÑ8 ex = 8, lim xÑ´8 ex = 0 lim xÑ8 qx = 8, lim xÑ´8 qx = 0 if q ą 1 lim xÑ8 qx = 0, lim xÑ´8 qx = 8 if 0 ă q ă 1 • The graph of 2x is given below. The graph of qx, for any q ą 1, is similar. 368 POWERS AND LOGARITHMS B.2 LOGARITHMS x y 1 2 3 −1 −2 −3 1 2 4 6 y = 2x B.2Ĳ Logarithms In the following, x and y are arbitrary real numbers that are strictly bigger than 0 (except where otherwise specified), p and q are arbitrary constants that are strictly bigger than one, and e is 2.7182818284, to ten decimal places. The notation ln x means loge x. Some people use log x to mean log10 x, others use it to mean loge x and still others use it to mean log2 x. • eln x = x, qlogq x = x • ln ex = x, logq qx = x for all ´8 ă x ă 8 • logq x = ln x ln q, ln x = logp x logp e , logq x = logp x logp q • ln 1 = 0, ln e = 1 logq 1 = 0, logq q = 1 • ln(xy) = ln x + ln y, logq(xy) = logq x + logq y • ln x y = ln x ´ ln y, logq x y = logq x ´ logq y • ln 1 y = ´ ln y, logq 1 y = ´ logq y • ln(xy) = y ln x, logq(xy) = y logq x • d dx ln x = 1 x, d dx logq x = 1 x ln q • ş ln x dx = x ln x ´ x + C, ş logq x dx = x logq x ´ x ln q + C, • lim xÑ8 ln x = 8, lim xÑ0+ ln x = ´8 lim xÑ8 logq x = 8, lim xÑ0+ logq x = ´8 • The graph of log10 x is given below. The graph of logq x, for any q ą 1, is similar. 369 POWERS AND LOGARITHMS B.2 LOGARITHMS x y 1 5 10 15 0.5 1.0 −0.5 −1.0 y = log10 x 370 TABLE OF DERIVATIVES Appendix C Throughout this table, a and b are constants, independent of x. F(x) F1(x) = dF dx a f (x) + bg(x) a f 1(x) + bg1(x) f (x) + g(x) f 1(x) + g1(x) f (x) ´ g(x) f 1(x) ´ g1(x) a f (x) a f 1(x) f (x)g(x) f 1(x)g(x) + f (x)g1(x) f (x)g(x)h(x) f 1(x)g(x)h(x) + f (x)g1(x)h(x) + f (x)g(x)h1(x) f (x) g(x) f 1(x)g(x)´f (x)g1(x) g(x)2 1 g(x) ´ g1(x) g(x)2 f g(x) f 1g(x) g1(x) 371 TABLE OF DERIVATIVES F(x) F1(x) = dF dx a 0 xa axa´1 g(x)a ag(x)a´1g1(x) sin x cos x sin g(x) g1(x) cos g(x) cos x ´ sin x cos g(x) ´g1(x) sin g(x) tan x sec2 x csc x ´ csc x cot x sec x sec x tan x cot x ´ csc2 x ex ex eg(x) g1(x)eg(x) ax (ln a) ax F(x) F1(x) = dF dx ln x 1 x ln g(x) g1(x) g(x) loga x 1 x ln a arcsin x 1 ? 1´x2 arcsin g(x) g1(x) ? 1´g(x)2 arccos x ´ 1 ? 1´x2 arctan x 1 1+x2 arctan g(x) g1(x) 1+g(x)2 arccsc x ´ 1 \|x\|? x2´1 arcsec x 1 \|x\|? x2´1 arccot x ´ 1 1+x2 372 TABLE OF INTEGRALS Appendix D Throughout this table, a and b are given constants, independent of x and C is an arbitrary constant. f (x) F(x) = ş f (x) dx a f (x) + bg(x) a ş f (x) dx + b ş g(x) dx + C f (x) + g(x) ş f (x) dx + ş g(x) dx + C f (x) ´ g(x) ş f (x) dx ´ ş g(x) dx + C a f (x) a ş f (x) dx + C u(x)v1(x) u(x)v(x) ´ ş u1(x)v(x) dx + C f y(x) y1(x) F y(x) where F(y) = ş f (y) dy a ax + C xa xa+1 a+1 + C if a ‰ ´1 1 x ln \|x\| + C g(x)ag1(x) g(x)a+1 a+1 + C if a ‰ ´1 373 TABLE OF INTEGRALS f (x) F(x) = ş f (x) dx sin x ´ cos x + C g1(x) sin g(x) ´ cos g(x) + C cos x sin x + C tan x ln \| sec x\| + C csc x ln \| csc x ´ cot x\| + C sec x ln \| sec x + tan x\| + C cot x ln \| sin x\| + C sec2 x tan x + C csc2 x ´ cot x + C sec x tan x sec x + C csc x cot x ´ csc x + C f (x) F(x) = ş f (x) dx ex ex + C eg(x)g1(x) eg(x) + C eax 1 a eax + C ax 1 ln a ax + C ln x x ln x ´ x + C 1 ? 1´x2 arcsin x + C g1(x) ? 1´g(x)2 arcsin g(x) + C 1 ? a2´x2 arcsin x a + C 1 1+x2 arctan x + C g1(x) 1+g(x)2 arctan g(x) + C 1 a2+x2 1 a arctan x a + C 1 x? x2´1 arcsec x + C (x ą 1) 374 TABLE OF TAYLOR EXPANSIONS Appendix E Let n ě be an integer. Then if the function f has n + 1 derivatives on an interval that contains both x0 and x, we have the Taylor expansion f (x) = f (x0) + f 1(x0) (x ´ x0) + 1 2! f 2(x0) (x ´ x0)2 + ¨ ¨ ¨ + 1 n! f (n)(x0) (x ´ x0)n + 1 (n + 1)! f (n+1)(c) (x ´ x0)n+1 for some c between x0 and x The limit as n Ñ 8 gives the Taylor series f (x) = 8 ÿ n=0 f (n)(x0) n! (x ´ x0)n for f. When x0 = 0 this is also called the Maclaurin series for f. Here are Taylor series expansions of some important functions. ex = 8 ÿ n=0 1 n!xn for ´8 ă x ă 8 = 1 + x + 1 2x2 + 1 3!x3 + ¨ ¨ ¨ + 1 n!xn + ¨ ¨ ¨ sin x = 8 ÿ n=0 (´1)n (2n + 1)!x2n+1 for ´8 ă x ă 8 = x ´ 1 3!x3 + 1 5!x5 ´ ¨ ¨ ¨ + (´1)n (2n + 1)!x2n+1 + ¨ ¨ ¨ cos x = 8 ÿ n=0 (´1)n (2n)! x2n for ´8 ă x ă 8 = 1 ´ 1 2!x2 + 1 4!x4 ´ ¨ ¨ ¨ + (´1)n (2n)! x2n + ¨ ¨ ¨ 1 1 ´ x = 8 ÿ n=0 xn for ´1 ď x ă 1 = 1 + x + x2 + x3 + ¨ ¨ ¨ + xn + ¨ ¨ ¨ 375 TABLE OF TAYLOR EXPANSIONS 1 1 + x = 8 ÿ n=0 (´1)nxn for ´1 ă x ď 1 = 1 ´ x + x2 ´ x3 + ¨ ¨ ¨ + (´1)nxn + ¨ ¨ ¨ ln(1 ´ x) = ´ 8 ÿ n=1 1 nxn for ´1 ď x ă 1 = ´x ´ 1 2x2 ´ 1 3x3 ´ ¨ ¨ ¨ ´ 1 nxn ´ ¨ ¨ ¨ ln(1 + x) = ´ 8 ÿ n=1 (´1)n n xn for ´1 ă x ď 1 = x ´ 1 2x2 + 1 3x3 ´ ¨ ¨ ¨ ´ (´1)n n xn ´ ¨ ¨ ¨ (1 + x)p = 1 + px + p(p ´ 1) 2 x2 + p(p ´ 1)(p ´ 2) 3! x3 + ¨ ¨ ¨ + p(p ´ 1)(p ´ 2) ¨ ¨ ¨ (p ´ n + 1) n! xn + ¨ ¨ ¨ 376 3D COORDINATE SYSTEMS Appendix F F.1Ĳ Cartesian Coordinates Here is a figure showing the definitions of the three Cartesian coordinates (x, y, z) y z x px, y, zq px, y, 0q x y z and here are three figures showing a surface of constant x, a surface of constant x, and a surface of constant z. y z x px, 0, 0q surface of constant x (a plane) y z x p0, y, 0q surface of constant y (a plane) y z x p0, 0, zq surface of constant z (a plane) Finally here is a figure showing the volume element dV in cartesian coordinates. 377 3D COORDINATE SYSTEMS F.2 CYLINDRICAL COORDINATES dx dy dz volume element dV “ dx dy dz F.2Ĳ Cylindrical Coordinates Here is a figure showing the definitions of the three cylindrical coordinates r = distance from (0, 0, 0) to (x, y, 0) θ = angle between the x axis and the line joining (x, y, 0) to (0, 0, 0) z = signed distance from (x, y, z) to the xy–plane px, y, zq px, y, 0q y z x z r θ The cartesian and cylindrical coordinates are related by x = r cos θ y = r sin θ z = z r = b x2 + y2 θ = arctan y x z = z Here are three figures showing a surface of constant r, a surface of constant θ, and a surface of constant z. y z x r surface of constant r (a cylindrical shell) y z x θ surface of constant θ (a plane) y z x p0, 0, zq surface of constant z (a plane) 378 3D COORDINATE SYSTEMS F.3 SPHERICAL COORDINATES Finally here is a figure showing the volume element dV in cylindrical coordinates. r dr dz r dθ volume element dV “ r dr dθ dz F.3Ĳ Spherical Coordinates Here is a figure showing the definitions of the three spherical coordinates ρ = distance from (0, 0, 0) to (x, y, z) φ = angle between the z axis and the line joining (x, y, z) to (0, 0, 0) θ = angle between the x axis and the line joining (x, y, 0) to (0, 0, 0) z y x p0, 0, zq px, y, 0q px, 0, 0q θ ϕ px, y, zq ρ ρ sin ϕ ρ cos ϕ ρ sin ϕ sin θ and here are two more figures giving the side and top views of the previous figure. z p0, 0, zq ρ cos ϕ px, y, 0q side view ρ sin ϕ ϕ ρ px, y, zq y x px, 0, 0q top view ρ sin ϕ sin θ ρ sin ϕ cos θ ρ sin ϕ θ px, y, 0q The cartesian and spherical coordinates are related by x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ ρ = b x2 + y2 + z2 θ = arctan y x φ = arctan a x2 + y2 z 379 3D COORDINATE SYSTEMS F.3 SPHERICAL COORDINATES Here are three figures showing a surface of constant ρ, a surface of constant θ, and a surface of constant φ. z y x ρ surface of constant ρ (a sphere) z y x θ surface of constant θ (a plane) z y x ϕ surface of constant ϕ (a cone) Finally, here is a figure showing the volume element dV in spherical coordinates z y x dρ ρ dϕ ρ sin ϕ dθ volume element dV “ ρ2 sin ϕ dρ dθ dϕ and two extracts of the above figure to make it easier to see how ρ dφ and ρ sin φ dθ arise. 380 3D COORDINATE SYSTEMS F.3 SPHERICAL COORDINATES z y x ρ dϕ ϕ dϕ dρ z y x ρ sin ϕ ϕ ρ dθ ρ sin ϕ dθ 381 ISO COORDINATE SYSTEM NOTATION Appendix G In this text we have chosen symbols for the various polar, cylindrical and spherical coor-dinates that are standard for mathematics. There is another, different, set of symbols that are commonly used in the physical sciences and engineering. Indeed, there is an interna-tional convention, called ISO 80000-2, that specifies those symbols1. In this appendix, we summarize the definitions and standard properties of the polar, cylindrical and spherical coordinate systems using the ISO symbols. G.1Ĳ Polar Coordinates In the ISO convention the symbols ρ and ϕ are used (instead of r and θ) for polar coordi-nates. ρ = the distance from (0, 0) to (x, y) ϕ = the (counter-clockwise) angle between the x axis and the line joining (x, y) to (0, 0) ρ px, yq x y φ Cartesian and polar coordinates are related by x = ρ cos ϕ y = ρ sin ϕ ρ = b x2 + y2 ϕ = arctan y x 1 It specifies more than just those symbols. See and The full ISO 80000-2 is available at https: //www.iso.org/standard/64973.html — for $$. 382 ISO COORDINATE SYSTEM NOTATION G.1 POLAR COORDINATES The following two figures show a number of lines of constant ϕ, on the left, and curves of constant ρ, on the right. x y lines of constant φ x y curves of constant ρ Note that the polar angle ϕ is only defined up to integer multiples of 2π. For example, the point (1, 0) on the x-axis could have ϕ = 0, but could also have ϕ = 2π or ϕ = 4π. It is sometimes convenient to assign ϕ negative values. When ϕ ă 0, the counter-clockwise angle ϕ refers to the clockwise angle \|ϕ\|. For example, the point (0, ´1) on the negative y-axis can have ϕ = ´π 2 and can also have ϕ = 3π 2 . x y ρ “ 1, φ “ ´π{2, φ “ 3π{2 3π{2 π{2 It is also sometimes convenient to extend the above definitions by saying that x = ρ cos ϕ and y = ρ sin ϕ even when ρ is negative. For example, the following figure shows (x, y) for ρ = 1, ϕ = π/4 and for ρ = ´1, ϕ = π/4. Both points lie on the line through ρ “ 1, φ “ π{4 ρ “ ´1, φ “ π{4 x y π{4 the origin that makes an angle of 45˝ with the x-axis and both are a distance one from the origin. But they are on opposite sides of the origin. The area element in polar coordinates is dA = ρ dρ dϕ 383 ISO COORDINATE SYSTEM NOTATION G.2 CYLINDRICAL COORDINATES dφ dρ ρ ρ dφ G.2Ĳ Cylindrical Coordinates In the ISO convention the symbols ρ, ϕ and z are used (instead of r, θ and z) for cylindrical coordinates. ρ = distance from (0, 0, 0) to (x, y, 0) ϕ = angle between the x axis and the line joining (x, y, 0) to (0, 0, 0) z = signed distance from (x, y, z) to the xy-plane px, y, zq px, y, 0q y z x z ρ φ The cartesian and cylindrical coordinates are related by x = ρ cos ϕ y = ρ sin ϕ z = z ρ = b x2 + y2 ϕ = arctan y x z = z Here are three figures showing a surface of constant ρ, a surface of constant ϕ, and a surface of constant z. y z x ρ surface of constant ρ (a cylindrical shell) y z x φ surface of constant φ (half a plane) y z x p0, 0, zq surface of constant z (a plane) 384 ISO COORDINATE SYSTEM NOTATION G.3 SPHERICAL COORDINATES Finally here is a figure showing the volume element dV in cylindrical coordinates. ρ dρ dz ρ dφ volume element dV “ ρ dρ dφ dz G.3Ĳ Spherical Coordinates In the ISO convention the symbols r (instead of ρ), ϕ (instead of θ) and θ (instead of ϕ) are used for spherical coordinates. r = distance from (0, 0, 0) to (x, y, z) θ = angle between the z axis and the line joining (x, y, z) to (0, 0, 0) ϕ = angle between the x axis and the line joining (x, y, 0) to (0, 0, 0) z y x p0, 0, zq px, y, 0q px, 0, 0q φ θ px, y, zq r r sin θ r cos θ r sin θ sin φ Here are two more figures giving the side and top views of the previous figure. z p0, 0, zq r cos θ px, y, 0q side view r sin θ θ r px, y, zq y x px, 0, 0q top view r sin θ sin φ r sin θ cos φ r sin θ φ px, y, 0q 385 ISO COORDINATE SYSTEM NOTATION G.3 SPHERICAL COORDINATES The cartesian and spherical coordinates are related by x = r sin θ cos ϕ y = r sin θ sin ϕ z = r cos θ r = b x2 + y2 + z2 ϕ = arctan y x θ = arctan a x2 + y2 z Here are three figures showing a surface of constant r, a surface of constant ϕ, and a surface of constant θ. z y x r surface of constant r (a sphere) z y x φ surface of constant φ (half a plane) z y x θ surface of constant θ (a cone) Finally, here is a figure showing the volume element dV in spherical coordinates z y x dr r dθ r sin θ dφ volume element dV “ r2 sin θ dr dθ dφ and two extracts of the above figure to make it easier to see how r dθ and r sin θ dϕ arise. 386 ISO COORDINATE SYSTEM NOTATION G.3 SPHERICAL COORDINATES z y x r dθ θ dθ dr z y x r sin θ θ r dφ r sin θ dφ 387 CONIC SECTIONS AND QUADRIC SURFACES Appendix H A conic section is the curve of intersection of a cone and a plane that does not pass through the vertex of the cone. This is illustrated in the figures below. An equivalent1 (and often circle ellipse parabola hyperbola used) definition is that a conic section is the set of all points in the xy–plane that obey Q(x, y) = 0 with Q(x, y) = Ax2 + By2 + Cxy + Dx + Ey + F = 0 being a polynomial of degree two2. By rotating and translating our coordinate system the equation of the conic section can be brought into one of the forms3 • αx2 + βy2 = γ with α, β, γ ą 0, which is an ellipse (or a circle), • αx2 ´ βy2 = γ with α, β ą 0, γ ‰ 0, which is a hyperbola, • x2 = δy, with δ ‰ 0 which is a parabola. 1 It is outside our scope to prove this equivalence. 2 Technically, we should also require that the constants A, B, C, D, E, F, are real numbers, that A, B, C are not all zero, that Q(x, y) = 0 has more than one real solution, and that the polynomial can’t be factored into the product of two polynomials of degree one. 3 This statement can be justified using a linear algebra eigenvalue/eigenvector analysis. It is beyond what we can cover here, but is not too difficult for a standard linear algeba course. 388 CONIC SECTIONS AND QUADRIC SURFACES The three dimensional analogs of conic sections, surfaces in three dimensions given by quadratic equations, are called quadrics. An example is the sphere x2 + y2 + z2 = 1. Here are some tables giving all of the quadric surfaces. name elliptic cylinder parabolic cylinder hyperbolic cylinder sphere equation in standard form x2 a2 + y2 b2 = 1 y = ax2 x2 a2 ´ y2 b2 = 1 x2 + y2 + z2 = r2 x = constant cross–section two lines one line two lines circle y = constant cross–section two lines two lines two lines circle z = constant cross–section ellipse parabola hyperbola circle sketch name ellipsoid elliptic paraboloid elliptic cone equation in standard form x2 a2 + y2 b2 + z2 c2 = 1 x2 a2 + y2 b2 = z c x2 a2 + y2 b2 = z2 c2 x = constant cross–section ellipse parabola two lines if x = 0 hyperbola if x ‰ 0 y = constant cross–section ellipse parabola two lines if y = 0 hyperbola if y ‰ 0 z = constant cross–section ellipse ellipse ellipse sketch 389 CONIC SECTIONS AND QUADRIC SURFACES name hyperboloid of one sheet hyperboloid of two sheets hyperbolic paraboloid equation in standard form x2 a2 + y2 b2 ´ z2 c2 = 1 x2 a2 + y2 b2 ´ z2 c2 = ´1 y2 b2 ´ x2 a2 = z c x = constant cross–section hyperbola hyperbola parabola y = constant cross–section hyperbola hyperbola parabola z = constant cross–section ellipse ellipse two lines if z = 0 hyperbola if z ‰ 0 sketch 390
188015	https://pmc.ncbi.nlm.nih.gov/articles/PMC10785355/	Unlocking the mysteries of VLDL: exploring its production, intracellular trafficking, and metabolism as therapeutic targets - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Lipids Health Dis . 2024 Jan 12;23:14. doi: 10.1186/s12944-023-01993-y Search in PMC Search in PubMed View in NLM Catalog Add to search Unlocking the mysteries of VLDL: exploring its production, intracellular trafficking, and metabolism as therapeutic targets Jingfei Chen Jingfei Chen 2 Reproductive Medicine Center, Department of Obstetrics and Gynecology, Research Institute of Blood Lipid and Atherosclerosis, The Second Xiangya Hospital, Central South University, Changsha, Hunan 410011 China Find articles by Jingfei Chen 2, Zhenfei Fang Zhenfei Fang 1 Research Institute of Blood Lipid and Atherosclerosis, the Second Xiangya Hospital, Central South University, Changsha, Hunan 410011 China 3 Department of Cardiovascular Medicine, The Second Xiangya Hospital, Central South University, Changsha, Hunan 410011 China Find articles by Zhenfei Fang 1,3, Qin Luo Qin Luo 1 Research Institute of Blood Lipid and Atherosclerosis, the Second Xiangya Hospital, Central South University, Changsha, Hunan 410011 China 3 Department of Cardiovascular Medicine, The Second Xiangya Hospital, Central South University, Changsha, Hunan 410011 China Find articles by Qin Luo 1,3, Xiao Wang Xiao Wang 4 State Key Laboratory of Membrane Biology, Peking University, Beijing, 100871 China Find articles by Xiao Wang 4, Mohamad Warda Mohamad Warda 5 Department of Biochemistry and Molecular Biology, Faculty of Veterinary Medicine, Cairo University, Giza, 12211 Egypt 6 Department of Physiology, Faculty of Veterinary Medicine, Ataturk University, Erzurum, 25240 Turkey Find articles by Mohamad Warda 5,6, Avash Das Avash Das 7 Department of Pathology, Beth Israel Deaconess Medical Center, Harvard Medical School, Boston, MA 02215-5400 USA Find articles by Avash Das 7, Federico Oldoni Federico Oldoni 8 Department of Molecular Genetics, University of Texas Southwestern Medical Center, Dallas, TX USA Find articles by Federico Oldoni 8, Fei Luo Fei Luo 1 Research Institute of Blood Lipid and Atherosclerosis, the Second Xiangya Hospital, Central South University, Changsha, Hunan 410011 China 3 Department of Cardiovascular Medicine, The Second Xiangya Hospital, Central South University, Changsha, Hunan 410011 China Find articles by Fei Luo 1,3,✉ Author information Article notes Copyright and License information 1 Research Institute of Blood Lipid and Atherosclerosis, the Second Xiangya Hospital, Central South University, Changsha, Hunan 410011 China 2 Reproductive Medicine Center, Department of Obstetrics and Gynecology, Research Institute of Blood Lipid and Atherosclerosis, The Second Xiangya Hospital, Central South University, Changsha, Hunan 410011 China 3 Department of Cardiovascular Medicine, The Second Xiangya Hospital, Central South University, Changsha, Hunan 410011 China 4 State Key Laboratory of Membrane Biology, Peking University, Beijing, 100871 China 5 Department of Biochemistry and Molecular Biology, Faculty of Veterinary Medicine, Cairo University, Giza, 12211 Egypt 6 Department of Physiology, Faculty of Veterinary Medicine, Ataturk University, Erzurum, 25240 Turkey 7 Department of Pathology, Beth Israel Deaconess Medical Center, Harvard Medical School, Boston, MA 02215-5400 USA 8 Department of Molecular Genetics, University of Texas Southwestern Medical Center, Dallas, TX USA ✉ Corresponding author. Received 2023 Sep 4; Accepted 2023 Dec 26; Collection date 2024. © The Author(s) 2024 Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit The Creative Commons Public Domain Dedication waiver ( applies to the data made available in this article, unless otherwise stated in a credit line to the data. PMC Copyright notice PMCID: PMC10785355 PMID: 38216994 Abstract Reducing circulating lipid levels is the centerpiece of strategies for preventing and treating atherosclerotic cardiovascular disease (ASCVD). Despite many available lipid-lowering medications, a substantial residual cardiovascular risk remains. Current clinical guidelines focus on plasma levels of low-density lipoprotein (LDL). Recent attention has been given to very low-density lipoprotein (VLDL), the precursor to LDL, and its role in the development of coronary atherosclerosis. Preclinical investigations have revealed that interventions targeting VLDL production or promoting VLDL metabolism, independent of the LDL receptor, can potentially decrease cholesterol levels and provide therapeutic benefits. Currently, methods, such as mipomersen, lomitapide, and ANGPTL3 inhibitors, are used to reduce plasma cholesterol and triglyceride levels by regulating the lipidation, secretion, and metabolism of VLDL. Targeting VLDL represents an avenue for new lipid-lowering strategies. Interventions aimed at reducing VLDL production or enhancing VLDL metabolism, independent of the LDL receptor, hold promise for lowering cholesterol levels and providing therapeutic benefits beyond LDL in the management of ASCVD. Keywords: Atherosclerotic cardiovascular disease, Very low-density lipoprotein, Low-density lipoprotein, LDL receptor-independent pathway Introduction Despite significant therapeutic advances, atherosclerotic cardiovascular disease (ASCVD) remains one of the major causes of mortality worldwide. High cholesterol and high triglyceride (TG) levels are considered important risk factors for the development of ASCVD . Triglyceride-rich lipoproteins (TRLs) are a type of lipoprotein that are rich in triglycerides . It consists mainly of very low-density lipoprotein (VLDL) produced by the liver and chylomicrons generated in the intestines. The TRLs play a vital role in lipid metabolism by transporting triglycerides and lipids to peripheral tissues for energy production or storage. VLDL facilitates fatty acid supply during fasting to muscles, and chylomicrons primarily provide fatty acids to adipose tissue. This regulation is predominantly governed by ANGPTL4/3/8 . TRLs are converted into smaller and denser particles called TRL remnants through the action of lipoprotein lipase and other enzymes. These TRL remnants have fewer TGs and phospholipids but are enriched in cholesteryl esters. Elevated levels of TGs are linked to a higher risk of ASCVD. Unfortunately, the existing therapies aimed at lowering TGs have shown limited effectiveness in reducing cardiovascular risk . TRLs have been identified as a causal risk factor for ASCVD [5, 6] and reported to be more atherogenic per particle than LDL cholesterol (LDL-C) [7, 8]. Moreover, elevated plasma TRL levels are dose-dependently associated with acute pancreatitis risk. In patients with persistently high triglyceride levels despite high-intensity statin therapy, guidelines often suggest considering adjunctive treatments such as fibrates, niacin, or long-chain omega-3 fatty acids . Diet and exercise play crucial roles in reducing TRLs, with clinical guidelines emphasizing them as foundational for TRL reduction . Recent studies have shown that VLDL-C plays a role in the etiology and progression of ASCVD. It has been estimated that the level of VLDL-C may account for 46% of the risk of apolipoprotein B (APOB)-related myocardial infarction [10, 11]. As a major source of TRL, especially during the fasting state, VLDL is a critical player in lipid metabolism. However, therapeutic strategies safely targeting the secretion or clearance of VLDL are lacking. A better understanding of the mechanism of ASCVD related to circulating VLDL can help identify new therapeutic targets for patients with a residual risk of ASCVD. Here, we summarize the latest research progress on VLDL metabolism and its potential as a lipid-lowering target, hoping to provide a reference for future research. Composition of VLDL Chylomicrons, VLDL, intermediate-density lipoprotein (IDL; VLDL remnants), LDL, and high-density lipoprotein (HDL) are the main categories of lipoproteins that can range in density from low to high . The density of VLDL is very low, ranging from 0.93 to 1.006 g/mL, and the size of the mass is approximately 30–80 nm. VLDL has APOB as its structural protein and contains other apolipoproteins, such as apolipoprotein C2 (APOC2), APOC3, apolipoprotein E (APOE), and apolipoprotein A5 (APOA5) (Fig.1). APOC3 is a smaller apolipoprotein consisting of 79 amino acid residues. It is predominantly located on TRLs and HDL in circulation . APOC3 functions as a lipoprotein lipase (LPL) inhibitor, impeding the breakdown of TRLs . Deficiency or inhibition of APOC3 can lead to decreased TG levels . In addition to protein, VLDL primarily carries TG while also containing cholesterol esters and phospholipids . However, the composition of VLDL particles is subjected to changes, and the amount of TG carried by each particle can also vary dramatically. Hydrophobic lipids (TG and cholesterol ester) form the core of VLDL. Phospholipids form a monolayer on the hydrophilic surface with unesterified cholesterol and exchangeable apolipoproteins. Fig. 1. Open in a new tab Schematic figure of apolipoproteins and the lipid composition of VLDL. Very low-density lipoprotein (VLDL) contains apolipoprotein B100 (APOB) on its surface as a structural protein and a surface monolayer of phospholipids; free cholesterol; and apolipoproteins such as apolipoprotein C2 (APOC2), APOC3, apolipoprotein E (APOE), and APOA5. The central core of VLDL contains cholesterol esters and triglycerides Biogenesis and lipidation of VLDL The biosynthesis of VLDL is a multistep process that begins in the rough endoplasmic reticulum (RER) (Fig.2). Hepatic VLDL assembly starts with the cotranslational translocation of APOB across the rough ER membrane, where microsomal triglyceride transfer protein (MTP) helps in the first lipid decoration of APOB . Without sufficient lipidation, nascent APOB is degraded . Thus, the biogenesis of VLDL relies on the availability of APOB, phospholipids and TGs. TGs include saturated and monoenoic fatty acids, while phospholipids constitute the most significant proportion of polyunsaturated fatty acids. Fig. 2. Open in a new tab Biogenesis, lipidation and intracellular trafficking of VLDL. The assembly of VLDL in the liver begins with the cotranslational translation of apolipoprotein B (APOB) across the rough endoplasmic reticulum (RER) membrane. Microsomal triglyceride transfer protein (MTP) plays an essential role in the initial lipidation of APOB by extracting phospholipids and triglycerides (TGs) from the endoplasmic reticulum (ER). Without sufficient lipidation, nascent APOB is degraded. After the initial lipidation, VLDL is further lipidated with a large amount of lipids to form mature VLDL. To exit the endoplasmic reticulum, VLDL is packaged into transport vesicles called coat protein complex II (COPII), which is initiated by the activation of the small Ras-like GTPase SAR1. With the availability of lipids, partially lipidated APOB (pre-VLDL) will become fully lipidated with the bulk of lipids. The lipidation of very low-density lipoprotein (VLDL) is a two-step process [14, 16, 17]. The first step of lipidation occurs in the lumen of RERs during the translation of APOB, where a small amount of lipid combines with APOB to form pre-VLDL with the assistance of MTP . The second step of VLDL lipidation involves pre-VLDL receiving a large number of lipids to form mature VLDLs. However, where this event occurs, only in the endoplasmic reticulum (ER) or in the Golgi apparatus, is controversial [18–20]. Previous studies have shown that transmembrane 6 superfamily member 2 (TM6SF2), a polytopic ER protein, plays a role in this second step of VLDL lipidation [16, 21]. The protein localizes to the smooth ER and the ER-Golgi intermediate compartment (ERGIC) [16, 21]. Researchers have proposed that two types of VLDL exist: triglyceride-rich VLDL1 and triglyceride-poor VLDL2 [22, 23]. Triglyceride-poor VLDL2 was formed after pre-VLDL acquired additional lipids [22, 23]. Triglyceride-poor VLDL2 can be directly secreted from hepatocytes or fused with APOB-free lipid droplets to form triglyceride-rich VLDL1. Other proteins are also reported to be involved in the assembly of VLDL, including cell death-inducing DFF45-like effector B (CIDEB), phospholipid transfer protein (PLTP) [24, 25], lysophosphatidylcholine acyltransferase 3 (LPCAT3) and transmembrane protein 41B (TMEM41B) . The most recent study reported that tissue-type plasminogen activator (tPA) can bind to apoB and inhibit the lipid transfer of MTP to APOB, thereby reducing the assembly of VLDL and plasma levels of APOB-lipoprotein cholesterol . Nevertheless, this is an area of burgeoning interest. Intracellular trafficking of VLDL VLDLs are generated in the ER and then transported to the Golgi for secretion (Fig.2). For precise delivery, exporting VLDLs from the ER to the Golgi may require efficient and specialized transport machinery. VLDLs are packaged into transport vesicles generated by coat complex II (COPII) to exit the ER . COPII assembly on the ER surface is initiated by activating the secretion-associated Ras-related GTPase 1 (SAR1) via its guanine nucleotide exchange factor (GEF) Sect.12 . SAR1 exposes its N-terminal amphipathic α-helix upon activation and inserts into the ER membrane . Activated SAR1 recruits the inner coat complex Sect.23/Sect.24 and, subsequently, the outer coat complex Sect.13/Sect.31 to form COPII-coated vesicles . The size of a typical COPII vesicle ranges from 60 to 70 nm in diameter , while lipid-containing APOB may be oversized for regular COPII vesicles . To accommodate this unique cargo, specialized factors may be needed to facilitate the ER export of VLDLs . Notably, germline mutations in human SAR1B, one of the two SAR1 paralogs, cause chylomicron retention disease (CMRD), an inborn metabolic defect in fat absorption due to the retention of chylomicrons (the APOB-containing lipoprotein produced in the gut) in the small intestine . Consistently, selective inactivation of murine Sar1b in the liver depletes fasting plasma lipids due to blockage of VLDL secretion . To bridge the lipid-bearing APOB particles in the ER lumen to the cytosolic COPII machinery, Surfeit 4 (SURF4) acts as the cargo receptor to selectively escort the ER-Golgi transport of lipoproteins in a physiological setting . Genetic ablation of hepatic Surf4 in mice depletes plasma lipids near zero, resulting from selective retention of lipoproteins in the ER . Importantly, a genome-wide analysis (GWAS) showed a strong association between plasma LDL levels and SURF4 in humans, further supporting the essential role of the specialized SURF4-mediated lipoprotein export pathway in systemic lipid homeostasis. In addition to the profound impacts of SAR1B and SURF4 on VLDL secretion, additional factors, including transport and Golgi organization protein 1 (TANGO1), TANGO1-like (TALI) and meningioma-expressed antigen 6 (Mea6) , are involved in the secretion of APOB-containing lipoproteins; however, how these auxiliary factors coordinate with each other has not been fully elucidated. Comprehensive reviews are cited for further reading on the biogenesis and intracellular trafficking of VLDL . Metabolism of VLDL in circulation Despite extensive research efforts, the metabolism of VLDL in the bloodstream has not been fully elucidated (Fig.3). Once released into circulation by the liver, VLDL triglycerides and phospholipids become susceptible to hydrolysis by various lipase families, resulting in the liberation of free fatty acids. These free fatty acids can be taken up by energy-demanding cells such as those of the heart and skeletal muscle or stored in adipose tissue. Additionally, the triglycerides on VLDL can be exchanged with cholesterol esters on HDL through the action of cholesteryl ester transfer protein (CETP) . Three crucial regulators of VLDL metabolism include lipase families: LPL, hepatic lipase, and endothelial lipase. As synthesized and secreted by the liver, hepatic lipase is anchored by sulfate proteoglycans (HSPGs) on the cell surface of hepatocytes and endothelial cells. Endothelial lipase is predominantly present in vascular endothelial cells of organs such as the liver, lung, kidney, and placenta . LPL synthesis occurs in the parenchymal cells of white adipose tissue and energy-consuming tissues (e.g., heart, skeletal muscle, and brown adipose tissue). It is then transported to the luminal surface of vascular endothelial cells via glycosylphosphatidylinositol-anchored high-density lipoprotein-binding protein 1 (GPIHBP1). Hepatic lipase catalyzes the hydrolysis of TG and phospholipids , while LPL, the rate-limiting enzyme, primarily catalyzes TG . Both hepatic lipase and endothelial lipase exhibit a strong affinity for phospholipids [41, 43]. Variants in LPL, hepatic lipase, and endothelial lipase have been linked to lipid traits in genome-wide association studies (GWASs) , underscoring their significant role in lipid metabolism. The activities of LPL, hepatic lipase, and endothelial lipase are regulated by various factors. Glycosylphosphatidylinositol anchored high-density lipoprotein binding protein 1 (GPIHBP1), APOC2, and APOA5 promote the activity of LPL, while its activity can be inhibited by members of the angiopoietin-like (ANGPTL) protein families (ANGPTL3, ANGPTL4, and ANGPTL8) and by APOC3 . The inhibitory effect of ANGPTL3 on LPL is particularly pronounced when ANGPTL3 forms a complex with ANGPTL8, which is activated upon refeeding [45–47]. ANGPTL3 can also inhibit the activity of endothelial lipase . Genetic cohort studies have paved the way for developing and clinically applying inhibitors targeting APOC3 and ANGPTL3, which have shown promising potential in reducing lipid levels. Fig. 3. Open in a new tab Metabolism of VLDL in circulation. After entering the bloodstream, VLDL undergoes hydrolysis by lipoprotein lipase (LPL), endothelial lipase, and hepatic lipase, leading to the release of free fatty acids from the core triglycerides and surface phospholipids. The activity of LPL is inhibited by apolipoprotein C3 (APOC3) and angiopoietin-like protein (ANGPTL)3/4/8, while it is activated by APOA5 and APOC2. When a large amount of triglycerides in VLDL is hydrolyzed, VLDL can be further metabolized into VLDL remnants, nearly half of which are absorbed by the liver, while the remaining VLDL remnants are further metabolized into low-density lipoprotein (LDL). LDL can bind to LDL receptors (LDLRs) and be taken up by the liver. Excessive VLDL remnants and LDL can deposit in the blood vessel walls, forming atherosclerotic plaques After LPL, a significant portion of triglycerides are removed from VLDL, causing a change in its composition and the transformation of triglycerides into a VLDL remnant. Approximately half of the VLDL remnants are recognized and taken up by liver cells through receptors such as LDLR, LDLR-related protein (LRP), the VLDL receptor, and heparan sulfate proteoglycan (HSPG) receptors (syndecan-1) [48, 49]. The remaining 50% of VLDL remnants lacking ApoE are LDL, which is internalized by cells via receptor-mediated endocytosis through LDLR [23, 50]. LDL cholesterol and other lipids can be stored or utilized by cells. The phospholipids found in VLDL can be hydrolyzed by hepatic lipase and endothelial lipase . However, the specific biological function of these phospholipids on the surface of VLDL is not yet fully understood. When there is an excess of APOB-containing lipoprotein particles in the bloodstream, including VLDL, VLDL remnants, and LDL, they can deposit on the vascular wall and contribute to the development of atherosclerosis . Therapies for the reduction of LDL-C and its challenge Numerous clinical studies have established that cholesterol, particularly LDL-C, is a causative risk factor for ASCVD. Over the past few decades, significant advancements have been made in lipid-lowering therapies. The primary lipid-lowering drugs currently used include statins, ezetimibe, and proprotein convertase subtilisin/kexin type 9 (PCSK9)-inhibiting monoclonal antibodies. Statins are the most frequently prescribed lipid-lowering medications due to their effectiveness in reducing LDL-C [53, 54]. Statins exhibit a modest triglyceride-lowering effect of only approximately 9–31% , depending on baseline triglyceride levels. Although PCSK9 inhibitors effectively reduce LDL-C levels, they have a limited impact on TG levels, typically reducing LDL-C levels by only approximately 10–20% [56, 57]. A reduction of 1 mmol/L (38.7 mg/dL) in both statins and nonstatin drugs has been shown to be associated with a significant 21% decrease in the risk of major cardiovascular events . In recent years, mounting evidence has suggested that an elevated TRL may contribute to this residual cardiovascular risk . In addition to TRLs, the enduring risk of ASCVD may be linked to elevated levels of lipoprotein (a) (Lp(a)) and chronic inflammation . Lp(a) has been identified as a contributing factor to cardiovascular risk, as substantiated by both epidemiological and genetic research . Ongoing clinical trials seek to validate whether diminishing Lp(a) levels can effectively reduce cardiovascular events, providing us with additional insights. Various anti-inflammatory medications have been used in individuals with ASCVD or those at risk. Notably, both canakinumab and colchicines have been shown to reduce the risk of ASCVD, with no significant impact on plasma lipid levels. VLDL, the precursor of LDL, serves as the primary source of TG and TRL during fasting. Targeting VLDL may reduce both TG and LDL levels. Inhibition of the biogenesis and lipidation of VLDL to lower blood lipids Under the action of lipoprotein lipase, VLDL TG is hydrolyzed and converted into VLDL remnants and LDL . The fasting plasma LDL-C and TG concentrations partially depend on the amount of VLDL secreted by the liver. Inhibiting the secretion of hepatic VLDL lipids can effectively reduce the level of circulating lipids through three main pathways: (1) reducing the amount of lipids needed for VLDL assembly, (2) inhibiting the lipidation of VLDL in the liver, and (3) inhibiting the transport and secretion of hepatic VLDL particles. 1) Targeting APOB APOB is the structural protein for VLDL, and inhibiting the synthesis of the APOB protein can significantly reduce the synthesis of hepatic VLDL. Current therapies lower plasma lipids by inhibiting APOB synthesis [64, 65]. Mipomersen, an oligonucleotide targeting hepatic APOB mRNA, showed an efficient lipid-lowering effect with a 50% reduction in APOB [64, 65]. However, due to its liver toxicity, this medicine is exclusively approved for use and administration to patients with homozygous familial hypercholesterolemia (HoFH). As expected, any treatment that inhibits the biogenesis of APOB directly may cause hepatic steatosis, which limits its usage. 2) Inhibition of VLDL lipidation As discussed above, the first step of VLDL lipidation requires MTP and PLTP, while the second step of VLDL lipidation is associated with TM6SF2 and other proteins [14, 16, 17, 21, 66]. A selective inhibitor of MTP, lomitapide, has been approved by both the FDA and EMA for use in treating HoFH . Although lomitapide successfully lowered LDL-C levels by 40–50%, it was also associated with hepatic steatosis and gastrointestinal side effects . MTP is necessary for the assembly of VLDL in hepatocytes by facilitating the incorporation of triglycerides into VLDL , and inhibition of MTP decreases the efficacy of lipids (mainly TG) export by hepatocytes, hence increasing hepatic steatosis. However, a recent clinical trial showed that lomitapide caused mild to moderate hepatic steatosis without affecting hepatic stiffness after more than nine years of follow-up . However, the safety of MTP inhibition still needs to be explored. There are currently no drugs targeting the second step of VLDL lipidation. Genetic evidence has shown that TM6SF2 loss-of-function mutations are linked to decreased plasma lipids and ASCVD , indicating an adequate lipid-lowering capacity and cardioprotective effect. However, genetic evidence and preclinical studies confirmed that inactivation of TM6SF2 resulted in hepatic steatosis and increased aminotransferase levels [21, 69]. Mechanisms that. Block secretion and thereby lipid flux out of the liver are likely nonviable mechanisms due to the development of steatosis. Targeting the intracellular trafficking of VLDL COPII-coated transport vehicles are thought to be responsible for transporting VLDL from the ER to the Golgi . Researchers have found that there are subtle mechanisms involved in ensuring the precise and selective delivery of VLDL . The membrane protein SURF4 shuttles between the ER and the Golgi apparatus, which can selectively assist nascent VLDL trafficking from the ER to the Golgi . Mice lacking SURF4 had a drastic decrease in plasma VLDL, although typical secretory proteins were unaffected . Inactivation of SURF4 resulted in a remarkable decrease of approximately 90% in plasma LDL-C and TG levels and entirely prevented the development of atherosclerotic plaques produced by a high-cholesterol diet coupled with overexpression of PCSK9 . Although liver cholesterol and TG levels were elevated in mice with complete deficiency of SURF4, significant protection against pathological dyslipidemia and atherosclerosis was observed in heterozygotes for Surf4 knockout without any observable hepatic damage [36, 70]. These data provide additional insight into the hypothesis that targeting VLDL secretion may be an effective method for decreasing the levels of atherogenic lipids. TANGO1, TALI , and Mea6 were also reported to regulate the ER-to-Golgi transport of VLDL by interacting with coat proteins of COPII. Hepatocyte-specific deletion of TANGO1 or Mea6 results in a significant defect in VLDL secretion and severe fatty liver [34, 37]. Due to their severe side effects, their potential to become lipid-lowering drug targets is limited. Strategies targeting VLDL metabolism in circulation VLDL is crucial for transporting TG and cholesterol from the liver to peripheral tissues through the bloodstream. The TG carried by VLDL can undergo hydrolysis, releasing free fatty acids taken up by tissues. Similarly, the cholesterol carried by VLDL is primarily utilized by tissues through the uptake of VLDL remnants. As mentioned earlier, the core triglycerides of VLDL can be hydrolyzed by lipases present in the circulation, leading to the formation of VLDL remnants. These remnants can be directly taken up by the liver or further metabolized into LDL. The activity of LPL and endothelial lipase can be inhibited by proteins such as ANGPTL3 and APOC3. Newly published research indicates that inhibiting ANGPTL3 and APOC3 can significantly reduce circulating levels of triglycerides and cholesterol . Clinical trials have also shown promising results, highlighting the potential of these agents as effective lipid-lowering agents. 1) Targeting ANGPTL3 ANGPTL3 is an inhibitor of both LPL and endothelial lipase . Genetic studies have shown that loss-of-function mutations in ANGPTL3 significantly reduce plasma TG and cholesterol levels and have significant cardiovascular protective effects [72, 73]. Multiple methods have been explored for inhibiting ANGPTL3, including monoclonal antibodies, antisense oligonucleotides (ASOs), and mRNA interference (mRNAi) . Preclinical studies have shown that inactivation of ANGPTL3 can significantly decrease TG levels and promote the clearance of VLDL remnants [45, 74–76]. Clinical studies have shown that ANGPTL3 inhibitors have very good lipid-lowering effects [77, 78]. Inhibition of ANGPTL3 in combination with statins induced a 47% reduction in LDL-C and a 55% reduction in TG in patients with familial hypercholesterolemia or refractory hypercholesterolemia [77, 78]. We hypothesize that changes in VLDL metabolism may be an important mechanism underlying the lipid-lowering effect of ANGPTL3 inhibitors . Studies have shown that ANGPTL3 inhibitors can lower plasma lipids through both LDLR-dependent and LDLR-independent pathways , but the exact underlying mechanisms are still unknown. ANGPTL3 has been shown to hasten VLDL processing through an LDLR-independent mechanism, with VLDL being removed before LDL production . LDLR-independent pathways also enable the combination of this lipid-lowering strategy with LDLR-dependent lipid-lowering strategies such as statins. Evinacumab, a monoclonal antibody that inhibits ANGPTL3, has gained approval for treating HoFH in the European Union (EU), United Kingdom (UK), and the United States (US). Conversely, the development of vupanorsen, an antisense oligonucleotide (ASO) aimed at reducing hepatic ANGPTL3 production, was halted due to hepatic side effects. 2) Targeting ANGPTL4 ANGPTL4 is produced by various cells and tissues, such as the liver and adipose tissue, and predominantly controls LPL activity during fasting . Genetic research has validated the role of ANGPTL4 in regulating plasma TG levels. Loss-of-function mutations in ANGPTL4 correlate with lower plasma TG levels and a reduced risk of ASCVD . Nonetheless, developing anti-ANGPTL4 strategies is challenging because complete ANGPTL4 inactivation across the body in mice can lead to severe clinical complications. Mice without ANGPTL4 exhibit mesenteric lymphadenopathy and undergo a significant acute phase response . 3) Targeting APOC3 APOC3 is an endogenous antagonist of LPL and hepatic lipase. APOC3 is a protein present on certain lipoproteins in the body, including VLDL and LDL. Researchers have found a correlation between APOC3 LOF mutations and lower plasma triglyceride levels as well as a decreased risk of cardiovascular disease that was 40% lower [84, 85]. Recent research has focused on developing drugs that target APOC3 as potential therapies for hypertriglyceridemia and other lipid-related disorders. Studies have shown that inhibiting APOC3 can reduce triglyceride levels by 70.9% and improve lipid profiles [86–88]. ASOs, monoclonal antibodies, and small molecule inhibitors are in development to target APOC3. Volanesorsen, an ASO directed at APOC3, has gained approval in Europe for treating familial chylomicronemia syndrome (FCS) patients. Although APOC3 inhibition has shown promise in both preclinical and clinical studies, additional research is needed to comprehensively grasp its therapeutic potential and long-term safety implications. Apolipoprotein C2 (APOC2), an exchangeable small apolipoprotein on TRL, activates LPL. Notably, APOC2 mimetic peptides reduce TG levels by displacing APOC3 from TRL, alleviating the inhibitory impact of APOC2 on LPL . Review strengths and limitations This review primarily focuses on clinical aspects and summarizes the research on the composition, production, metabolism, and other aspects of VLDL. The possibility of intervening in VLDL by regulating these aspects has also been investigated. Not only does it provide insights from a clinical perspective, but it also includes fundamental research that can benefit both clinical and basic researchers. However, our review has several limitations. Due to the constraints of article length, our description of the underlying mechanisms may not be thorough enough. Our focus is on elucidating the metabolism of VLDL in circulation and its potential targets, and reviews of its composition and generation may not be thorough enough. Challenge and future directions Current clinical guidelines and research on lipid-lowering effects on reducing ASCVD risk primarily revolve around LDL, with limited attention given to VLDL metabolism and clearance. The significance of VLDL as a precursor to TRL, a major source of remnant cholesterol in the fasting state, has yet to be fully appreciated. A burgeoning body of literature reveals that VLDL-C may account for nearly half of APOB-related myocardial infarctions, piquing interest in the potential role of VLDL-C in reducing the residual risk of ASCVD . Remnant cholesterol has been proposed as a new opportunity for reducing residual cardiovascular risk [6, 66], indicating the importance of targeting VLDL. Recent studies have also demonstrated that modulating the composition of VLDL through lipases such as LPL, hepatic lipase, or endothelial lipase can considerably enhance both LDLR-dependent and nondependent clearance pathways for VLDL remnants [86–88, 90]. These strategies can effectively reduce triglycerides, remnant cholesterol, and LDL-C by targeting VLDL. However, despite their potential, numerous challenges and difficulties remain. Existing approaches to inhibit VLDL secretion can lead to hepatic lipid accumulation, which may limit their future applications. Therefore, developing better targeted therapies to reduce VLDL secretion without causing hepatic lipid accumulation is crucial. Inhibiting SURF4 to decrease VLDL secretion and circulating lipids has been shown to be dose dependent and safe , suggesting that it has the potential to lower plasma lipids without inducing hepatic steatosis. Furthermore, novel lipid-lowering strategies are needed to inhibit VLDL assembly and, in combination, prevent intracellular lipid excess by promoting free fatty acid oxidation or reducing triglyceride synthesis. Promising avenues reside in strategies promoting VLDL metabolism and clearance, such as ANGPTL3 inhibitors . In patients with HoFH, ANGPTL3 inhibitors can lead to a nearly 50% reduction in LDL-C , offering hope for FH patients and improving lipid-lowering options for individuals with high residual cardiovascular risk or statin intolerance. Research has established a causal relationship between VLDL remnants and cardiovascular risk , highlighting the importance of targeting VLDL in lipid-lowering strategies to address the current challenges in lipid-lowering treatment. Currently, there is a need for simpler and more reliable methods for monitoring circulating VLDL in patients, which can circumvent difficulties in assessing the therapeutic efficacy of interventions targeting VLDL. Therefore, further research is needed to explore and develop accessible and cost-effective methods for detecting circulating VLDL. Acknowledgements We thank Dr. Helen Hobbs, Dr. Yan Wang and Dr. Xiao Liu for helpful discussion and suggestions. Abbreviations APOB Apolipoprotein B ANGPTL Angiopoietin-like protein APOA5 Apolipoprotein A5 APOC2 Apolipoprotein C2 APOE Apolipoprotein E ASCVDs Atherosclerotic cardiovascular disease ASO Antisense oligonucleotide CIDEB Cell death-inducing DFF45-like effector B CETP Cholesteryl ester transfer protein CMRD Chylomicron retention disease COPII Coat complex II ER Endoplasmic reticulum ERGIC ER-Golgi intermediate compartment GWAS Genome-wide analysis study HDL High-density lipoprotein HMG-CoA 3-hydroxy-3-methylglutaryl coenzyme A HoFH Homozygous familial hypercholesterolemia IDL Intermediate-density lipoprotein GPIHBP1 Glycosylphosphatidylinositol anchored high-density lipoprotein binding protein 1 LDL Low-density lipoprotein LPCAT3 Lysophosphatidylcholine acyltransferase 3 LPL Lipoprotein lipase LRP LDLR-related protein Mea6 Meningioma-expressed antigen 6 MTP Microsomal triglyceride transfer protein PCSK9 Proprotein convertase subtilisin/kexin type 9 PLTP Phospholipid transfer protein TMEM41B Transmembrane protein 41B RER Rough endoplasmic reticulum SAR1 Secretion-associated Ras-related GTPase 1 VLDL Very low-density lipoprotein TALI TANGO1-like TANGO1 Transport and Golgi organization protein 1 TG Triglyceride TM6SF2 Transmembrane 6 superfamily member 2 TRL Triglyceride-rich lipoprotein Author contributions FL conceived the idea. JC, ZF, FL wrote the manuscript; JC. ZF, QL, XW, MW, AD, FO read through and corrected the manuscript. All the authors have read and approved the final manuscript. Funding This project was supported by grants from the National Natural Science Foundation of China [No. 82100495 to F. Luo, No. 82201879 to J.F. Chen]; the Hunan Provincial Natural Science Foundation of China [No. 2021JJ40852 to F. Luo, No. 2022JJ40675 J.F. Chen]; the Scientific Research Project of Hunan Provincial Health Commission [No. 202203014009 to F. Luo, No. 202305037231 to J.F. Chen]; and the Scientific Research Launch Project for new employees of the Second Xiangya Hospital of Central South University (to F. Luo and J.F. Chen). China Postdoctoral Science Foundation (No. 331046 and 2023T160738 to F. Luo and J.F. Chen), Key Research and Development Program of Hunan Province of China (grant number 2021SK2004 to Z.F.). Data availability No data were generated or analyzed for this manuscript. 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Articles from Lipids in Health and Disease are provided here courtesy of BMC ACTIONS View on publisher site PDF (2.6 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction Composition of VLDL Biogenesis and lipidation of VLDL Intracellular trafficking of VLDL Metabolism of VLDL in circulation Therapies for the reduction of LDL-C and its challenge Inhibition of the biogenesis and lipidation of VLDL to lower blood lipids Targeting the intracellular trafficking of VLDL Strategies targeting VLDL metabolism in circulation Review strengths and limitations Challenge and future directions Acknowledgements Abbreviations Author contributions Funding Data availability Declarations Footnotes References Associated Data Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
188016	https://quizlet.com/114780473/unit-test-1-topic-review-activity-flash-cards/	Unit Test 1 Topic Review Activity Flashcards \| Quizlet hello quizlet Study tools Subjects Create Log in Science Chemistry Inorganic Chemistry Unit Test 1 Topic Review Activity 5.0 (5 reviews) Save The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 mc010-1.jpg Fe2(SO4)3 + 2Na3PO4 What is the theoretical yield of Fe2(SO4)3 if 20.00 g of FePO4 reacts with an excess of Na2SO4? Click the card to flip 👆 26.52 g Click the card to flip 👆 1 / 9 1 / 9 Flashcards Learn Test Blocks Match Created by ethangibson4 Students also studied Flashcard sets Study guides Chemistry Vocab, Section 1 13 terms sandovalsilv20294501 Preview Naming and creating formulas 21 terms alh88830 Preview Diatomic Molecules 7 terms hayyym1 Preview Chemistry elements set 2 12 terms jwsnow09 Preview Chemistry Chapters 3-4 41 terms Laceymayj7 Preview chrome-extension://bpmcpldpdmajfigpchkicefoigmkfalc/views/app.html 63 terms abdulai_gassama Preview Covalent Bonds 10 terms Mckenzie_Buzzard Preview 7.1 chem 25 terms Haley_Finta Preview chem exam guide 80 terms z_martinez2027 Preview Chemistry Module 6 21 terms barneybloodhound9 Preview science chapters 10-13 study guide 8 terms jamesmcctrainer Preview CHE131 Test 1 58 terms kyong_and_jin Preview Unit 2 Review 10 terms OrangeCat85022 Preview H.bio 1.2 33 terms Seungran_Lim Preview IB Chemistry Chapter 1 59 terms kelsee3929 Preview Pre-unit 1 Review 30 terms duchesseke Preview OAT General Chemistry: Solutions 57 terms Gabby_Ward46 Preview Exam 2 Chemistry 21 terms danicaa633 Preview CHEM3432 Exam 4 176 terms Jenna_Fette3 Preview Chemistry final 13 terms Brock_Merrill Preview Bio 2.1 2.2 49 terms gabbygrandone Preview Electronegativity 13 terms superIdksomething Preview Classify the following as an element, compound, homogenous, or heterogeneous mixture? 17 terms brooklynndyson Preview CHM 210- Exam 1 58 terms lakingiager1 Preview chem unit 7 test 15 terms lnassef27 Preview Group 1A 23 terms hunterlreeves06 Preview Chemistry regents 74 terms arien_60 Preview d orbitals 5 terms snwitt Preview Terms in this set (9) The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 mc010-1.jpg Fe2(SO4)3 + 2Na3PO4 What is the theoretical yield of Fe2(SO4)3 if 20.00 g of FePO4 reacts with an excess of Na2SO4? 26.52 g Consider the equation below. N2 + 3H2 mc011-1.jpg 2NH3 What is the mole ratio of hydrogen to ammonia? 3:2 The equation represents the combustion of sucrose. C12H22O11 + 12O2 mc007-1.jpg 12CO2 + 11H2O If there are 10.0 g of sucrose and 8.0 g of oxygen, how many moles of sucrose are available for this reaction? 0.029 mol Consider the reaction below. 2H2O mc030-1.jpg 2H2 + O2 How many moles of hydrogen are produced when 6.28 mol of oxygen form? 12.6 mol Consider the balanced equation. 2HCl + Mg mc031-1.jpg MgCl2 + H2 If 40.0 g of HCl react with an excess of magnesium metal, what is the theoretical yield of hydrogen? 1.11 g A chemical equation can be used to solve stoichiometry problems provided that it includes the coefficients that are needed for balancing the equation. The chemical equation below shows the formation of aluminum oxide (Al2O3) from aluminum (Al) and oxygen (O2). 4Al + 3O2 mc029-1.jpg 2Al2O3 The molar mass of O2 is 32.0 g/mol. What mass, in grams, of O2 must react to form 3.80 mol of Al2O3? 182 The chemical equation below shows the burning of magnesium (Mg) with oxygen (O2) to form magnesium oxide (MgO). 2Mg + O2 mc009-1.jpg 2MgO The molar mass of O2 is 32.0 g/mol. What mass, in grams, of O2 is required to react completely with 4.00 mol of Mg? 64.0 The following balanced equation shows the formation of water. 2H2 + O2 mc020-1.jpg 2H2O How many moles of oxygen (O2) are required to completely react with 27.4 mol of H2? 13.7 mol Learn More About us About Quizlet How Quizlet works Careers Advertise with us For students Flashcards Test Learn Study groups Solutions Modern Learning Lab Quizlet Plus Study Guides Pomodoro timer For teachers Live Blog Be the Change Quizlet Plus for teachers Resources Help center Sign up Honor code Community guidelines Terms Privacy California Privacy Your Privacy/Cookie Choices Ads and Cookie Settings Interest-Based Advertising Quizlet for Schools Parents Language Get the app Country United States Canada United Kingdom Australia New Zealand Germany France Spain Italy Japan South Korea India China Mexico Sweden Netherlands Switzerland Brazil Poland Turkey Ukraine Taiwan Vietnam Indonesia Philippines Russia © 2025 Quizlet, Inc. Students Flashcards Learn Study Guides Test Expert Solutions Study groups Teachers Live Blast Categories Subjects Exams Literature Arts and Humanit... Languages Math Science Social Science Other Flashcards Learn Study Guides Test Expert Solutions Study groups Live Blast Categories Exams Literature Arts and Humanit... Languages Math Science Social Science Other
188017	https://chem.libretexts.org/Courses/Santa_Barbara_City_College/SBCC_Chem_101%3A_Introductory_Chemistry/02%3A_Measurement_and_Problem_Solving/2.04%3A_Significant_Figures_in_Calculations	2.4: Significant Figures in Calculations - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not 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MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Tue, 25 Feb 2020 01:50:51 GMT 2.4: Significant Figures in Calculations 208077 208077 Janice Lawson { } Anonymous Anonymous User 2 false false [ "article:topic", "significant figures", "showtoc:no", "transcluded:yes", "source-chem-47449" ] [ "article:topic", "significant figures", "showtoc:no", "transcluded:yes", "source-chem-47449" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. 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SBCC Chem 101: Introductory Chemistry 5. 2: Measurement and Problem Solving 6. 2.4: Significant Figures in Calculations Expand/collapse global location SBCC Chem 101: Introductory Chemistry Front Matter 1: The Chemical World 2: Measurement and Problem Solving 3: Matter and Energy 4: Atoms and Elements 5: Molecules and Compounds 6: Chemical Composition 7: Chemical Reactions 8: Gases 9: Electrons in Atoms and the Periodic Table 10: Chemical Bonding 11: Liquids, Solids, and Intermolecular Forces 12: Solubility & Reaction Types 13: Solutions 14: Acids and Bases 15: Radioactivity and Nuclear Chemistry Back Matter 2.4: Significant Figures in Calculations Last updated Feb 25, 2020 Save as PDF 2.3: Significant Figures - Writing Numbers to Reflect Precision 2.5: The Basic Units of Measurement picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report Submit a Peer Review View on CommonsDonate Page ID 208077 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. Rounding 3. Multiplication and Division 1. 1. Example 2.4.1 2. Solution Addition and Subtraction Example 2.4.2 Solution a Exercise 2.4.2 Calculations Involving Multiplication/Division and Addition/Subtraction Example 2.4.3 Solution a. Perform multiplication first. Then, perform the addition. Round the final answer. b. Perform multiplication first. Then, perform the addition. Round the final answer. c. Perform division first. Perform subtraction next. Round the final answer. Exercise 2.4.3 Summary Learning Objectives Use significant figures correctly in arithmetical operations. Rounding Before dealing with the specifics of the rules for determining the significant figures in a calculated result, we need to be able to round numbers correctly. To round a number, first decide how many significant figures the number should have. Once you know that, round to that many digits, starting from the left. If the number immediately to the right of the last significant digit is less than 5, it is dropped and the value of the last significant digit remains the same. If the number immediately to the right of the last significant digit is greater than or equal to 5, the last significant digit is increased by 1. Consider the measurement 207.518 m. Right now, the measurement contains six significant figures. How would we successively round it to fewer and fewer significant figures? Follow the process as outlined in Table 2.4.1. \| Number of Significant Figures \| Rounded Value \| Reasoning \| --- Table 2.4.1: Rounding examples\| 6 \| 207.518 \| All digits are significant \| \| 5 \| 207.52 \| 8 rounds the 1 up to 2 \| \| 4 \| 207.5 \| 2 is dropped \| \| 3 \| 208 \| 5 rounds the 7 up to 8 \| \| 2 \| 210 \| 8 is replaced by a 0 and rounds the 0 up to 1 \| \| 1 \| 200 \| 1 is replaced by a 0 \| Notice that the more rounding that is done, the less reliable the figure is. An approximate value may be sufficient for some purposes, but scientific work requires a much higher level of detail. It is important to be aware of significant figures when you are mathematically manipulating numbers. For example, dividing 125 by 307 on a calculator gives 0.4071661238… to an infinite number of digits. But do the digits in this answer have any practical meaning, especially when you are starting with numbers that have only three significant figures each? When performing mathematical operations, there are two rules for limiting the number of significant figures in an answer—one rule is for addition and subtraction, and one rule is for multiplication and division. In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the least precise operation. An answer is no more precise than the least precise number used to get the answer. Multiplication and Division For multiplication or division, the rule is to count the number of significant figures in each number being multiplied or divided and then limit the significant figures in the answer to the lowest count. An example is as follows: The final answer, limited to four significant figures, is 4,094. The first digit dropped is 1, so we do not round up. Scientific notation provides a way of communicating significant figures without ambiguity. You simply include all the significant figures in the leading number. For example, the number 450 has two significant figures and would be written in scientific notation as 4.5 × 10 2, whereas 450.0 has four significant figures and would be written as 4.500 × 10 2. In scientific notation, all significant figures are listed explicitly. Example 2.4.1 Write the answer for each expression using scientific notation with the appropriate number of significant figures. 23.096 × 90.300 125 × 9.000 Solution a \| Explanation \| Answer \| --- \| \| The calculator answer is 2,085.5688, but we need to round it to five significant figures. Because the first digit to be dropped (in the tenths place) is greater than 5, we round up to 2,085.6. \| 2.0856×10 3 \| b \| Explanation \| Answer \| --- \| \| The calculator gives 1,125 as the answer, but we limit it to three significant figures. \| 1.13×10 3 \| Addition and Subtraction How are significant figures handled in calculations? It depends on what type of calculation is being performed. If the calculation is an addition or a subtraction, the rule is as follows: limit the reported answer to the rightmost column that all numbers have significant figures in common. For example, if you were to add 1.2 and 4.71, we note that the first number stops its significant figures in the tenths column, while the second number stops its significant figures in the hundredths column. We therefore limit our answer to the tenths column. We drop the last digit—the 1—because it is not significant to the final answer. The dropping of positions in sums and differences brings up the topic of rounding. Although there are several conventions, in this text we will adopt the following rule: the final answer should be rounded up if the first dropped digit is 5 or greater, and rounded down if the first dropped digit is less than 5. Example 2.4.2 13.77 + 908.226 1,027 + 611 + 363.06 Solution a \| Explanation \| Answer \| --- \| \| The calculator answer is 921.996, but because 13.77 has its farthest-right significant figure in the hundredths place, we need to round the final answer to the hundredths position. Because the first digit to be dropped (in the thousandths place) is greater than 5, we round up to 922.00 \| 922.00=9.2200×10 2 \| b \| Explanation \| Answer \| --- \| \| The calculator gives 2,001.06 as the answer, but because 611 and 1027 has its farthest-right significant figure in the ones place, the final answer must be limited to the ones position. \| 2,001.06=2.001×10 3 \| Exercise 2.4.2 Write the answer for each expression using scientific notation with the appropriate number of significant figures. 217 ÷ 903 13.77 + 908.226 + 515 255.0 − 99 0.00666 × 321 Answer a:0.240=2.40×10−1Answer b:1,437=1.437×10 3Answer c:156=1.56×10 2Answer d:2.14=2.14×10 0 Remember that calculators do not understand significant figures. You are the one who must apply the rules of significant figures to a result from your calculator. Calculations Involving Multiplication/Division and Addition/Subtraction In practice, chemists generally work with a calculator and carry all digits forward through subsequent calculations. When working on paper, however, we often want to minimize the number of digits we have to write out. Because successive rounding can compound inaccuracies, intermediate rounding needs to be handled correctly. When working on paper, always round an intermediate result so as to retain at least one more digit than can be justified and carry this number into the next step in the calculation. The final answer is then rounded to the correct number of significant figures at the very end. Video 2.4.1: Significant figures in mixed operations ( Video 2.4.2: In the worked examples in this text, we will often show the results of intermediate steps in a calculation. In doing so, we will show the results to only the correct number of significant figures allowed for that step, in effect treating each step as a separate calculation. This procedure is intended to reinforce the rules for determining the number of significant figures, but in some cases it may give a final answer that differs in the last digit from that obtained using a calculator, where all digits are carried through to the last step. Example 2.4.3 2(1.008 g) + 15.99 g 137.3 s + 2(35.45 s) 118.7⁢g 2−35.5⁢g Solution a. \| Explanation \| Answer \| --- \| \| 2(1.008 g) + 15.99 g = ##### Perform multiplication first. 2 (1.008 g 4 sig figs) = 2.01 6 g4 sig figs The number with the least number of significant figures is 1.008 g; the number 2 is an exact number and therefore has an infinite number of significant figures. ###### Then, perform the addition. 2.01 6 gthousandths place + 15.9 9 g hundredths place (least precise) = 18.006 g ###### Round the final answer. Round the final answer to the hundredths place since 15.99 has its farthest right significant figure in the hundredths place (least precise). \| 18.01 g (rounding up) \| b. \| Explanation \| Answer \| --- \| \| 137.3 s + 2(35.45 s) = ###### Perform multiplication first. 2(35.45 s 4 sig figs) = 70.90 s4 sig figs The number with the least number of significant figures is 35.45;the number 2 is an exact number and therefore has an infinite number of significant figures. ###### Then, perform the addition. 137.3 s tenths place (least precise)+ 70.90 shundredths place= 208.20 s ###### Round the final answer. Round the final answer to the tenths place based on 137.3 s. \| 208.2 s \| c. \| Explanation \| Answer \| --- \| \| 118.7⁢g 2−35.5⁢g = ###### Perform division first. 118.7⁢g 2 4 sig figs= 59.35 g4 sig figs The number with the least number of significant figures is 118.7 g; the number 2 is an exact number and therefore has an infinite number of significant figures. ###### Perform subtraction next. 59.35 ghundredths place− 35.5 g tenths place (least precise)= 23.85 g ###### Round the final answer. Round the final answer to the tenths place based on 35.5 g. \| 23.9 g (rounding up) \| Exercise 2.4.3 Complete the calculations and report your answers using the correct number of significant figures. 5(1.008s) - 10.66 s 99.0 cm+ 2(5.56 cm) Answer a-5.62 sAnswer b110.2 cm Summary Rounding If the number to be dropped is greater than or equal to 5, increase the number to its left by 1 (e.g. 2.9699 rounded to three significant figures is 2.97). If the number to be dropped is less than 5, there is no change (e.g. 4.00443 rounded to four significant figures is 4.004). The rule in multiplication and division is that the final answer should have the same number of significant figures as there are in the number with the fewest significant figures. The rule in addition and subtraction is that the answer is given the same number of decimal places as the term with the fewest decimal places. 2.4: Significant Figures in Calculations is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. 2.4: Significant Figures in Calculations by Henry Agnew, Marisa Alviar-Agnew is licensed CK-12. Original source: Toggle block-level attributions Back to top 2.3: Significant Figures - Writing Numbers to Reflect Precision 2.5: The Basic Units of Measurement Was this article helpful? 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188018	https://secure.medicalletter.org/TML-article-1493a	Friday, September 19, 2025 Login/Register Subscribe \| Donate \| Search Because the source matters. Not a subscriber My Account My Exams My Orders Logout The Medical Letter on Drugs and Therapeutics FROMISSUE1493 April 25, 2016 View Complete Issue Table of Contents 2019 In Brief: New Recommendations for Use of Metformin in Renal Impairment Download PDF: US English Show Related TermsHide Related Terms View Complete Issue Send Article Feedback Med Lett Drugs Ther. 2016 Apr 25;58(1493):51 Disclosures Objective(s) The FDA has required labeling changes that replace serum creatinine (SCr) with estimated glomerular filtration rate (eGFR) as the parameter used to determine the appropriateness of treatment with the biguanide metformin (Glucophage, and others) in patients with renal impairment. These changes will allow more patients with mild to moderate renal impairment to receive metformin, which is generally the first drug prescribed for treatment of type 2 diabetes. Metformin was previously contraindicated in women with a SCr level ≥1.4 mg/dL and in men with a SCr level ≥1.5 mg/dL, but use of SCr as a surrogate indicator tends to underestimate renal function in certain populations (e.g., younger patients, men, black patients, patients with greater muscle mass). The calculation of eGFR takes into account age, race, and sex, as well as SCr level, providing a more accurate assessment of kidney function. A literature review summarized in an FDA Drug Safety Communication concluded that, based on eGFR, metformin is safe to use in patients with mild renal impairment and in some patients with moderate renal impairment.1 The eGFR should be calculated before patients begin treatment with metformin and at least annually thereafter. Metformin is now contraindicated in patients with an eGFR <30 mL/min/1.73 m2, and starting treatment with the drug in patients with an eGFR between 30 and 45 mL/min/1.73 m2 is not recommended. If the eGFR falls below 45 mL/min/1.73 m2 in a patient already taking metformin, the benefits and risks of continuing treatment should be assessed. Metformin should be not be administered for 48 hours after an iodinated contrast imaging procedure in patients with an eGFR <60 mL/min/1.73 m2 or a history of liver disease, alcoholism, or heart failure, or in those receiving intra-arterial contrast, and the eGFR should be re-evaluated before treatment is restarted. FDA Drug Safety Communication: FDA revises warnings regarding use of the diabetes medicine metformin in certain patients with reduced kidney function. Available at: www.fda.gov. Accessed April 14, 2016. EMAIL THIS ARTICLE In Brief: New Recommendations for Use of Metformin in Renal Impairment Published: April 25, 2016 The FDA has required labeling changes that replace serum creatinine (SCr) with estimated glomerular filtration rate (eGFR) as the parameter used to determine the appropriateness of treatment with the biguanide metformin (Glucophage, and others) in patients with renal... © The Medical Letter, Inc. All Rights Reserved. The Medical Letter, Inc. does not warrant that all the material in this publication is accurate and complete in every respect. The Medical Letter, Inc. and its editors shall not be held responsible for any damage resulting from any error, inaccuracy, or omission. This article has been freely provided. Previous articleNext article
188019	https://physics.stackexchange.com/questions/692322/potential-energy-density-formula-for-waves-on-a-string	forces - Potential energy density formula for waves on a string - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Potential energy density formula for waves on a string Ask Question Asked 3 years, 7 months ago Modified3 years, 7 months ago Viewed 135 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. For a string whose position is labelled by the x coordinate and the purely transverse displacement labelled by the y coordinate, then according to this handout, the potential energy density is u p(x)=1 2 T(∂y∂x)2 u p(x)=1 2 T(∂y∂x)2 where T T is the constant tension in the string. In the derivation for this formula, it was assumed that ∂y∂x<<1,∂y∂x<<1, i.e. the string is close to equilibrium. My question is, when is such a formula useful? For what waves in the real world can we use this formula? The condition ∂y∂x<<1∂y∂x<<1 seems to be quite a strong one such that the wave is nearly non-existent. forces energy waves potential Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications asked Feb 2, 2022 at 15:44 TaeNyFanTaeNyFan 4,396 1 1 gold badge 15 15 silver badges 73 73 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. That's a variation of the small angle approximation. It means the slope of the tangent line to the curve is much smaller than one. This means tan x≈x tan⁡x≈x. That's good to one part in 10 if \|x tan x−1\|<0.1\|x tan⁡x−1\|<0.1. Examples of this use is really any system approximately in equilibrium. For example the actual equation of motion for a pendulum is d 2 θ d t 2+g l sin θ=0 d 2 θ d t 2+g l sin⁡θ=0. While its non-linear, setting sin θ→θ sin⁡θ→θ yields an easier equation to solve. That approximation for this problem is typically considered pretty good for initial displacements less than ten degrees. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Feb 2, 2022 at 16:06 R. RomeroR. Romero 2,931 8 8 silver badges 14 14 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. 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188021	https://www.youtube.com/watch?v=Uwts72SSvWw	ES 5 Math Subtract Decimals Using a Number Line _ Academics DropBox _ Staff - Academics 1170 subscribers Description 497 views Posted: 19 Jul 2022 Transcript: here we will take a look at subtracting decimals using a number line this standard extends students previous experiences with adding and subtracting whole numbers and their understanding of place value with decimals this standard requires that students utilize models drawings and strategies based on place value rather than relying on algorithms here we're going to look at subtracting 8 and 752 thousandths from twelve and four tenths to begin with we'll make an estimate so that we could determine whether our actual answer is reasonable or not twelve and four tenths i'm just going to go ahead and round that to the nearest whole number for my estimate i know that it is just four tenths away from 12 but six tenths away from 13 so it is a little closer to 12. so i can estimate this to be 12 minus well i know that 8 and hundred fifty-two thousandths is close to nine wholes closer to nine than it is to eight so i will round that to nine and i know that twelve minus nine is three and so this lets me know that our answer should be around three i actually know it will end up being a little bit more than three since i rounded eight and seventy five eight and seven hundred fifty two thousands up and i rounded twelve and four tenths down just a little bit so i'm going to use a number line i'm just going to create an open number line here at the bottom and then i'm going to start with our twelve and four tenths here down to the right hand side so that i can show my subtraction um and move to the left on my number line um as i subtract so i'm going to subtract 8 and 752 thousandths from 12 and 4 tenths not all at once though i'm going to decompose this eight and seven hundred fifty two thousands so first thing i'm going to do is just work with the whole number eight so i'm going to go ahead and subtract eight wholes so i'll show a jump where i'm going to subtract eight and i know now i would be left with four and four tenths now i when i look at the number of tenths in the seven hundred fifty two thousands i see i have seven tenths some of you may be able to mentally subtract seven tenths from four and four tenths others may find it helpful to decompose that i'm going to decompose that and just go ahead and take away four tenths at first so i can get to the friendly number of four so i'm going to subtract four tenths and now i would be at four wholes exactly well i still have three tenths to subtract until i've taken away the entire seven tenths so i'm going to go ahead and subtract three tenths now and again you may be able to do this mentally in one jump or you might have also needed to break it up just one tenth at a time so i know if i subtracted one tenth from four i would be at three and nine tenths if i subtracted another tenth i would be at three and eight tenths and then if i subtracted one more tenth i would be at three and seven tenths so again you might have been able to just subtract those three tenths altogether and get right to that three and seven tenths or you might have needed to do that one tenth at a time either way it's perfectly fine so now i can see that i have already subtracted so our 8 and 752 thousandths i've subtracted the 8 wholes and i've now subtracted the 7 10 in those two steps now i'm going to focus on the five hundredths well what i'm looking at three and seven tenths it can be helpful to think about that as three and seventy hundredths which it is equivalent to that can help me subtract the five hundredths that i need to subtract i know 70 hundredths minus five hundredths would be sixty five hundredths i'm going to go ahead and subtract the five hundredths now and i would be at three and sixty-five hundredths so now i've subtracted the five hundredths and what i have left to subtract are the two thousandths there so again it might help to think about this and not just three and sixty-five hundredths but think about it as three and six 650 thousandths because we need to take two thousandths away from that well i know that six hundred fifty thousandths minus two thousandths would actually be six hundred forty eight thousandths so i can go ahead and subtract two thousandths and i would be at three and 648 thousandths and that is our final answer that is the difference between twelve and four tenths and eight and seven hundred fifty-two thousands so the difference is three and six hundred forty-eight thousands i'll compare that to our estimated answer which was three uh three and six hundred forty-eight thousands is pretty close to three and if you remember what we said in the beginning was we actually expected our answer would be a little more than three just based on how we had rounded those numbers for our estimate so again this is how we can subtract decimals using a number line
188022	https://www.k5learning.com/free-math-worksheets/fifth-grade-5/algebra/expressions-2-variables	Skip to main content Breadcrumbs Worksheets Math Grade 5 Algebra Expressions with 2 variables Expressions with 2 variables Evaluate expressions with x an y Students are given algebraic expressions with two variables (x,y) and evaluate them for given values of the variables. Add/subtract only: Worksheet #1 Worksheet #2 Worksheet #3 4 operations: Worksheet #4 Worksheet #5 Worksheet #6 With exponents: Worksheet #7 Worksheet #8 Worksheet #9 Become a Member These worksheets are available to members only. Join K5 to save time, skip ads and access more content. Learn More Join Now What is K5? K5 Learning offers free worksheets, flashcards and inexpensive workbooks for kids in kindergarten to grade 5. Become a member to access additional content and skip ads. Help us give away worksheets Our members helped us give away millions of worksheets last year. We provide free educational materials to parents and teachers in over 100 countries. If you can, please consider purchasing a membership ($24/year) to support our efforts. Members skip ads and access exclusive features. Learn about member benefits Join Now
188023	https://shop.elsevier.com/books/head-and-neck-imaging-2-volume-set/som/978-0-323-05355-6	Head and Neck Imaging - 2 Volume Set - 5th Edition \| Elsevier Shop We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. For more information, see our Cookie Policy. Accept all cookies Cookie settings Books Journals Browse by subject Back Discover Books & Journals by subject ### Life Sciences Agricultural & Biological Sciences Drug Discovery Immunology Life Sciences Microbiology & Virology Neuroscience Pharmaceutical Sciences Pharmacology Toxicology All ### Physical Sciences & Engineering Astronomy, Astrophysics, Space Science Built Environment Chemical Engineering Chemistry Computer Science Earth & Planetary Sciences Energy & Power Engineering & Technology Environmental Sciences Mathematics Materials Science Physics All ### Social Sciences & Humanities Arts & Humanities Business, Management & Accounting Decision Sciences Economics & Finance Forensics Psychology Social Sciences All ### Health Dentistry Health Professions Medicine Nursing & Midwifery Veterinary Science & Veterinary Medicine All Search Open search View Cart US My account Open main site navigation Home Books Subjects Medicine Head and Neck Imaging - 2 Volume Set Head and Neck Imaging - 2 Volume Set Expert Consult- Online and Print 5th Edition - April 11, 2011 Latest edition Imprint: Mosby Authors:Peter M. Som, Hugh D. Curtin Language: English Hardback ISBN:9780323053556 9 7 8 - 0 - 3 2 3 - 0 5 3 5 5 - 6 eBook ISBN:9780323086936 9 7 8 - 0 - 3 2 3 - 0 8 6 9 3 - 6 eBook ISBN:9780323087926 9 7 8 - 0 - 3 2 3 - 0 8 7 9 2 - 6 eBook ISBN:9780323248938 9 7 8 - 0 - 3 2 3 - 2 4 8 9 3 - 8 Head and Neck Imaging, by Drs. Peter M. Som and Hugh D. Curtin, delivers the encyclopedic and authoritative guidance you’ve come to expect from this book – the expert guidanc…Read more Purchase options Info / Buy BACK-TO-SCHOOL Fuel your confidence! Up to 25% off learning resources Shop now Description Head and Neck Imaging, by Drs. Peter M. Som and Hugh D. Curtin, delivers the encyclopedic and authoritative guidance you’ve come to expect from this book – the expert guidance you need to diagnose the most challenging disorders using today’s most accurate techniques. New state-of-the-art imaging examples throughout help you recognize the imaging presentation of the full range of head and neck disorders using PET, CT, MRI, and ultrasound. Enhanced coverage of the complexities of embryology, anatomy, and physiology, including original color drawings and new color anatomical images from Frank Netter, help you distinguish subtle abnormalities and understand their etiologies. Access to the complete book’s contents is available online at www.expertconsult.com, which allows you to compare its images onscreen with the imaging findings you encounter in practice. Key features Compare your imaging findings to thousands of crystal-clear examples representing every type of head and neck disorder…both inside the book and onscreen at www.expertconsult.com. Gain an international perspective from global authorities in the field. Find information quickly with a logical organization by anatomic region. Readership Radiology Product details Edition: 5 Latest edition Published: April 11, 2011 Imprint: Mosby Language: English About the authors PS Peter M. Som Affiliations and expertise Professor of Radiology, Otolaryngology, and Radiation Oncology; Chief of Head and Neck Radiology, Mount Sinai Medical Center, New York, NY HC Hugh D. Curtin Affiliations and expertise Professor of Radiology, Harvard Medical School; Chief of Radiology, Department of Radiology, Massachusetts Eye and Ear Infirmary, Boston, MA Related books product Useful links Book awardsBook bestsellersBook imprintsBook series(opens in new tab/window)Flexible eBook solutionsNew book releasesUpcoming book releases Quick help eBook format help(opens in new tab/window)My account(opens in new tab/window)Returns & refunds(opens in new tab/window)Shipping & delivery(opens in new tab/window)Subscriptions & renewals(opens in new tab/window)Support & contact(opens in new tab/window)Tax exempt orders(opens in new tab/window) Solutions ClinicalKey(opens in new tab/window)ClinicalKey AI(opens in new tab/window)Embase(opens in new tab/window)Evolve(opens in new tab/window)Mendeley(opens in new tab/window)Knovel(opens in new tab/window)Reaxys(opens in new tab/window)ScienceDirect(opens in new tab/window) About About Elsevier(opens in new tab/window)Careers(opens in new tab/window)Newsroom(opens in new tab/window) Copyright © 2025 Elsevier, its licensors, and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies. Terms & conditions Privacy policy Accessibility Cookie Settings Support & contact
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LOGIN SHOPPING BASKET LOGIN BUY A BLOOD TEST Quick test finder Best Sellers Full Body MOT Women's Health Men's Health Hormone Tests Fertility Sexual Health Sports and Fitness Wellness and Nutrition Vitamins A-Z list Tests For Children Bundles Symptoms search Gift cards BOOK A CLINIC APPOINTMENT Find your nearest clinic London Clinics London Clinics Canary Wharf Chiswick Croydon Liverpool Street Victoria East and South East Brighton and Hove Cambridge Chelmsford Crawley Croydon Guildford Maidstone Milton Keynes Oxford Southampton Tunbridge Wells West and South West Bristol Cheltenham Exeter Oxford Plymouth Southampton Midlands Birmingham North, North East and North West Leeds Liverpool Manchester Newcastle York Wales Cardiff Scotland Edinburgh Glasgow SERVICES Weight loss medication Online GP Consultations Fertility Specialist Consultation Menopause Specialist Consultation Longevity Medicine GP Consultation HOW IT WORKS Frequently Asked Qs Health Dashboard Our clinics Home kits Subscriptions GP consultations Fertility consultations Menopause consultations Children tests Customer stories Custom test request Wellness Library Wellness Library Women's Health Men's Health Health advice from our GPs PARTNERSHIPS Partner Services Affiliates Employee Health Checks ABOUT US Meet the team Contact us Custom test request About us About us Meet the team Contact us Most popular tests Health Dashboard Customer Stories Wellness Library Partner Services Affiliates LOGIN Close Buy a blood test Quick test finder Best Sellers Full Body MOT Women's Health Men's Health Hormone Tests Fertility Sexual Health Sports and Fitness Wellness and Nutrition Vitamins A-Z list Tests For Children Bundles Symptoms search Gift cards Book a clinic appointment Find your nearest clinic London Clinics Canary Wharf Chiswick Croydon Liverpool Street Victoria East and South East Brighton and Hove Cambridge Chelmsford Crawley Croydon Guildford Maidstone Milton Keynes Oxford Southampton Tunbridge Wells West and South West Bristol Cheltenham Exeter Oxford Plymouth Southampton Midlands Birmingham North, North East and North West Leeds Liverpool Manchester Newcastle York Wales Cardiff Scotland Edinburgh Glasgow Services Weight loss medication Online GP Consultations Fertility Specialist Consultation Menopause Specialist Consultation Longevity Medicine GP Consultation How it works Frequently Asked Qs Health Dashboard Our clinics Home kits Subscriptions GP consultations Fertility consultations Menopause consultations Children tests Customer stories Custom test request Wellness Library Women's Health Men's Health Health advice from our GPs Partnerships Partner Services Affiliates Employee Health Checks About us Meet the team Contact us Custom test request July 15, 2024 Dr. Adam Staten Monitoring Low and High Progesterone Levels Progesterone is a crucial hormone for women's health, playing a vital role in the menstrual cycle and pregnancy. It regulates the menstrual cycle, supports pregnancy, and maintains overall hormonal balance. Imbalances in progesterone levels can lead to a range of symptoms and health issues, making it essential to monitor and manage these levels effectively. One Day Tests offers a comprehensive Progesterone Blood Test that can help determine whether ovulation is occurring and if any hormonal imbalances are contributing to menstrual irregularities. With quick turnaround times and advanced health dashboards, One Day Tests helps you stay informed and proactive about your hormonal health. The Important Role of Progesterone in the Body In women, progesterone is produced in the ovaries by the corpus luteum after ovulation. If pregnancy occurs, the placenta takes over its production to sustain the pregnancy. Both men and women produce smaller quantities of progesterone in the adrenal glands. This hormone is essential for maintaining a balanced endocrine system, supporting reproductive functions, and influencing overall health. By regulating menstrual cycles in women and contributing to testosterone production in men, progesterone plays a crucial role in overall well-being. What Is the Normal Level of Progesterone? Progesterone levels vary depending on the timing of the test. Levels start to rise midway through the menstrual cycle, continuing to increase for about 6 to 10 days before falling if fertilisation does not occur. During early pregnancy, levels continue to rise. Normal ranges based on different phases and conditions are as follows: Female (pre-ovulation): Less than 1 ng/mL or 3.18 nmol/L Female (mid-cycle): 5 to 20 ng/mL or 15.90 to 63.60 nmol/L Male: Less than 1 ng/mL or 3.18 nmol/L Postmenopausal: Less than 1 ng/mL or 3.18 nmol/L Pregnancy 1st trimester: 11.2 to 90.0 ng/mL or 35.62 to 286.20 nmol/L Pregnancy 2nd trimester: 25.6 to 89.4 ng/mL or 81.41 to 284.29 nmol/L Pregnancy 3rd trimester: 48 to 150 ng/mL or more, or 152.64 to 477 nmol/L or more Day 21 Progesterone Level Testing Day 21 testing plays a vital role in fertility assessments by measuring progesterone levels at a key point in the menstrual cycle. Typically conducted around day 21 in a 28-day cycle (about 7 days before the next period), this test coincides with peak progesterone production, confirming ovulation and evaluating the luteal phase. While the length of the follicular phase can vary among women, the luteal phase consistently lasts about 14 days. Your peak progesterone level should occur seven days after ovulation and seven days before your period starts. Adequate progesterone is essential for preparing the uterine lining for embryo implantation and supporting early pregnancy. Monitoring progesterone allows healthcare providers to identify and address potential fertility issues, improving the chances of conception. This test also helps determine if progesterone levels are sufficient to maintain a potential pregnancy, as low levels can impede implantation and overall fertility. Symptoms of Low Progesterone Levels Symptoms in Women Low progesterone levels can lead to a variety of symptoms that affect menstrual and reproductive health. These symptoms include irregular menstrual cycles, where periods may become erratic or infrequent. Women might also experience heavy bleeding and premenstrual syndrome (PMS) symptoms such as mood swings, bloating, and breast tenderness. Additionally, low progesterone can make it difficult to maintain a pregnancy, increasing the risk of miscarriage and fertility issues. Illnesses Associated with Low Progesterone Levels Low progesterone levels in women can lead to several health conditions: Infertility: Progesterone is essential for preparing the uterus for the implantation of a fertilised egg. Low levels can prevent successful implantation, leading to infertility. Miscarriage: Adequate progesterone is crucial for maintaining pregnancy. Insufficient levels can lead to early pregnancy loss. Polycystic Ovary Syndrome (PCOS): Women with PCOS often have hormonal imbalances, including low progesterone, which can cause irregular menstrual cycles and infertility. Menstrual Disorders: Low progesterone can cause heavy bleeding, irregular periods, and severe PMS symptoms. Illnesses Associated with High Progesterone Levels High progesterone levels in women can be linked to several health conditions: Ovarian Cysts: Elevated progesterone can contribute to the formation of ovarian cysts, which can cause pelvic pain, bloating, and menstrual irregularities. Congenital Adrenal Hyperplasia (CAH): This genetic disorder affects the adrenal glands and can lead to excessive progesterone production, disrupting normal hormonal balance. Monitoring and maintaining balanced progesterone levels is crucial for overall health and well-being. One Day Tests offers a comprehensive progesterone blood test to help you keep track of your hormone levels and address any imbalances effectively. Our Progesterone Blood Test Progesterone Blood Test £ £ 37.00 Measure Your Progesterone Levels With One Day Tests Regular blood testing for progesterone levels is crucial for maintaining hormonal balance and overall health. By monitoring these levels, you can detect imbalances early and address potential health issues before they become severe. One Day Tests offers a comprehensive progesterone blood test, providing insights on whether ovulation is occurring and if there are any hormonal imbalances contributing to menstrual irregularities. Get in touch by phone at 0845 527 07 67, email us at hello@onedaytests.com, or message us on WhatsApp for more information. Previous article A Guide to Anti-Müllerian Hormone (AMH) Blood Testing Next article Understanding Progesterone Levels: Importance, Health Issues, and Testing COMPARE
188025	https://math.stackexchange.com/questions/2458538/gcda-b-is-unique-up-to-units-in-a-unique-factorization-domain	elementary number theory - gcd(a,b) is unique up to units in a unique factorization domain - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more gcd(a,b) is unique up to units in a unique factorization domain Ask Question Asked 7 years, 11 months ago Modified7 years, 11 months ago Viewed 2k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Im trying to prove that for some a,b a,b (not both zero) in a unique factorization domain then the gcd(a,b a,b) is of the form {g u:u g u:u is a unit} where g g=gcd(a,b a,b). I'm looking for a hint. What I know is: gcd(a,b a,b)= g=m a+n b g=m a+n b for some m,n m,n in my domain. g\|a g\|a and g\|b g\|b. a=a 1 a 2⋅⋅⋅a n a=a 1 a 2⋅⋅⋅a n and b=b 1 b 2⋅⋅⋅b m b=b 1 b 2⋅⋅⋅b m I'd appreciate any help. Thanks in advance. elementary-number-theory Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Oct 5, 2017 at 6:25 ABCABC 379 1 1 silver badge 8 8 bronze badges Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. Using the definition of gcd, if you take two of them, they divide each other, so are associate Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 5, 2017 at 6:47 TancrediTancredi 1,562 9 9 silver badges 13 13 bronze badges Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Bézout's identity does not necessarily hold in a unique factorization domain. In a general domain, a greatest common divisor of a a and b b is defined to be any element d d such that d∣a d∣a and d∣c d∣c for all c c, if c∣a c∣a and c∣b c∣b, then c∣d c∣d In a unique factorization domain a gcd of two elements is proved to exist (via the middle school rule), but in any domain we can prove that if a gcd of a a and b b exists, then it is determined up to invertible elements; more precisely if d d and e e are greatest common divisors of a a and b b, then there exists an invertible element u u such that e=d u e=d u. Proof. By property 1 (applied to e e), e∣a e∣a and e∣b e∣b; by property 2 (applied to d d), e∣d e∣d. Similarly, d∣e d∣e. Hence d=e u d=e u and e=d v e=d v, so e=e u v e=e u v and either e=0 e=0 or u v=1 u v=1. If u v=1 u v=1, then u u is invertible; if e=0 e=0, we can take u=1 u=1.□◻ Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 5, 2017 at 8:53 egregegreg 246k 21 21 gold badges 155 155 silver badges 353 353 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions elementary-number-theory See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 2Why isn't gcd(x 2+3 x+2,x 2+x)=(x+1)gcd(x 2+3 x+2,x 2+x)=(x+1)? [unit normalization of gcds] Related 0Use the unique factorization for integers theorem and the definition of logarithm to prove that log 3(7)log 3⁡(7) is irrational. 2Prime factorization problem... 2Proof that if gcd(a,b)=1 gcd(a,b)=1 and a∣n a∣n and b∣n b∣n, a b∣n a b∣n 2On gcd(a,x)=gcd(b,x)=k⟹g c d(a b,x)=k gcd(a,x)=gcd(b,x)=k⟹g c d(a b,x)=k Hot Network Questions Exchange a file in a zip file quickly Do we need the author's permission for reference Is there a way to defend from Spot kick? How to convert this extremely large group in GAP into a permutation group. в ответе meaning in context Is direct sum of finite spectra cancellative? Are there any world leaders who are/were good at chess? 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188026	https://www.ncbi.nlm.nih.gov/books/NBK547748/	Embryology, Uterus - StatPearls - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Search term Embryology, Uterus Mariana Moncada-Madrazo; Cristobal Rodríguez Valero. Author Information and Affiliations Authors Mariana Moncada-Madrazo 1; Cristobal Rodríguez Valero 2. Affiliations 1 Tecnológico de Monterrey 2 TecSalud Tecnológico de Monterrey Last Update: July 25, 2023. Go to: Introduction The development of the female reproductive system requires an elaborate process. In females, reproductive organs divide into three main groups: gonads, reproductive ducts, and external genitalia. The female reproductive system derives from four origins: mesoderm, primordial germ cells, coelomic epithelium, and mesenchyme. The uterus forms during Mullerian organogenesis accompanied by the development of the upper third of the vagina, the cervix, and both fallopian tubes. Knowledge of the embryology of the female reproductive tract provides insight into congenital pathologies that are related to these organs. The objective of this activity is to review uterine embryology and its clinical significance. Go to: Development Up to the fifth and sixth week of fetal life, the genital system remains indifferent. Two pairs of genital ducts are present at this time: the mesonephric (Wolffian duct) and paramesonephric (Mullerian duct). In females, the absence of anti-Mullerian hormone (AMH) and SRY gene conditions the regression of Wolff ducts and further differentiation of Mullerian ducts. The upper third of the vagina, the cervix, both fallopian tubes, and the uterus derive from the paramesonephric ducts. During the seventh week, paired paramesonephric ducts arise from focal invaginations of the coelomic epithelium that is found on the upper pole of each mesonephros, shortly after this the Mullerian ducts grow caudally and laterally to the urogenital ridges. In the eighth week, a vertical fusion of paramesonephric ducts occurs. The fused cranial end gives origin to the left and right parts of what will ultimately become the uterus. This structure contains mesoderm that will form the endometrium and myometrium. The unfused cranial ends of the Mullerian ducts will develop into the fallopian tubes, the fimbrial portion of the fallopian tubes derives from the tip of this structure that remains open and acquires a funnel shape. The caudal end of the fused ducts will form the upper third of the vagina. At this stage, a midline septum is present along these structures, and within the uterine cavity, this septum usually reabsorbs completely around 20 weeks, but it can persist and produce a septate uterus. Regarding uterine ligaments, both the round ligament and the ovarian ligament develop from the gubernaculum, and undifferentiated mesenchymal tissue is attached to the ovary in the female fetus. The round ligament must attach to both the ovary and uterus for the ovary to be in place. By the end of the first trimester development of the uterus and the other structures derived from the Mullerian ducts is complete. Go to: Molecular Level For the first 10 weeks, the human fetus has the potential to become either female or male. The final phenotype depends on genetic information that influences differentiation in the embryonic structures. A female fetus will classically develop if there is the presence of a XX genotype. For a male fetus to develop there must be the presence of a Y chromosome that codes for SRY protein that enables testicular, epididymis, ductus deferens, ejaculatory duct, and seminal vesicles differentiation and secretion of anti-Mullerian hormone (AMH) from the Sertoli cells which will inhibit the further differentiation of paramesonephric ducts and condition their regression. In cases where this characteristic does not occur, an immature female fetus, or an intersex fetus will develop. Go to: Function The primary function of the uterus is reproductive. The principal elements of uterine physiology are the endometrium and myometrium. The uterus accepts the ovum after fertilization, holds and provides nutrients and oxygen for the fetus and during birth, and it contracts to cause delivery. The uterus is a hormone-sensitive organ: differentiation, proliferation, exfoliation of the endometrium, and contraction during childbirth get regulated by the interaction between itself and the hypothalamus, pituitary gland, and ovaries. Go to: Testing Assessment of uterine anatomy through imaging is a priority when Mullerian abnormalities are suspected. Magnetic resonance imaging (MRI) has proven to be highly sensitive and specific. Therefore it is considered the gold standard for these pathologies. Additionally to its diagnostic accuracy, more often than not, Mullerian remnants are present. MRI permits the identification of endometrial activity within the Mullerian structures. This imaging technique can also demonstrate urinary tract, gastrointestinal, and skeletal abnormalities that can accompany these entities. Regarding treatment decisions, MRI can aid in the differentiation between surgically correctable abnormalities and inoperable forms. Go to: Pathophysiology Deviation from normal development of reproductive ducts can result in congenital structural anomalies referred to as Mullerian anomalies. Mullerian anomalies may result from arrested development of the paramesonephric ducts, failure of fusion of the paramesonephric ducts, or failure of resorption of the medial septum. The development of the gastrointestinal and urinary system occurs closely in time and space to this phenomenon and anomalies in the development of these organ systems may also affect the female reproductive system and vice-versa. Go to: Clinical Significance Mullerian anomalies can be either clinically asymptomatic and missed in routine gynecological examinations or manifest with infertility and amenorrhea. These pathologies represent unique challenges for establishing reproductive health. The American fertility association has classified this anomaly in seven categories : Class I: Hypoplasia/uterine hypoplasia. (Mayer Rokitansky Kuster Hauser syndrome) Class II: Unicornuate uterus Class III: Uterus didelphys. Class IV: Bicornuate uterus Class V: Septate uterus Class VI: Arcuate uterus Class VII: T-shaped uterus resulting from the exposure to Diethylstilbestrol in fetal life Mayer Rokitansky Kuster Hauser syndrome has a prevalence of 1 in 4000 to 5000 births and is one of the most frequent Mullerian abnormalities. It characteristically presents with uterine and vaginal agenesis or hypoplasia and can be accompanied by renal and bone abnormalities. Patients usually arrive at care due to primary amenorrhea with normal secondary sexual characteristics. Treatment of vaginal aplasia consists of the creation of a neovagina surgically or by dilation. Depression, anxiety, and female identity issues often occur in these patients, thus seeking counseling and peer support groups are recommended activities before and during treatment. Alternatives for having children include adoption and gestational surrogacy; this merits discussion during the consult. Also of clinical relevance after fetal life rudimentary structures derived from paramesonephric ducts can persist and be encountered in clinical practice: Testicular appendage: From the cranial end of the fussed paramesonephric ducts in males. Prostatic utricle: A small sac found in the prostatic urethra in males. Morgagni hydatid in females from the cranial end of the paramesonephric duct that does not contribute to the fallopian tube Go to: Review Questions Access free multiple choice questions on this topic. Comment on this article. Figure Uterus Embryology. Contributed by O Chaigasame, MD Go to: References 1. Kobayashi A, Behringer RR. Developmental genetics of the female reproductive tract in mammals. Nat Rev Genet. 2003 Dec;4(12):969-80. [PubMed: 14631357] 2. WITSCHI E. Embryology of the uterus: normal and experimental. Ann N Y Acad Sci. 1959 Jan 09;75:412-35. [PubMed: 13845458] 3. Robbins JB, Broadwell C, Chow LC, Parry JP, Sadowski EA. Müllerian duct anomalies: embryological development, classification, and MRI assessment. J Magn Reson Imaging. 2015 Jan;41(1):1-12. [PubMed: 25288098] 4. Guioli S, Sekido R, Lovell-Badge R. The origin of the Mullerian duct in chick and mouse. Dev Biol. 2007 Feb 15;302(2):389-98. [PubMed: 17070514] 5. Warne GL, Kanumakala S. Molecular endocrinology of sex differentiation. Semin Reprod Med. 2002 Aug;20(3):169-80. [PubMed: 12428197] 6. Chaudhry SR, Chaudhry K. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Jul 24, 2023. Anatomy, Abdomen and Pelvis: Uterus Round Ligament. [PubMed: 29763145] 7. Roly ZY, Backhouse B, Cutting A, Tan TY, Sinclair AH, Ayers KL, Major AT, Smith CA. The cell biology and molecular genetics of Müllerian duct development. Wiley Interdiscip Rev Dev Biol. 2018 May;7(3):e310. [PubMed: 29350886] 8. de Ziegler D, Pirtea P, Galliano D, Cicinelli E, Meldrum D. Optimal uterine anatomy and physiology necessary for normal implantation and placentation. Fertil Steril. 2016 Apr;105(4):844-54. [PubMed: 26926252] 9. Chandler TM, Machan LS, Cooperberg PL, Harris AC, Chang SD. Mullerian duct anomalies: from diagnosis to intervention. Br J Radiol. 2009 Dec;82(984):1034-42. [PMC free article: PMC3473390] [PubMed: 19433480] 10. Thomas DFM. The embryology of persistent cloaca and urogenital sinus malformations. Asian J Androl. 2020 Mar-Apr;22(2):124-128. [PMC free article: PMC7155797] [PubMed: 31322137] 11. Troiano RN, McCarthy SM. Mullerian duct anomalies: imaging and clinical issues. Radiology. 2004 Oct;233(1):19-34. [PubMed: 15317956] 12. Morcel K, Camborieux L, Programme de Recherches sur les Aplasies Müllériennes. Guerrier D. Mayer-Rokitansky-Küster-Hauser (MRKH) syndrome. Orphanet J Rare Dis. 2007 Mar 14;2:13. [PMC free article: PMC1832178] [PubMed: 17359527] 13. Sajjad Y. Development of the genital ducts and external genitalia in the early human embryo. J Obstet Gynaecol Res. 2010 Oct;36(5):929-37. [PubMed: 20846260] Disclosure:Mariana Moncada-Madrazo declares no relevant financial relationships with ineligible companies. Disclosure:Cristobal Rodríguez Valero declares no relevant financial relationships with ineligible companies. 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188027	https://ntrs.nasa.gov/api/citations/19870003970/downloads/19870003970.pdf	NASA Technical Memorandum 89409 [NASA-TM-89409) WATER FACILITIES IN N87-13403 EETHOSPECT ASD PECSPEET: AN I L L U M I N A T I N G TOOL EOR VEHICLE D E S X G N (NASA) 31 p C S C L 01A Unclas G3/02 43618 Water Facilities in Retrospect and Prospect -an Illuminating Tool for Vehicle Design Gary E. Erickson, David J. Peake, John Del Frate, Andrew M. Skow, and Gerald N. Malcolm November 1986 NASA National Aeronautics and Space Administration NASA Technical Memorandum 89409 Water Facilities in Retrospect and Prospect -an Illuminating Tool for Vehicle Design Gary E. Erickson, David J. Peake, John Del Frate, Ames Research Center, Moffett Field, California Andrew M. Skow, Gerald N. Malcolm, Eidetics International, Torrance, California November 1986 NASA National Aeronautics and Space Administration Ames Research Center Moffett Field, California 94035 WATER FACILITIES IN RETROSPECT AND PROSPECT-- AN ILLUMINATING TOOL FOR VEHICLE DESIGN Gary E. Erickson. David J. Peake. and John Del Frate NASA Ames Research Center Moffett Field, California 94035 USA and Andrew M. Skow and Gerald N . Malcolm Ei detics International Torrance, California 90505 USA S U M A R Y Water facilities play a fundamental role in the design of air, ground, and marine vehicles by provid- ing a qualitative, and sometimes quantitative, description of complex flow phenomena. channels, and tow tanks used as flow-diagnostic tools have experienced a renaissance in recent years in response to the increased complexity of designs suitable for advanced technology vehicles. These vehicles are frequently characterized by large regions of steady and unsteady three-dimensional flow separation and ensuing vortical flows. The visualization and interpretation of the complicated fluid motions about iso- lated vehicle components and complete configurations in a time- and cost-effective manner in hydrodynamic test facilities is a key element in the development of flow control concepts and. hence. improved vehicle designs. Water tunnels, This paper presents a historical perspective of the role of water facilities in the vehicle design The application of water facilities to specific aerodynamic and hydrodynamic flow problems is process. discussed, and the strengths and limitations of these important experimental tools are emphasized. 1. INTRODUCTION The successful development of vehicles suitable for aerodynamic and hydrodynamic applications requires an understanding of the configuration flow field at design and off-design conditions. facilities (tunnels, channels, and tow tanks) have historically provided valuable information on the fundamental fluid mechanics of two-dimensional (2-0) and three-dimensional (3-0) aerodynamic and hydrody- namic shapes operating at low speeds. The application of hydrodynamic facilities to basic aerodynamic research problems that was pioneered by Ludwig Pranti in the early days of flight continues to the present day. moted a surge of interest in water facilities as a diagnostic tool. The utilization of water tunnels, channels. and tow tanks to visualize the 3-0 separated and vortical flow fields about advanced, highly swept, military aircraft and slender missile configurations operating at extreme attitudes (angles of attack and sideslip) has been demonstrated by many researchers. attained by correlating study results with aerodynamic facility and flight results has led to a substan- tial increase in the applications of relatively simple water facilities to the complex aerodynamic and hydrodynamic shapes of interest for present- and future-generation vehicles. zation and nonintrusive measurement techniques have been developed, modified, and implemented to provide qualitative and quantitative information on the steady and unsteady flow fields about isolated vehicle components and complete configurations. direct proportion to the advancement of air, ground. and marine vehicle technologies. As a consequence, water facilities are now an important element in the design process, yielding insight into the complicated vortical fluid mechanisms and structures that are characteristic of flows about advanced-technology vehicles. Water The advent of high-performance flight vehicles that incorporate flow separation by design has pro- The increased level of confidence Concurrently, flow visuali- In general, the use of hydrodynamic facilities has risen in The visualization of a flow phenomenon represents a major advancement toward understanding and, sub- sequently. controlling the fluid mechanism. Flow visualization is easily performed in water and, as a consequence, aerodynamic as well as hydrodynamic problems have long been studied in water facilities. flow properties of water are similar to those in air, provided the flow simulations are restricted to the incompressible regime. The selection of water as a flow-visualization medium is largely based on the 800-fold increase in the density relative to air and, consequently, the excellent light-reflecting charac- teristics of tracers injected into the flow field. num powder, dye, or hydrogen bubbles. for example) provides direct visualization of steady and unsteady flows. ever, cost, complexity, and facility space are factors that constrain the test section size; hence, the model scale is generally relatively small. ft/sec in order to preserve the clarity of the flow structure, to avoid excessive model loads, and to preclude cavitation when it is an undesired phenomenon. The combination of small model scale and low fluid velocities results in Reynolds numbers that are orders of magnitude less than those achievable on larger-scale models in wind tunnels or o n full-scale air, ground, and marine vehicles operating in their respective environments. The Suitable illumination of the tracer particles (alumi- At the same scale and speed. the Reynolds number is 15 times greater in water than in air. How- In addition, the fluid velocities are typically less then 1 0 As a consequence of this Reynolds number mismatch. the fluid motion under 1 consideration must be the kind t h a t i s insensitive t o changes i n the Reynolds number. fundamental structure of the flow must be similar, regardless o f the Reynolds number. It i s fortunate t h a t many o f the complex flow phenomena, particularly 3-0 motions with vortices o r predominantly separated regimes, associated w i t h advanced-technology vehicles, lend themselves t o q u a l i t a t i v e (and sometimes quan- t i t a t i v e ) evaluation i n these f a c i l i t i e s . water tunnels. channels, and towing tanks a t low Reynolds numbers are topics o f considerable debate i n the technical conunity. the strengths and l i m i t a t i o n s of hydrodynamic t e s t f a c i l i t i e s applied t o flow problems i n aeronautical and maritime f l u i d dynamics. number o f water f a c i l i t i e s t h a t have been i n operation over the years and that remain i n operation today, the discussion w i l l Center on representative f a c i l i t i e s . The applications o f water f a c i l i t i e s t o vehicle design, encompassing the interests of the various sessions o f t h i s meeting. are discussed i n Section 3. O f necessity. only the flavor of the myriad basic and applied aerodynamic and related hydrodynamic studies w i l l be provided t o demonstrate the u t i l i t y and current potential o f water f a c i l i t i e s . A t the least, the The interpretation and application o f the r e s u l t s obtained i n This meeting i s both timely and useful as it provides a forum f o r the discussion o f Section 2 o f t h i s Paper w i l l present a historical review o f water f a c i l i t i e s . Because o f the large 2. HISTORICAL PERSPECTIVE 2.1 From Antiquity t o the Renaissance The visualization o f complex flow phenomena i n a water medium spans millennia. Observations and speculations on v o r t i c a l flows i n nature go back t o prehistoric times (Ref. 1). Stone Age a r t i f a c t s depicted the s p i r a l m t i o n s that were frequently observed i n a water medium. Over 2.000 years ago, the writings o f A r i s t o t l e (384-322 8.C.) described the observations of whirlpools a t sea and the resultant loss of ships caught i n t h i s powerful f l u i d motion (Ref. 1). Von Karman (Ref. 2) wrote o f an early paint- ing i n Bologna, I t a l y . of S t . Christopher walking through a flowing stream t h a t showed alternating vor- tices, o r Karman "vortex streets." behind the saint's foot. I n the centuries t o follow, the nonscientific observations o f flows i n water persisted. w r i t i n g s and elaborate drawings of Leonardo da Vinci (1452-1519) (Ref. 3). (Ref. 1). Leonardo's a r t i s t i c descriptions of nature single out a particular phenomenon, such as a vortex, from a global flow f i e l d and are. therefore, i n conon w i t h s c i e n t i f i c experiment. f l o w sketches are shown i n Fig. 1 taken from Ref. 1 . f l u i d motions i n water and a i r . According t o the account o f Truesdell (Ref. 4). such a conclusion could come only from d i r e c t observation o f the f l u i d motions i n both media. I n fact, Leonardo was the f i r s t t o discuss water and a i r as f l u i d s , and he designed the f i r s t water tunnel flow-visualization f a c i l i t y as shown i n Fig. 2 (from Ref. 5). water medium due t o the e f f e c t s o f compressibility. 2.2 From 1600 t o 1800 The e a r l i e s t documentation of flow mechanisms i n water f o r s c i e n t i f i c purposes appears t o be i n the As pointed out by Lugt Examples o f h i s f l u i d Leonardo frequently discussed the s i m i l a r i t y of Leonardo also recognized the l i m i t a t i o n s o f simulations of a i r f l o w i n a A century a f t e r Leonardo's passing, S i r Isaac Newton (1642-1727) claimed that h i s so-called "sine- square law o f a i r resistance" applied as w e l l to water. where the forces were proportional t o the respec- t i v e densities (Ref. 2). mental r e s u l t s i n water t o motion i n a i r and vice versa. o f Edme Mariotte (1620-1684). who measured the force acting on a f l a t p l a t e submerged i n a stream o f water. ous shapes, which were put i n motion by means of a rotating, o r whirling, arm. (1717-1783). Antoine Condorcet (1743-1794). and Charles Bossut (1730-1814) towed ship models i n s t i l l water (see Ref. 2). technique that i s so prevalent today. vortices (Ref. 1). 2.3 From 1800 t o 1900 Newton's statement was noteworthy i n t h a t it promoted the application o f experi- Newton's influence was present i n the experiment Jean Charles de Borda (1733-1799) performed experiments in a water f a c i l i t y using bodies o f vari- Jean Le Rond d'Alembert These investigations represented perhaps the f i r s t application o f the towing-tank I n 1780, J. C. Wilke used a water f a c i l i t y t o study atmospheric Sporadic experiments using water as the working f l u i d continued i n t o the 19th century. I n 1839, Hagen (Ref. 6) conducted studies o f water flowing through c y l i n d r i c a l tubes. where he observed the transi- t i o n from laminar t o turbulent flow. Hagen subsequently conducted more detailed testing of flow s t a b i l i t y i n tubes and documented h i s r e s u l t s i n 1854 (Ref. 7). experiments (Ref. 8) i n which he demonstrated that f l o w t r a n s i t i o n occurred when a parameter, now called the Reynolds number, exceeded a c e r t a i n value. rather than photographs. o r i g i n a l apparatus a t the University o f Manchester and representative r e s u l t s are documented by Van Dyke (Ref. 9). I n 1897, Hele Shaw performed experiments (Ref. 10) i n a t h i n tank t o study 2-D highly-viscous flows. The experiments of researchers spanning four centuries, s t a r t i n g with the work o f Leonardo, con- t r i b u t e d t o the development of new flow-visualization f a c i l i t i e s and techniques which l e d t o major advances i n f l u i d mechanics. I n 1883, Osborne Reynolds conducted h i s classic His flow-visualization results were documented by sketches Interestingly, Reynolds' experiments were repeated nearly 100 years l a t e r i n h i s 2 2.4 From 1900 t o 1935 hydrodynamic problems was pioneered by Ludwig Prandtl i n Germany beginning i n the early 1900s. Prandtl conducted experiments on 2-0 shapes i n a water channel where the flow was visualized on the free surface using aluminum powder. His flow-visualization method was f i r s t documented i n 1904 i n Ref. 11. A compila- t i o n o f Prandtl's 2-0 flow-visualization results obtained a t the Kaiser Wilhelm I n s t i t u t e (KWI) for Flow Research i n Gottingen. Germany was provided by Prandtl and Tietjens (Ref. 12) i n 1934. included (1) the propagation of turbulence i n boundary layers, (2) vortex shedding downstream of a plate, (3) vortex development behind a nonrotating cylinder and the development of Karman vortices, (4) laminar and turbulent boundary layers, (5) flow development around a r o t a t i n g cylinder, (6) s t a r t i n g vortex gener- ated by an a i r f o i l . (7) flow i n a diffuser with and without suction at the walls, and (8) cavitation phenomena. mechanics through the use of a water f a c i l i t y and the intuitiveness of the experimenter i s exemplified by item (3) above. Karl Hiemenz. a student o f Prandtl, b u i l t a water channel i n 1911 t o observe the flow separation behind a cylinder. The o s c i l l a t o r y vortex shedding t h a t he observed was unexpected and the phenomenon was a t f i r s t attributed t o model and tunnel wall asymmetries. The flow s i t u a t i o n was always repeatable, however, despite careful checks of the model and t e s t f a c i l i t y . This attracted the attention of von Karman (Ref. 2) who speculated t h a t there may be a natural and i n t r i n s i c reason f o r the phenom- non. now-famous Karman "vortex street .'I The u t i l i z a t i o n of water f a c i l i t i e s as tools t o study a wide range of fundamental aerodynamic and His studies Examples of Prandtl's results are shown i n Fig. 3 (from Ref. 12). The advancement o f f l u i d He calculated the s t a b i l i t y o f such a system o f vortices, which led t o the understanding of the The e f f o r t s o f Prandtl and h i s colleagues confirmed that, under c e r t a i n conditions, the flow patterns i n the v i c i n i t y of a body are s i m i l a r i n a i r , water, o r other l i q u i d o r gaseous f l u i d s . Prandtl demonstrated t h a t the f l u i d dynamic characteristics of a body exposed t o a f l o w i n one medium could be predicted from experiments i n a different medium. Prandtl's work overlapped the development of f l i g h t vehicles f o r m i l i t a r y and commercial applications. gated i n hydrodynamic f a c i l i t i e s . t e r i s t i c s i n any f l u i d , including a i r at low Mach numbers. Since the p r a c t i c a l design o f a i r c r a f t d i d not seriously account for compressibility during the f i r s t 20-30 years o f i t s h i s t o r y because of the low f l i g h t speeds, water f a c i l i t i e s were applied t o f l i g h t - v e h i c l e design, a l b e i t on a limited basis. problems were o f a 2-0 nature and emphasized the boundary-layer behavior and f low-separation characteris- t i c s o f a i r f o i l shapes suitable for the r e l a t i v e l y unswept wings o f t h i s period. The KWI i n Gottingen, Germany, maintained i t s r o l e as a leader i n the study o f laminar and turbulent boundary layers and the drag of aerodynamic shapes (Ref. 12). More frequently, however, water channel f a c i l i t i e s were employed f o r hydrodynamic f l o w problems r e l a t i n g t o cavitation phenomena and the resistance o f surface ships and submersibles. Hoerner (Ref. 13) provides an extensive l i s t of references o f water tunnel, channel. and towing-tank investigations o f marine vehicles conducted during t h i s period. 2.5 From 1935 t o 1950 On t h i s basis, and airplane wings were, accordingly, investi- It i s noted that hydrodynamic can r e f e r t o imcompressible flow charac- From approximately 1910 t o 1935 the majority o f water-facil i t y investigations o f aerodynamic flow The contributions o f water f a c i l i t i e s t o vehicle design increased during the period 1935-1950, l a r g e l y because o f the global p o l i t i c a l climate that would lead t o World War I 1 and the ensuing demands f o r m i l i t a r y a i r and sea superiority. industry t o improve the design o f propeller and j e t a i r c r a f t during World War 11. Reichardt (Ref. 14) described the r e s u l t s o f hydrodynamic tests t o develop e f f i c i e n t low-speed aerodynamic shapes using a c a v i t a t i o n method. The basis o f t h i s study was that gaseous flows a t sonic speed had a c e r t a i n resem- blance t o f l u i d motions a t cavitation. The t e s t models were streamlined i n the water tunnel t o delay c a v i t a t i o n onset. The resultant shapes provided a rough approximation of the desired geometries i n a i r t h a t would e x h i b i t delayed onset o f l o c a l l y sonic flow on the surface. ing, and wing-body t r a n s i t i o n on the M e 262 were modified i n t h i s manner. water f a c i l i t y t o a i d i n the development o f marine vehicles. oped, i n part, from towing-tank investigations (Ref. 13). uous shape, a streamlined conning tower, and guns integrated i n t o the tower. The water f a c i l i t i e s i n Gottingen were used by the German a i r c r a f t The nacelles, nacelle-wing f a i r - The Germans also u t i l i z e d a An advanced submarine type " X X I " was devel- The t e s t results l e d t o a h u l l having a contin- The A l l i e s also used water f a c i l i t i e s during World War I1 f o r the improved design o f marine and f l i g h t vehicles. Notable water tunnel, channel, and towing-tank f a c i l i t i e s were located a t the National Physical Laboratory (NPL) and Admiralty Research Laboratory (ARL) i n England and a t the California I n s t i t u t e o f Technology (CalTech). National Advisory Committee f o r Aeronautics (NACA). and David Taylor Model &asin (DTMB) i n the United States. The majority o f water f a c i l i t y applications t o vehicle design during t h i s period pertained t o marine c r a f t . t i o n phenomena associated with ship propellers and hydrofoils (ship rudders and submarine planes. f o r example) and the drag/resistance o f hydrofoils, displacement hulls. submersibles, torpedoes. surface- piercing struts, seaplane floats/skis, and planing c r a f t (Ref. 13). addressed and solved i n water f a c i l i t i e s included t h a t o f submarine periscope vibration, associated with periodic vortex shedding; t h i s problem was alleviated by means o f guidevanes. drag a t water entry was achieved by improved nose shapes developed from w a t e r - f a c i l i t y testing (Ref. 13). forebody and afterbody shapes and f i n arrangements was performed i n 1944 by Knapp (Ref. 15) i n the CalTech High Speed Water Tunnel. p r o j e c t i l e and bomb designers during the l a t t e r stages o f the war. Typical studies i n these f a c i l i t i e s addressed the cavita- Hydrodynamic problems that were The reduction o f missile A parametric study of the flow separation characteristics of p r o j e c t i l e s having various The detailed flow-field observations from t h i s study were intended t o a i d the Aircraft wings were s t i l l r e l a t i v e l y 3 unswept and o f high aspect r a t i o . AS a consequence, a i r f o i l s were generally tested, and the r e s u l t s were corrected f o r 3-0 effects. A common problem was manifested i n the experimental investigations of marine and f l i g h t vehicles during t h i s period, namely, how t o transfer the model r e s u l t s t o f u l l - s c a l e operation. sions and flow speeds o f water tunnel, channel, and towing-tank i n s t a l l a t i o n s generally resulted i n Reynolds numbers i n model testing that were typically two t o three orders of magnitude less than those of the f u l l - s c a l e operation. nents, s k i n f r i c t i o n and wave resistance, that were governed by d i f f e r e n t s i m i l a r i t y laws. Unless f u l l - scale dimensions were used, it was not possible to simulate correctly the full-scale conditions i n towing tanks. It became comnon practice, then, t o Satisfy the dynamic s i m i l a r i t y (Froude number), producing the proper wave pattern, and t o correct the skin-friction component on the basis o f the Reynolds number. The Reynolds number could be increased by heating the water. However, the d i s p a r i t y between model- and f u l l - scale conditions was s t i l l large. The Scaling of cavitation phenomena observed i n w a t e r - f a c i l i t y t e s t i n g o f ship propellers and hydrofoils was a Persistent cha1)enge. p a r t i c u l a r l y under conditions o f i n c i p i e n t and f u l l y developed cavitation. The introduction o f jet-powered a i r c r a f t toward the end of World War I 1 and the ensuing emphasis on high-speed f l i g h t posed p a r t i c u l a r problems for water-facility simulations owing t o the s i g n i f i c a n t Reynolds number gap and the effects of compressibility (Mach number). higher speeds resulted i n undesired cavitation phenonena t h a t r e s t r i c t e d the speed up t o which a flow pattern i n water could be expected t o represent that around the same shape i n air. Cavitation could occur w i t h i n the f l u i d a t some distance from the model surface. i n the cores o f vortices from a i r c r a f t propeller blade t i p s o r i n the separated flow past blunt o r bluff bodies, f o r example, which precluded the correla- t i o n o f the water-facility data t o the conditions i n a i r . Closed-section f a c i l i t i e s could be pressurized t o delay c a v i t a t i o n onset: however. t h i s was done at increased f a c i l i t y complexity and cost. As a r e s u l t o f these considerations. constraints were necessarily imposed on the use o f water f a c i l i t i e s as a design tool, p a r t i c u l a r l y w i t h regard t o f l i g h t vehicles. 2.6 From 1950 t o 1960 The l i m i t e d dimen- For example. the drag of displacement h u l l s consisted o f two predominant compo- Operation o f water f a c i l i t i e s a t The work o f German researchers during World W a r I 1 and o f R. T. Jones i n the United States during the same period l e d t o wing sweep as a means of delaying transonic flow e f f e c t s (see Ref. 2). The f l u i d mechanics associated w i t h these wings was not well-understood owing t o the 3-0 nature o f the flow. A t off-design conditions, the swept separation l i n e s and ensuing vortices were important features o f the 3-0 flow f i e l d . The e a r l y 1950s also marked the advent o f first-generation supersonic transport (SST) a i r - c r a f t featuring thin, slender wings. l i f t f o r improved takeoff and landing performance. This was a s i g n i f i c a n t departure from the time-honored attached-flow wing designs. The complexity o f the wing v o r t i c a l flow f i e l d and the associated aerodynamic nonlinearities were demonstrated i n the :ate 1940s and early 195Cs by researchers i n England, France, the United States, Canada. and Sweden (Refs. 16-20, f o r example). These experimental investigations underscored the need f o r a flow-visualization t o o l i n order t o understand and control the powerful f l u i d motions. The early studies suggested t h a t once flow separation occurred everywhere along the leading edge o f a swept wing, the fundamental character o f the vortex was insensitive t o the Reynolds number (Ref. 21). structure also remained much the same over a wide range o f the Mach number (Ref. 22). pravided the wing leading edge was swept w i t h i n the Mach cone and shock waves d i d not interact with the vortices. o f w a t e r - f a c i l i t y applications t o a i r c r a f t design therefore emerged and was characterized by extensive t e s t i n g of slender wings suitable f o r comnercial and m i l i t a r y a i r c r a f t configurations w i t h a requirement f o r supersonic operation. the NPL i n England, the Office National d'Etudes e t de Recherches Aerospatiales (ONERA) i n France, the National Aeronautical Establishment (NAE) i n Canada, the NACA and CalTech i n the United States, and Kungl Tekniska Hogskolan (KTH) i n Sweden. The pump-driven horizontal water tunnels at the NPL and NAE. the v e r t i c a l water tunnel a t ONERA operating by gravity discharge, the towing tank a t the NACA, the free- surface water channel and pressurized high-speed water tunnel a t CalTech. and the water tank a t the KTH represented several d i f f e r e n t f a c i l i t y designs that were used f o r a comnon purpose. namely, t o develop a flow-visualization data base on a i r c r a f t and aircraft-related configurations. v i s u a l i z a t i o n from experiments conducted i n these installations established water f a c i l i t i e s as a useful diagnostic t o o l t o study the flow about 3-0 aerodynamic shapes operating w i t h i n an expanded f l i g h t enve- lope. For example, the water tunnels a t the NPL and ONERA contributed t o the understanding o f the con- t r o l l e d flow separations and v o r t i c a l motions on the SST Concorde developed j o i n t l y by the English and the French. The water tank a t the KTH was u t i l i z e d i n Sweden's pioneering studies o f canard-wing a i r c r a f t . The water tunnel a t the NAE deserves special note owing t o i t s c o l o r f u l history. a t the Aerodynamische Versuchsantalt i n 1939 i n Gottingen, Germany. and, subsequently, was shipped t o Canada a f t e r World War I1 (Ref. 23). the design of cockpit canopy shapes for the CF-100 and CF-103 a i r c r a f t by generating a c a v i t a t i o n bubble whose shape conformed t o the constant-pressure contour (Ref. 24). t o Reichardt's (Ref. 14). whose work i n Germany was discussed e a r l i e r . formed of the l a t e r a l i n s t a b i l i t y caused by asymmetric breakdown of wing leading-edge vortices. subject remains of great interest t o modern-day fighter a i r c r a f t configurations. BY design, the configurations u t i l i z e d leading-edge vortex-induced The vortex flow A new era The p r i n c i p a l advocates o f w a t e r - f a c i l i t y applications t o air-vehicle design i n the 1950s included The c l a r i t y o f the flow The f a c i l i t y was b u i l t One of the early investigations conducted a t the NAE was related t o This experimental approach was s i m i l a r A n undocumented study was per- The NAE f a c i l i t y was This 4 also u t i l i z e d t o study the e f f e c t s o f the exhaust from jet-powered a i r c r a f t on the f l o w about t a i l surfaces. Water f a c i l i t i e s maintained t h e i r r o l e i n the marine-vehicle design process i n the post-World War I1 years. Major contributions t o the design of surface ships, submersibles, and marine propulsion systems were made by researchers i n many countries. Primary contributors t o marine vehicle technology included the NACA, DTMB, Iowa I n s t i t u t e of Hydraulic Research, S t . Anthony F a l l s Hydraulic Laboratory (University of Minnesota), and CalTech i n the United States; the Ship Model Basin in the Netherlands; the Supramar company i n Switzerland; and the State Shipbuilding i n Sweden. Reference 25 l i s t s many o f the research i n s t a l l a t i o n s involved i n maritime t e s t i n g during t h i s period. began i n the l a t e 1950s. the p o s s i b i l i t y of high speeds a t sea were due t o the greatly increased understanding o f the flow past hydrofoils. I n a manner s i m i l a r t o the f i e l d of aeronautics, water f a c i l i t i e s played a key r o l e i n the advancement o f hydrofoi 1 technology. 2.7 From 1960 t o 1970 An upsurge i n the use o f hydrofoil c r a f t As noted by Acosta (Ref. 26). the successful achievements o f hydrofoil c r a f t and Numerous slender-wing a i r c r a f t configurations emerged during the 1960s t h a t were characterized by leading-edge vortex formation a t off-design conditions. These a i r c r a f t included the F-111, YF-12. and XB-70 i n the United States; the Mirage 111 and I V i n France; the HP-115 experimental a i r c r a f t i n England; the Concorde i n a j o i n t English/French effort; and the Viggen i n Sweden. highly maneuverable transonic f i g h t e r a i r c r a f t began during t h i s period i n the United States, and t h i s work would culminate i n the F-16 and F-18 fighters i n the 1970s. wing-body strakes o r leading-edge extensions (LEXes) t o generate concentrated vortices f o r enhanced l i f t a t takeoff and landing and a t subsonic/transonic maneuvering conditions. The water tunnels a t the NPL i n England, ONERA i n France, and NAE i n Canada continued t o be u t i l i z e d successfully i n the study of slender-wing vortices. A copy o f the NPL tunnel was b u i l t a t the University o f Southhampton, England, and s t a t i c and dynamic testing o f slender-wing vortex flows was performed (Ref. 27). d e l t a wings and specific a i r c r a f t configurations such as the Concorde and the Douglas F-50 (Ref. 29) i n water tunnels, wind tunnels, and f l i g h t . These results established confidence that, f o r the special case o f leading-edge vortices on thin, highly swept surfaces, water f a c i l i t i e s could be used as a diagnostic tool despite the Reynolds number gap. ONERA assembled a laboratory consisting o f two v e r t i c a l water tunnels and a water tank t h a t was dedicated t o the study o f aerodynamic and hydrodynamic flow phenomena. The French were a t the forefront of flow-visualization technology and they applied water f a c i l i t i e s t o a wide range o f aerodynamic flow situations. boundary-layer separation on 2-0 and 3-D shapes, vortex-dominated flows on slender bodies and wings, j e t mixing and i n t e r a c t i o n phenomena. blowing for boundary-layer control, vortex enhancement by spanwise blowing, ground effects on the flow about the Concorde using a moving ground board, models w i t h internal flows (engine intakes, f o r example), and a helicopter r o t o r i n t r a n s l a t i o n and hover (Ref. 30). laboratory became the standard o f excellence i n flow v i s u a i i z a t i o n and was the forebearer o f numerous other water f a c i l i t i e s throughout the world. In addition, the design o f From the outset, these designs employed Excellent correlations were obtained by Werl6 (Ref. 28) a t ONERA o f the vortex behavior on These included laminar and turbulent boundary layers, The ONERA The NAE water tunnel was a consistent source of useful f l o w - f i e l d information on complex flow phenom- ena i n the 1960s. The 3-0 separated flow-characteristics about numerous aerodynamic shapes were i n v e s t i - oated. f o i l s , flaps, etc. Other experiments featured l i f t i n g propellers a t high angles o f inclination, ducted fans, and fan-in-wing arrangements. increased sophistication of w a t e r - f a c i l i t y experiments o f a i r c r a f t models and o f the information gained from these studies. observed i n order t o assess the flow e f f e c t s leading t o changes i n the configuration forces and moments. The circulation-control concept f o r enhanced, h i g h - l i f t aerodynamics was studied on round a i r - The wing-submerged l i f t i n g - f a n investigations were indicative o f the I n t h i s case, the f a n - a i r f o i l interactions and wing-fan e f f l u x interactions were The heavy commercial and m i l i t a r y transport a i r c r a f t that appeared during t h i s period generated pow- e r f u l t r a i l i n g vortex systems that posed a f l i g h t safety hazard t o t r a i l i n g a i r c r a f t . f a c i l i t i e s were well suited t o study the vortex patterns and methods of wake a l l e v i a t i o n , and the National Aeronautics and Space Administration (NASA), the Douglas A i r c r a f t Company, and the Lockheed-Georgia Company i n the United States employed such f a c i l i t i e s f o r t h i s purpose. experiments of the vortex flows shed from upswept rear fuselages similar t o those of rear-loading a i r - c r a f t . The vortices were found t o promote adverse e f f e c t s on the p i t c h s t a b i l i t y and cruise drag and t o produce undesirable loadings on rear cargo doors (Ref. 31). 2.8 From 1970 t o 1980 Water towing-tank The NAE conducted water tunnel Vehicle designers i n the 1970s were once again confronted w i t h a large experimental data gap caused by the emerging requirement f o r advanced t a c t i c a l missile and f i g h t e r a i r c r a f t configurations t o operate i n a controlled manner a t extreme attitudes. wings of highly maneuverable f l i g h t vehicles were not well understood. The flow-visualization and flow- measurement techniques i n wind tunnels and i n f l i g h t were inadequate f o r the detailed d e f i n i t i o n o f the h i g h l y 3-0 flow f i e l d s t h a t were o f t e n characterized by multiple vortex development. vortex interactions. and vortex breakdown. i t i e s i n the vehicle design process. The vortex-dominated flows shed from the slender bodies and As a consequence. the 1970s marked a s i g n i f i c a n t upsurge i n the use o f water f a c i l - 5 To improve the understanding of the structure o f vortex core breakdown, experiments were conducted by Sarpkaya i n the United States using a water f a c i l i t y (Ref. 32). Water t o which s w i r l was imparted by upstream vanes flowed through a s l i g h t l y divergent tube and the forms o f vortex bursting were observed. The experimental r e s u l t s also supported the development o f computational methods t o predict vortex core i n s t a b i l i t i e s . Water tunnels. channels, and towing tanks were i n operation i n many countries, and studies i n support o f vehicle design were performed i n the United States, Canada, France, England, Belgium, Germany, Switzerland, the Netherlands. Sweden, Russia, Australia, Japan, and China. The confidence i n water f a c i l - i t i e s as a flow-diagnostic t o o l was reflected by the d i v e r s i t y o f research subjects encompassing air. ground, and marine vehicles. high angles of attack was Northrop Corporation i n the United States. of Werl6 (ONERA) i n France, M. S. Cahn and G. R. Hall l e d the e f f o r t a t Northrop i n the mid-1970s t o develop a water f a c i l i t y t h a t would augment the aircraft design process (Ref. 33). Of a Small p i l o t tunnel modeled after the ONERA gravity-discharge f a c i l i t y was the visualization o f the LEX vortex flows on a small-scale model o f the Northrop YF-17 lightweight f i g h t e r configuration. v i v i d d e f i n i t i o n o f the YF-17 v o r t i c a l flows stinulated s u f f i c i e n t support t o construct a larger water tunnel t h a t i s s t i l l i n operation today. This f a c i l i t y was used extensively i n a l l o f the m i l i t a r y a i r - c r a f t programs a t Northrop t o understand and control the forebody and wing vortex flows, vortex interac- t i o n s and breakdown, and vortex-empennage interactions. United States inventory and o f numerous foreign m i l i t a r y a i r c r a f t configurations were tested i n the Northrop i n s t a l l a t i o n . The r o l e o f t h i s tunnel rapidly expanded t o include the study o f 2-0 nozzle exhaust e f f e c t s on afterbody flow separation, nozzle exhaust plumes and j e t mixing processes. forebody and wing vortex control by active and passive means, hot-gas reingestion phenomena on V/STOL a i r c r a f t i n ground proximity, vortex f l o w management f o r improved performance o f top-mounted inlets, thrust reverser plume t r a j e c t o r i e s and the e f f e c t s on wing and t a i l flow fields, the structure o f swirling jets, s e l f - induced l a t e r a l o s c i l l a t i o n s (wing rock) o f slender planforms, deflected wing leading- and trailing-edge f l a p e f f e c t s on vortex s t a b i l i t y . o s c i l l a t i n g wing control surfaces f o r flow control. vortex shedding on an a i r c r a f t model i n a f l a t spin, and canard-wing, forward-swept wing. and oblique wing flow f i e l d s (Ref. 34). ONERA i n France and the NAE i n Canada continued t o excel i n t h e i r high-quality and diverse applica- t i o n s o f water f a c i l i t i e s t o vehicle design. The confidence gained from years o f experience i n hydrody- namic t e s t i n g l e d t o the use o f the ONERA and NAE f a c i l i t i e s t o study ground vehicle configurations such as high-speed trains, trucks, automobiles, and snowmobiles. regions o f 3-0 flow separation and t o develop aerodynamic "fixes" t o improve the vehicle performance. Marine vehicle investigations were also undertaken i n these tunnels t o study the separated flow f i e l d s about the superstructure o f surface ships and highly maneuverable submarines. The trend toward high- performance a i r , ground, and marine vehicles f a c i l i t a t e d the acceptance o f water f a c i l i t i e s as a design t o o l , owing t o the complicated f l u i d motions that were o f t e n vortex-dominated. i n s t a l l a t i o n was u t i l i z e d t o study the vortex formation on the upper surface o f an o s c i l l a t i n g p r o f i l e , which simulated the c y c l i c variation i n p i t c h of a helicopter r o t o r blade (Ref. 30). vortex-shedding characteristics o f a spinning fighter model were investigated. e;cllla:lon t e s t i n g of a modern a i r c r a f t configuration t o i d e n t i f y the e f f e c t s o f the body vortices on the s t a t i c and dynamic cross derivatives (Ref. 35). the dynamic s t a l l behavior of o s c i l l a t i n g a i r f o i l s (Ref. 36) a t the U . S . Army Research and Technology Laboratories (AVRADCOM) a t NASA Ames Research Center. was the measurement of the force and moment t i m e h i s t o r i e s i n combination with v i v i d off-body flow visualization. A leader i n the application of water tunnels t o the simulation o f f i g h t e r a i r c r a f t flow f i e l d s a t Influenced i n large part by the work An early application The Models o f v i r t u a l l y every f i g h t e r a i r c r a f t i n the The water f a c i l i t i e s were used t o i d e n t i f y The French and Canadians were leaders i n the investigation o f unsteady v o r t i c a l motions. The ONERA I n addition, the The NAE performed forced I n t e r e s t i n the unsteady aerodynamics o f helicopter r o t o r blades l e d t o numerous investigations o f A unique feature o f the water tunnel experiments The Flow Research Company towing tank and the Tracor Hydronautics Ship Model Basin i n the United States emerged as important water f a c i l i t i e s for aerodynamic flow simulations. The Flow Research f a c i l i t y was used i n the general research o f unsteady aerodynamics; separated flows; and laminar, transitional, and turbulent boundary layers. t r a i l i n g vortex systems generated by models of wide-body commercial transport a i r c r a f t such as the Boeing 747. International (Ref. 37) and the McDonnell-Douglas Corporation (Ref. 38). v i s u a l i z a t i o n of ground-effect phenomena associated with m u l t i j e t arrangements. The Tracor model basin was used extensively by NASA f o r measurements o f the The continued interest i n V/STOL a i r c r a f t prompted the development o f water f a c i l i t i e s a t Rockwell These were dedicated t o the The U.S. A i r Force Wright Aeronautical Laboratories (AFWAL) constructed a small water tunnel operat- i n g by g r a v i t y discharge. f i g h t e r models, including several forward-swept-wi ng configurations. comnunities. i n Gottingen. Germany, was applied t o the study o f wing leading-edge vortices (Ref. 39). This f a c i l i t y proved useful i n visualizing the vortex flows about advanced Water f a c i l i t i e s also experienced a renaissance i n the European government, industry, and university The towing tank a t Deutsche Forschungs- und Versuchsanstalt f u r Luft- und Raumfahrt (DFVLR) A water tunnel 6 b u i l t a t Messerschmitt-Bolkow-Blohm (MBB) in Munich. Germany, became a useful a i d i n recognizing and solv- ing flow problems during the a i r c r a f t development phase (Refs. 40 and 41). w i t h a h i s t o r y o f water f a c i l i t y experience dating back t o 1950, continued i t s fundamental f l u i d mechanics research. The behavior o f slender missile vortices a t extreme attitudes was investigated i n a water tunnel a t the B r i t i s h Aerospace M i l i t a r y A i r c r a f t Division (Ref. 42). s e n s i t i v i t y o f d e l t a wing vortex breakdown was performed i n a water tunnel by Svenska Aeroplan Aktiebolaget (SAAB) i n Sweden (Ref. 43). f a c i l i t y a t the von Karman I n s t i t u t e ( V K I ) i n Belgium (Ref. 44). tained a leadership r o l e i n hydrodynamic t e s t i n g of displacement h u l l s and marine propulsion systems. tematic experiments o f vortex flows i n the mid-1970s using water towing tank and water tunnel i n s t a l l a - tions. tank and the experimental trends pertaining t o the vortex structure and dissipation were i n q u a l i t a t i v e agreement with available wind tunnel and f l i g h t test r e s u l t s (Ref. 45). The structure and breakdown o f the leading-edge vortices shed from slender delta and cranked wing planforms were observed i n the water tunnel (Ref. 46). 2.9 From 1980 t o the Present The University o f Stuttgart, A study o f the Reynolds number Leading-edge vortex flow studies were also conducted i n a water The Netherlands Ship Model Basin main- The Aeronautical Research Laboratories (ARL) of the Australian Department o f Defence conducted sys- The t r a i l i n g - v o r t e x system generated by a rectangular planform wing was investigated i n the towing Water f a c i l i t i e s have gained general acceptance throughout the world as valuable diagnostic tools t o a i d i n the vehicle design process. motions about complicated aerodynamic and hydrodynamic shapes has been u t i l i z e d i n vehicle development programs i n several countries. shown i n the sketches i n Fig. 4 (from Ref. 1). aerodynamic shapes t h a t are dominated by vortex flows. 3-0 separated regimes i s an important element i n air, ground, and marine vehicle performance optimization. The unique a b i l i t y o f these i n s t a l l a t i o n s t o visualize 3-0 v o r t i c a l Vortical flows have become a "way o f l i f e " on a l l classes o f vehicles as Clearly, the i d e n t i f i c a t i o n and control of these The emphasis on low-observable f l i g h t vehicles has led t o The sophistication o f water f a c i l i t i e s and flow v i s u a l i z a t i o n and measurement techniques has increased i n concert w i t h the advancement i n vehicle technology. I n addition t o the q u a l i t a t i v e informa- t i o n gained from water-flow experiments, e f f o r t s are now under way t o extract more quantitative data than previously possible. handling high-resolution images has made it possible t o combine flow-visualization and digital-image pro- cessing techniques t o obtain quantitative information. f a c i l i t y installations. iments o f v o r t i c a l motions using 2-0 laser velocimetry (Ref. 48). generate an intense sheet o f l i g h t t o enhance the visualization o f the flow structure i n a r b i t r a r y planes (Ref. 49). A laser-induced fluorescence visualization technique can provide more detailed information on the structure o f complex flows (Ref. 50). The interest i n unsteady aerodynamics has l e d t o more sophisti- cated model support apparatus and instrumented models. As indicated by Gad-el-Hak (Ref. 47). the advent o f advanced computers capable o f Lasers have become a key element i n many water Laser optics have also been used t o Quantitative f l o w - f i e l d information has been obtained i n recent water-flow exper- The water f a c i l i t y "standard bearers" o f the 1970s continue t o make major contributions i n t h i s decade t o the understanding and control o f the flow about advanced vehicles. Northrop has been placed on the establishment o f a flow visualization data base on present- and future- generation f i g h t e r a i r c r a f t configurations t h a t are characterized by highly coupled forebody and wing vortex systems a t extreme attitudes (see Ref. 51). ONERA i n France and the NAE i n Canada continue a long- standing t r a d i t i o n o f excellence i n water f a c i l i t y applications t o a i r , ground, and marine vehicle design (Refs. 52 and 53. respectively). Several w a t e r - f a c i l i t y i n s t a l l a t i o n s i n the United States have assumed leadership positions i n aero- nautical and related hydrodynamic research i n recent years. The water tunnel a t NASA Ames-Dryden F l i g h t Research F a c i l i t y , modeled a f t e r the Northrop tunnel. has become a "workhorse" f o r NASA since i t s incep- t i o n i n the early 1980s. other U.S. government agencies t o increase the experimental data base on advanced m i l i t a r y a i r c r a f t con- figurations. Force supermaneuverability program t o study the unsteady aerodynamics o f slender wings and bodies undergo- ing p i t c h i n g o s c i l l a t i o n s (Refs. 54 and 55). maneuvers o f a slender generic f i g h t e r model (Ref. 56). constructed a f a c i l i t y with a v e r t i c a l t e s t section s i m i l a r t o the NASA Ames-Dryden tunnel and i ; i n i t i a t - ing a number of projects t o support t h e i r aeronautics research programs. The success o f the p i l o t water tunnel a t AFWAL a t Wright-Patterson A i r Force Base has l e d t o the i n s t a l l a t i o n o f a larger ONERA-type water tunnel w i t h a v e r t i c a l 24- by 24-in. t e s t section. This tunnel i s undergoing operational checkout t e s t s and i s expected t o be i n use as a research f a c i l i t y i n November 1986. completed a new water tunnel/channel w i t h a 15- by 20-in. horizontal t e s t section incorporating a unique downstream viewing window t o permit flow visualization i n the cross-flow plane as well as the usual plan- form and side views. construction and i s intended p r i m a r i l y t o provide a c a p a b i l i t y f o r performing both s t a t i c and dynamic experiments a t higher angles of attack ( t o 90") t o support technology advancements r e l a t e d t o f i g h t e r a i r c r a f t . 24- by 24-in. t e s t section was recently i n s t a l l e d a t the General Dynamics Corporation i n F o r t Worth t o support t h e i r in-house research programs i n advanced f i g h t e r technology. In recent years. emphasis a t Considerable work has been done i n cooperation with industry, universities. and The Flow Research Company towing tank has most recently been used i n support o f the A i r The Tracor model basin was used t o simulate the pitch-up Several other organizations have recently acquired water tunnels. NASA's Langley Research Center Eidetics International has A larger version o f the same tunnel with 24- by 36-in. t e s t section i s also under A water tunnel designed by the Visual Aerodynamics Division o f Eidetics International w i t h a 7 The i n s t a l l a t i o n s c i t e d above have equally s i g n i f i c a n t counterparts i n Europe. I n addition t o ONERA, the B e r t i n and Company water tunnel i n France was u t i l i z e d recently t o t e s t a canard-wing arrangement f o r which flow-visualization. pressures, velocities, and forces were obtained (Ref. 57). Numerous i n s t a l l a - t i o n s are i n operation i n England and are used f o r diverse aerodynamic and hydrodynamic flow problems. The No. 2 Ship Tank and Rotating Arm i n the Maneuvering Tank a t the Admiralty Research Establishment, which has been I n operation f o r decades, demonstrated the powerful vortex flows shed from the h u l l o f a modern. highly maneuverable submarine model (Ref. 58). obtained on slender missile configurations i n water tunnel and wind tunnel f a c i l i t i e s was recently made a t the B r i t i s h Aerospace M i l i t a r y A i r c r a f t Division (Ref. 59). Q u a l i t a t i v e and quantitative testing o f para- chute canopies o f various shape and porosity w a s conducted i n the Southhampton (England) Towing Channel t o evaluate p i t c h s t a b i l i t y characteristics (Ref. 60). I M I Summerfield i n England studied the internal flow characteristics o f a ramrocket combustion chamber (Ref. 61). and the flow-visualization results led t o an improved f u e l supply design. Unsteady flow phenomena were investigated i n the towing tank o f DFVLR i n Germany, where a i r c r a f t models underwent prescribed accelerations and decelerations (Ref. 62). University o f Stuttgart constructed two new water f a c i l i t i e s . and recent r e s u l t s o f research work con- ducted i n these i n s t a l l a t i o n s are provided i n Ref. 63. c r a f t configurations and isolated airframe components have been tested i n the MBB water tunnel (Ref. 41). A systematic comparison of vortex positions The Models o f complete m i l i t a r y and comnercial a i r - Water f a c i l i t i e s are an important element i n vehicle design i n Asian countries. A high-speed water tunnel (up t o 10 ft/sec) designed by the Visual Aerodynamics Division o f Eidetics International was i n s t a l l e d i n a research laboratory a t the Aero Industry Development Center i n Taiwan i n 1984 t o support the a i r c r a f t development programs i n t h a t country. This tunnel has a 24- by 24-in. horizontal t e s t sec- tion. The Peking I n s t i t u t e i n the People's Republic of China has employed a water tunnel t o visualize the vortex flows about strake-wing planforms (Ref. 64). Japanese researchers have made extensive use o f water f a c i l i t i e s t o understand the f l u i d flows about the various aerodynamic and hydrodynamic shapes. o f t h e i r work are provided i n Ref. 9. Mitsubishi Heavy Industries employed a towing tank t o study surface ship designs. A study (Ref. 65) a t Mitsubishi led t o the solution o f a v i b r a t i o n problem by improved h u l l design obtained through systematic water f a c i l i t y testing. The burgeoning applications o f water f a c i l i t i e s t o the design o f vehicles are apparent from the pre- ceding h i s t o r i c a l review. The following section w i l l discuss specific investigations i n water f a c i l i t i e s t o demonstrate t h e i r r o l e i n the vehicle design process. These representative investigations w i l l demon- s t r a t e the strengths and l i m i t a t i o n s o f water f a c i l i t i e s i n aeronautical and related hydrodynamic research. Examples 3. SPECIFIC APPLICATIONS OF W A T E R FACILITIES TO AERONAUTICAL AND RELATED HYDRODYNAMIC PROBLEMS Experiments performed i n the NAE. CalTech, Northrop, and NASA Ames-Dryden water f a c i l i t i e s are described i n t h i s section t o provide a flavor of t h e myriad applications o f these i n s t a l l a t i o n s t o a i r , ground, and marine vehicle design. NAE and CalTech have long-standing t r a d i t i o n s i n the application o f water tunnels t o aeronautical and related hydrodynamic problems. The Northrop and NASA Ames-Dryden f a c i i t i e s represent a "new generation" o f water tunnels t h a t have made s i g n i f i c a n t contributions t o vehicle design. Empha s i s w i l l be placed on experiments pertinent t o current and f u t u r e comnercial and m i l i t a r y vehicles. 3.1 National Aeronautical Establishment, Ottawa, Canada Collectively. these i n s t a l l a t i o n s represent over 100 years of w a t e r - f a c i l i t y experience. The NAE water tunnel (Fig. 5) i s a continuous-flow, closed-circuit design having a 10- by 13-in. The horizontal t e s t section. diverse experiments conducted i n t h i s f a c i l i t y encompass steady and unsteady, and attached and separated flows about air, ground, and marine vehicles. Dobrodzicki (Ref. 23). The f l u i d tracers reveal the fan efflux, recirculation region, and fountain effect. jet-wing f l o w interactions were useful i n interpreting the anomalies i n the l i f t , drag, and pitching moment characteristics obtained i n wind tunnel tests. i n t e r e s t t o advanced fighter a i r c r a f t configurations w i t h a requirement f o r short takeoff and v e r t i c a l landing (STOVL) capability. regarding the jet-induced e f f e c t s on the configuration aerodynamics and potential hot-gas reingestion problems. s l o t t e d trailing-edge f l a p arrangement. t h a t was applied t o the McDonnel-Douglas YC-15 m i l i t a r y transport configuration. e f f e c t induced by the high-velocity j e t i s clearly i l l u s t r a t e d i n the flow-visualization photograph. i s noted t h a t the boundary-layer separation characteristics o f the unpowered wing and f l a p combination are not accurately represented i n the water tunnel owing t o the subscritical Reynolds number and the subse- quent laminar separation. However, the a b i l i t y o f the j e t exhaust t o reattach the flow t o the wing and f l a p surface i s q u a l i t a t i v e l y represented. t i a l flow f i e l d are simulated i n a quantitative sense. Flow velocities i n the working section can vary from 0.2 t o 10 ft/sec. A detailed description o f t h i s i n s t a l l a t i o n i s provided by Figure 6 i l l u s t r a t e s the flow about a wing w i t h a submerged l i f t i n g fan i n proximity t o the ground. The fan-wing and The simulation o f t h i s flow s i t u a t i o n i s o f current For such configurations. water flow v i s u a l i z a t i o n could provide insight Figure 7 shows the interaction o f the simulated exhaust from a wing-mounted engine with a wing and This i s representative o f the externally blown-flap (EBF) concept The supercirculation It A t higher j e t momentum. the jet-induced e f f e c t s on the poten- 8 The boundary-layer flow-separation characteristics of a m i l i t a r y transport model are shown i n Fig. 8. t y p i c a l o f water tunnel operation. Used judiciously, however, flow visualization can indicate regions on the a i r c r a f t surface t h a t may be susceptible t o f l o w separation. A related study concerns the flow separation from the upswept r e a r fuselage t y p i c a l o f a rear-loading transport a i r c r a f t . The vortex p a i r shed frm the a f t fuselage section i s shown i n Fig. 9. The strength and location o f the vortices w i l l vary w i t h the Reynolds number because o f the lack o f a fixed l i n e o f boundary-layer separation. Despite t h i s , the water flow simulation was used t o i d e n t i f y the source o f degraded performance and s t a b i l i t y problems as well as unsteady loads on cargo doors. This flow s i t u a t i o n resembles the vortex formation on displacement hulls and submersible vehicles. The aerodynamic cross-coup1 ing effects associated w i t h an o s c i l l a t i n g generic f i g h t e r model were studied i n water tunnel flow-visualization experiments. water flow simulation revealed a l a t e r a l o s c i l l a t i o n o f the forebody vortices caused by an o s c i l l a t i o n i n p i t c h and. as a result, the vortices were observed t o s h i f t from one side t o the other o f a top-mounted v e r t i c a l fin. The r e l a t i v e l y simple experiment provided a plausible flow mechanism that would lead t o secondary l a t e r a l aerodynamic forces i n response t o a primary pitching maneuver. This study indicated that the unsteady, separated flow f i e l d about a slender configuration could be studied i n a q u a l i t a t i v e sense i n a water f a c i l i t y despite the s e n s i t i v i t y of t h e forebody v o r t i c a l motions t o the Reynolds number. The modulation of the f low-separation characteristics t o improve ground vehicle performance was e f f e c t i v e l y demonstrated i n the water tunnel. t r a c t o r - t r a i l e r due t o the i n s t a l l a t i o n of a cab deflector. This modification led t o reduced drag and increased s t a b i l i t y and i s now a standard " f i x " on most ground transport vehicles o f t h i s class. r e s u l t s from t h i s investigation indicate t h a t small-scale model t e s t i n g i n a water f a c i l i t y operating a t low Reynolds number can y i e l d substantial design improvements on vehicles that are Reynolds-number- sensitive and nonvortex-dominated. Results o f such t e s t s must be carefully interpreted because o f the low Reynolds number conditions A representative r e s u l t i s shown i n Fig. 10. The Figure 1 1 reveals a large improvement i n the flow about a The 3.2 California I n s t i t u t e of Technology, Pasadena, California The Hydrodynamics Laboratory a t CalTech consists o f the Free Surface Water Tunnel (FSWT) and the High Speed Water Tunnel (HSWT). The F S W T depicted i n Fig. 12 i s a closed-circuit c i r c u l a t i o n system l y i n g i n a v e r t i c a l plane. The horizontal t e s t section i s 20 in. wide by 30 in. deep and flow speeds up t o 25 ft/sec can be obtained. The top. or free, surface i s an air-water interface. This arrangement allows investiga- t i o n s t o be performed on bodies acting on or a t a prescribed distance below the water surface, such as a ventilated hydrofoil. interchangeable 2-0 and axisymmetric working sections. The 2-0 section. used p r i n c i p a l l y t o obtain sec- t i o n a l characteristics of hydrofoils, i s 6 in. by 3 0 in. by 50 in. long, whereas the axisymnetric section i s 14 in. i n diameter and 46 in. long. Flow speeds up t o 100 ft/sec and pressures from 100 psig t o the vapor pressure o f water are achievable. moments and t o obtain quantitative f l o w - f i e l d information using a laser doppler velocimeter (LDV). detailed account o f the CalTech f a c i l i t i e s i s provided by Ward (Ref. 66). The HSWT i s also a closed-circuit design l y i n g i n a v e r t i c a l plane and features I n addition t o flow visualization, the capabilities e x i s t t o measure steady and unsteady forces and A Applications o f the CalTech water f a c i l i t i e s to vehicle design have t y p i c a l l y pertained t o marine c r a f t . Studies performed i n the H5WI demonstratea rne importance of tne bounadry i d y e r i r i cav iiaiiuri iricepiiuri. This was accomplished using a Schlieren system t o visualize the o r i g i n and migration o f c a v i t a t i o n bubbles within the boundary layer on a b l u f f body and their subsequent entrainment i n t o the mainstream. sentative r e s u l t from a study o f c a v i t a t i o n phenmena i s shown i n Fig. 13. which depicts the c a v i t a t i n g flow over a 2-0 wedge a t high angle o f attack. The inception and scaling o f c a v i t a t i o n on hydrodynamic shapes have been studied extensively. A repre- Recent emphasis has been placed on the development o f i n l e t s suitable f o r water-jet propulsion sys- Sophisticated i n l e t models have been tested i n the HSWT and F S W T that featured translating and tems. r o t a t i n g lips, variable-geometry walls, and auxiliary o r secondary inlets. content, the thickness o f the approaching boundary layer. and numerous boundary-layer control devices on the inception o f c a v i t a t i o n and i n l e t recovery efficiency have been determined i n these investigations. Hydrofoil development projects have addressed the e f f e c t s o f flaps i n cavity flow, the study o f hydrofoil sections having good performance i n f u l l y wetted and c a v i t y flow, and the performance o f venti- l a t e d f o i l s near a f r e e surface. (Ref. 67) t o predict the forces on supercavitating o r ventilated hydrofoils o f f i n i t e aspect r a t i o , arbi- t r a r y shape, and variable submergence. faces were undertaken i n the FSWT. measured using an LDV system and the data were used t o confirm a theory f o r the structure and decay r a t e o f a t r a i l i n g vortex. t o y i e l d q u a l i t a t i v e and quantitative surface and off-body f l o w - f i e l d information t o a i d i n theory devel- opment and marine and f l i g h t vehicle design. The influence o f upstream a i r The l a t t e r experiment provided v e r i f i c a t i o n o f a theory by Furuya Aeronautical research projects pertaining t o the t r a i l i n g vortex systems generated by 1 i f t i n g sur- The a x i a l and tangential v e l o c i t y p r o f i l e s i n the wake region were The experiments performed i n the CalTech installations exemplify the c a p a b i l i t i e s o f water f a c i l i t i e s 9 3.3 Northrop Corporation, A i r c r a f t Division, Hawthorne, California The Northrop Corporation water tunnel i s a continuous-flow. closed-circuit f a c i l i t y having 16- by 24-in. v e r t i c a l t e s t section (Fig. 14). This i n s t a l l a t i o n has been used as a diagnostic t o o l t o a i d i n the a i r c r a f t design process since i t s inception i n 1977. It was preceded by a small p i l o t water tunnel having a 6- by 6-in. v e r t i c a l t e s t section that could operate i n g r a v i t y discharge and continuous-flow modes. The Northrop i n s t a l l a t i o n s marked the advent of water f a c i l i t i e s t h a t were dedicated t o the study o f the vortex flows developed on advanced tactical/fighter a i r c r a f t operating a t extreme attitudes. These i n s t a l l a t i o n s s a t i s f i e d the need f o r a visualization t o o l t o improve the understanding and control o f the complex v o r t i c a l flows t h a t have become characteristic o f highly maneuverable m i l i t a r y a i r c r a f t beginning w i t h the General Dynamics YF-16 and the Northrop YF-17 lightweight f i g h t e r s i n the 1970s. A t a very e a r l y stage o f i t s operation. the p i l o t tunnel demnstrated the u t i l i t y o f a water f a c i l i t y operating a t very low speed (0.25 ft/sec) and low Reynolds number (10,00O/ft) t o v i v i d l y depict the v o r t i - cal motion about a small-scale model of a complete f i g h t e r configuration. Figure 15 shows the vortex a r i s i n g from flow separation along the sharp edge o f a wing LEX on a 1/72-scale YF-17 a t an angle of attack o f 20". results, bear l i t t l e resemblance t o the r e a l f l o w and are p r i n c i p a l l y o f public r e l a t i o n s value. This i s hardly the case, however, because t h i s single flow-visualization photograph demonstrates several important f l o w - f i e l d features t h a t have been observed i n wind tunnels and i n f l i g h t (Ref. 34). interpretation o f the nonlinear forces and moments and vortex-induced surface pressures i s f a c i l i t a t e d by the detailed 3-D flow visualization. the wing surface and the favorable LEX vortex-induced effects on the wing flow-separation characteristics correlate well w i t h the nonlinear l i f t increase a t moderate and high angles o f attack. the vortex core over the wing surface l i m i t s the maximum vortex-induced l i f t increment and can promote p i t c h i n s t a b i l i t y . The proximity o f the unburst v o r t i c a l flows t o suitably placed twin v e r t i c a l t a i l s enhances the s t a b i l i z e r effectiveness. ing near the t a i l surfaces can induce a severe b u f f e t environment leading t o structural failure. through the boundary-layer bleed s l o t s a t the juncture o f the LEX and fuselage was observed t o have an adverse e f f e c t on the LEX vortex core s t a b i l i t y a t high angles o f attack, thereby reducing the maximum l i f t i n comparison t o the configuration w i t h slots closed. Clearly. the a v a i l a b i l i t y o f off-body flow- f i e l d trends through simple water-facility experiments i s o f great value i n the a i r c r a f t design process. To establish confidence i n the water flow simulations, flow-visualization studies were performed i n the p i l o t and larger-scale water tunnels o f the leading-edge vortex behavior on thin, sharp-edged delta wings encompassing a wide range of the leading-edge sweep angle. The vortex positions determined from the water tunnel t e s t i n g were i n reasonable agreement w i t h r e s u l t s obtained a t higher Reynolds numbers i n wind tunnels (Ref. 34). This was due t o the i n s e n s i t i v i t y of the primary flow separation a t the sharp leading edge t o changes i n the Reynolds number. The f a c t t h a t theoretical methods which ignore viscous effects can reasonably predict vortex flow aerodynamics i s one indication of the Reynolds number i n s e n s i t i v i t y o f such phenomenon. tunnel i s limited, however. because o f the viscous e f f e c t s near the wing surface. The upper-surface boundary-layer f l o w separates near the leading edge, generating a secondary vortex having a sense o f rota- t i o n which i s opposite t o t h a t o f the primary vortex. The secondary separation l i n e and the strength and location o f the secondary vortex vary with the Reynolds number. The location o f the primary vortex core w i i i be affected by the state of rne boundary k y e r ana w i i i be somewndi inbodra ana nigner o f f the wing surface when s u b c r i t i c a l (laminar) separation occurs. Wortmann has commented i n a recent paper (Ref. 68) t h a t this, and s i m i l a r water f a c i l i t y I n addition, the For example, the development of a concentrated v o r t i c a l flow above The breakdown of Under certain conditions, however, the occurrence o f vortex burst- The flow The agreement between the vortex positions determined i n the water tunnel and the wind The vortex breakdown characteristics compared favorably w i t h s i m i l a r observations made i n wind tunnels and i n f l i g h t , as shown i n Fig. 16. conditions the r e l a t i v e importance o f i n e r t i a and viscous and pressure terms was simulated i n the water tunnel. Experience suggests t h a t the adverse pressure gradient i n the external potential flow f i e l d i s the dominant parameter affecting the vortex breakdown a t high angles o f attack. the i n v i s c i d Euler methods t o "capture" the vortex breakdown phenomenon on t h i s class o f aerodynamic shape i s based on s i m i l a r reasoning (Ref. 69). a sharp leading edge, a water flow simulation i s expected t o provide an acceptable representation o f the size and structure of the wake shed from a thin wing a t a high angle o f attack and, consequently, the pressure f i e l d through which a vortex core must traverse. The water tunnel photograph o f Fig. 17 i l l u s - t r a t e s t h i s flow s i t u a t i o n on a small-scale model o f an advanced f i g h t e r featuring a sharp, highly swept LEX. n a l noise i n t e n s i t y which i s associated with the forward advance o f the vortex breakdown p o s i t i o n on e i t h e r side of the canopy. Another water tunnel-to-flight correlation o f a q u a l i t a t i v e nature i s shown i n Fig. 18, which depicts LEX vortex breakdown on a current f i g h t e r a i r c r a f t at high angle o f attack. water tunnel model (an inexpensive, p l a s t i c model) e x h i b i t s vortex bursting over the wing panel a t a loca- t i o n approximating the burst p o s i t i o n on the full-scale a i r c r a f t . o f the adverse pressure gradient i n the external p o t e n t i a l flow f i e l d - - t h a t w a t e r - f a c i l i t y vortex- breakdown r e s u l t s can be applied t o higher-Reynolds-number phenomena i n a i r a t high angles o f attack. o f these preliminary investigations. years demonstrated the strengths and limitations o f w a t e r - f a c i l i t y simulations o f f i g h t e r a i r c r a f t flow This c o r r e l a t i o n indicated that under c e r t a i n r e s t r i c t i v e The apparent success o f Provided flow separation occurs simultaneously everywhere along A t a s i m i l a r angle o f attack, p i l o t s o f this a i r c r a f t have detected a sudden increase i n the exter- The It i s upon t h i s premise--the dominance The water tunnel became an important element i n the major a i r c r a f t programs a t Northrop as a r e s u l t The ensuing flow-visualization experiments spanning the next several 10 fields. vortex-dominated f l o w f i e l d were insensitive t o the Reynolds number. experiments. modifications f o r which wind tunnel and f l i g h t test data were available. and area modifications shown i n Fig. 19 increased the maximum l i f t and improved the s t a t i c l a t e r a l - directional s t a b i l i t y characteristics a t high angles o f attack. enhanced s t a b i l i t y o f the LEX v o r t i c a l f l o w at angles of attack near maximum l i f t , a delay t o higher angle of attack o f the pronounced vortex breakdown asymnetry i n sideslip, and increased dynamic pressure i n the v i c i n i t y o f the centerline v e r t i c a l t a i l arising from the delayed breakdown o f the windward LEX vortex. It was determined from the flow-visualization t e s t i n g of these models t h a t faired-over engine i n l e t s Emphasis was placed on studies a t high angles of attack where the phenomenological aspects o f the The F-5E, F-5F, and F-20A f i g h t e r configurations were the subject of extensive flow-visualization The water tunnel was used primarily t o obtain off-body flow-field information on airframe For example, the LEX planform The water tunnel t e s t i n g revealed could y i e l d misleading results regarding the vortex behavior a t high angles o f attack. LEX vortex breakdown characteristics indicated that influx i n t o the side-mounted i n l e t s a t moderate t o high angles of attack induced a local upwash near the LEX apex so that the e f f e c t i v e was approximately 2" higher i n comparison t o the blocked-inlet case. f i g h t e r models were tested w i t h flowing inlets. f i g h t e r configurations w i t h top-mounted inlets. The water tunnel flow visualization. i n combination w i t h wind tunnel t e s t results, showed t h a t careful integration of the i n l e t on the fuselage t o take advantage of the LEX vortex-induced sweeping action could effectively control the i n l e t pressure recovery and dynamic d i s t o r t i o n a t high angles o f attack (Ref. 70). The F-5 configuration was also used as a test bed t o evaluate propulsive lift-enhancement concepts. A representative study featured the application of spanwise blowing t o the wing upper surface t o reener- gize the LEX vortex a t high a. A t sufficiently high blowing rates, a discrete vortex from the wing lead- ing edge was also apparent, as shown i n Fig. 20. required i n water tunnel simulations t o effect a p a r t i c u l a r change i n the flow f i e l d was somewhat higher than t h a t i n wind tunnel t e s t i n g performed a t higher Reynolds numbers. blowing rates t h a t were necessary t o energize the laminar boundary layer on the wing upper surface. o f the F-5F a t zero s i d e s l i p was responsible f o r the large aerodynamic asymnetries a t high angles of attack t h a t were encountered i n wind tunnel and f l i g h t testing. studies i s shown i n Fig. 21. Although the primary separation along the forebody sides was sensitive t o the Reynolds number, the forebody vortex asymnetry was promoted by an i n v i s c i d hydrodynamic i n s t a b i l i t y associated w i t h a crowding together o f the vortices near the nose (Ref. 71). nism was amenable t o study i n the water tunnel. Because o f the laminar separation. however, the vortices were more widely spaced along the forebody i n comparison t o the case o f turbulent separation. Therefore, the angle o f attack f o r onset of the vortex asymmetry was t y p i c a l l y a few degrees higher than the onset o f aerodynamic asymmetries determined from wind tunnel and f l i g h t tests. For example, the i n t h i s region a As a consequence, a l l ensuing The a b i l i t y t o visualize the vortex-engine i n l e t interactions subsequently l e d t o detailed studies of In general, it was found t h a t the blowing momentum This was due t o the increased Water tunnel studies confirmed t h a t the asymnetric shedding of the vortices from the slender forebody A representative r e s u l t from these As a result, the flow mecha- The a b i l i t y of the F-5 "shark nose" depicted i n Fig. 22 t o a l l e v i a t e the vortex asymnetry was demon- strated i n water tunnel flow-visualization tests. the nose increased the l a t e r a l spacing o f the vortices and, therefore, reduced the s u s c e p t i b i l i t y t o flow- f i e l d asymmetries. Larit decredse i n tne aerodynamic asynmecries aiong w i r n improveo oeparrure/spin resistance. Similarly, the nose shape on the reconnaissance version of the F-5F w i t h i t s forward-looking oblique window (RECCE nose) was shown i n water flow studies t o reduce the asymmetric vortex shedding a t high a. apex region and the a b i l i t y of a water f a c i l i t y to visualize the f l o w - f i e l d changes. study of numerous other active and passive methods o f asymmetric sideload a l l e v i a t i o n i n the water tunnel. These included nose strakes. h e l i c a l separation t r i p s , and normal and tangential blowing on the forebody. The flow-visualization experiments provided a rapid assessment of the a b i l i t y o f the flow- control devices t o reduce/el iminate/reverse the body vortex asymmetry. The devices that were i d e n t i f i e d as promising flow modulators i n the water tunnel proved e f f e c t i v e i n c o n t r o l l i n g the 6-zero asymnetries i n wind tunnel t e s t i n g o f subscale F-5 models. The broader planform and flattened cross section near Wind tunnel and f l i g h t testing o f the F-5F/shark nose combination revealed a s i g n i f i - The l a t t e r studies demonstrated the s e n s i t i v i t y o f the vortex flows t o the geometry o f the forebody This l e d t o the The water tunnel was a useful t o o l i n analyzing the directional s t a b i l i t y trends obtained i n wind tunnels and i n f l i g h t since at high angles of attack the forebody can strongly a f f e c t the s t a t i c direc- t i o n a l s t a b i l i t y . An example includes the F-5F forebody. which develops an unusual vortex orientation i n s i d e s l i p as shown i n Fig. 23, O n the basis of water tunnel/wind t u n n e l / f l i g h t correlations, the forebody vortices and t h e i r unique orientation were identified as the primary source o f s t a t i c directional s t a b i l - i t y a t high angles o f attack. As a consequence, the e f f e c t o f forebody modifications on the yaw s t a b i l i t y could generally be surmised from water tunnel testing by observing any changes t o the forebody vortex structure and location. The water tunnel t e s t i n g of the F-5 and related f i g h t e r models revealed several l i m i t a t i o n s t o the water f a c i l i t y simulations. tunnel i s influenced by the wake region produced by laminar separation from the rear p o r t i o n o f the wing. A t low angles of attack, generally less than 10". the vortex core i n a water As shown in Figs. 24 and 25, t h i s a l t e r s the vortex path and also produces premature dissipation o f 11 the vortex because of entrainment o f turbulent f l u i d from the separated wake. o f the vortex behavior i n wind tunnels and i n f l i g h t a t s i m i l a r angles o f attack indicated t h a t the vortex does not burst over the wing (see Fig. 25). conforms closely t o the curvature o f the wing, p a r t i c u l a r l y when leading- and trailing-edge flaps are deflected, and can i n t e r a c t with downstream airframe components. ary-layer separation a t the subcritical conditions i n the water tunnel masks the quantitative e f f e c t s o f deflected leading- and trailing-edge devices on the vortex s t a b i l i t y . behavior on wings with thickness, camber, twist, and/or leading-edge bluntness i s also inadequate a t low-a conditions. Investigations o f these configurations, therefore, are 1 imited t o high angles o f attack where flow separation occurs everywhere along the wing leading edge. Under such conditions, the fundamental character o f the vortex-dominated flow i s s i m i l a r t o t h a t observed a t higher Reynolds number. (Ref. 72). t o which the water-facility results can be extrapolated t o higher Reynolds number. from wing leading-edge snags and lower and upper surface fences. f o r example, generally proved d i f f i c u l t t o simulate i n the water tunnel because o f the strong interaction between the separated wing boundary layer and the vortical motions. An attempt t o represent the interaction o f the vortex generated from the nacelle strake, o r engine "ear," o f a comnercial transport model w i t h the wing upper surface proved f u t i l e . the wing upper surface and closely followed the curvature o f the wing without breakdown. vortex interactions, and breakdown on f i g h t e r a i r c r a f t featuring r e l a t i v e l y large LEXes i n proximity t o the forebody. angle-of-attack range owing t o the persistence o f the v o r t i c a l motions and t h e i r proximity t o one another. F-18 i s shown i n Fig. 26. s i d e s l i p and the entrainment o f t h i s vortex p a i r by the dominant LEX v o r t i c a l flows. i n t e r a c t i o n was very sensitive t o small changes i n the s i d e s l i p angle. forebody vortex orientation i n s i d e s l i p was found. under c e r t a i n conditions. t o influence the wing s t a l l behavior and, hence. the l a t e r a l s t a b i l i t y characteristics (Ref. 73). For example, the forebody vortices were resistant t o asymmetric orientation i n sideslip when radome strakes were i n s t a l l e d a t 40" above the maximum half-breadth. windward side o f the a i r c r a f t . This flow s i t u a t i o n was unsteady, however, as the leeward body vortex would periodically pass underneath the windward v o r t i c a l flow. vortex interactions resulted i n powerful vortex-induced downwash and sidewash on the windward wing panel, thereby delaying complete wing s t a l l t o angles o f attack greater than 40". This e f f e c t was also observed i n wind tunnel smoke-flow visualization. The increased r o l l s t a b i l i t y a r i s i n g from the strake i n s t a l l a - t i o n was confirmed i n subscale wind tunnel and f u l l - s c a l e f l i g h t testing. I n addition. the wind tunnel model i n s t a l l e d on a f r e e - t o - r o l l r i g and the f u l l - s c a l e vehicle i n f l i g h t revealed modest strake-induced l a t e r a l oscillations, or wing rock, which was consistent w i t h the unsteady forebody-LEX vortex interac- t i o n s observed i n the water tunnel and wind tunnel flow-visualizations. The investigations demonstrated how changes t o the forebody flow could be amplified downstream t o a f f e c t the wing aerodynamics and how the understanding o f complicated flow interactions could be improved through water tunnel experiments. A t the higher angles o f attack, the water tunnel provided useful f l o w - f i e l d r e s u l t s regarding the effects o f the canard downwash on the wing flow f i e l d . For example, the delay o f leading-edge flow separation on the wing, the progressive development o f the wing vortex w i t h increased angle o f attack, and the enhance- ment o f the l a t t e r a t high a i n the presence o f the canard downwash f i e l d were demonstrated i n water flow experiments. The flow-visualization photograph o f Fig. 27 shows a discrete wing vortex on a small- scale model o f the Swedish Viggen a i r c r a f t a t o = 30". which i s w e l l beyond the angle o f attack f o r s t a l l o f the isolated wing. The l a t e r a l s e n s i t i v i t y a t high angles o f attack associated w i t h asymmetric vortex breakdown i n side- s l i p i s an inherent characteristic o f any f i g h t e r a i r c r a f t employing large amounts o f vortex lift. Flow- v i s u a l i z a t i o n investigations were conducted t o improve the f l o w s i t u a t i o n depicted i n Fig. 28 by suitable modifications t o the LEX planform and addition o f LEX fences and s l o t s t o modulate the vortex core break- down behavior i n sideslip. low-speed wind tunnel data trends. Water tunnel studies of a slender hypersonic research configuration unconstrained i n r o l l revealed a self- induced, bounded l a t e r a l o s c i l l a t i o n s i m i l a r t o that observed i n the wind tunnel. the o s c i l l a t o r y leading-edge vortex core and breakdown phenomena provided insight i n t o possible triggering and sustaining mechanisms o f t h i s single-degree-of-freedom o s c i l l a t i o n . Water tunnel tests were performed on a 2-D ejector nozzle t o study the e f f e c t s o f s w i r l on the exhaust plume characteristics. reduced the primary nozzle potential core and, hence. the mixing shroud length as shown i n Fig. 29. results, which confirmed a theory developed by Chu (Ref. 74), were useful f o r such applications as j e t noise reduction. I n contrast, observations Rather, the discrete vortex core exhibits a t r a j e c t o r y that For s i m i l a r reasons, the laminar bound- The simulation of the vortex This was the j u s t i f i c a t i o n f o r water tunnel studies o f the Space Shuttle Orbiter by Lorincz The scale of the v o r t i c a l motions r e l a t i v e t o the boundary-layer thickness w i l l determine the degree The vortices generated I n f l i g h t , however, the strake vortex observed under conditions o f natural condensation traversed A major contribution o f the water tunnel pertained t o the visualization o f m u l t i p l e vortex flows, Such conf igurations were shown t o develop strong flow interactions throughout the extended A n example o f the strong coupling o f forebody and LEX vortices on a small-scale model o f the The flow f i e l d i s characterized by symnetric forebody vortex shedding a t zero This multiple vortex Furthermore. modulation o f the A t small sideslip angles, the body vortex system was actually biased toward the The strake e f f e c t s on the forebody-LEX The simulation of m u l t i p l e vortex interactions was extended t o canard-wing f i g h t e r configurations. Excellent correlations were obtained between the f l o w - f i e l d observations and The occurrence o f wing rock i s comnon t o slender-wing a i r c r a f t a t high angles o f attack. The v i s u a l i z a t i o n of The flow-visualization experiments indicated t h a t s w i r l dramatically The 12 Another application o f the water tunnel t o a nonvortex-dominated flow f i e l d pertained t o thrust reversers on an advanced fighter configuration i n and out of ground proximity. f r o m these experiments i s shown i n Fig. 30. ranges o f the nozzle geometry. orientation. and j e t v e l o c i t y r a t i o . Emphasis was placed on the j e t block- age and entrainment effects on the v e r t i c a l and horizontal s t a b i l i z e r flow f i e l d s . It was d i f f i c u l t t o i d e n t i f y the pertinent flow mechanisms. however. and the correlations w i t h t a i l loads information obtained i n wind tunnel t e s t s were necessarily limited. 3.4 NASA Ames Research Center. Dryden F l i g h t Research F a c i l i t y , Edwards, California The NASA Ames-Dryden Flow Visualization System (FVS) (Fig. 31) i s a single-return f a c i l i t y with a 16- A representative r e s u l t The reverser plume shape and t r a j e c t o r y were observed for by 24-in. v e r t i c a l t e s t section. operation since 1983. surement o f the separated flow f i e l d s about advanced m i l i t a r y a i r c r a f t configurations. NASA in-house research. numerous cooperative ventures have been undertaken with industry, universities, and other government agencies. developing laser-enhanced visualization (LEV) and 2-D laser velocimeter (LV) techniques f o r the water flow studies. In addition. successful attempts have been made t o obtain quantitative information using instru- mented models. This installation, modeled a f t e r the Northrop water tunnel, has been i n The NASA f a c i l i t y has been used almost exclusively f o r the visualization and mea- In addition t o The NASA f a c i l i t y has expanded on the Northrop tunnel c a p a b i l i t i e s by Many o f the flow-visualization studies have been made i n support o f the NASA High-Alpha Technology Flow-field information has been obtained on small-scale models o f the F/A-18, as shown i n Program. Fig. 32, t o i d e n t i f y effective passive and active vortex flow-control concepts f o r future wind tunnel and f l i g h t testing. A related study sponsored by the Navy featured quantitative measurements o f the F/A-18 twin v e r t i c a l t a i l b u f f e t characteristics i n the presence o f L E X vortex breakdown. combination with e x i s t i n g wind tunnel and f l i g h t t e s t data, provided an improved understanding o f vortex- empennage interactions a t high angles of attack that can lead t o severe t a i l buffeting. The surface hot- f i l m anemometer r e s u l t s showed high turbulence a c t i v i t y on the f i n s a t conditions coincident w i t h vortex bursting observed from flow visualization. cies from the water tunnel tests correlated well with wind tunnel tests a t higher Reynolds number (Ref. 75). A generic study (Ref. 48) was made of a concept t o improve the v e r t i c a l - t a i l b u f f e t environment. This t e s t featured the generation of a "free vortex" w i t h an imposed downstream pressure gradient t o pro- mote core bursting. surements showed t h a t the active flow control significantly reduced the turbulence intensity. i n f e r r e d from these preliminary results that core blowing applied t o the F/A-18 configuration would have a favorable e f f e c t on the vortex-empennage interactions a t high angles o f attack. t o evaluate potential dynamic l i f t benefits a t angles o f attack beyond the s t a t i c maximum lift. v i s u a l i z a t i o n revealed a lag i n the flow-field response t o the a i r c r a f t motion and a delay o f the LEX vortex bursting i n comparison t o the s t a t i c case. The l a t t e r phenomenon was very transient. however, as the f l o w f i e l d r a p i d l y assumed i t s steady-state condition upon termination of the maneuver. I n contrast t o t h e r e s u l t s obtained a t NAE on a slender generic f i g h t e r model, there was no s i g n i f i c a n t l a t e r a l o s d l - l a t i o n o f the vortex flows due t o the pitching maneuver. vortices which emanated from fixed lines o f separation and therefore were resistant t o large l a t e r a l excursions. The water tunnel investigation, i n The vortex frequencies, vortex bursting, and dominant frequen- Blowing along the core was then i n i t i a t e d t o delay the onset o f bursting. L V mea- I t can be A q u a l i t a t i v e study was made o f an F/A-18 model undergoing p i t c h o s c i l l a t i o n s and ramp-type motions The flow This was a t t r i b u t e d t o the dominance o f the LEX The Space Shuttle Orbiter configuration was tested t o evaluate the vortex-shedding patterns on the t h i c k cranked wing a t high angles o f attack. used t o develop a LEV c a p a b i l i t y (Ref. 49) and the r e s u l t a n t technique was then applied t o several other configurations. including a powered AV-8A Harrier model i n ground proximity and a drag-reduction concept featuring a t r a i l i n g disk behind the base o f a cylinder. plane f r o m the l a t t e r study i s shown i n Fig. 34. Douglas F-15 and F-4. the Grumman/DARPA X-29, the General Dynamics F-16XL. and the NASA/General Dynamics F-106 with vortex flaps. The vortex flow behavior on the NASA/LTV/Rockwell F-8 oblique wing t e s t a i r c r a f t configuration was visualized as shown i n Fig. 35. O f p a r t i c u l a r i n t e r e s t was the asymmetric vortex forma- t i o n and breakdown on t h i s skewed wing arrangement and the interaction o f the v o r t i c a l flows w i t h the v e r t i c a l t a i l . able levels o f s t a t i c lateral-directional s t a b i l i t y a t moderate and high angles o f attack. A representative r e s u l t i s shown i n Fig. 33. This model was A t y p i c a l flow vlsualization i n a streamwise Other a i r c r a f t configurations t h a t have been studied i n the NASA i n s t a l l a t i o n include the McOonnell- The understanding and control of these phenomena are essential i n order t o obtain accept- In addition t o experiments on specific a i r c r a f t configurations, basic aerodynamic research programs have also been supported by water tunnel experiments i n the NASA Ames-Dryden f a c i l i t y . One example i s a recent study (Ref. 76) performed by Eidetics International f o r the A i r Force t o investigate methods o f vortex control t o enhance aerodynamic control on f i g h t e r a i r c r a f t a t high attitudes. The aim o f t h i s flow-visualization study was t o explore methods of a l t e r i n g the hatural state of the forebody and LEX vortices by i n t e r j e c t i n g i n t o the flow f i e l d either small surfaces o r blowing jets. 13 4. S U H M A R Y A review has been made of the r o l e of water f a c i l i t i e s i n vehicle design. The use o f water as a flow-visualization medium began very early. i n the 15th century with the observations and sketches o f Leonard0 da Vinci. who also designed the f i r s t water tunnel. Leonard0 hypothesized t h a t flows i n water and a i r were similar, which was o f great impor- tance t o the advancement of f l u i d mechanics. The s c i e n t i f i c application o f water flow v i s u a l i z a t i o n began In the centuries t o follow, sporadic experiments on simple shapes were performed i n water, p r i m a r i l y f o r marine c r a f t applications. The Pioneering research o f Ludwig Prandtl and h i s colleagues and the dawn- ing o f the era o f f l i g h t i n the early 20th century marked an upsurge i n the use o f water f a c i l i t i e s f o r vehicle design. I n the ensuing decades up through World War 11. water tunnels. channels, and towing tanks yielded useful qualitative. and sometimes quantitative, information on various aerodynamic and hydrody- namic shapes suitable f o r f l i g h t and marine vehicles. The trend toward increased vehicle size and speed posed a scaling problem f o r water f a c i l i t y simula- tions. The matching of the Froude number i n marine c r a f t t e s t i n g was generally a t the expense o f a large Reynolds number gap. The scaling of c a v i t a t i o n phenomena was a continual challenge. tions were necessarily r e s t r i c t e d t o the incompressible flow regime and could not represent the phenomena encountered on high-speed a i r c r a f t t h a t emerged during World War 11. match was also present i n the testing o f f l i g h t vehicles i n water f a c i l i t i e s . These problems continue t o the present day. The u t i l i z a t i o n o f controlled f l o w separations and vortex flows by design t o improve vehicle perfor- mance emerged i n the 1950s. This increased the u t i l i t y o f water f a c i l i t i e s as a vehicle design tool largely because the fundamental structure of these flows was insensitive t o the Reynolds number. The r a p i d advancement of vehicle technology since that time has resulted i n expanded operating envelopes and corresponding increase i n the flow-field complexity. 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Goodman, A. and Brown, C.. "An Experimental Study t o Determine the Vortex-Flow and the Subsonic and Dynamic S t a b i l i t y Characteristics o f A i r c r a f t Operating a t High Angles-of-Attack," presented a t the A G A R D F l u i d Dynamics Panel Symposium on Aerodynamic and Related Hydrodynamic Studies Using Water F a c i l i t i e s , 20-23 Oct. 1986, Monterey. Calif. 56. 57. Chezlepretre. 8. and Brocard. Y., "Qualification d'un Tunnel Hydrodynamique Pour Des Pesees de Marquettes Aeronautiques," presented a t the A G A R 0 F l u i d Dynamics Panel Symposium on Aerodynamic and Related Hydrodynamic Studies Using Water Facilities. 20-23 Oct. 1986, Monterey. C a l i f . Lloyd, A. R. J. M. and Campbell, I. F.. "Experiments t o Investigate the Vortices Shed From a Submarine-Like Body o f Revolution," presented a t t h e AGARD F l u i d Dynamics Panel Symposium on Aerody- namic and Related Hydrodynamic Studies Using Water F a c i l i t i e s , 20-23 Oct. 1986, Monterey, C a l i f . Davies, A. G., "A Comparative Study o f Vortex Flows i n Wind and Water Tunnels," presented a t the A G A R D F l u i d Dynamics Panel Symposium on Aerodynamic and Related Hydrodynamic Studies Using Water F a c i l i t i e s , 20-23 Oct. 1986. Monterey. Calif. Cockrell. 0. J., Polpitiye, S. J.. Vavuz. T., and Shen, C.. "Measurements of Aerodynamic Forces on Unsteadily-kving Bluff Parachute Canopies," presented a t the AGARD F l u i d Dynamics Panel Symposium on Aerodynamic and Related Hydrodynamic Studies Using Water F a c i l i t i e s . 20-23 Oct. 1986, Monterey, C a l i f . 5 8 . 59. 60. 61. Boszko, P. J. and Owen, G. S., "Water Flow Visualization o f a Ramrocket Combustion Chamber," pre- sented a t the AGARD F l u i d Dynamics Panel Symposium on Aerodynamic and Related Hydrodynamic Studies Using Water F a c i l i t i e s , 20-23 Oct. 1986. Monterey. Calif. 16 62. Bippes, H . , "The Use of a Water Towing Tank f o r Aerodynamic Testing and a Method f o r Quantitative Evaluation of Stereo Photographs," presented a t the AGARD F l u i d Dynamics Panel Symposium on Aerody- namic and Related Hydrodynamic Studies Using Water F a c i l i t i e s , 20-23 Oct. 1986. Monterey. Calif. Strunz, M.. "A New Laminar Water Tunnel t o Study the Transition Process i n a Blasius Layer and a Separation Bobble." presented a t t h e AGARD F l u i d Dynamics Panel Symposium on Aerodynamic and Related Hydrodynamic Studies Using Water Facilities. 20-23 Dct. 1986, Monterey, C a l i f . 64. Wenhan, S . . Mouji, L.. Bocheng, Zhou, Chenghao, Q., and Shanwen. X., "An Experimental Investigation o f Leading-Edge Spanwise Blowing," Intern. Counc. of Aeron. Sci ., ICAS-82-6.6.2, Aug. 1982. Fujita. T., "Flow "Visualization o f Ship Model i n Towing Tank." Flow Visualization 11, Proc. 2nd Intern. Symp. on Flow Visualization, September 9-12, 1980, Bochum, W. Germany (Ed. by W. Merzkirch). Ward, T. M.. "The Hydrodynamics Laboratory a t the California I n s t i t u t e o f rechnology - 1976," Trans. ASME, 3. F l u i d Eng., Dec. 1976, pp. 740-748. Furuya. D., "Three-Dimensional Theory on Supercavitating Hydrofoils Near a Free Surface." J. F l u i d Mech.. Vol. 71. Pt. 2. 1987. pp. 339-359. Wortmann. A., "On Reynolds Number Effects i n Vortex Flow Over A i r c r a f t Wings." AIAA Paper 84-0037. Jan. 1984. 63. 65. 66. 67. 6 8 . 69. Hitzel. S. M., "Vortex Breakdown o f Leading-Edge Vortices Above Swept Wings." Dornier Report, Aktenvermerk BF30-2576/84. Friedrichshafen. Jan. 1980. 70. Williams, T. L. and Hunt, B. L.. "Top Mounted I n l e t System F e a s i b i l i t y f o r Transonic-Supersonic Fighter Applications," AIAA Paper 80-1809, Aug. 1980. Keener, E. R. and Chapman, G. T., "Similarity i n Vortex Asymnetries Over Slender Bodies and Wings," AIAA J., Vol. 15. No. 9. Sept. 1977, pp 1370-1372. 71. 72. Lorincz, D. J., "Space Shuttle Orbiter Flow Visualization Study." NASA CR-163092. Feb. 1980. 73. Erickson, G. E. and Gilbert. W. P., "Experimental Investigation o f Forebody and Wing Leading-Edge Vortex Interactions a t High Angles of Attack," AGARD-CP-342. July 1983. Chu, C.-W., Der, J . , Jr.. Ortiz. V. M., and Widynski, T. C., "Effect o f S w i r l on the Potential Core i n Two-Dimensional Ejector Nozzles," J. A i r c r a f t , Vol. 20. No. 2. Feb. 1983, pp. 191-192. Lan. C. E., Lee, 1. G.. and Wentz, W. H., "Investigation o f Empennage Buffeting." Technical Report CRINC-FRL-714-1, F1 ight Research Laboratory, Univ. Kansas Center f o r Research, Inc., July 1986. 76. Malcolm. G. N. and Skow. A. M., "Enhanced C o n t r o l l a b i l i t y Through Vortex Manipulation on Fighter A i r c r a f t a t High Angles o f Attack," A I A A Paper 86-2277-CP. Aug. 1986. 74. 75. 17 1 3 2 4 Figure 1. Vinci (from Ref. 1). Fluid flow sketches by Leonard0 d a Figure 2. glass." read from right to left as was Leonardo's habit, represents the f i r s t historic example o f a water tunnel (from Ref. 5). The box with the inscription "white (a) Flow past a knife edge. ( b ) Flow along aft portion of b l u n t body. (c) KBrmhn vortex street downstream of cylinder. (d) Flow in a sharply diverging channel with wall suction. Figure 3. Representative water flow-visualization results of Prandtl (from Ref. 12). 18 (a) Leading-edge vortex on a slender wing. (c) Wingtip vortices on a comnercial transport a i r c r a f t . (e) Bilge and stern vortices on displacement hulls. (b) Vortices on the lee side of a missile. (d) Side-edge vortices on an automobile. (f) Vortex flows on an a i r c r a f t c a r r i e r f l i g h t deck. (9) Vortex flows about a maneuvering submarine. Figure 4. Sketches of vortex flows shed from air, ground, and marine vehicles (from Ref. 1). 19 Figure 5 . NAE water tunnel. Figure 6. tunnel). Wing-submerged l i f t i n g fan (NAE water Figure 7. EBF arrangement (NAE water tunnel). Figure 8. (NAE water tunnel). Surface flow on a C-5 transport model Figure 9. Vortex wake o f l i f t i n g fuselage, side view (NAE water tunnel). Figure 10. angle o f attack 45"--vortices asymmetrical (NAE water tunnel). Modern a i r c r a f t with long forebody-- 20 ORiG-IFShl FxGE I S OF POOR QUALITY (a) Original version. (b) With deflector on cab. Figure 1 1 . Truck t r a i l e r (NAE water tunnel). Figure 12. Sketch of the CalTech FSWT. Figure 13. Cavitating flow about 2-D wedge at high angle o f attack (CalTech FSWT). Figure 14. Northrop Corporation water tunnel. 2 1 REYNOLDS NO. (BASED ON CENTER- LINE CHORD) 0 D u1 G c1 0 b 1 Y u 4 c U . 40 10 55 0 WATER TUNNEL 4.1 x lo4 8 WATER TUNNEL 9.8 x 103 + WATER TUNNEL 1.0 x lo4 0 WINDTUNNEL 1.5 X lo6 A WINDTUNNEL 1.3 X lo6 d i 1 60 WIND TUNNEL WIND TUNNEL WIND TUNNEL FLIGHT WATER TUNNEL WIND TUNNEL WIND TUNNEL WATER TUNNEL WATER TUNNEL 65 70 WING SWEEP - 9.0 x 105 2.0 x 106 1 . 0 8.0 x 104 2.0 x 106 1.0 x 106 3.0 x io4 3.0 x 104 1.4&1.7X106 40.0 X lo6 L ~ . DEG 1 85 Figure 15. Vortex flows about YF-17 model a t a = 20" (Northrop water tunnel). Figure 16. number on delta wing vortex breakdown at the trailing edge (from Ref. 34). Effects of wing sweep and Reynolds LEX VORTEX CORE ' VORTEX BREAKDOWN Figure 17. Vortex breakdown on a small-scale model of an advanced fighter configuration (Northrop water tunnel). 22 (a) Northrop water tunnel--dye injection. (b) F1 ight--natural condensation. Figure 18. Correlation of vortex breakdown on a current fighter aircraft at high angle of attack. 0 W6LEX 40 0 W 0 I 0 + Y 30 2 : : 20 a Y -I u z a 10 n 0 0.20 0.40 0.60 0.80 1.00 VORTEX BREAKDOWN LOCATION, XIC, Figure 19. attack (from Ref. 34). Effect of LEX planform modification on the progression of vortex bursting with the angle of 23 " t 4 I i separated flow II ' ' ,c- (a) Blowing o f f . (b) Blowing on. Figure 20. (Northrop water tunnel). E f f e c t o f wing upper surface spanwise blowing on the leading-edge vortex behavior a t ( I = 24" VORTEX ASYMMETRY VORTEX ASYMMETRY Figure 21. Asymmetric forebody vortex shedding a t Figure 22. Close-up o f vortex flows developed on zero sideslip (Northrop water tunnel). forebody w i t h shark nose (Northrop water tunnel). 24 I WIND VOR PREMATURE CORE INSTABILITY NEAR TRAILING EDGE 0 Figure 23. Forebody vortex orientation i n Figure 24. Vortex f l o w a t low angle o f attack s i d e s l i p (Northrop water tunnel). (Northrop water tunnel ). Figure 25. Comparison o f vortex flow behavior i n Figure 26. Forebody-wing vortex flow interactions water and wind tunnel f a c i l i t i e s (Northrop). on an advanced f i g h t e r model (Northrop water tunnel). L3 Figure 27. Vortex flows on a canard-wing fighter Figure 28. Asymmetric vortex breakdown i n model (Northrop water tunnel). sideslip (Northrop water tunnel). PREDIC TlON NO SWIRL 22 5 SWIRL 1 Figure 29. Effect of swirl on 2-D ejector nozzle Figure 30. Thrust-reverser effect i n ground flow (Northrop water tunnel). proximity (Northrop water tunnel). 26 Figure 31. NASA Ames-Dryden FVS. Figure 32. o f the F/A-18 (NASA Ames-Dryden water tunnel). Vortex flows about a small-scale model Figure 33. o f the Space Shuttle O r b i t e r (NASA Arnes-Dryden water tunnel ). Vortex flows about a small-scale model Figure 34. flow about a cylinder with t r a i l i n g disk (NASA Ames-Dryden water tunnel). Laser-enhanced v i s u a l i z a t i o n o f the 27 h (a) Low angle of attack. (b) High angle o f attack. Figure 35. Ames-Dryden water tunnel). Flow visualization o f a small-scale model of the F-8 oblique wing aircraft configuration (NASA 28 2. Government Accession No. 1. Report No. NASA TM-89409 4. Title and Subtitle WATER FACILITIES IN RETROSPECT AND PROSPECT--AN ILLUMINATING TOOL FOR VEHICLE DESIGN 3. Recipient's Catalog No. 5. Report Date 6. Performing Organization Code November 1986 Ames Research Center 7. Author(sJ and Andrew M. Skow and Gerald N. Malcolm (Eidetics International, Torrance, CA 90505) 9. Performing Organization Name and Address Gary E. Erickson, David J. Peak, and John Del Frate, 11. Contract or Grant No. 8. Performing Organization Report No. A-87021 10. Work Unit No. Moffett Field, CA 94035 13. Type of Report and Period Covered 12. Sponsoring Agency Name and Address National Aeronautics and Space Administration Washington, DC 20546 Technical Memorandum 14. Sponsoring Agency Code 505-6171 6 Abstract Water facilities play a fundamental role in the design of air, ground, and marine vehicles by Water providing a qualitative, and sometimes quantitative, description of complex flow phenomena. tunnels, channels, and tow tanks used as flow-diagnostic tools have experienced a renaissance in recent years in response to the increased complexity of designs suitable for advanced technology vehicles. three-dimensional flow separation and ensuing vortical flows. The visualization and interpretation of the complicated fluid motions about isolated vehicle components and complete configurations in a time- and cost-effective manner in hydrodynamic test facilities i s a key element in the development of flow control concepts and, hence, improved vehicle designs. These vehicles are frequently characterized by large regions of steady and unsteady This paper presents a historical perspective of the role of water facilities in the vehicle design process. emphasized. The application of water facilities to specific aerodynamic and hydrodynamic flow prmhlams is discussed, and the st;engths 2nd :imitations of 4.L.--- ~ I I C W Z ~ I I I ~ U ~ --- - L - - L L ~ H L aapar ---- - - I - - 1111er1t~.i tvuiv m e 7. Key Words (Suggested by Author(sJ1 Water tunnels Aircraft configuration Flow fields Vortical flows 18. Distribution Statement Unclassified-Unlimited Subject Category - 02 1 9 . Security Classif. (of this report) 20. Security Classif. (of this prgeJ 21. No. of Pages Unclassified Unclassified 2 8 22. Rice' A03
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This comprehensive guide offers you an in-depth understanding of their definitions, importance, applications, and structure. You'll also discover the mathematical formula behind these devices, illustrated by practical examples and real-world applications. Furthermore, the crucial characteristics of the parallel inductors and how they affect their usage will be elucidated. Finally, learn how to add inductors in parallel effectively with a step-by-step guide, including potential challenges and tips to surpass them, so that you can maximise the advantages of this fundamental physics phenomenon. Get started Millions of flashcards designed to help you ace your studies Sign up for free + Add tag Immunology Cell Biology Mo Why is the property of shared voltage across parallel inductors important in practical applications such as power grids? Show Answer + Add tag Immunology Cell Biology Mo What are some practical implications of using inductors in series or in parallel? Show Answer + Add tag Immunology Cell Biology Mo What are some potential challenges when adding inductors in parallel and how can you overcome them? Show Answer + Add tag Immunology Cell Biology Mo What happens to the total inductance when inductors are connected in a series configuration? Show Answer + Add tag Immunology Cell Biology Mo How is a parallel arrangement of inductors characterised? Show Answer + Add tag Immunology Cell Biology Mo Can you explain about the voltage and current in a parallel setup of inductors? Show Answer + Add tag Immunology Cell Biology Mo How is the total inductance in a parallel arrangement calculated? Show Answer + Add tag Immunology Cell Biology Mo What is the most prominent characteristic of parallel inductors? Show Answer + Add tag Immunology Cell Biology Mo What happens to the overall inductance when inductors are connected in a parallel configuration? Show Answer + Add tag Immunology Cell Biology Mo What does the term 'Inductors in Parallel' refer to in the realm of electricity and magnetism? Show Answer + Add tag Immunology Cell Biology Mo What is an important real-world application of 'Inductors in Parallel'? Show Answer + Add tag Immunology Cell Biology Mo What is the most prominent characteristic of parallel inductors? Show Answer + Add tag Immunology Cell Biology Mo What happens to the overall inductance when inductors are connected in a parallel configuration? Show Answer + Add tag Immunology Cell Biology Mo What does the term 'Inductors in Parallel' refer to in the realm of electricity and magnetism? Show Answer + Add tag Immunology Cell Biology Mo What is an important real-world application of 'Inductors in Parallel'? Show Answer + Add tag Immunology Cell Biology Mo Why is the property of shared voltage across parallel inductors important in practical applications such as power grids? Show Answer + Add tag Immunology Cell Biology Mo What are some practical implications of using inductors in series or in parallel? Show Answer + Add tag Immunology Cell Biology Mo What are some potential challenges when adding inductors in parallel and how can you overcome them? Show Answer + Add tag Immunology Cell Biology Mo What happens to the total inductance when inductors are connected in a series configuration? Show Answer + Add tag Immunology Cell Biology Mo How is a parallel arrangement of inductors characterised? Show Answer + Add tag Immunology Cell Biology Mo Can you explain about the voltage and current in a parallel setup of inductors? Show Answer + Add tag Immunology Cell Biology Mo How is the total inductance in a parallel arrangement calculated? 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Save ArticleSave Article Jump to a key chapter Understanding Inductors in Parallel Unpacking the Inductors in Parallel Formula Comparing Inductors in Series and Inductors in Parallel Key Characteristics of Inductors in Parallel How to Add Inductors in Parallel Inductors in Parallel - Key takeaways Similar topics in Physics Related topics to Electricity and Magnetism Flashcards in Inductors in Parallel Learn faster with the 15 flashcards about Inductors in Parallel Frequently Asked Questions about Inductors in Parallel About StudySmarter Test your knowledge with multiple choice flashcards 1/3 Why is the property of shared voltage across parallel inductors important in practical applications such as power grids? A. The shared voltage property reduces the total inductance in the system, a feature which isn't particularly valuable in practical applications like power grids. B. The shared voltage property helps to increase the power output of the system, a feature not utilized in power grids. C. The shared voltage property ensures each inductor in a parallel arrangement experiences the same current, which is not important in practical applications such as power grids. D. The shared voltage property ensures each inductor in a parallel arrangement experiences the same voltage regardless of their individual inductance values or current flowing in their branch. This is useful in systems like power grids, where all components need to operate at a specified voltage level. 1/3 What are some practical implications of using inductors in series or in parallel? A. Inductors in series and parallel have no practical implications and serve only theoretical purposes. B. Inductors in series are often used when a high level of inductance is required, and also in noise filtering. Inductors in parallel are used where a lower inductance is needed and are frequently found in power supplies, AC generators and radio frequency tuning circuits. C. Inductors, regardless of being in series or parallel, are only used in high-frequency circuits. D. Inductors in series are often used when a low level of inductance is required, and also in power supplies. Inductors in parallel are used where a higher inductance is needed and are frequently found in noise filtering. 1/3 What are some potential challenges when adding inductors in parallel and how can you overcome them? A. Challenges include maintaining accuracy in calculation due to decimal values, differentiating between series and parallel, mutual inductance, and ensuring good quality of inductor. These can be overcome by careful attention to detail, accurate calculations, correct identification of configuration, and sourcing good-quality components. B. Challenges include achieving a higher total inductance, inability to connect inductors at the same two points, achieving uniform voltage across each inductor and overcoming the reciprocal nature of capacitors. Calculation errors, wrong sourcing and lack of configuration options can't be avoided. C. The main challenge is adding inductors of different materials. Overcoming this situation involves limiting to homogenous inductors, avoiding capacitor coupling, eliminating magnetic interference and ensuring constant current. These are all inaccurate. D. Challenges include adding too many inductors in a parallel arrangement, coupling between capacitors, achieving uniform magnetic field and absence of resistance. Solutions include limiting the number of inductors, restricting capacitor coupling, ensuring distinct magnetic fields and integrating resistors. Score That was a fantastic start! You can do better! Sign up to create your own flashcards Access over 700 million learning materials Study more efficiently with flashcards Get better grades with AI Sign up for free Already have an account? Log in Good job! Keep learning, you are doing great. Don't give up! Next Open in our app Understanding Inductors in Parallel Physics is rife with terms and principles that serve as the bedrock of many technological advancements. A key concept in the realm of electricity and magnetism is that of 'Inductors in Parallel'. Embrace each layer of this potent term as you unlock deep layers of physical understanding. Thanks to this ad, StudySmarter remains free.Sign up for a free account to access a better learning experience. Basic Definition of Inductors in Parallel Inductors in parallel refer to a configuration where two or more inductors share the same voltage across their ends. They are commonly used in circuits such as AC generators and transformers. By using the term 'Parallel', it is implied that these inductors are placed side by side, hence each one experiences the same voltage. The total inductance for inductors in a parallel configuration is provided by the formula: 1 L Total=1 L 1+1 L 2+1 L 3+⋯ Surprisingly, just like resistors, only connected in a parallel manner does the total inductance tend to decrease. Importance and Application of Inductors in Parallel Inductors in Parallel are cornerstone elements in many practical applications, particularly in signal processing, power supply filters, and even your everyday radio transmission. Their properties of storing energy and limiting current make them invaluable in the real world. Thanks to this ad, StudySmarter remains free.Sign up for a free account to access a better learning experience. Structure of Inductors in Parallel Brimming with physical import, parallel inductors have a unique configuration, setting them apart from series or solo inductor arrangements. In a parallel inductor circuit, multiple inductors are connected across the same two points, resulting in the same voltage drop across each inductor. However, the current in each parallel branch may vary depending on the inductance of each inductor. Refer to the table beneath for an shed light on such an arrangement: In parallel setup The voltage across each inductor The sum of currents flowing through each inductor In a parallel connection of inductors Remains the same Adds up to give the total current in the circuit Components and Arrangement of Inductors in Parallel Think of a simple example where you have three inductors with inductances of L 1=3 m H, L 2=6 m H and L 3=9 m H connected in parallel across a voltage source with V=12 V. The voltage across each inductor will be the same as that of the source, hence 12 V. The total inductance L Total can be evaluated using the rule of parallel inductors as: 1 L Total=1 3+1 6+1 9⇒L Total=1.5 m H In conclusion, understanding the structure, components, and arrangement of Inductors in Parallel is crucial for making sense of myriad real-world appliances and systems. They are one more example of physics' ubiquitous influence on our day-to-day living. Your browser does not support the video tag. Thanks to this ad, StudySmarter remains free.Sign up for a free account to access a better learning experience. Unpacking the Inductors in Parallel Formula When it comes to understanding inductors in parallel, the formula for the total inductance is crucial. The formula demonstrates an interesting and often surprising characteristic: unlike resistors where total resistance increases in a series connection and decreases in a parallel one, for inductors, it's quite the opposite. Team up with friends and make studying fun Sign up for free Mathematical Explanation of the Inductors in Parallel Formula To comprehend the Inductors in Parallel formula better, it's vital to grasp its structure. As with capacitors, the mathematics behind parallel inductors is based on reciprocation. The inverse of the total inductance is the sum of the inverses of each individual inductor's inductance. As previously mentioned, the formula is as follows: 1 L Total=1 L 1+1 L 2+1 L 3+⋯ Here each L 1, L 2, L 3 and so on represents the inductance of the individual inductors. The equation is formulated this way because the voltage in a parallel arrangement is constant, while the current through each branch can vary. As inductors influence the current, the parallel formula reflects these characteristics. Let's look at a scenario for clarity: Suppose we have three parallel connected inductors with inductance values of 2 H, 3 H, and 6 H. The total inductance would be: 1 L Total=1 2+1 3+1 6 L Total=1 H This result is counter-intuitive to most people given that three components, each with an inductance greater than 1, combine to create a total inductance of just 1. Applying the Inductors in Parallel Formula Applying the formula for inductors in parallel is a matter of following the mathematical steps, taking into account the inductance of each inductor involved. Remember, since the configuration is parallel, the voltage remains the same across all inductors. Factors that affect the calculation include the inductance of each inductor and their arrangement. Here is a succinct list of the steps to apply this formula: Determine the inductance for each individual inductor in the circuit. Find the reciprocal of each individual inductance value. Add together all these reciprocal values. Take the reciprocal of the result from the previous step to find the total inductance. Real-world Applications of the Inductors in Parallel Formula The formula for inductors in parallel is not just a mathematical curiosity; it has real-world applications in many areas of technology. Particularly in electronics, it plays a pivotal role. Here are a few examples: Impedance Matching:Parallel inductors are used in impedance matching, important for maximising signal transmission or power transfer in radio antennas. Power Supply Filters:In power supplies, inductors in parallel help filter out high-frequency noise, improving function. Resonant Circuits:Parallel configurations are part of resonant or tuned circuits, used in applications ranging from radio tuning to medical imaging. Using the right tool for the right job is a crucial part of any technician’s role. From straight physics and component-level circuit design to system-level power applications, knowing how to apply the formula for inductors in parallel helps leverage the interesting properties of these components. Remembering that the total inductance value in a parallel setup decreases opens doors to optimisation possibilities in design and application. Stay organized and focused with your smart to do list Sign up for free Comparing Inductors in Series and Inductors in Parallel Transitioning from discussing solely about inductors in parallel, let's now delve into a comparison with their counterparts, 'inductors in series'. These two scenarios may seem intertwined, however, both configurations impact electrical circuits differently, altering the overall inductance and effectively shaping the course of the current. Main Differences Between Inductors in Series and Inductors in Parallel Though both inductors in series and in parallel configurations are common in electronic circuits, their behaviour and responses to voltage and current differ substantially. Understanding these differences is pivotal to grasping their individual roles in circuit performance, design, and functionality. Series Arrangement of Inductors: When two or more inductors are connected in series, they share the same current, with voltage varying across each one. It may seem surprising, but the total inductance in such a situation increases. The formula for total inductance in a series arrangement is a straightforward addition of individual inductor values: L Total=L 1+L 2+L 3+⋯Parallel Arrangement of Inductors: As discussed earlier, in a parallel configuration, all inductors share the same voltage, with currents in different branches likely to vary. Contrary to the intuitive approach, the total inductance value decreases in a parallel setting, with the formula given by: 1 L Total=1 L 1+1 L 2+1 L 3+⋯ The following table summarises the key differences between inductors in series and in parallel: CharacteristicInductors in SeriesInductors in Parallel Current flow Same across all inductors Can vary across different branches Voltage drop Can vary across each inductor Same across all inductors Total Inductance Increases Decreases Studying Examples of Inductors in Series and in Parallel Let's look at an example each for inductors in series and in parallel to augment our theoretical insights with concrete setups: Example of Inductors in Series: Suppose three inductors of inductance 1 H, 2 H, and 3 H respectively, are connected in series with a voltage source of 12V. The total inductance is the sum of individual inductances: L Total=1 H+2 H+3 H=6 HExample of Inductors in Parallel: Now, consider the same three inductors, but this time connected in parallel with a 12V source. The total inductance is calculated using the parallel connection formula: 1 L Total=1 1 H+1 2 H+1 3 H This gives us L Total=0.545 H, which is less than the inductance of any single inductor. Access millions of flashcards designed to help you ace your studies Sign up for free Practical Implications of Using Inductors in Series or in Parallel Both series and parallel arrangements of inductors have numerous practical applications. Making the choice between the two largely depends on the requirements of the circuit or system in question. Implications of Inductors in Series: As connecting inductors in series increases the total inductance, this configuration is often used when a high level of inductance is required, but space or supply limitations necessitate the use of smaller inductors. Typical applications include noise filtering where a higher inductance is favourable for effective elimination of unwanted frequency signals. Implications of Inductors in Parallel: The primary practical implication of inductors in parallel is related to their property of retaining the same voltage while allowing different currents in distinct branches. They can also be used where a lower inductance is required but only larger-value inductors are available. They are frequently deployed in power supplies, AC generators and radio frequency tuning circuits. One way to visualise the influence of these configurations is by picturing water flowing through pipes. Inductors in series are akin to a single narrow pipe allowing water (current) to flow through, while parallel inductors resemble multiple pipes, allowing variable amounts of water to flow, but subject to the same water pressure (voltage). Conceptualise the requirements of your circuit - based on the nature of the inductors, the circuit requirements and the practical constraints, you can determine whether a series or parallel configuration makes the most sense for your particular use-case. Don't let perceived complexity hamper your apprehension; even the vast and varied landscape of physics can be navigated successfully with the right analytical tools at your disposal. Key Characteristics of Inductors in Parallel Just like resistors and capacitors, inductors too can be arranged in a parallel configuration. This arrangement gives rise to certain notable characteristics which define how these inductors interact and influence the overall performance and function of a circuit. Find relevant study materials and get ready for exam day Sign up for free In-depth Analysis of Inductors in Parallel Characteristics Understanding the characteristics of inductors in parallel is pivotal to analysing how they behave within an electrical circuit. 1: Shared Voltage: Perhaps the most prominent characteristic of parallel inductors is that all inductors in a parallel arrangement share the same voltage across them. This is a result of their common connection to the voltage input source. 2: Reciprocal Nature of Total Inductance: Unlike resistors, for inductors in a parallel arrangement, the total inductance is calculated differently. Instead of adding the inductances of each inductor, we have to take the inverse of the sum of the inverses of each inductor's inductance. The formula to calculate the total inductance (L Total) is: 1 L Total=1 L 1+1 L 2+1 L 3+⋯ Thus, the reciprocal nature of total inductance in parallel is a defining characteristic. 3: Different Currents: While all parallel inductors share the same voltage, the current passing through each branch can vary. The value of the current is dependent on the inductance and other properties of each individual inductor. Real-life Examples of Inductors in Parallel Characteristics Let's take the example of power distribution lines, a common real-world operation where inductors are involved in a parallel arrangement. In this application, transformers function as inductors playing an essential role in stepping up and stepping down voltages as per requirements. Simultaneously, the multiple distribution lines effectively create a parallel connection. With the transformers (inductors) all connected to the same voltage, the characteristic of shared voltage across parallel inductors is clearly demonstrated. The variation in distribution loads leads to variable current through the multiple transformers. This, in another example of different currents flowing through parallel inductors. Further, if additional power lines (inductors) are added, the total inductance value effectively decreases, despite the inclusion of more inductors. This clearly demonstrates the reciprocal nature of total inductance in a parallel arrangement. Effects of the Characteristics of Inductors in Parallel on Their Usage The unique characteristics of parallel inductors not only shape their behaviour but also influence how and where they are used. Voltage Equalisation: The shared voltage characteristic ensures each inductor in a parallel arrangement experiences the same voltage regardless of their individual inductance values or the current flowing in their branch. This factor comes handy in systems like power grids, where all connected components need to operate at a specified voltage level. It's also critical in impedance matching in signal transmission systems. Current Distribution: The ability to carry different currents through different branches despite a shared voltage is another useful feature of parallel inductors. This property is exploited in applications like distribution transformers, where different transformers carry different loads even though they are connected to the same supply voltage. Lowering Total Inductance: The reciprocal nature of total inductance in a parallel arrangement allows system designers to decrease the overall inductance of an assembly by adding inductors. This comes into play in radio frequency tuning circuits or power generator systems, where precise control of inductance can optimise performance. To capitalise on these properties fully, it's necessary to understand these characteristics thoroughly. The complex realms of electronics and electricity can ensure many surprises, but through comprehensive understanding, you too can harness the power of these exciting phenomena. How to Add Inductors in Parallel Combining inductors in a parallel configuration is a key aspect of designing and analysing electronic circuits. The overall inductance in this arrangement is smaller than the smallest inductor in the group. Let's go step-by-step on how to add inductors in parallel and illustrate this concept with suitable examples. Step-by-step Guide on How to Add Inductors in Parallel Adding inductors in parallel might seem like a formidable task at first, but with a systematic approach and a good understanding of the fundamental principles, you can easily achieve it. Here is a step-by-step guide on how to do so: Step 1: Understand the Circuit Configuration: Recognise that a parallel arrangement of inductors exists when all the inductors are linked at the same two points, causing them to share the same voltage. Step 2: Identify the Inductance Values: Note the individual inductance values of each inductor in the parallel configuration. These values are typically denoted in Henries (H). Step 3: Use the Formula for Parallel Inductors: Apply the formula for combining inductors in parallel: 1 L Total=1 L 1+1 L 2+1 L 3+⋯ Here, L Total is the total inductance, and L 1,L 2,L 3,… signify the inductance of individual inductors. Step 4: Solve for the Total Inductance: Perform the calculation to find the total inductance. Note that the result may be a decimal or fraction, which is normal given the reciprocal nature of the formula. Example Scenarios of Adding Inductors in Parallel VBearing all the steps in mind, let's illustrate them with an example. Consider we have three inductors with values of 1 H, 2 H, and 3 H in a parallel arrangement. We can calculate the total inductance in the following way: 1 L Total=1 1 H+1 2 H+1 3 H⇒L Total≈0.545 H This result shows that the total inductance of the system is lower than that of any individual inductor, which is a distinctive characteristic of parallel inductors. Potential Challenges in Adding Inductors in Parallel and Tips to Overcome These While following the process of combining inductors in parallel may seem straightforward, it can present a few challenges. Here are a few potential issues and some advice on how to navigate them: Accuracy in Calculation: When dealing with the formula for adding inductors in parallel, it's easy to make a mistake, particularly because of its reciprocal nature. Be careful with the decimal values to maintain accuracy. Differentiating Series and Parallel: Make sure you correctly identify whether the inductors are configured in series or parallel. Pay close attention to how the inductors are connected. Mutual Inductance: In specific configurations and close proximity, inductors can create a magnetic field that interacts with neighbouring inductors, leading to a phenomenon called mutual inductance. To overcome this issue, ensure the physical layout of circuit components minimises such interactions. Quality of Inductor: The quality of the inductor used can affect circuit performance. Ensure you source your components from reliable suppliers and calculate their specifications accurately. All of these challenges can be overcome with a thorough understanding of the underlying principles of electricity and magnetism and by paying careful attention to the details. Remember, proper planning and meticulous execution are key to successful circuit design and analysis. Inductors in Parallel - Key takeaways Understanding the structure, components, and arrangement of Inductors in Parallel is crucial for making sense of real-world appliances and systems. For inductors, the total inductance decreases when arranged in parallel as opposed to resistors where total resistance decreases in a parallel connection. The formula for the total inductance in a parallel inductors arrangement is: 1 L Total=1 L 1+1 L 2+1 L 3+⋯, where each L 1, L 2, L 3 and so on represents the inductance of the individual inductors. Real-world applications of inductors in parallel include impedance matching (maximising signal transmission or power transfer in radio antennas), filtering high-frequency noise in power supply filters and resonant or tuned circuits used in applications ranging from radio tuning to medical imaging. Comparing inductors in series and in parallel configurations: In a series arrangement, they share the same current with voltage varying across each one and total inductance increases with the addition of individual inductor values. In a parallel configuration, all inductors share the same voltage with currents likely varying across different branches and contrary to intuition, the total inductance value decreases. Key characteristics of inductors in parallel include shared voltage across them owing to their common connection to the voltage input source, reciprocal nature of total inductance and different currents passing through each branch depending on the inductance and other properties of each individual inductor. Adding inductors in parallel: Determine the inductance for each individual inductor in the circuit, find the reciprocal of each individual inductance value, add together all these reciprocal values, and then take the reciprocal of the result to find the total inductance. 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The shared voltage property ensures each inductor in a parallel arrangement experiences the same voltage regardless of their individual inductance values or current flowing in their branch. This is useful in systems like power grids, where all components need to operate at a specified voltage level. What are some practical implications of using inductors in series or in parallel? Inductors in series are often used when a high level of inductance is required, and also in noise filtering. Inductors in parallel are used where a lower inductance is needed and are frequently found in power supplies, AC generators and radio frequency tuning circuits. What are some potential challenges when adding inductors in parallel and how can you overcome them? Challenges include maintaining accuracy in calculation due to decimal values, differentiating between series and parallel, mutual inductance, and ensuring good quality of inductor. These can be overcome by careful attention to detail, accurate calculations, correct identification of configuration, and sourcing good-quality components. What happens to the total inductance when inductors are connected in a series configuration? The total inductance increases. The formula for total inductance in a series arrangement is a straightforward addition of individual inductor values: L_Total = L_1 + L_2 + L_3 + … How is a parallel arrangement of inductors characterised? A parallel arrangement of inductors exists when all the inductors are linked at the same two points, causing them to share the same voltage. Can you explain about the voltage and current in a parallel setup of inductors? In a parallel connection of inductors, the voltage across each inductor remains the same whereas the sum of currents flowing through each inductor adds up to give the total current in the circuit. Learn faster with the 15 flashcards about Inductors in Parallel Sign up for free to gain access to all our flashcards. Sign up with Email Already have an account? Log in Frequently Asked Questions about Inductors in Parallel What is the formula for calculating the total inductance of inductors in parallel? The total inductance of inductors in parallel (L) is calculated using the formula: 1/L = 1/L1 + 1/L2 + 1/L3 + ... , where L1, L2, L3, etc. are the inductances of individual inductors. How does the principle of superposition apply to inductors in parallel? The principle of superposition applies to inductors in parallel by simply adding together their individual inductances to get the total inductance. This is similar to the method used for resistors in series. What factors can affect the overall inductance when inductors are arranged in parallel? The overall inductance of parallel inductors can be affected by the individual inductance values of each inductor, the number of inductors in the configuration, and any mutual inductance (interaction) between them. What happens to the magnetic fields of inductors when they are connected in parallel? When inductors are connected in parallel, their magnetic fields independently store energy. There is no direct interaction or merging between their magnetic fields unless the inductors are sufficiently close to each other to cause magnetic field coupling. Can mutual inductance occur between inductors arranged in parallel and how does it affect the total inductance? Yes, mutual inductance can occur between inductors arranged in parallel if their magnetic fields interact. It may increase or decrease the total inductance, depending on whether the magnetic coupling is aiding (fields add) or opposing (fields cancel). Save Article How we ensure our content is accurate and trustworthy? 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188029	https://cacr.uwaterloo.ca/techreports/2006/cacr2006-29.pdf	1 New Binary Sequences with Optimal Autocorrelation Magnitude Nam Yul Yu, Student Member, IEEE and Guang Gong, Member, IEEE Department of Electrical and Computer Engineering University of Waterloo, Canada Abstract New binary sequences of period for even are found. These sequences can be described by a interleaved structure. The new sequences are almost balanced and have four-valued autocorrelation, i.e., , which is optimal with respect to autocorrelation magnitude. Complete autocorrelation distribution and exact linear complexity of the sequences are mathematically derived. From the simple implementation with a small number of shift registers and a connector, the sequences have a beneﬁt of obtaining large linear complexity. Index Terms Binary sequences, interleaved sequences, linear complexity, optimal autocorrelation. I. INTRODUCTION Binary pseudorandom sequences with optimal autocorrelation play important roles in many areas of communication and cryptography. In code-division multiple access (CDMA) commu-nication systems, the sequences are needed to acquire accurate timing information of received signals. In cryptography, on the other hand, the sequences are used to generate key streams in stream cipher encryptions. For binary sequences of period , binary -sequences are traditionally well-known sequences with ideal two-level autocorrelation of . Due to their good randomness proper-ties and simple implementation , -sequences have been widely used for communication systems. Besides -sequences, several other inequivalent classes of binary sequences with ideal two-level autocorrelation have been constructed and discovered . 2 For binary sequences of even period , on the other hand, Lempel, Cohn, and Eastman showed that autocorrelation must have at least two distinct out-of-phase values and a difference between any two autocorrelation values is divisible by . Therefore, optimal auto-correlation is if , and or if . Several classes of binary sequences of even period with optimal autocorrelation are known. Initially, Lempel, Cohn, and Eastman presented a class of the balanced binary sequences of period for odd prime . (It is known that this has been already described in and called as the Sidelnikov sequences.) Then, No, Chung, Song, Yang, Lee, and Helleseth gave another class of the binary sequences of period for odd prime using a polynomial over a ﬁnite ﬁeld. From a group division structure by the Chinese Remainder theorem, Ding, Helleseth, and Martinsen also presented several families of the binary sequences of period for odd prime which correspond to almost difference sets. Using known cyclic difference sets, Arasu, Ding, Helleseth, Kumar, and Martinsen constructed four classes of almost difference sets which give inequivalent classes of the binary sequences of period . These sequences generally contain the binary sequences of period constructed from the product method in . Recently, Zhang, Lei, and Zhang presented an almost difference set corresponding to the binary sequence of period by adding two indices to one class of the almost difference set in where the corresponding cyclic difference set is from the Legendre sequences. For a period , the autocorrelation or is optimal from the Lempel, Cohn, and Eastman’s assertion in the sense that it has the two out-of-phase values with the smallest magnitudes. If we allow three out-of-phase values with the smallest magnitudes, on the other hand, then optimal autocorrelation should be , where the autocorrelation is optimal with respect to its magnitude. In practical applications, we believe that it has the same meaning as conventional optimal autocorrelation. Consequently, the autocorrelation of is also considered as optimal in this correspondence. In , Gong introduced the interleaved structure of sequences which is indeed a good method not only for understanding a sequence structure, but also for constructing new sequences of an interleaved form . In this correspondence, we show that a binary sequence of period shown in can be represented by a interleaved structure. We also show that a binary product sequence of period with optimal autocorrelation can 3 be represented as a interleaved structure. Inspired by these interpretations, we discover a new construction of binary sequences of period with autocorrelation by the interleaved method. In details, we use a interleaved structure deﬁned by a perfect binary sequence of period and a binary -sequence of period . In the interleaved structure, a sequence deﬁned over is used as a shift sequence. The new sequences are almost balanced, i.e., a difference between the numbers of zeros and ones in a period is , and optimal with respect to autocorrelation magnitude. Complete autocorrelation distribution and exact linear complexity of the sequences are mathematically derived. From the simple implementation with a small number of shift registers and a connector, the sequences have a beneﬁt of obtaining large linear complexity. This correspondence is organized as follows. In Section II, we give preliminary concepts and deﬁnitions on binary sequences for understanding this correspondence. Interleaved structures of known binary sequences of period with optimal autocorrelation are presented in Section III. In Section IV, new binary sequences of period for even with autocorrelation are constructed using a interleaved structure, and the autocorrelation distribution is mathematically derived. In Section V, linear complexity of the sequences is investigated and the implementation is discussed. Concluding remarks are given in Section VI. II. PRELIMINARIES Following notation will be used throughout this correspondence. is a ring of integers modulo , and . is a ﬁnite ﬁeld with elements and is a multiplicative group of . For a binary sequence , . is a complement of , or where the addition is computed modulo . For a sequence over and an integer , where the addition is computed modulo . For positive integers and , let . A trace function from to is denoted by , i.e., or simply as if and the context is clear. 4 A. Equivalence of Sequences Let and be two periodic sequences. Then, they are called cyclically equivalent if there exists an integer such that for all Otherwise, they are called cyclically distinct. B. Balance and Almost Balance Properties Let be a binary sequence of period . Then is called balanced if the number of zeros is nearly equal to the number of ones in a period, i.e., where denotes a difference between the numbers of zeros and ones of a binary sequence in a period. For odd , is balanced if and only if , and for even , it is balanced if and only if . On the other hand, if is even and , then is called almost balanced . C. Autocorrelation (Periodic) autocorrelation of a binary sequence of period is deﬁned by where is a phase shift of and the indices are computed modulo . For a sequence of period , it is implied that occurs only at . is called optimal autocorrelation if it satisﬁes 1) if , or 2) if , or 3) if , or 4) or if for all . In particular, case 1) is called ideal two-level autocorrelation and a binary sequence of case 1) corresponds to a cyclic difference set . Complete classes of all the known inequivalent binary sequences of period with ideal two-level autocorrelation and the corresponding cyclic difference sets are summarized in . 5 Binary sequences of cases 2) - 4) are described by almost difference sets and the other methods. Several classes of binary sequences of cases 2) and 3) are described by the correspon-ding almost difference sets in and , respectively. On the other hand, four classes of binary sequences of case 4) and the corresponding almost difference sets are presented in where the sequences generally contain the binary sequences constructed from the product method in . Another almost difference set corresponding to a binary sequence of case 4) is presented in by adding two indices to one class of the almost difference sets in . From a ﬁnite ﬁeld approach, furthermore, binary sequences of period for odd prime corresponding to cases 3) and 4) are also described in and , respectively. For a survey of binary and quadriphase sequences with optimal autocorrelation, see . In this correspondence, if for , we consider that it is also optimal in the sense that its autocorrelation magnitude is identical to that of case 4). D. Perfect Sequences Let be a binary sequence of period . If autocorrelation is equal to for all , then is called a perfect sequence. A perfect sequence is also deﬁned for a nonbinary sequence by extending the deﬁnition of its autocorrelation . For nonbinary cases, a few polyphase perfect sequences are known in and . However, the only known perfect binary sequence is or its complement . For a period of , no perfect binary sequences are discovered , and it is conjectured in that no other perfect binary sequences exist except for . E. Product Sequences Let and be binary sequences of periods and , respectively, where . Then a product sequence of period is deﬁned by a component-wise addition of where the addition is computed modulo . Autocorrelation of the product sequence is given by (1) where the indices of a sequence are computed modulo its own period . 6 F. Almost Difference Set (ADS) Sequences of Period In , Arasu, Ding, Helleseth, Kumar, and Martinsen presented binary sequences of period with optimal autocorrelation. Let be a binary sequence of period with ideal two-level autocorrelation and a matrix be deﬁned by (2) where is any integer in and the indices are computed modulo . A binary sequence of period is deﬁned by where and (3) Then has optimal autocorrelation of for every by providing the corresponding almost difference set . Throughout this correspondence, is called an ADS sequence. G. Interleaved Sequences Let be a binary sequence of period where both and are not equal to . Then, we can arrange by an matrix, i.e., . . . . . . . . . If each column is either a cyclic shift of a binary sequence of period or a zero sequence, then is called a binary interleaved sequence . According to the deﬁnition, where denotes a cyclic left shift operation, and or if . Here, a transpose notation is omitted because we consider it as a sequence. In the interleaved sequence, and are called a base sequence and a shift sequence of , respectively. In this correspondence, the matrix is used for an array form of , denoted by ! . 7 In ! , the interleaved structure is preserved by adding a binary sequence of period . Let " . In ", is used as an indicator sequence where if , or otherwise. H. Interleaved Structure of Binary -sequences Let and be a binary -sequence of period represented by # where # is a primitive element of . (As a binary -sequence, we consider a sequence satisfying the constant-on-cosets property .) Then, is represented by a interleaved sequence , i.e., ! . In its array form , a base sequence is a binary -sequence of period represented by $ % where $ # is a primitive element of . Also, a shift sequence is given by # In other words, is a zero sequence of length , and is a cyclic shift of a binary -sequence of period . III. INTERLEAVED STRUCTURES OF KNOWN BINARY SEQUENCES WITH OPTIMAL AUTOCORRELATION In this section, we examine interleaved structures of binary sequences of period with optimal autocorrelation. First, a interleaved structure of the ADS sequence is given using a binary sequence of period with ideal two-level autocorrelation and the constant-on-cosets property as a base sequence. Then, a product sequence with optimal autocorrelation is also examined by a interleaved structure where each column is a cyclic shift of a perfect binary sequence of period . A. ADS Sequences In the ADS sequence deﬁned by (3), let and be a binary sequence of period with ideal two-level autocorrelation and the constant-on-cosets property. If we represent 8 by a interleaved structure, then . . . . . . . . . . . . where % and is an integer, . From the constant-on-cosets property of and , we have , , , where the indices are computed modulo . From this, we have the following interleaved structure of . Property 1: Let be the ADS sequence of period . Then, has a interleaved structure, i.e., ! where 1) is a binary sequence of period with ideal two-level autocorrelation and the constant-on-cosets property, 2) is a shift sequence, 3) is a perfect binary sequence where is an integer, . From the interleaved structure, is cyclically distinct for each . If , in particular, it is pointed out in that is equivalent to a product sequence of and , i.e., . Thus, a product sequence of period with optimal autocorrelation is a special case of the ADS sequence. Example 1: For , let be a binary -sequence of period , i.e., . 9 If , then . Then, the corresponding ADS sequence is given by B. Product Sequences In Section III-A, a product sequence of period with optimal autocorrelation is represented by a interleaved structure as a special case of the ADS sequence ( ). Here, we show that it is also represented by a interleaved structure with different base and shift sequences. Let be a binary product sequence of period where is a perfect binary sequence of period and a binary sequence of period with ideal two-level autocorrelation. Then has optimal autocorrelation from (1). Note that and for any integer & . We ﬁrst consider a interleaved structure ' associated with of length . Then, is given by % where the index is computed modulo . From and , we have (4) Thus, the product sequence has the following interleaved structure. Property 2: Let be a binary product sequence of period with optimal autocorrelation. Then, it has a interleaved structure, i.e., ! where 1) is a perfect binary sequence, 2) is a shift sequence deﬁned over where the -th element is given by (5) 10 3) is a binary sequence of period with ideal two-level autocorrelation. In the interleaved structure of , its -th column is given by if , or otherwise. Example 2: For , a product sequence of period with optimal autocorrelation is given by where , a perfect binary sequence of period , and , a binary -sequence of period . From (5), is deﬁned by Then, is given by ! Now, we focus on a shift sequence over of the product sequence of period for an integer & with optimal autocorrelation. For , the shift sequence shown in (5) can also be represented by a interleaved structure, i.e., . . . . . . . . . . . . . . . . . . Interestingly, the interleaved structure of is given by where % % (6) 11 Note that and both expressions are used throughout this correspondence. In next section, we will give a new construction of binary sequences of period for even with optimal four-valued autocorrelation by modifying the shift sequence of the interleaved structure of a product sequence. IV. NEW BINARY SEQUENCES WITH OPTIMAL FOUR-VALUED AUTOCORRELATION In this section, we present a new construction of binary sequences of period for even with optimal four-valued autocorrelation, i.e., for any . A. Construction Construction 1: Let & be a positive integer. A new binary sequence of period is deﬁned by a interleaved structure of ! where is a perfect binary sequence of period , is a binary -sequence of period with the constant-on-cosets property, is a sequence deﬁned over of period , and represented by a interleaved structure, i.e., where % Æ if % if (7) where % and Æ or . In fact, the new sequence is obtained by modifying the shift sequence of (6) in the interleaved structure of a product sequence. Remark 1: With Æ and cyclically distinct binary -sequences of , Construction 1 gives cyclically distinct binary sequences , where ( is the Euler-totient function. Theorem 1: Let ! be the binary sequence from Construction 1. Then is almost balanced. Proof: In the interleaved structure of ! , note that the )-th column is a cyclic shift of either or , which is determined by . In 12 other words, each column of has ones if , or ones if . Since is a binary -sequence of period with the balance property , the number of ones in is given by Hence, a difference between the numbers of zeros and ones in a period is , i.e., is almost balanced. B. Autocorrelation To compute the autocorrelation function of , we ﬁrst consider Proposition 1. Proposition 1: Let ! be the binary sequence from Construction 1. Let . Then the autocorrelation function is given by (8) where , , and is autocorrelation of a base sequence . In , the index is computed modulo . As in equation (12) of , on the other hand, if ) (9) Proof: From ! , it is immediate from Lemma 2 of . In Construction 1, note that the indicator sequence and the shift sequence have the interleaved structures. Then, ) in Proposition 1 is represented by a difference array + , , where arrays , and , represent and , respectively. From the shift-and-add property of the binary -sequence , it is clear that + also represents a binary -sequence of period . Also, the -th column is either a cyclic shift of a binary -sequence of period or a zero sequence from the interleaved structure of binary -sequences. In the following lemma, we further study a structure of the array +. From now on, we use following notation in all lemmas and theorems in Section IV. where (10) 13 Lemma 1: Let , and , be arrays of and , respectively, where . In the difference array + , , , the -th column has the following properties. 1) If , then a zero column exists only at . 2) If , then a zero column exists at exactly one for with and where ‘’ is computed modulo . Proof: In , , is a zero column and is a cyclic shift of a binary -sequence from Section II-H. In another array , of , on the other hand, is a zero column and is a cyclic shift of a binary -sequence. In the difference + , , , therefore, is still a zero column if . If , on the other hand, neither nor can be a zero column because both and are nonzero columns. Instead, a zero column exists at another column index with and because there should be exactly one zero column in the difference array + which represents a binary -sequence. This completes a proof of Lemma 1. In Proposition 1, ) can also be represented by a interleaved structure . To obtain , we need the following two lemmas. Lemma 2: In the array structure of in (7), -shift of is given by (11) where is deﬁned in Construction 1 and extended to for . Proof: If we arrange in an array form, then we have . . . . . . . . . . . . . . . . . . . . . . . . (12) 14 where the ‘’ addition is from (9) and computed modulo . From the deﬁnition of in (7), % (13) Also, (14) where Æ if , or otherwise. From (13) and (14), we see that in (12), every element in a difference vector between the % -th row and the %-th row for % is . Thus, we can write (12) as For , if is deﬁned by , then is given by (11) from . Lemma 3: With the notation of Proposition 1 and Lemma 2, let ) % where ) , % , and . Then, is given by if and Æ if and Æ if and if and (15) In a array of , each column has constant elements given as follows. If , then % for all (16) If , then Æ % Æ % % for and (17) where ‘’ is computed modulo . 15 Proof: A array structure of is given by where is added to all elements of the array. From Lemma 2, we see that the -th column vector of is given by % From ) % , it is clear that is the %-th element of , i.e., (18) Together with (18) and (7), (15) follows immediately. The assertions of (16) and (17) are from (15). Now, we are ready to compute . Theorem 2: Let ! be the binary sequence of period for an integer & from Construction 1. Then it has four-valued optimal autocorrelation, i.e., for any . Precisely, its complete autocorrelation is given by if if and or - . or / if - or / or / if / where , , and for some - , / , and . . Proof: To compute , we use (8) in Proposition 1. Since for nonzero , nonzero is determined by cases of in (8). We have the following three cases. Recall (10) and the arrays + and of and in Lemmas 1 and 3, respectively. Case 1. : If in this case, then and , a trivial in-phase autocorrelation. If , on the other hand, and then in (8) because for all nonzero . Thus, we have if if and (19) 16 Case 2. and : From (16), is a constant array where each element is and + is balanced from the difference property of -sequences, i.e., . If , then for all ) . Hence, from (8). If , on the other hand, and for all ) and hence . From (10), note that if , then . Therefore, if - if - . (20) where - and . are some elements in and , respectively. Case 3. and : Let ) % . From Lemma 3, we have three distinct ’s in , i.e., 0 0 where 0 . (Note that Æ in (7).) 0 corresponds to ’s of with and . For such ’s, is a constant column of 0 and there exists one such that is a zero column from Lemma 1. On the other hand, 0 corresponds to and , respectively, and both and are nonzero -sequences in +. 0 : In this case, all ’s of with and are zero columns. On the other hand, and are nonzero. Let and be the numbers of zeros and ones in ’s for with and . Then, From (8), . 0 : In this case, either or is a zero column for given 0 and Æ. On the other hand, all other columns are nonzero. If and are the numbers of zeros and ones in or , then Thus, . 0 : In this case, no columns are zero in . Thus, from . From (10), note that 0. Combining , and , we have if / if / or / if / (21) 17 where / is some element in . In (21), note that implies . If we combine (19), (20), and (21), then the proof is completed. Remark 2: In Remark 1, we have many cyclically distinct sequences of according to Æ and . In Theorem 2, however, the distinction disappears regarding their autocorrelations and consequently all the sequences from Construction 1 have identical autocorrelation distribution regardless of Æ and . Theorem 3: With the notation in Theorem 2, complete distribution of is given by time times times times Proof: From Theorem 2, a trivial in-phase autocorrelation occurs only once. Hence, we count the other exclusive cases of Theorem 2. Case 1. : - : In this case, possible ’s are . Thus its number of occurrences is . / or / : Note that . By the Chinese Remainder theorem, we have a unique solution of for / with given / . Thus, the number of distinct solutions of in for / is for all / . From the isomorphism , the number of ’s in for / is . Considering the exclusive cases of / and / , the number of such ’s is . Combining and , the number of occurrences of is 1 . Case 2. : This corresponds to / . By a similar approach to Case 1-, the number of such ’s is equal to 1 2 . Case 3. : The number of such ’s is 1 1 1 . From Cases 1 - 3, the proof is completed. Example 3: For , consider a new sequence in Construction 1 with Æ . In its interleaved structure ! , a base sequence is and an indicator 18 TABLE I AUTOCORRELATION VALUES OF IN EXAMPLE 3 sequence is , a binary -sequence of period . A shift sequence is deﬁned by (7), and and are represented by arrays, respectively, i.e., Then, a new sequence of period is given by Consider . From , we have from (10). + and are given by + Hence, we see that Lemmas 1 and 3 are true in this case. From + and at , we can easily compute . The autocorrelation function from Theorem 2 is shown in Table I. From Theorem 3, the complete autocorrelation distribution is given by time times times times Both the autocorrelation function and the distribution are veriﬁed from computer experiments. 19 V. OTHER ASPECTS OF NEW BINARY SEQUENCES In this section, we derive exact linear complexity of the new binary sequences and show that large linear complexity can be obtained from the sequences. Implementation of the sequences requires only a small number of shift registers and a simple logic. Linear complexity of a sequence is deﬁned as the shortest length of a shift register which generates the sequence, or equivalently a degree of the minimal polynomial of the sequence . Before examining linear complexity of the sequence from Construction 1, we consider the following lemmas. Lemma 4: Let or . Let 3 be a binary sequence of period such that 3 if if (22) where is a positive integer and is an integer, . Then, the minimal polynomial of is . Proof: Let 3 3 3 3 . From (22), is given by if if From , the minimal polynomial of is given by for both cases of . Lemma 5: Let ! be the binary sequence from Construction 1. Then, is represented by , i.e., 3 (23) where 3 is the binary sequence from Lemma 4. Proof: Assume that where 4 . Since at , 4 for (24) 20 For with , let % % where % and % . Then, % % % (25) where % and % correspond to row and column indices of the interleaved structure of , respectively. Recall the interleaved structure of a sequence in (4). Then, is represented as 4 4 (26) at . From Construction 1, on the other hand, Æ (27) at . From (25), note that % . Also, and . Then, from (26) and (27), 4 4 Æ Æ Æ 0 (28) where Æ and the indices are computed modulo . Let 0 . Since has a period , and have periods and from (28), respectively. Also, it is easily known from (28) that if Æ or if Æ which is identical to from Lemma 4. Thus, 4 for (29) From (24) and (29), in (22), and hence (23) is true. Linear complexity of the binary sequences from Construction 1 is presented by Theorem 4. Theorem 4: Let be the binary sequence of period for an integer & from Construction 1. Then, linear complexity of is given by Proof: By deﬁnition of the minimal polynomials, , the minimal polynomial of is determined by the least common multiple (lcm) of the minimal polynomials of , , and in Lemma 5. Let , and be the minimal polynomials of , and , respectively. 21 It is easily known that and is a primitive polynomial of degree . From Lemma 4, ) 3 ) 3 ) 3 Note that and if & , . Finally, ) 3) 3 ) 3 where a degree of or linear complexity of is given by which completes the proof. Remark 3: In Theorem 4, we requested that & . For the case of , the new binary sequence of period also has optimal autocorrelation for any . However, its linear complexity is . It is because in the proof of Theorem 4, is a factor of , and thus the minimal polynomial of is given by ) 3 . Table II shows linear complexities of the three different binary sequences of period with optimal autocorrelation. For the product and the ADS sequences in Section III adopting -sequences of period as an indicator or a base sequence in their interleaved structures, their linear complexities are given by and , respectively, which will be discussed in Appendix I. In Table II, the new binary sequences from Construction 1 provides much larger linear complexity than the other two classes of sequences. Linear complexities of Table II are conﬁrmed by computer experiments using the Berlekamp-Massey algorithm . Figure 1 shows implementation of the new binary sequence of period with Æ . (If Æ , we only need to change the initial state of the 3-stage LFSR from to .) In Fig. 1, the 3-stage linear feedback shift register (LFSR) is enabled and connected to other LFSRs only at time . In the implementation, is generated by combining only shift registers and a simple connector. From Theorem 4, 22 TABLE II LINEAR COMPLEXITY OF BINARY SEQUENCES OF PERIOD WITH OPTIMAL AUTOCORRELATION ADOPTING A BINARY -SEQUENCE Period Product ADS New Sequence Sequence Sequence however, its actual linear complexity is for 2, where we obtain the large linear complexity with the low implementation cost. Remark 4: Let ) be linear complexity of a binary sequence of period with ideal two-level autocorrelation which is used as an indicator or a base sequence of the product or the ADS sequences in Section III. Then, linear complexities of the product and the ADS sequences are ) and at most ) , respectively, which will be proved in Appendix I. If ) , then the linear complexities can be larger than that of the new sequences from Construction 1. In this case, however, we need as many numbers of shift registers as the linear complexities for their implementation, which requires the larger implementation cost. VI. CONCLUSION From a interleaved structure, we have constructed new binary sequences of period for even with four-valued autocorrelation which is optimal with respect to autocorrelation magnitude. Complete autocorrelation distribution and exact linear complexity of the sequences have been mathematically derived. Only with shift registers and a simple connector, the sequences are implemented to give large linear complexity. 23 Connected at -stage LFSR Fig. 1. Implementation of a new binary sequence of period with optimal autocorrelation Æ APPENDIX I In Appendix I, we show two lemmas on linear complexities of the product and the ADS sequences of period in Section III. Lemma 6: Let be the product sequence of period with optimal autocorrelation shown in Section III, where is a perfect binary sequence of period and a binary sequence of period with ideal two-level autocorrelation. Then, its linear complexity is given by ) where ) is linear complexity of . Proof: The minimal polynomial of a perfect binary sequence is . Let be the minimal polynomial of . In general, where is a primitive polynomial over . We assume without loss of generality. Then the minimal polynomial of is given by ) 3 . Therefore, a degree of or linear complexity of is given by ) where ) is a degree of or linear complexity of . Lemma 7: Let be the ADS sequence of period deﬁned by (3) with a binary two-level autocorrelation sequence of period and a matrix in (2). If linear complexity of is ), then linear complexity of is at most ) , i.e., ) . In particular, the equality is achieved if is a binary -sequence and . 24 Proof: From (2) and (3), is represented by where (30) where is a perfect binary sequence of period and is a binary sequence deﬁned by if if (31) If , we have and . Thus, the linear complexity of is given by ) from Lemma 6. For a nontrivial ADS sequence with , it is clear that is a binary sequence of period . From (31), is represented by a interleaved structure where its ﬁrst column is and the second column is . Since each column sequence is a form of a shift-and-decimation of , its minimal polynomial is identical to , the minimal polynomial of . Let be the minimal polynomial of . From the interleaved structure of , we have (32) from Lemma 1 of . Similar to the proof of Lemma 6, we assume and thus . From (30), the minimal polynomial of is given by ) 3 ) 3 (33) where is the minimal polynomial of a perfect binary sequence . From (32) and (33), ) ) where ) is a degree of . In particular, if is a binary -sequence, then is a primitive polynomial of degree ) and thus we have or from (32). If , and thus . Hence, and consequently, ) from (33). REFERENCES K. T. Arasu, C. Ding, T. Helleseth, P. V. Kumar, and H. Martinsen, “Almost difference sets and their sequences with optimal autocorrelation,” IEEE Trans. Inform. Theory, vol. 47, no. 7, pp. 2934-2943, Nov. 2001. L. D. Baumert, Cyclic Difference Sets, Berlin, Springer-Verlag, 1971. E. R. Berlekamp. Algebraic Coding Theory. Aegean Park Press, CA, Revised ed. 1984. 25 J. F. Dillon and H. Dobbertin, “New cyclic difference sets with Singer parameters,” Finite Fields and Their Applications 10, pp. 342-389, 2004. C. Ding, T. Helleseth, and K. Y. Lam, “Several classes of binary sequences with three-level autocorrelation,” IEEE Trans. Inform. Theory, vol. 45, no. 7, pp. 2606-2612, Nov. 1999. C. Ding, T. Helleseth, and H. 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Gordon, W. H. Mills, and L. R. Welch, “Some new difference sets,” Canadian J. Math., vol. 14, no. 4, pp. 614-625, 1962. T. Helleseth and P. V. Kumar, Sequences with Low Correlation, a chapter in Handbook of Coding Theory, edited by V. Pless and C. Huffmann, Elsevier Science Publishers, 1998. D. Jungnickel and A. Pott, “Difference sets: An introduction,” in Difference Sets, Sequences and their Correlation Properties, A. Pott, P. V. Kumar, T. Helleseth, and D. Jungnickel, Eds., 1999, vol. 542, NATO Science Series C, pp. 259-296. A. Lempel, M. Cohn, and W. Eastman, “A class of balanced binary sequences with optimal autocorrelation property,” IEEE Trans. Inform. Theory, vol. IT-23, no. 1, pp. 38-42, Jan. 1977. R. Lidl and H. Niederreiter, “Finite Fields,” in Encyclopedia of Mathematics and Its Applications, vol. 20, Reading, MA: Addison-Wesley, 1983. H. D. L¨ uke, “Sequences and arrays with perfect periodic correlation,” IEEE Trans. Aerosp. Electron. Syst., vol. 24, no. 3, pp. 287-294, May 1988. H. D. L¨ uke, H. D. Schotten, and H. Hadinejad-Mahram, “Binary and quadriphase sequences with optimal autocorrelation properties: A survey,” IEEE Trans. Inform. Theory, vol. 49, no. 2, pp. 3271-3282, Dec. 2003. A. Maschietti, “Difference sets and hyperovals,” Designs, Codes and Cryptography, vol. 14, pp. 89-98, 1998. J. L. Massey, “Shift-register synthesis and BCH decoding,” IEEE Trans. Inform. Theory, vol. IT-15, no. 1, pp. 122-127, Jan. 1969. J. S. No, H. Chung, and M. S. Yun, “Binary pseudorandom sequences of period with ideal autocorrelation generated by the polynomial ,” IEEE Trans. Inform. Theory, vol. 44, no. 3, pp. 1278-1282, May 1998. J. -S. No, H. Chung, H. -Y. Song, K. Yang, J. -D. Lee, and T. Helleseth, “New construction for binary sequences of period with optimal autocorrelation using ,” IEEE Trans. Inform. Theory, vol. 47, no. 4, pp. 1638-1644, May 2001. 26 J. S. No, S. W. Golomb, G. Gong, H. K. Lee, and P. 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188030	https://www.sciencedirect.com/science/article/pii/S2211568415002508	Nipple discharge: The role of imaging - ScienceDirect Skip to main contentSkip to article Journals & Books ViewPDF Download full issue Search ScienceDirect Outline Abstract Keywords The different types of discharge The clinical examination Smear of the discharge Imaging Advantages Limitations The approach to nipple discharges Conclusion Clinical case Disclosure of interest References Show full outline Cited by (32) Figures (18) Show 12 more figures Tables (1) Table 1 Diagnostic and Interventional Imaging Volume 96, Issue 10, October 2015, Pages 1017-1032 Continuing education program: focus… Nipple discharge: The role of imaging Author links open overlay panel N.Lippa, G.Hurtevent-Labrot, S.Ferron, M.Boisserie-Lacroix Show more Outline Add to Mendeley Share Cite rights and content Under an Elsevier user license Open archive Referred to by Les écoulements mamelonnaires Journal de Radiologie Diagnostique et Interventionnelle, Volume 96, Issue 5, October 2015, Pages 434-450 N. Lippa, G. Hurtevent-Labrot, S. Ferron, M. Boisserie-Lacroix Abstract Nipple discharge is a common symptom in breast medicine. It is usually benign in origin (papillomas and galactophore duct ectasia) although it is essential not to miss the risk of an underlying malignant lesion (5%) mostly due to in situ carcinomas. Clinical examination is essential in the management, distinguishing benign “physiological” discharge from discharge suspected of being “pathological” in which further investigations with mammography and ultrasound are required. When the conventional imaging assessment for pathological nipple discharge is normal, breast MRI is gradually replacing galactography although this is still an emerging and invalidated indication. In this context and if the whole imaging assessment is normal, surgery is no longer the only solution for patients, who can now be offered regular monitoring. Previous article in issue Next article in issue Keywords Nipple discharge Breast symptom Mammography Ultrasound MRI Nipple discharge is a common symptom with poorly defined management. It is the third most common reason for consulting in breast medicine after mastodynia and palpable masses and has a prevalence of 5 to 10% , . Approximately 80% of women will develop at least one episode of nipple discharge during their fertile lives most of which are benign in origin and involve papillo-adenoma (35–56%) and/or galactophore duct ectasia (6–59%) . It is essential, however, not to miss the risk of an underlying malignant lesion, which is not that uncommon but varies greatly between series from 5 to 23%. The lesions concerned are mostly in situ ductal carcinoma (ISDC) . The different types of discharge In everyday clinical practice, several types of discharge are seen which have very different clinical features from each other. The following types can be distinguished: • galactorrhea: milky discharge (Fig. 1a) sometimes greenish in color and more or less abundant, persistent, bilateral and/or from multiple pores and usually seen post-partum. In this situation they may last from 1 to 2 years ; 1. Download: Download full-size image Figure 1. Different types of nipple discharges. a: milky discharge; b: yellowish±greenish multiple pore discharge in a context of fibrocystic breast disease; c: straw-colored single pore discharge; d: bloody single pore discharge. •discharge in pregnant women: uni- or bilateral and usually occurring during the second trimester of pregnancy. This is occasionally bloody even if no underlying breast disease is present and may continue for up to two years after pregnancy or breast-feeding ; •purulent discharge: this occurs on a background of infection or inflammation (complicated ductal ectasia) and is often accompanied by a more or less putrid smell. Associated symptoms are redness, pain, warmth, edema,±pyrexia; •multiple pore discharge: this is yellowish, greenish or multicolored and often occurs bilaterally due to breast dysfunction (galactophore duct ectasia, fibrocystic breast disease) or periductal mastitis (Fig. 1b); •single or pauci-pore discharge: this is often spontaneous and recurrent, varying in color but usually serous or bloody and secondary to underlying duct disease (Fig. 1c, d). The clinical examination This is essential, as it determines the need for further investigations and distinguishes between benign “physiological” discharge from suspicious “pathological” discharge. It begins with the usual general breast examination and is then focused on the nipples ideally using a magnifying lamp. It is based on three distinct phases all of which are essential: the clinical enquiry, inspection and palpation. The clinical enquiry This begins with the usual aspects of the clinical enquiry in breast medicine (age, hormone status, past medical history, past breast or ovarian, personal and family history, etc.). It should then seek to retrace the history and optimally describe the features of the discharge, recording its date of onset, duration, frequency, whether or not it is spontaneous (staining the bra) or provoked (by being expressed from the nipple or pressure on a “trigger” point) and its abundance. Finally, issues which may be related to the symptoms should be investigated: date of last pregnancy, recent breast-feeding, receipt of medications (anticoagulants or psychotropic agents such as the neuroleptics), history of recent trauma, smoking habit, etc. Secondly, descriptive aspects of the discharge are supplemented by inspection and palpation which also establish whether or it is isolated or associated with another clinical abnormality. Signs are looked for specifically pointing towards a malignant or benign cause which influences the diagnostic management: a palpable mass or recent nipple retraction, change in the areola or signs of inflammation, etc. Inspection Inspection should ideally be carried out using a magnifying lamp (Fig. 2a), and is initially designed to distinguish “true” nipple discharge from its differential diagnoses of pseudodischarge. These latter diagnoses include weeping associated with disease of the areola (eczema, dermatosis, erosive adenomatosis of the nipple), exteriorization of secretions secondary to pronounced nipple invagination or discharge from the peri-areolar region and not from the nipple, occurring in a girl (secondary to a para-areolar cyst evacuating itself through a Montgomery tubercle). 1. Download: Download full-size image Figure 2. Different stages of the clinical examination. a: inspection of the discharge using a magnification lamp; b: examination of a color of the discharge by placing a few drops on a dressing; c: technique for reproducing a discharge not present on the day of the consultation by gentle pressure and separation of the nipple-areolar region between thumb and index figure. In this case, the discharge is single pore and straw-colored (black arrows). Once the differential diagnoses have been excluded, the site of the discharge should then be established (uni-, pauci- or multi-orifice, uni- or bilateral) together with its color (milky or white, clear or serous or “gin clear”, yellow, orange, more or less dark green, chestnut, sero-sanginous, blood red or black) (Fig. 1a–d) which is better assessed by placing a few drops on a dressing (Fig. 2b). Palpation Palpation is performed using the usual breast examination technique in full daylight with the patient seated and then lying down. Palpation is carried out quadrant by quadrant with centripetal expression of the gland looking for a mass or “trigger point” inducing the discharge. If no discharge is present or only small amounts are produced on the day of the discharge, an attempt should be made to reproduce it by pressing gently on the areola, and not on the nipple, with thumb and index finger, pulling it apart (Fig. 2c). This is an essential stage as it can identify the watch hands of a single discharge from the nipple . Clinical summary The clinical examination should therefore endeavor to best describe the features of the discharge and to retrace its history, recording its date of onset, duration, frequency, whether it is spontaneously provoked, abundance, site and color. Conventionally, a bilateral provoked longstanding or intermittent discharge from multiple pores which is milky or dark green in color is considered to be non-suspicious or “physiological”. Conversely, a unilateral, single pore, spontaneous persistent discharge which is not green or milky (serous or clear, yellow, orange or bloodstained) is considered to be “pathological” as it is associated with an increased risk of underlying disease (Table 1). Table 1. Clinical classification of discharges. \| Empty Cell \| Non-suspicious=physiological \| Suspicious=pathological \| --- \| \| Laterality \| Bilateral \| Unilateral \| \| Number of pore(s) concerned \| Multiple pore \| Single pore \| \| Discharge production \| Provoked \| Spontaneous \| \| Frequency \| Long standing, intermittent \| Persistent \| \| Color \| Milky, creamy, chestnut-colored or dark green \| Clear, serous, bloody \| It is important to note, however, that controversies remain about the criteria used to describe a discharge, which vary between studies , , , and also on those used to define physiological and pathological discharges , . The clinical examination is also essential to guide breast ultrasound investigation by identifying a possible “trigger point” which induces the discharge or the watch hands of the discharge on the nipple which allows the radiological examination to be focused on a precise area. Smear of the discharge The role of a discharge smear in the diagnostic strategy for discharges is controversial and this is no longer recommended routinely in clinical practice . A discharge smear is a simple investigation to perform which begins by spreading the discharge onto slides. The discharge is obtained either by direct imprint from the nipple onto a slide if spontaneous discharge is present or by gentle pressure on the areola (Fig. 3a). After smearing, the slide is fixed and stained either by drying in open air and staining with May-Grünwald-Giemsa or by fixing in alcohol, or by spraying on a lacquer and staining with Papanicolaou (Fig. 3b). The macroscopic appearance and color of the discharge should be completed on the laboratory form, as the color of the discharge has been shown in some studies to be more sensitive than the cytological examination itself . 1. Download: Download full-size image Figure 3. The discharge smear. a: illustration of technique for performing the smear; b: cytology of a hemorrhagic single pore nipple discharge after staining. A clump of papillary cells are present with a few atypical changes on a blood stained background. A smear is recommended for a single pore bloody discharge but is of no benefit in bilateral multiple pore discharges . Conversely, it is debated for non-bloody single pore discharge regardless of color: •advantages: it is easy to perform, painless and may guide towards an etiologic diagnosis; •disadvantages: it offers average or even poor sensitivity which varies greatly depending on the study sometimes needing it to be repeated to increase its sensitivity . It also requires an experienced breast cytopathologist to be available and there is lack of close correlation between cytological results and possible underlying organic lesion ; •regardless, it is only valid if the result is malignant because of a false negative rate of over 50% for malignant lesions , . Imaging Most series in the literature have examined the utility of imaging investigations either in detecting papillomas (with or without a context of discharge) or in detecting a malignant/at risk lesion on a background of discharge. In practice, imaging investigations are used to detect a lesion –benign or malignant– which explains the origin of the discharge, hence the limitation of published findings on this subject. Mammography Mammography is the investigation of first line for a pathological discharge but is still limited, particularly because of its poor sensitivity of around 20–25% , , . A negative mammograph in the context of nipple discharge does not therefore exclude any underlying disease. It can be used to investigate for the cause of a discharge by detecting: • calcifications (Fig. 4a–c): ∘macrocalcifications which are benign in appearance and round or “eggshell” like, located behind the areola and suggestive of a papilloma, or as “rods” due to calcifications of galactophore duct ectasia, ∘microcalcifications: these do not always distinguish benign from malignant duct diseases. As an example, microcalcifications due to papillomas are usually suspicious in appearance , . On the other hand, ductal or segmental microcalcifications classified as BI-RADS grade 5 suggest a malignant duct disease from the outset, mostly due to ISDC; 1. Download: Download full-size image Figure 4. Different types of calcifications found on mammography for discharge. a: “eggshell” calcifications in the retro-areolar region due to a central papillo-adenoma; b: bilateral “rod shaped” calcifications which are pathognomonic of calcifications from galactophore duct ectasia (red arrows), in this case in the right breast; c: fine polymorphic microcalcifications arranged segmentally due to an ISDC with microinfiltration on a background of hemorrhagic single pore nipple discharge. • masses (Fig. 5a, b): when these are analyzed using the Breast Imaging-Reporting And Data System (BI-RADS) descriptive criteria they may be due to a papilloma, a papillary carcinoma or an infiltrating ductal carcinoma (IDC). A round-shaped mass associated to a greater or lesser extent with calcifications and located behind the areola is suggestive in the context of a papilloma; 1. Download: Download full-size image Figure 5. Different masses found on mammography in the context of nipple discharge. a: oval mass with partially masked circumscribed outlines in the external retro-areolar region due to a central papillo-adenoma; b: irregular shaped mass with indistinct outlines due to a papillary carcinoma (red arrow). • others: galactophore duct ectasia (Fig. 6a, b), focal density asymmetry. 1. Download: Download full-size image Figure 6. Galactophore duct ectasia visible on mammography on a background of nipple discharge (red arrows). a: in a patient with brownish discharge occurring at night; b: in a 41-year-old patient in the context of galactorrhoea after breast-feeding. There is no information in the literature about technical details for performing and reading mammographs in this situation. The retro-areolar region, which is often more difficult to examine, however, must be studied carefully adding a localized compression film or magnified films centered on this region if the least doubt is present. Ultrasound Ultrasound is performed immediately after the mammograph even if the mammograph is normal and if needed secondarily as a “second-look” ultrasound if additional breast MRI investigation has been performed. Details for performing this technically have been described by Stavros (Fig. 7). Ultrasound offers better sensitivity than mammography to detect intraductal lesions responsible for a pathological discharge, with a sensitivity in the region of 65% and specificity of between 75 and 85% depending on the study , , . 1. Download: Download full-size image Figure 7. Diagram showing the technique for carrying out ultrasound for nipple discharge according to Stavros. In this situation, it is used to investigate for: • mammary duct (or galactophore duct) ectasia (Fig. 8a–c), defined by a duct diameter of over 3 mm, which is usually associated with green or chestnut-colored discharge. This must be examined along its long axis to better study its content. An echoluscent content, however, is not necessarily reassuring as this is seen in half of the cases of papillomas and in 14% of cases of intraductal carcinomas ; 1. Download: Download full-size image Figure 8. Ductal or galactophore duct ectasia on ultrasound. a: galactophore duct ectasia containing thick “beaded” secretions which is not vascularized seen on color Doppler ultrasound-mode; b: galactophore duct ectasia with a completely echogenic content filled with thick secretions; c: galactophore duct ectasia containing a small budding endoductal mass with a central vascular pedicle on color Doppler ultrasound-mode due to a small papilloma. • a dilated duct with occasionally tortuous echogenic contents (Fig. 9), which is not specific for any cause (thick secretions, ductal hyperplasia or papilloma), but in the context and in conjunction with clinical examination findings (watch hands of a single pore discharge) should lead to histological samples being taken; 1. Download: Download full-size image Figure 9. Dilated tortuous duct with an echogenic content on ultrasound (red arrows) in a case of bloody discharge with erosive adenomatosis of the nipple due to a bulky central papillo-adenoma located in the nipple and extending to several ducts. • a complex or solid mass (Fig. 10a–f), which is usually endoductal, and should be examined according to the BI-RADS criteria. In this situation, however, difficulties arise from the many differential diagnoses as the image may be due to a papilloma, a papillary carcinoma, an intraductal carcinoma or a complex lesions of fibrocystic breast disease; 1. Download: Download full-size image Figure 10. Different masses seen on ultrasound in nipple discharge. a: rounded endoductal mass with a small central vascular pedicle on color Doppler-mode representing a small papilloma; b: budding echogenic mass dilating a galactophore duct representing a peripheral papillo-adenoma; c: heterogeneous hypoechogenic attenuating plaque on ultrasound, suspicious in appearance and representing a sclerosing papilloma; d: budding echogenic mass with intense vascularisation within the lesion in color Doppler-mode in a galactophore duct ectasia representing a central papillo-adenoma colonized by an intermediary grade ISDC; e: complex intensely vascularized endoductal mass in the retro-areolar region representing a papilloma colonized by papillary carcinoma lesions; f: BI-RADS 4 solid mass in a case of hemorrhagic single pore nipple discharge due to an infiltrating ductal carcinoma. •an attenuating “non mass” occasionally due to ISDC lesions. Ultrasound however is very limited in detecting in situ carcinomas as it is negative in almost 80% of cases . Ultrasound can also show signs of fibrocystic breast disease when it is performed in the context of multiple pore discharge. The limitations of ultrasound in this situation include: the discontinuous visualization of the duct system; the difficulty in detecting lesions lying peripherally, particularly if these are small in size (e.g. peripheral papillo-adenomas); the lack of specific ultrasound criteria to distinguish papillo-adenomas from papillary carcinomas. On the other hand, various authors highlight its high added value after MRI as a “second-look” ultrasound , . Galactography This was formerly the method of choice in the diagnostic approach to a pathological single pore nipple discharge with no clinical mammography or ultrasound pointers. It could be used to detect and locate the lesion responsible as galactophore duct ectasia, an intraluminal defect (filling defect), single or multiple, stenosis, complete obstruction or wall irregularity (Fig. 11). 1. Download: Download full-size image Figure 11. Diagnostic galactography image. The EUSOMA working group in 2010 , however, highlighted the many disadvantages of this investigation. It is occasionally technically impossible because of failed catheterization of the pore due to the need to reproduce the discharge on the day of the investigation. It is invasive with a risk of extravasation and complications due to allergy to the iodinated contrast medium, or mastitis. It cannot establish the cause of the pathological discharge and differentiate malignant from benign lesions; it is occasionally incomplete, with an incomplete galactography rate of around 15% . At present it is important to distinguish diagnostic from “topographic” galactographies. There is greater use of diagnostic galactography in France , although the investigation is inadequate in this situation particularly because of the high false negative rate (≈20%) and poor sensitivity of 50% or less depending on the study , . Lorenzon et al. also referred to it as an investigation belonging “in the past”, highlighting the fact that it was invasive and time-consuming . Son et al. described it as being unpleasant or even painful for the patient and noted that many centers had replaced galactography with other imaging techniques. On the other hand depending on the centre, “topographic” galactography is still performed by the radiologist preoperatively or by the surgeon in the operating theatre (after injecting blue dye through the pore involved) because of its good localizing value helping to minimize the volume of breast excised , , . MRI There are few studies in the literature examining the role of MRI in the diagnostic strategy for nipple discharge. This is currently an emerging indication proposed by the learned societies but not validated by published findings. According to the HAS, “MRI is a specialist investigation which can provide further information in some cases of discharge where it is presumed there is a proliferative intra-galactophore duct lesion” (level of evidence C) . According to the EUSOMA (European Society of Breast Cancer Specialists) working group, “the scientific evidence for the utility of MRI in the management of pathological discharge is inadequate to justify it being performed in usual practice”. However, “MRI-galactography” may be thought of as an alternative to investigate a suspicious nipple discharge, i.e. one from a single pore which is unilateral, if galactography has failed or the patient refuses the investigation . In usual practice, we perform a breast MRI for pathologic nipple discharge which has a normal breast ultrasound assessment. This opinion is also held by Lorenzon et al. who consider that the investigation should be performed when the cause of the discharge is unexplained on conventional imaging . Advantages Firstly, MRI offers excellent sensitivity for pathological nipple discharge compared to conventional imaging, which varies according to the most recent studies from 88 to 95% , . Similarly, it has a high negative predictive value of around 90%, false negatives mostly involving low grade ISDC or IDC around a millimeter in size , . When an underlying pathological lesion is responsible for the discharge this investigation can be used for a more detailed assessment of the spread of the lesion by mammography and ultrasound, allowing better quality, surgical excision by optimalizing preoperative land marking for preservative treatment. Finally it guides the “second-look” ultrasound, which facilitates detecting the lesion responsible which is not seen on the primary investigation and taking percutaneous samples . Limitations The major disadvantage of MRI is that it often detects additional images or “false positives” which result in MRI monitoring or biopsies being taken which are unrelated to the pathological nipple discharge . It appears to be more difficult with this technique to characterize an endoductal lesion an therefore to guide the diagnosis towards a benign or malignant lesion, which makes “second-look” ultrasound even more essential . Furthermore, absence of an image abnormality on morphological T1- and T2-weighted sequences, on MR-galactography and lack of contrast enhancement does not however completely exclude a malignant lesion . MRI offers a specificity of around 75% in the different published series which some authors consider to be only moderate. What to look for on MRI? The most common abnormality seen with a pathological nipple discharge is enhancement without a mass (EWM) for malignant, atypical or papillary lesions , (Fig. 12a–f). In decreasing order of frequency, these are ductal EWM (Fig. 12a), focal regions (Fig. 12b) followed by segmentally or regionally distributed EWM (Fig. 12c). It is important to bear in mind that a ductal or segmental EWM can be seen in both papillomas and papillomatoses . In this situation, the segmental spatial distribution and micronodular or annular nature of the lesion are the EWM features which have the greatest positive predictive value for malignancy. 1. Download: Download full-size image Figure 12. Different abnormalities seen on MRI in cases of pathological nipple discharge without abnormality on the conventional mammography-ultrasound assessment. a: branched enhancement without ductal mass at the union of the internal quadrants and nipple in the left breast in a case of single pore bloody discharge associated with erosive adenomatosis of the nipple: bulky central papillo-adenoma carcinoma located in the nipple and extending to several ducts; b: enhancement without mass appearing as a heterogenous local area 40×25×15 mm in the lower part of the SIQ of the left breast (MRI BI-RADS grade 4) in a patient with single pore straw-colored discharge from her left breast and a normal mammography-ultrasound assessment which was followed by MRI-guided macrobiopsies: ISDC; c: MIP reconstructions of micronodular segmental enhancement without mass in the supero-internal retro-areolar region of the right breast in a case of spontaneous persistent single pore bloody discharge from the right breast: high grade ISDC with microinfiltration; d: small round mass with smooth outlines in the infra-areola region of the left breast (red arrow) showing a type 3 enhancement curve (with “wash-out”) in a case of spontaneous persistent single pore serous discharge: central papillo-adenoma; e: relatively extensive endoductal mass in a case of persistent single pore serous discharge from the left breast: papillo-adenoma associated with at risk ILN 1 and ILN 2 lesions; f: T1-weighted image without enhancement showing galactophore duct ectasia in the right breast with unenhanced T1 hyperintensity in a case of persistent nocturnal multiple pore bloody nipple discharge. Masses are far less commonly seen and when these are due to a papilloma they are usually small, round or oval masses with regular outlines, which are microlobulated or finely speculated and are hypervascularized with falsely suspicious type 2 or 3 enhancement curves (Fig. 12d, e). Finally, galactophore duct ectasia is also relatively common and often appears as a T1-weighted hyperintensity on unenhanced images due to protein-rich secretions or blood contained within the duct (Fig. 12f). This is sometimes even the only abnormality seen on MRI. MRI protocol The MRI protocol is standardized and should at least include T1- and T2-weighted MR images together with a dynamic image both without and then with IV administration of gadolinium chelate using sections of 3 mm or less in order to be able to detect small abnormalities . Examination of abnormalities is facilitated by producing multiplaner reconstructions (MPR). Specific MRI sequences for investigating a discharge Indirect ducto-MRI is a recent technique , based on the principle of hydrography, i.e. using a very highly weighted T2 sequence (equivalent to a cholangio-MRI sequence). The benefit of ducto-MRI is that it establishes the intraductal site of a mass or enhancement without mass (Fig. 13a–c). This is extremely useful information as by locating the lesion within the duct, it can confirm the relationship between the lesion and the discharge. Identification of the pathological duct is not however always straightforward as the duct may be filled with fluid and the section plane must be tangential to the long axis of the duct in order to be able to examine it over a greater length. In addition, if the introductal fluid contains blood or protein the duct may appear iso-intense on ducto-MRI images. 1. Download: Download full-size image Figure 13. Further investigation with breast MRI and indirect ducto-MRI image in a patient with a spontaneous right breast single pore sero-sanginous discharge for several weeks: central papillo-adenoma. a: ultrasound: galactophore duct ectasia with an echogenic content occupied proximally by non-vascularized echogenic material in color Doppler ultrasound-mode; b: breast MRI with dynamic images: unenhanced image, 2nd dynamic stage after enhancement and subtraction in the same stage. Appearances of internal retro-areolar galactophore duct ectasia appearing as an unenhanced hyperintensity on T1-weighted imaging (yellow arrows) combined with homogeneous “branched” proximal ductal enhancement without mass (red arrows); c: indirect galacto-MRI view in the sagittal plane showing galactophore duct ectasia interrupted by an intraductal tissue structure. The most recent studies on the subject however including the study by Boisserie-Lacroix et al. failed to establish any conclusion about the actual utility of indirect ducto-MRI compared to a conventional MRI protocol in this situation. Direct ducto-MRI described by a few authors involves direct intraductal injection of the MRI contrast medium diluted according to the same principle as in galactography. This technique appears to produce identical diagnostic results to those of galactography although the technique is not approved in France. The approach to nipple discharges The first stage in any nipple discharge is the clinical examination, in order to describe the discharge, investigate for related signs (mass, nipple retraction, etc.) and then distinguish between “physiological” and “pathological” discharge in order to establish whether it is necessary to perform further imaging investigations. In all case of pathological discharge in patients over 35 years old, further investigations including mammography and then ultrasound are required which remain unquestioned despite their limitations in this situation . The approach to a multiple pore or bilateral discharge In this situation the causes responsible are galactorrhoea (due to primary or secondary hyperprolactinemia) or abnormalities associated with breast dysfunction such as secretory galactophore duct ectasia and fibrocystic breast disease. Theoretically this type of discharge does not require a further imaging assessment . This recommendation, however, requires a more subtle interpretation and imaging investigations should be performed if a bloody multiple pore discharge is present . The approach to a single pore discharge The standard investigation to be performed for single pore discharges is breast ultrasound. After this investigation their abnormalities found are classified using the BI-RADS criteria. Percutaneous histological samples should of course be taken for BI-RADS grade 4 or 5 abnormalities and also for BI-RADS 3 lesions when their sites are consistent with clinical examination findings (such as with a watch hands appearance for a single pore discharge or a “trigger point”) as the investigations have been triggered by a clinical symptom and clinical findings should take priority over imaging. In the specific case in which a single pore discharge is green or green-chestnut in color and the conventional assessment is normal or BI-RADS grade 2 and further investigations are compatible with galactophore duct(s) ectasia or fibrocystic breast disease, clinical monitoring should then be offered until the discharge dries up, which may take 2 to 36 months . In the occasional cases when patients are in too much discomfort, surgery can then be considered, attempting to identify the duct responsible for the discharge with blue dye. In the case of periductal mastitis, the clinical enquiry must focus on smoking habit and smoking cessation. What to do for a pathological nipple discharge with a normal conventional assessment? Although there is still inadequate scientific evidence, most authors agree (and the utility of this is confirmed in our own experience) that further investigations should be performed using breast MRI, with or without a galacto-MRI sequence. Although according to EUSOMA working group, this indication is not currently approved , MRI is however more sensible than a conventional assessment and may show contrast enhancement corresponding to the lesion responsible for the discharge. In this situation a “second-look” ultrasound should then be performed to attempt to find an ultrasound reflection of the MRI contrast enhancement. If an ultrasound appearance is present which is consistent with the MRI abnormality, percutaneous samples should be taken, preferably under ultrasound guidance and in most cases a clip should then be applied in order to confirm the consistency between MRI contrast enhancement and the ultrasound abnormality particularly if the histological result is benign. If the MRI contrast enhancement involves enhancement without mass located in the same quadrant as the pathological single pore nipple discharge and no finding is seen on the “second-look” ultrasound which is consistent with the MRI abnormality it is then useful to take enlarged postero-anterior and lateral mammography views in this territory looking for microcalcifications which correlate with the MRI contrast enhancement . If microcalcifications are seen which may be related to the enhancement without mass, percutaneous histological samples are then taken by the stereotactic macrobiopsy procedure followed by application of a clip. Finally, if no microcalcifications are seen on the additional mammography films and if the MRI contrast enhancement in question is classified as BI-RADS grade 4 or 5, an MRI biopsy should be performed after being confirmed in a multidisciplinary team meeting if this is available within the centre. Failing this, pyramidectomy can be considered after identifying the pathological duct with blue dye. What is the approach to a pathological nipple discharge with a normal full diagnostic assessment? Historically, surgical excision/pyramidectomy was the standard practice for any single pore bloody discharge or even any “pathological” discharge because of the risk of cancer which was considered to be significant. The assumption was to deem that the negative clinical examination apart from the pathological discharge, cytology and imaging assessment (mammography+ultrasound+MRI) did not exclude a malignant lesion , , , although this approach was far too invasive. Firstly, discovery of ISDC from the single symptom of breast discharge with no associated imaging abnormality is extremely rare and secondly, an invasive cancer very rarely presents with discharge, and in this situation is always associated with a predominant intraductal component. More recently, Ashfaq et al. and Sabel et al. , have shown that the risk of an occult malignant lesion is extremely low if the clinical examination (excluding the discharge) and the imaging assessment are normal and that most of these cases are low grade ISDC or very small IDC. In addition, the malignant lesions associated with pathological nipple discharge often carry a good prognosis . According to these authors, if the imaging assessment is normal in a pathological nipple discharge short term monitoring would appear to be reasonable. In practice, Ashfaq et al. proposed that regular patient monitoring be set up every 6 months for 2 years or until the discharge resolves spontaneously, which occurs in 81% of cases over two years . The following monitoring protocol is suggested: •clinical monitoring every 6 months; •ultrasound monitoring every 6 months; •mammography monitoring every 12 months. If the patient refuses this monitoring and the discomfort causes by the discharge is excessive or if it is still present after 2 years, pyramidectomy can be considered. The approach to the diagnostic management of pathological nipple discharge is summarized in the diagnostic decision algorithm shown below (Fig. 14). 1. Download: Download full-size image Figure 14. Proposed diagnostic decision algorithm for the investigation of nipple discharge. The approach to persistant galactorrhea The first reflex action should be to confirm the patient is not pregnant or breast-feeding, if necessary with a blood β-HCG measurement . The combination of galactorrhoea and amenorrhea should immediately raise the question of hyperprolactinemia and an evident secondary cause should be excluded on the clinical enquiry: either drug-induced iatrogenic effects (antidepressants, neuroleptics, H2 antagonists, antiemetics and antihypertensives) or a cause due to a specific clinical situation (severe hypothyroidism or end stage renal or hepatocellular renal failure). If no cause is found a blood prolactin measurement is required. If hyperprolactinemia is present, the first approach is to exclude hypothyroidism (by measurement of TSH) or renal impairment (measurement of renal function) and then, if no laboratory abnormalities are present, to perform a hypothalamic and pituitary MRI looking for a pituitary adenoma or another organic lesion (a sellar or suprasellar lesion). The other causes of galactorrhoea are diagnoses of exclusion but ones which although far rarer must not however be missed: drugs (cannabis, marijuana), amphetamines, excess caffeine consumption, repeated rubbing from an unsuitable bra and bronchogenic carcinoma (ectopic prolactin production). Treatment then depends on the cause: •prolactinoma: dopamine agonists (bromocriptine, cabergoline) and occasional use of surgery and radiotherapy; •iatrogenic effects: stop and replace the drug in question with another compound belonging to the same family but which has a lesser prolactin raising effect; •idiopathic galactorrhoea (normal prolactin concentration): reassure the patient and prescribe low doses of dopaminergic agonists in incapacitating cases. What do to for a nipple discharge associated with infection? The treatment for infectious causes is mostly medical with antipyretics and antibiotic therapy although far more rarely surgery is used to treat any abscess present. Vigilance is required in inflammatory presentations of discharge in order not to miss a possible underlying breast carcinoma. What to do when a nipple discharge occurs in a man? This symptom should always be considered to be suspicious in a man because of the high incidence of underlying breast carcinoma which has been found to be 23% in this situation . Breast nipple discharge may even be the only presenting sign of an ISDC. In terms of further investigations, mammography and ultrasound are both essential although mammography is more sensitive than ultrasound (100% vs 83.3%). Conclusion The diagnostic management of nipple discharge is based on a thorough analysis of clinical examination findings to distinguish benign “physiological” from suspicious “pathological” discharges, which require a conventional mammography and ultrasound assessment. If no abnormality is present on this assessment, diagnostic galactography has been gradually replaced by breast MRI (whether or not combined with galacto-MRI) images which needs however to be validated by published findings in order to justify its use in everyday practice. Although in this situation, benign causes are by far more common, there is a significant risk of an underlying malignant lesion. According to recent studies, however, the risk of an occult malignant lesion is extremely low if the whole imaging assessment is normal. In addition, malignant lesions associated with pathological nipple discharge often carry a good prognosis. The approach of operating on any single pore bloody discharge or even any “pathological” discharge with a normal imaging assessment because of the risk of underlying cancer, therefore appears to be too invasive and in the most recent publications has given way to regular short term monitoring with the patient's agreement. Refusal for monitoring, persistent symptoms or overly severe however, represent a reason for carrying out pyramidectomy. Take-home messages •Nipple discharge is a common symptom. •Benign causes are most common: papillomas and galactophore duct ectasia. •It is important not to miss the risk of an underlying malignant lesion (≈5%) which is mostly an in situ ductal carcinoma. •Clinical examination is essential in the diagnostic approach, distinguishing benign “physiological” from suspicious “pathological” discharges which require further conventional imaging investigations (mammography and ultrasound). •Theoretically, mulipore and/or bilateral discharges do not require a further imaging assessment. •There is no longer an indication for diagnostic galactography in France although “topographic” galactography is still performed preoperatively in some centers because of its good localizing value. •Breast MRI is an emerging indication which may be useful in cases of pathological nipple discharge when the conventional imaging assessment is normal, although its use in everyday practice has not yet been validated by the learned societies or the findings in the literature. •Surgery is not the only solution for patients with a pathological nipple discharge and normal imaging assessment. They can now be offered regular monitoring. Clinical case At the end of October 2014, a gynecologist requested an opinion about a left nipple discharge which had developed two months previously in a 48-years-old woman. The discharge was single pore, spontaneous, bloody, isolated and troubled the patient because of its amount. The patient has been amenorrheic on the contraceptive pill. She has no significant past personal history although she has a family history of breast cancer in two paternal aunts after the age of 50 years old. The conventional assessment performed outside of the centre (2 view standard mammography combined with breast ultrasound) is normal. The gynecologist performed a smear of the discharge on slides which showed “foamy galactophore duct cells and microclumps of epithelial cells with slight atypia which could be consistent with proliferative sclerocystic breast disease”. Questions 1.Do the features of this discharge allow investigations to be stopped at this point? What would you propose? Justify your approach. 2. MR examination is performed in November. The discharge has become intermittent and clear. You have a very highly weighted T2 sagittal image (Fig. 15), transverse T1-weighted axial images T1- (Fig. 16) and T2-weighted images together with dynamic T1-weighted images with fat saturation after contrast enhancement (Fig. 17a–c). Do you think the protocol is optimal? How would you interpret the atypical enhancement between the two breasts? What would you propose? 1. Download: Download full-size image Figure 15. Sagittal “galacto-MRI” view. 1. Download: Download full-size image Figure 16. T1-weighted MRI view without fat saturation. 1. Download: Download full-size image Figure 17. Dynamic T1-weighted MRI views with fat saturation after gadolinium chelate enhancement. 3. The patient was seen the next week for a “second-look” assessment. The further mammography films are normal. Before the ultrasound she tells you that she feels a projection in the left internal para-areolar region on which the ultrasound is centered. As no abnormality was found further investigations were performed in this region using ShearWave elastography (Fig. 18). What is the approach now? 1. Download: Download full-size image Figure 18. Ultrasound image with further investigation using ShearWave elastography centered on the union of the internal quadrants of the left breast. Answers 1.The discharge has all of the features of a “pathological” discharge which requires investigations to be continued, particularly as the smear showed “slight atypia”. There is currently no longer an indication to perform diagnostic galactography and the decision is made to carry out breast MRI. 2.“Galacto-MRI” was performed optimally followed by a standard protocol. This did not show duct ectasia. Micronodular enhancement over a 40-mm segmental region along its transverse axis classified as BI-RADS grade 5 is found after enhancement in the retro-areolar region at the junction of the internal quadrants for which a “second-look” mammography assessment (magnified orthogonal film centered on the region) and centered ultrasound is proposed. 3.A careful “second-look” ultrasound shows no abnormality although on elastography, the palpable projection has an elastography score of 100 kPa, which is locally higher than the neighboring tissue. There is a good topographical location between the clinical abnormality, ShearWave elastography findings and suspicious MRI contrast enhancement. Ultrasound guided 10 G diameter macrobiopsies are therefore proposed with aspiration (as this is a subtle non-mass image) with application of a clip at the end of the procedure. Four out of the 6 samples contain high grade in situ ductal carcinoma (ISDC) lesions and in December 2014, the patient underwent lumpectomy with landmarking and sentinel lymph node biopsy. Ultimately, surgery found an ISDC with microinfiltration. Revision surgery was carried out as the excision margins were not clear. Disclosure of interest The authors declare that they have no conflicts of interest concerning this article. Special issue articles Recommended articles References L. Chen, W.-B. Zhou, Y. Zhao, X.-A. Liu, Q. Ding, X.-M. Zha, et al. Bloody nipple discharge is a predictor of breast cancer risk: a meta-analysis Breast Cancer Res Treat, 132 (1) (2012), pp. 9-14 Google Scholar M. Lorenzon, C. Zuiani, A. Linda, V. Londero, R. Girometti, M. Bazzocchi Magnetic resonance imaging in patients with nipple discharge: should we recommend it? Eur Radiol, 21 (5) (2011), pp. 899-907 CrossrefView in ScopusGoogle Scholar M. Morrogh, A. Park, E.B. Elkin, T.A. 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Rivin del Campo Value of mammography and breast ultrasound in male patients with nipple discharge Eur J Radiol, 82 (3) (2013), pp. 478-484 View PDFView articleView in ScopusGoogle Scholar Cited by (32) Is Ductography Still Warranted in the 21st century? 2019, Breast Journal ### Retrospective statistical analysis on the diagnostic value of ductography based on lesion pathology in patients presenting with nipple discharge 2019, Breast Journal ### Common breast problems 2019, American Family Physician ### Nipple discharge: Imaging variability among U.S. radiologists 2018, American Journal of Roentgenology ### Ultrasonographic evaluation of women with pathologic nipple discharge 2017, Ultrasonography ### Diagnostic performance of MRI versus galactography in women with pathologic nipple discharge: A systematic review and meta-analysis 2017, American Journal of Roentgenology View all citing articles on Scopus Copyright © 2015 Published by Elsevier Masson SAS Part of special issue 96/10-2015 -FMC CME "Women's imaging" Download full issue Other articles from this issue Acute pelvic pain in females in septic and aseptic contexts October 2015 E.Pages-Bouic, …, P.Taourel View PDF ### Chronic pelvic pain: An imaging approach October 2015 V.Juhan View PDF ### Breast pain and imaging October 2015 C.Balleyguier, …, C.Dromain View PDF View more articles Recommended articles No articles found. 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188031	http://lampz.tugraz.at/~hadley/ss1/materials/thermo/gp/gp/Fermi-function.html	Properties of the Fermi function In a system of non interacting fermions the Fermi function (f(E)) gives the propability that an energy level $E$ is occupied at the absolut temperature $T$. For low temperatures almost all fermions are in states with energies lower than the chemical potential (\mu). For theses temperatures the Fermi function looks like a step function. $$ f(E) = \frac{1}{\exp{\big(\frac{E-\mu}{k_BT}\big)}+1} $$ \| \| \| --- \| \| [ \begin{equation} \phantom{[eV^-1]} f(E)\; \end{equation} ] \| \| \| [ \begin{equation} \quad{} (E - \mu) \; [\text{eV}] \end{equation} ] \| It should be noted, that the Fermi function does not actually apply to fermion systems like metals as there is interaction between the fermions in these systems. However, it is still a good first approximation, when calculating their thermodynamical properties. Derivative of the Fermi function As the Fermi function mostly changes only in the close vicinity of the chemical potential, its derivative is mostly zero. Close to the chemical potential the derivative shows a negative peak, which in the limit of (T \rightarrow 0) becomes a dirac delta function. $$ \frac{df(E)}{dE} = -\frac{1}{k_BT}\frac{\exp{\big(\frac{E-\mu}{k_BT}\big)}}{\big(\exp{\big(\frac{E-\mu}{k_BT}\big)}+1\big)^2} $$ \| \| \| --- \| \| [ \begin{equation} \frac{df(E)}{dE}\; [\text{eV}^{-1}] \end{equation} ] \| \| \| [ \begin{equation} \quad{} (E - \mu) \; [\text{eV}] \end{equation} ] \| Because of this behaviour, thermodynamic integrals that contain the Fermi function are often first integrated by parts. $$ \int\limits_{-\infty{}}^{\infty{}}H(E)f(E)\;{}dE = K(\infty)f(\infty) - K({-\infty})f({-\infty}) - \int\limits_{-\infty{}}^{\infty{}}K(E)\frac{df}{dE}\;{}dE\; ,$$ with (K(E)) being $$ K(E) = \int\limits_{-\infty{}}^{E}H(E')\;{}dE'\; .$$ The border terms vanish as the Fermi function becomes zero for (E\rightarrow \infty) and (K(-\infty)=0). One ends up with an integral which only needs to be evaluated numerically for a finite interval around (\mu). $$\int\limits_{-\infty{}}^{\infty{}}H(E)f(E)\;{}dE = - \int\limits_{-\infty{}}^{\infty{}}K(E)\frac{df}{dE}\;{}dE\; \approx - \int\limits_{\mu - 8k_BT}^{\mu + 8k_BT}K(E)\frac{df}{dE}\;{}dE\;.$$ Integral of the Fermi function In the case of some thermodynamic properties like the grand potential (\phi) or the entropy (s) the integrals that need to be calculated contain the integral of the Fermi function in the form, $$F(E)= \int\limits_{E}^{\infty}\frac{1}{\exp{\left(\frac{E'-\mu}{k_BT}\right)}+1}dE'.$$ Let $x = \frac{E'-\mu}{k_BT}$ and multiply the integrand by $\exp (-x)/\exp (-x)$, $$ F(E) = k_BT\int\limits_{\frac{E-\mu}{k_BT}}^{\infty}\frac{\exp(-x)}{1+\exp(-x)}dx.$$ The integrand has the form $\frac{du}{u}$ so the integral is $\ln(u)$. $$ F(E) = k_BT\ln\left(1+\exp(-x)\right)\Big{\|}_{\frac{E-\mu}{k_BT}}^{\infty}=k_BT\left(\ln\left(1+\exp(-\infty)\right)-\ln\left(1+\exp(-\frac{E-\mu}{k_BT})\right)\right).$$ The first term on the right vanishes because $\ln(1)=0$, $$ F(E) = -k_BT\ln{\Bigg[\exp{\bigg({-\frac{E-\mu}{k_BT}}\bigg)}+1\Bigg]} $$ \| \| \| --- \| \| [ \begin{equation} \phantom{^-1}F(E)\; [\text{eV}] \end{equation} ] \| \| \| [ \begin{equation} \quad{} (E - \mu) \; [\text{eV}] \end{equation} ] \| One can often convert the integrals involving the integral of the Fermi function (F(E)) into one involving the Fermi function by integrating by parts. The border terms vanish as (F(E)) also goes to zero rather quickly for energies above (\mu). $$\int\limits_{-\infty{}}^{\infty{}}H(E)F(E)\;{}dE = - \int\limits_{-\infty{}}^{\infty{}}K(E)f(E)\;{}dE\; $$ $$ K(E) = \int\limits_{-\infty{}}^{E}H(E')\;{}dE'\; .$$ One can now apply a similar scheme to the one shown in the section above. For example, the grand potential (\phi) was calculated in this way (Calculate (\phi)). Second integral of the Fermi function There is an interesting property regarding the grand potential (\phi) in the case of a 2D fermi gas. A solution can be found in terms of a dilogarithm fuction which can be used to check the numerical calculation. $$ \phi = -k_BT\int\limits_{-\infty{}}^{\infty{}}D(E)\ln{\Bigg[\exp{\bigg({-\frac{(E-\mu)}{k_BT}}\bigg)}+1\Bigg]}\;{}dE = \int\limits_{-\infty{}}^{\infty{}}D(E)F(E)\;{}dE$$ As the density of states is a constant function starting from some E0 the integral reduces to $$\phi = D_0\int\limits_{E_0}^{\infty{}}F(E)\;{}dE = D_0\bigg[F_2(E)\bigg]_{E_0}^\infty \; ,$$ with $F_2(E)$ being the second integral of the Fermi function. In contrast to the first integral, the second integral does not have a closed form: $$F_2(E) = -(k_BT)^2\text{Li}_2(x)$$ $$x = -\exp{\bigg(-\frac{E-\mu}{k_BT}}\bigg)$$ Here Li2(x) denotes the dilogarithm. \| \| \| --- \| \| [ \begin{equation} F_2(E)\; [\text{eV}^{2}] \end{equation} ] \| \| \| [ \begin{equation} \quad{} (E - \mu) \; [\text{eV}] \end{equation} ] \| The dilogarithm and therefore $F_2(E)$ vanishes for $E\rightarrow\infty$. The grand potential then takes the form: $$\phi{}(T) = -D_0F_2(E_0,T)$$ Usually the lowest occupied energy $E_0$ in case of a free fermi gas is chosen as the zero point and the chemical potential $\mu$ has a positive value. In the plots below, like in the cases before, a presentation with $\mu$ as the zero point is better suited. Therefore $E_0$ is negative and a change of the chemical potential will result in a shift of $E_0$. On the left the negative second integral of the Fermi function is plotted against the energy $E$. The value of the chemical potential (and therefore the position of $E_0$ relative to it) depends in case of the 2D fermi gas just on the electron density $n$ and the mass of the particle (we choose free electrons $m=m_e$). The same goes for the constant density of states $D_0$. $$\mu \approx E_F = \frac{\pi{\hbar{}}^2}{m}n$$ $$ D_0 = \frac{m}{\pi{\hbar{}}^2}$$ The value of $-F_2(E_0)$ at the energy $E_0$ is marked with a cursor. When regulating the temperature $T_0$, $F_2$ changes its shape. The point for a fixed electron density aka fixed $E_0$ moves then up and down. When looking only at the movement of this point, one obtains the temperature dependancy of the grand potential density $\phi(T)$ of a 2D fermi gas with exactly that electron density. \| \| \| \| --- \| [ \begin{equation} -F_2(E) \; [\text{eV}^2] \end{equation} ] \| \| [ \begin{equation} \phi(T)\; [\text{eV}] \end{equation} ] \| \| [ \begin{equation} \quad{} (E - \mu) \; [\text{eV}] \end{equation} ] \| [ \begin{equation} \quad{} T \; [\text{K}] \end{equation} ] \| \| \| T0 = 600 K \| For a fixed temperature $T_0$, when changing the electron density, the position of $E_0$ on the $-F_2$ curve changes. This results in a increase/decrease of $-F_2(E_0)$ and therfore of $\phi$. The increase/decrease depends hereby on the slope of $F_2$, which itself depends on the temperature $T_0$. Therefore the points of $\phi$ on the right change differently depending on their temperatures (small change for low temperatures and big change for high temperatures → shift + compressing/streching of graph).
188032	https://grail.eecs.csuohio.edu/~somos/rattri.html	Rational Triangles Definition Define a Rational Triangle as a triangle in the Euclidean plane such that all three sides measured relative to each other are rational. Once, it was thought that all triangles were rational. The discovery of counterexamples is attributed to the Pythagoreans. Any triangle similar to a rational triangle is rational also. Take as a unit the greatest common measure of the three sides. Then the length of the sides are positive integers whose greatest common measure is unity. All rational triangles can be uniquely constructed in this way from three positive integers with greatest common divisor unity and each less than the sum of the other two (triangle inequality). Right Triangles Define a Rational Right Triangle as a right triangle which is a rational triangle. A right triangle is a rational triangle if and only if all six trigonometrical ratios of the two complementary acute angles are rational. It can be proved, using the inscribed circle, that the tangent of its half angles are rational numbers. Define a Heronian Angle as an angle such that the tangent of its half is a rational number. Conversely, any rational number between zero and one is the tangent of half an acute angle of a rational right triangle. The rational number associated with one acute angle has numerator and denominator both odd, but for the other angle they are of different parity. All rational right triangles are constructed in this way from a rational number between zero and one in both ways. Heronian Triangles Define a Heronian Triangle as a rational triangle with area rational relative to the square of any side. The name refers to the formula for the area of a triangle given the sides known as Hero's formula. This remarkable formula is one-fourth the square root of the product of four factors, one of which is the perimeter of the triangle (the sum of the three sides), and the other three factors are gotten by subtracting one side from the the sum of the other two sides. The simplest triangle is an equilateral triangle which is a rational triangle, but is not Heronian because the area is not rational relative to the square of a side. Next simplest triangle is an isoceles right triangle which has area one half the square of any leg, but the hypoteneuse is not rational relative to a leg as the Pytagoreans discovered, and so the triangle is not Heronian. The simplest example of a Heronian triangle is a right triangle with ratio of sides 3:4:5. In fact, any rational right triangle is a Heronian triangle. The area of any triangle is half the perimeter times radius of the inscribed circle. Thus, the inradius of a Heronian triangle is rational relative to the sides. A radius from the incenter to any side splits it into two segments of length the semiperimeter diminished by the non-adjacent side length. It follows that the tangent of any bisected angle of the triangle is a rational number and therefore the original angle is a Heronian angle. Thus, a Heronian triangle is the same as a triangle having three Heronian angles. However, it suffices to require two Heronian angles for the following reason. The area of any triangle is half of base times height. Thus, all the altitudes of a Heronian triangle are rational relative to the sides. Any altitude of a Heronian triangle splits the triangle into two Heronian right triangles. The original triangle is the sum or difference of the two right triangles depending on if the altitude is internal or external. Lattice Triangles There is a formula for the area of a triangle computed as a quadratic function of the coordinates of its vertices. This formula is one-half of a sum of three terms, each of which is a difference of cross-products of the x and y coordinates of two points (a two by two determinant) in circular order. Thus, a triangle with rational coordinates has rational area, and, if the sides are rational, then the triangle is Heronian. Conversely, a Heronian triangle's vertices can be given rational coordinates. This is proven by an analysis of the reduction to the case of integer sides which leads to the triangle realization as an integer lattice triangle. Literature Since the study of arithmetical properties of triangles goes back at least to Pythagoras the published literature is ancient and well worked. There is no universal terminology or notation so I have decided to use my own variation of commonly used terms. One useful source is a slim book Pythagorean Triangles by Waclaw Sierpinski. It was published in 1962 and Dover Publications reprinted it in 2003. Another source is chapter four of Mathematical Recreations by Maurice Kraitchik reprinted by Dover Publications in 1953. A recent journal article is "On basic Heronian triangles" by Kozhegeldinov in Mathematical Notes 55 (1994) pp. 151-156 [MR 95c:51016]. There are scattered articles in the American Mathematical Monthly. One article is in the February 1997 issue on "An Infinite Set of Heron Triangles with Two Rational Medians" by Ralph H. Buchholz and Randall L. Rathbun. A more recent article on the same subject is in the March 2024 issue of The Mathematical Intelligencer by Andrew N.W. Hone titled "Heron Triangles and the Hunt for Unicorns". He also gave a talk "An Infinite Sequence of Heron Triangles With Two Rational Medians" at Colorado State University which is available to view on YouTube. Links There is some discussion of rational triangles in the Geometry Junkyard. There is an entry for Heronian Triangle in MathWorld. The subject of triangles is extensive. Another angle on the subject is Clark Kimberling's Triangle Centers. Tables Here is my Heronian Triangle Table Here is my Pythagorean Triple Table Back to my home page Last Updated Feb 19 2025 Michael Somos michael.somos@gmail.com Michael Somos "
188033	https://discovery.ucl.ac.uk/id/eprint/10150082/1/HPOct22.pdf	Helly-type Problems Imre B´ ar´ any and Gil Kalai October 27, 2021 Abstract In this paper, we present a variety of problems in the interface between combinatorics and geometry around the theorems of Helly, Radon, Carath´ eodory, and Tverberg. Through these problems we describe the fascinating area of Helly-type theorems, and explain some of its main themes and goals. 1 Helly, Carath´ eodory, and Radon theorems In this paper, we present a variety of problems in the interface between com-binatorics and geometry around the theorems of Helly, Radon, Carath´ eodory, and Tverberg. Helly’s theorem [Hel23] asserts that for a family {K1, K2, . . . , Kn} of con-vex sets in Rd where n ≥d + 1, if every d + 1 of the sets have a point in common, then all of the sets have a point in common. The closely related Carath´ eodory theorem [Car07] states that for S ⊂Rd, if x ∈conv S, then x ∈conv R for some R ⊂S, \|R\| ≤d + 1. The more general colorful Carath´ eodory theorem [B´ ar82] says the follow-ing. Let S1, S2, . . . , Sd+1 be d + 1 sets (or colors if you wish) in Rd. Suppose that x ∈Td+1 i=1 conv Si. Then there is a transversal T = {x1, . . . , xd+1} of the system S1, . . . , Sd+1, meaning that x1 ∈S1, x2 ∈S2, . . . , xd+1 ∈Sd+1 such that x ∈conv T. A transversal is also called a rainbow set when S1, . . . , Sd+1 are considered as colors. The uncolored version, that is, when S1 = S2 = . . . = Sd+1, is the classic result of Carath´ eodory. There is a closely related colorful version of Helly’s theorem due to Lov´ asz that ap-peared in [B´ ar82]. 1 Tverberg’s theorem [Tve66] states the following: Let x1, x2, . . . , xm be points in Rd with m ≥(r−1)(d+1)+1. Then there is a partition S1, S2, . . . , Sr of {1, 2, . . . , m} such that Tr j=1 conv {xi : i ∈Sj} ̸= ∅. This was a conjecture by Birch who also proved the planar case in a slightly diﬀerent form. The bound of (r −1)(d+1)+1 in the theorem is sharp as can easily be seen from the conﬁguration of points in a suﬃciently general position. The case r = 2 is Radon’s theorem [Rad21], another classic from 1921, which was used by Radon to prove Helly’s theorem. Helly’s original proof (published later) was based on a separation argument. Sarkaria [Sar92] gave a simple proof of Tverberg’s theorem based on the colorful Carath´ eodory theorem. This paper describes the fascinating area of Helly-type theorems, and ex-plains some of its main themes and goals through a large and colorful bouquet of problems and conjectures. Some of these problems are very precise and clear-cut, for instance, Sierksma’s conjecture (Conjecture 4.1), the cascade conjecture (Conjecture 5.1), and Problem 3.2 about volumes of intersections. Some of them are rather vague, for instance, Problem 2.1 about intersection patterns of Euclidean convex sets, and Problem 3.9 about the mutual po-sition of convex sets, and Problem 5.5 about topological conditions for the existence of Tverberg partitions. We hope to see the answers to many of the questions presented here in the near future. Often, results from convexity give a simple and strong manifestation of theorems from topology: Helly’s theorem manifests the nerve theorem from algebraic topology, and Radon’s theorem can be regarded as an early “linear” version of the Borsuk–Ulam theorem. One of our main themes is to further explore these connections to topology. Helly-type theorems also oﬀer complex and profound combi-natorial connections and applications that represent a second theme of this paper. For a wider perspective and many other problems we refer the reader to survey papers by Danzer–Gr¨ unbaum–Klee [DGK63], Eckhoﬀ[Eck79] and [Eck93], Tancer [Tan13], De Loera–Goaoc–Meunier–Mustafa [DLGMM19], and the forthcoming book of B´ ar´ any [B´ ar21a]. Here is a quick summary of the paper. Section 2 deﬁnes the nerve that records the intersection pattern of convex sets in Rd, describes some of its combinatorial and topological properties, and considers various extensions of Helly’s theorem, such as the fractional Helly theorem, which asserts that if a fraction α of all sets in a family of convex sets have a non-empty intersection, then there is a point that belongs to a fraction β(α, d) of the sets in the 2 family. Section 3 considers various reﬁnements and generalizations of Helly theorems such as the study of dimensions of intersections of convex sets, and the study of Helly-type theorems for unions of convex sets. Section 4 presents various extensions and reﬁnements of Tverberg’s theorem, starting with Sierksma’s conjecture on the number of Tverberg partitions. Section 5 studies the cascade conjecture about the dimensions of the Tverberg points and considers several connections with graph theory including a speculative connection with the four-color theorem. Section 6 deals with other Tverberg-type problems. Section 7 brings problems related to Carath´ eodory theorem and weak-epsilon nets, and Section 8 gives a glance at common transversals; rather than piercing a family of sets by a single point or a few points we want to stab them with a single or a few j-dimensional aﬃne spaces. Final conclusions are drawn in the last section. 2 Around Helly’s theorem 2.1 Nerves, representability, and collapsibility We start this section with the following basic deﬁnition: for a ﬁnite collection of sets F = {K1, K2, . . . , Kn}, the nerve of F is a simplicial complex deﬁned by N(F) = {S ⊂[n] : \ i∈S Ki ̸= ∅}. Helly’s theorem can be seen as a statement about nerves of convex sets in Rd, and nerves come to play in many extensions and reﬁnements of Helly’s theorem. A missing face S of a simplicial complex K is a set of vertices of K that is not a face but every proper subset of S is a face. Helly’s theorem asserts that a d-representable complex does not have a missing face with more than d + 1 vertices. A simplicial complex is d-representable if it is the nerve of a family of convex sets in Rd. Problem 2.1 Explore d-representable simplicial complexes. We refer the reader to the survey on d-representable complexes by Tancer [Tan13]. 3 Let K be a simplicial complex. A face F ∈K is free if it is contained in a unique maximal face. An elementary d-collapse step is the removal from K of a free face F with at most d vertices and all faces containing F. A simplicial complex is d-collapsible if it can be reduced to the empty complex by a sequence of elementary d-collapse steps. Wegner proved [Weg75] that every d-representable complex is d-collapsible. The converse does not hold even for d = 1: 1-representable complexes are the clique complex of interval graphs and 1-collapsible complexes are the clique complexes of chordal graphs. Here, a clique complex of a graph G is a simplicial complex whose faces correspond to the sets of vertices of complete subgraphs of G. Chordal graphs are graphs with no induced cycles of length greater than three. Intersection patterns of intervals (which are the same as 1-representable complexes) were completely characterized by Lekkerkerker and Boland [LB62]. They proved that interval graphs are characterized by being chordal graphs with the ad-ditional property that among every three vertices, one is a vertex or adjacent to a vertex in any path between the other two. They also described interval graphs in terms of a list of forbidden induced subgraphs. 2.2 The upper bound theorem For a ﬁnite collection of sets F = {K1, K2, . . . , Kn}, n ≥d + 1, in Rd, let N = N(F) be the nerve of F. We put fk(N) = \|{S ∈N : \|S\| = k + 1}\|. (The vector (f0(N), f1(N), . . . ) is called the f-vector of N, and is sometimes referred to also as the f-vector of F and written as f(F).) Helly’s theorem states that if fn−1(N) = 0 then fd(N) < n d+1 , or , with the f(F) notation, fn−1(F) = 0 implies fd(F) < n d+1 A far-reaching extension of Helly’s theorem was conjectured by Katchalski and Perles and proved by Kalai [Kal84b] and Eckhoﬀ[Eck85]. Theorem 2.1 (upper bound theorem) Let F be a family of n convex sets in Rd, and suppose that every d + r + 1 members of F have an empty intersection. Then, for k = d, . . . , d + r −1, fk(N(F)) ≤ d+r−1 X j=k j −d k −d n −j + d −1 d . The theorem provides best upper bounds for fd(F), . . . , fd+r−1(F) in terms of f0(F) provided fd+r(F) = 0. The proofs rely on d-collapsibility. 4 There is a simple case of equality: the family consists of r copies of Rd and n −r hyperplanes in general position. Theorem 2.1 is closely related to the upper bound theorem for convex polytopes of Peter McMullen [McM70]. In fact, a common proof was given by Alon and Kalai in [AK95]. Problem 2.2 Study cases of equality for the upper bound theorem. A place to start would be to understand 2-representable complexes K with f3(K) = 0 and f2(K) = n−1 2 . Theorem 2.1 implies the sharp version of the fractional Helly theorem of Katchalski and Liu [KL79]. The sharp version is due to Kalai [Kal84b]. Theorem 2.2 Let K be a d-representable complex. If fd(K) ≥α n d+1 , then dim(K) ≥βn, where β = β(d, α) = 1 −(1 −α)1/d+1. In other words, if F = {K1, . . . , Kn} is a family of convex sets in Rd (n ≥d + 1) and at least α n d+1 of the d + 1 tuples in F intersect, then F contains an intersecting subfamily of size βn. This is a result of central importance around Helly’s theorem. The existence of β(α) is referred to as the fractional Helly property, and if β →1 when α →1 this is referred to as the strong fractional Helly property. We note that a complete characterization of f-vectors of d-representable complexes was conjectured by Eckhoﬀand proved by Kalai [Kal84a, Kal86]. 2.3 Helly numbers and Helly orders It is useful to consider the following abstract notions of Helly numbers and Helly orders. Let F be a family of sets. The Helly number h(F) of F is the minimal positive integer h such that if a ﬁnite subfamily K ⊂F satisﬁes T K′ ̸= ∅for all K′ ⊂K of cardinality ≤h, then T K ̸= ∅. The Helly order ho(F) of F is the minimal positive integer h such that if a ﬁnite subfamily K ⊂F satisﬁes (1) every ﬁnite intersection of sets in K belongs to F and (2) T K′ ̸= ∅for all K′ ⊂K of cardinality ≤h, then T K ̸= ∅. Of course, when we consider families of sets closed under intersection, the Helly number and the Helly order coincide. So, for example, the topological Helly theorem, to be mentioned next in Section 2.4, asserts that the Helly order of topologically trivial sets in Rd is d + 1, and Amenta’s 5 theorem (Section 3.3 below) asserts that the family of unions of k pairwise disjoint convex sets in Rd has the Helly order k(d + 1). Let F be a family of sets. The fractional Helly number g(F) of F is the minimal positive integer g such that there is a function f(α) > 0, deﬁned for α > 0, with the following property: for every family K ⊂F of cardinality n, if at least α n g of the g-tuples in F intersect, then F contains an intersecting subfamily of size f(α)n. 2.4 Topological Helly theorem and Leray complexes Helly himself proved a topological version of his theorem [Hel30]. A good cover is a family of compact subsets of Rd such that every intersection of sets in the family is either empty or topologically trivial. (By “topologically triv-ial” we mean “contractible,” but it is suﬃcient to assume that all homology groups vanish.) Theorem 2.3 (topological Helly) If in a good cover of n subsets of Rd, n ≥d+1, every intersection of d+1 sets is non-empty, then the intersection of all the sets in the family is non-empty. A simplicial complex K is d-Leray if Hi(K′) = 0 for every induced sub-complex K′ of K and for every i ≥d. The well-known nerve theorem from algebraic topology asserts that if K is a ﬁnite family of sets that form a good cover then the nerve of K is topologically equivalent to S K. (The notion of “topologically equivalent” corresponds to the notion of “topologically trivial” in the deﬁnition of good covers.) It follows from the homological version of the nerve theorem that d-representable complexes are d-Leray. It is also easy to see that d-collapsible complexes are d-Leray. Remark: The nerve theorem played an important role in algebraic topology in the ’40s and ’50s, e.g., in showing that the de Rham homol-ogy coincides with other notions of homology. Helly’s topological theorem is remarkable since it came earlier than these developments. In Section 4.2 we will mention that Radon’s theorem can be seen as an early incarnation of the Borsuk–Ulam theorem in topology. Topological extensions of Helly-type theorems are an important part of the theory. Often, such extensions are considerably more diﬃcult to prove, but in a few cases the topological proofs are the only known ones even for the geometric results. There are 6 also a few cases where natural topological extensions turned out to be in-correct. The survey paper [DLGMM19] of De Loera, Goaoc, Meunier, and Mustafa emphasizes connections with combinatorial theorems closely related to the Brouwer ﬁxed-point theorem, starting with the Sperner lemma and the Knaster–Kuratowski–Mazurkiewicz theorem. A general problem is the following. Problem 2.3 (i) Find ﬁner and ﬁner topological and combinatorial proper-ties of d-representable complexes. (ii) Extend Helly-type theorems to good covers, Leray complexes, and be-yond. (iii) Find weaker topological conditions that suﬃce for the topological Helly theorem to hold. There is much to say about part (ii) of Problem 2.3. In several cases the way to go about it is to extend properties of d-representable complexes to d-Leray complexes. We will come back to such extensions later but we note that the upper bound theorem (Theorem 2.1), as well as the full characterization of their f-vectors, extends to d-Leray complexes; [Kal02]. This is also closely related to Stanley’s characterization [Sta75] of f-vectors of Cohen–Macaulay complexes. Figure 1: TK5 Regarding part (i) of Problem 2.3, we ﬁrst note a very easy connection with embeddability: if G is a graph we denote by TG the graph where for every edge e, we add a new vertex ve that is adjacent to the endpoints of e, and remove e itself; see Figure 1. If G is not planar, e.g., when G = T5, then 7 TG is not 2-representable. (Note however that Kn itself is 2-representable for every n.) White [Whi21] deﬁned the class of d-Matouˇ sek simplicial complexes that are related to topological invariants for embeddability as follows. Let K be an abstract simplicial complex with vertices V (K) = [n] = {1, 2, . . . , n}. We deﬁne the dual simplicial complex K′, with vertices V (K′) = {J ∈K : J is inclusion maximal}, and faces K′ = {α ⊆V (K′) : T J∈α J ̸= ∅}. We say that K is d-Matouˇ sek, if the Z2-index of the space ˆ K = {(x, y) ∈\| \| K′ \| \| 2 : \ supp(x) ∩ \ supp(y) / ∈K} is less than d. Here supp(x) denotes the support of x in K′, which is the inclusion-minimal face of K′ containing x. It is straightforward to verify that K is d-representable iﬀthere exists a linear map f : K′ →Rd, such that for every set I ⊆V (K) not in K, we have T i∈I f(αi) = ∅, where αi = {J ∈V (K′) : i ∈J}. This implies the existence of a Z2-map from ˆ K to Sd−1; thus any d-representable complex is also d-Matouˇ sek. White proved that nerves of good covers in Rd are d-Matouˇ sek and also showed that being 1-Matouˇ sek is equivalent to being 1-representable. Regarding part (iii), Debrunner [Deb70] showed that for the statement of topological Helly it suﬃces to assume that the (reduced) homology of intersections of k sets in the family 1 ≤k ≤d + 1 vanishes at and below dimension d−k, and even more general conditions were found by Montejano [Mon14]. 2.5 Conditions for the fractional Helly property Problem 2.4 (i) Find geometric, topological, and combinatorial conditions that imply the fractional Helly property. (i) Find geometric, topological, and combinatorial conditions that imply the strong fractional Helly property. We will mention here two conjectures regarding the fractional Helly prop-erty and two related theorems. A class of simplicial complexes is hereditary if it is closed under induced subcomplexes. Recall that for a simplicial complex K, fi(K) is the number of i-faces of K and b(K) is the sum of (reduced) Betti numbers of K. In connection with the fractional Helly theorem, Kalai and Meshulam [Kal10] formulated the following conjecture. 8 Conjecture 2.5 (Kalai and Meshulam) Let C > 0 be a positive number. Let F be the hereditary family of simplicial complexes deﬁned by the property that for every simplicial complex K ∈F with n vertices, b(K) ≤Cnd. Then for every α > 0 there is β = β(d, C) > 0 such that K ∈F and fd(K) ≥α n d+1 imply dim(K) ≥βn. The conclusion of the conjecture is referred to as the fractional Helly prop-erty of degree d. Kalai and Meshulam further conjectured that the conclusion holds even if one replaces b(K) with \|χ(K)\|, where χ(K) is the Euler charac-teristic of K. When d = 0, this conjecture is about graphs and it was proved (in its strong version) in [CSSS20]; see also [SS20]. Conjecture 2.6 (Kalai and Meshulam) Let U be a family of sets in Rd. Suppose that for every intersection L of m members of U, b(L) ≤γmd+1. Then U satisﬁes the fractional Helly property of order d. In some special cases the fractional Helly property has been established. For instance, Matouˇ sek [Mat04] showed that families of sets with a bounded VC dimension in Rd satisfy the fractional Helly property of order d. Another case is the so-called convex lattice sets. These are sets of the form Zd ∩C where Zd is the lattice of integer points in Rd and C is a convex sets in Rd. A result of B´ ar´ any and Matouˇ sek [BM03] asserts that families of convex lattice sets in Rd satisfy the fractional Helly property of order d+1. In both of these theorems the fractional Helly number is considerably smaller than the Helly number. For example, let F be the family of all convex lattice sets in Rd. The Helly number of F, h(F), is equal to 2d, as shown by Doignon [Doi73], while the fractional Helly number is d + 1; [BM03]. Problem 2.7 Does the assertion of the Radon theorem imply the fractional Helly property? An aﬃrmative solution to one interpretation of this question was recently given by Holmsen and Lee [HL21], who showed that for abstract convexity spaces, the ﬁnite Radon number r implies that the fractional Helly number is bounded by some function m(r) of r. Problem 2.8 Estimate m(r). 9 Convex sets are sets of solutions of systems of linear inequalities, and we can consider systems of polynomial inequalities of higher degrees. Conjecture 2.9 The family Bd k of sets of solutions in Rd of polynomial in-equalities of degree ≤k has the fractional Helly property. It is known [Mot55] (and is an easy consequence of Helly’s theorem itself) that the class Ad k of sets in Rd of common zeroes of systems of polynomial inequalities of degree ≤k has the Helly number d+k k . And we can even ask if this formula gives the precise fractional Helly number for the class Bd k. We conclude this section by mentioning an interesting recent abstract no-tion of convexity described by Moran and Yehudayoﬀ[MY20], which seems relevant to various problems raised in this paper and, in particular, to Prob-lem 2.7. In this notion of abstract convexity, which we call MY-convexity, we assume that every “convex set” is the intersection of “halfspaces.” We assume further that the VC dimension of the class of halfspaces is at most D. The class Bd k is an example of an MY-convexity space where the hafspaces are the sets of solutions of a single polynomial inequality of degree k. Problem 2.10 Consider an MY-convexity space X where the VC dimension of the class of halfspaces is at most D. (i) Does X have the fractional Helly number f(D) for some function f of D? (ii) Does X have the fractional Helly number D? 2.6 The (p, q)-property The conclusion of Helly’s theorem is that the family is intersecting; i.e., there is a point Rd that is included in all sets in the family. Problem 2.11 What conditions guarantee that the family is t-pierceable, meaning that there are t points such that every set in the family contains at least one? In the language of nerves, what conditions guarantee that the set of ver-tices of the nerve can be expressed as the union of t faces? A family of sets has the (p, q)-property if for every p members of the family some q have a non-empty intersection. Note that here we assume p ≥q > d. (For nerves this says that every set of p vertices spans a face with q vertices, and this is closely related to Tur´ an’s problem for hypergraphs.) Hadwiger and Debrunner [HD57] introduced the (p, q)-property and proved 10 Theorem 2.4 If a ﬁnite family of convex sets in Rd has the (p, q)-property and (d −1)p < (q −1)d, then it is p −q + 1-pierceable. A family of sets has the (p, q)r property if in its nerve every p vertices span at least r faces with q vertices. This was introduced by Montejano and Sober´ on [MS11] and further studied by Keller and Smorodinsky [KS18]. Montejano and Sober´ on proved (among other results) Theorem 2.5 A family of convex sets in Rd with the (d+2, d+1)d property is 2-pierceable. Hadwiger and Debrunner [HD57] conjectured in 1957 and Alon and Kleit-man [AK92b, AK92a] proved the following important theorem. Theorem 2.6 ((p, q)-theorem) For all p ≥q > d there exists f(d, p, q) such that if a family of convex sets in Rd has the (p, q)-property, then it is f(d, p, q)-pierceable. The bound on f(d, p, q) given in [AK92b] is enormous. The ﬁrst open case is d = 2 and p = 4, q = 3. It is known that f(2, 4, 3) is between 3 and 9, the lower bound is from [KGT01], and the upper bound is a recent result of McGinnis [McG20] who brought down the upper bound of 13 of [KGT01] to 9. Substantial improvements for the general case were given by Keller, Smorodinsky, and Tardos [KST18] and by Keller and Smorodinsky [KS20]. Problem 2.12 Improve further the bounds on f(2, 4, 3) and, more generally, on f(d, p, q). Alon, Kalai, Matouˇ sek, Meshulam [AKMM02] proved the following result that implies that the Alon–Kleitman theorem extends to good covers and Leray complexes (but with worse bounds). Theorem 2.7 For every q > d + 1 there exists C(d, q) with the following property: let F be a hereditary class of simplicial complexes satisfying the fractional Helly property of degree d. If a simplicial complex K ∈F has the property that every q vertices span a d-dimensional face, then the vertices of K can be covered by C(d, q) faces. See Eckhoﬀ[Eck03] for a survey on (p, q)-theorems. 11 2.7 A Ramsey type question Conjecture 2.13 For integers d ≥1 and r > 1 + ⌈d/2⌉, there is α = α(d, r) > 0 such that the following holds: let F be a family of n convex sets in Rd. Then F contains nα(d,r) sets such that either every r has a point in common or no r has a point in common. There is a large literature on this and related questions starting with a theorem of Larman, Matouˇ sek, Pach, and T¨ or˝ ocsik [LMPT94] that proves the case d = 2 and r = 2. Subsequent works are [APP+05] and [FPT11]. When r = d + 1 this conjecture holds with α = 1/(d + 1). This was observed by Keller and Smorodinsky (private communication) and follows from their improved (p, q)-theorems. The general phenomenon here (with several inter-esting manifestations) is that graphs and hypergraphs arising in geometry satisfy much stronger forms of Ramsey’s theorem than arbitrary graphs and hypergraphs. 2.8 Colorful, fractional colorful, and matroidal Helly theorems The colorful Helly theorem of Lov´ asz (see [B´ ar82]) asserts the following. Assume that C1, . . . , Cd+1 are ﬁnite families of convex sets in Rd with the property that every transversal K1, . . . , Kd+1 is intersecting, then T Ci ̸= ∅ for some i ∈[d+1]. Here transversal means that Ki ∈Ci for every i ∈[d+1]. The colorful version implies the original one when C1 = . . . = Cd+1. The analogous colorful version of the fractional Helly theorem says that if an α fraction of all transversals of the system C1, . . . , Cd+1 is intersecting, then one of the families, say Ci, contains an intersecting subfamily of size β\|Ci\|. Here α > 0, of course, and β = β(d, α) has to be positive. Such a theorem (with β = α/(d + 1)) was proved and used ﬁrst in [ABB+09]. The dependence of β was improved by Kim [Kim17], who showed in particular that β →1 as α →1. The optimal dependence of β on α and d is a recent result of Bulavka, Goodarzi, and Tancer [BGT20]. They use Kalai’s algebraic shifting technique [Kal84b] and raise the following interesting conjecture. Conjecture 2.14 Let K be a d-Leray simplicial complex whose vertex set V is partitioned into sets V1, . . . , Vd+1, called colors, and \|Vi\| = ni for i ∈[d+1]. Assume that K contains at least α Qd+1 1 ni colorful d-faces for some α > 0. 12 Then there is i ∈[d + 1] such that the dimension of the restriction of K to Vi is at least (1 −(1 −α)1/(d+1))ni −1. Kalai and Meshulam [KM05] extended the assertion of the colorful Helly theorem to the topological setting and also considered a matroidal version. A matroidal complex is the complex consisting of the independent sets of a matroid. Equivalently, M is a matroidal complex if and only if every induced subcomplex is pure, i.e., if all its maximal faces have the same cardinality. Theorem 2.8 Let X be a d-Leray complex on the vertex set V . Suppose that M is a matroidal complex on the same vertex set V with rank function ρ. If M ⊂X, then there exists τ ∈X with ρ(V \τ) ≤d. This theorem gives the colorful Helly property when M is a transversal matroid and it suggests a general way to extend results about colorings. We will encounter this idea in Section 4.3 below where we try to move from col-orful versions of Tverberg’s theorem to matroidal versions. Theorem 2.8 has interesting connections with advances in topological combinatorics related to Hall’s marriage theorem and “rainbow” matchings; see [AB06] and [AB09]. 3 More around Helly’s theorem 3.1 Dimensions of intersections: Katchalski’s theorems Let g(d, k) be the smallest integer with the following property: for every family of n convex sets in Rd, n ≥g(d, k), such that the dimension of in-tersection of every g(d, k) sets in the family is at least k, the dimension of intersection of all members of the family is at least k. Helly’s theorem as-serts that g(d, 0) = d + 1. In 1971 Katchalski [Kat71] proved the following interesting result. Theorem 3.1 g(d, 0) = d+1, g(d, k) = max{d+1, 2(d−k+1)} if 1 ≤k ≤d. Given a family K = {K1, K2, . . . , Kn} of convex sets in Rd and J ⊂[n], set K(J) = T j∈J Kj and write d(J) = dim K(J). A further remarkable result of Katchalski [Kat78] “reconstructs” the dimension of the intersection: Theorem 3.2 Let K = {K1, K2, . . . , Kn} and K′ = {K′ 1, K′ 2, . . . , K′ n} be two families of compact convex sets in Rd. If dK(J) = dK′(J) for every J, \|J\| ≤d + 1, then dK(J) = dK′(J) for every J. 13 Katchalski actually proved a stronger statement, namely, that the condi-tion dK(J) = dK′(J) for every J with (d + 1) −⌊d/2⌋≤\|J\| ≤d + 1 suﬃces for the conclusion of Theorem 3.2. More generally he proved that for every r ≥1 if dK(J) = dK′(J) for every J with (d + r) −⌊d/(r + 1)⌋≤\|J\| ≤d + r, then dK(J) = dK′(J) for every J. Deﬁne the D-nerve of a ﬁnite set of convex sets as its nerve K where every face S ∈K is labeled by the dimension of T i∈S Ki. We can regard the D-nerve as a nested collection of simplicial complexes that correspond to intersections of dimension ≥j. Problem 3.1 Explore combinatorial and topological properties of D-nerves of families of compact convex sets in Rd. 3.2 Helly with volume Theorems about volumes of intersections are closely related to theorems about dimensions of intersections. The natural question is, given a ﬁnite family F of convex sets in Rd, what condition guarantees that the intersec-tion T F not only is non-empty but also has volume at least one, say. The ﬁrst result in this direction is in [BKP82] of B´ ar´ any, Katchalski, and Pach. Theorem 3.3 (Helly with volume) Assume that F is a ﬁnite family of convex sets in Rd, \|F\| ≥2d, such that the intersection of any 2d sets from F has volume at least one. Then vol(T F) ≥d−2d2. The example of the 2d halfspaces in Rd whose intersection is the unit cube shows that the number 2d is the best possible in this theorem. In other words, 2d is the Helly number for volumes. However, the bound d−2d2 is not sharp and was improved ﬁrst by Nasz´ odi [Nas16] to (cd)−2d and later by Brazitikos [Bra17] to (cd)−1.5d. In both estimates, c > 0 is a universal constant. The following conjecture is still open. Problem 3.2 Show that under the conditions of Theorem 3.3 vol(T F) ≥ (cd)−d/2 where c > 0 is a constant. A similar result was established in [BKP82] for the diameter of the in-tersection. The Helly number is again 2d. So if the intersection of any 2d sets from the family F has diameter at least one, then diam F ≥cd−d/2. 14 This lower bound was improved in a series of recent papers: ﬁrst by Braz-itikos [Bra17] to cd−11/2, then by Ivanov and Nasz´ odi [IN21] to (2d)−3, and most recently by Almendra-Hern´ andez, Ambrus, and Kendall [AHAK21] to (2d)−2. This leads to the next problem. Problem 3.3 Assume that F is a ﬁnite family of convex sets in Rd, \|F\| ≥ 2d, such that the intersection of any 2d sets from F has diameter at least one. Then diam T F ≥cd−1/2. Recently, several further quantitative Helly-type results have appeared; see for instance [DFN21] and [DS21]. 3.3 Unions of convex sets: Around the Gr¨ unbaum– Motzkin conjecture Nina Amenta [Ame94] proved a Helly-type result on unions of disjoint convex sets. Theorem 3.4 Let F be a family of sets in Rd such that every member in F is the union of k disjoint compact convex sets. Suppose further that every intersection of members of F is also a union of k disjoint convex sets. If every k(d + 1) sets in F has a point in common, then T F ̸= ∅. In the language of Section 2.3, Theorem 3.4 asserts that the Helly order of the family of disjoint unions of k compact convex sets in Rd is (d + 1)k. This was conjectured by Gr¨ unbaum and Motzkin [GM61] who proved the case k = 2; Larman [Lar68] proved their conjecture for k = 3 and Amenta in its full generality. It is easy to see that this family has no ﬁnite Helly number. Kalai and Meshulam [KM08] proved that Amenta’s theorem extends topo-logically. They consider the following setting. Let K and L be simplicial complexes with a map from V (K) to V (L) such that the inverse image of every i face in L is the union of at most k i-faces of L. If K is d-Leray, then the Leray number of L is at most dk + k −1. Eckhoﬀand Nischke [EN09] showed that Amenta’s theorem extends com-binatorially. In the setting of the previous paragraph they proved that if K has no missing face of size d + 1 or larger, then L has no missing face of size k(d + 1) or larger. 15 3.4 More on families of unions of convex sets We may consider sets in Rd that can be represented as unions of k convex sets but delete the disjointness assumption. In this case Alon and Kalai [AK95] and Matouˇ sek [Mat97] proved the following result. Theorem 3.5 Assume that F is a ﬁnite family of sets in Rd such that every member in F is the union of k compact convex sets. Then F has a ﬁnite Helly order. Let us mention a recent topological Helly-type theorem by Goaoc, Pat´ ak, Pat´ akov´ a, Tancer, and Wagner [GPP+17] that strengthens Theorem 3.5. Theorem 3.6 For every γ > 0 there is h(γ, d) with the following property: let U be a family of sets in Rd. Suppose that for every intersection L of some members of U and every i ≤⌈d/2⌉−1, we have bi(L) ≤γ. Then, if every h(γ, d) members of U have a point in common, then all sets in U have a point in common. We note that Theorem 3.6 implies Theorem 3.5. In fact, its proof relies on the method developed by Matouˇ sek in [Mat97]. His method, connecting topological obstructions for embeddability to Helly-type theorems, is the basis of White’s notion [Whi21] of d-Matouˇ sek complexes. In connection with this we mention the following curious question. Conjecture 3.4 The Helly order of families of unions of two disjoint non-empty sets in Rd is d + 1. This is known to be false if “two” is replaced by a large integer even when d = 2. We say that two compact sets intersect nicely if the long Meyer–Vietoris exact sequence splits into short exact sequences dimensionwise. Problem 3.5 Let K = {K1, K2, . . . , Kn} be a ﬁnite family of compact sets such that for every set of indices I ⊂[n], K(I) is topologically equivalent to a ﬁxed topological space Z, and for every two sets of indices I, J ⊂[n], K(I) and K(J) intersect nicely. Then S K is topologically equivalent to a ﬁber bundle over N(K) with ﬁbers topologically equivalent to Z. A positive answer to Problem 3.5 would imply Conjecture 3.4 because a pair of disjoint unions of non-empty convex sets whose intersection is also a disjoint union of non-empty convex sets always intersect nicely. 16 3.5 A conjecture by Gao, Landberg, and Schulman Here is an interesting Helly-type conjecture by Gao, Langberg, and Schul-man [GLS08]. For a convex set K in Rd an ϵ enlargement of K is K+ϵ(K−K) (where K −K = {x −y : x, y ∈K}). Conjecture 3.6 For every d, k, and ϵ there is some h = h(d, k, ϵ) with the following property. Let F be a family of unions of k convex sets. Let Fϵ be the family obtained by enlarging all the involved convex sets by ϵ. If every h members of F have a point in common, then all members of Fϵ have a point in common. Of course, for k = 1 we can take ϵ = 0 by Helly’s theorem. 3.6 Boxes and products Problem 3.7 Let d1, d2, . . . , dr be a partition of d. Study Helly-type theo-rems for families of Cartesian products K1 × K2 × . . . × Kr of convex sets where dim Ki = di. The case of standard boxes, namely when d1 = d2 = · · · = dd = 1 is of special interest. Standard boxes have Helly number 2, and therefore their nerves are determined by their graphs. Eckhoﬀproved an upper bound the-orem for standard boxes [Eck88], and studied the extremal families [Eck91]. It is easiest to describe the families where the upper bound is attained. If fd+r = 0 (that is, the largest non-empty intersection is for d + r sets), then the family consists of r copies of Rd and roughly the same number of parallel copies of each of the d coordinate’s hyperplanes. Let K be the nerve of a family of standard boxes in Rd. Then K is a d-Leray complex and has the further property that if S is a set of vertices such that every pair of vertices in S form an edge, then S is a face of K. This property of the nerve corresponds to Helly number 2 for the original family and we refer to it as Helly number 2. Problem 3.8 Extend Eckhoﬀ’s upper bound theorem to the class of d-Leray complexes with no missing faces of size greater than 3 (namely, those corre-sponding to Helly number 2). 17 3.7 Mutual position of convex sets The study of nerves of convex sets is the study of intersection patterns of families of convex sets. When we start with a family of convex sets in Rd we can go further and consider intersection patterns of the convex hulls of all subfamilies. (We can go even further by alternating between taking con-vex hulls and intersections and by considering statements regarding k-ﬂat transversals rather than plain intersections.) (a) (b) (c) (d) Figure 2: Mutual positions of three convex sets. Figure 2 shows various possible positions of three convex sets in the plane: (a) the convex hull of every two sets intersects the third set, (b) the convex hull of any two sets is disjoint from the third set, but all pairwise convex hulls have a point in common, (c) the three convex hulls of pairs of sets have no point in common, (d) the convex hull of two sets intersect the third set. Statements in this wider language can be regarded as the study of mu-tual positions of convex sets and they are, of course, of interest even for conﬁgurations of points, which we discuss in the next sections. Problem 3.9 Are there interesting things to say about the mutual position of convex sets? 18 3.8 Order types for points and sets To conclude this section and prepare for the next, we brieﬂy mention the notion of order types (aka oriented matroids). These objects arise from conﬁgurations of points (or of hyperplanes) in real vector spaces, and can also be associated with directed graphs. Consider a sequence Y = (y1, y2, . . . , yn) of n points in Rd that aﬃnely span Rd. The order type described by Y can be seen as the set of all minimal Radon partitions. There is a more general axiomatic deﬁnition of order types that roughly requires that the restriction to every d + 3 points be an order type of d + 3 points in a real space. For general order types there is a topological representation that replaces the linear description of order types that correspond to point conﬁgurations. Another equivalent way to describe the order type is as follows: for every set J of subscripts i1, . . . , id+1 with 1 ≤i1 < . . . < id+1 ≤n, let sg(J, Y ) be the sign of the determinant of the (d + 1) × (d + 1) matrix yi1 yi2 · · · yid+1 1 1 · · · 1 . (1) Two sequences Y = (y1, y2, . . . , yn) and Z = (z1, z2, . . . , zn) of n points in Rd are equivalent (or have the same order type) if sg(J, Y ) = sg(J, Z) for all J ⊂[n] of size d + 1. For more on oriented matroids see [BLVS+93]. Returning to families of convex sets we note that one way to record the mutual position of n convex sets K1, K2, . . . , Kn in Rd is by listing all order types of sequences y1 ∈K1, y2 ∈K2, . . . , yn ∈Kn. Goodman and Pollack’s notion of allowable sequences for conﬁgurations [GP85] is a very useful way to study order types of planar conﬁgurations. The more general notion of interval sequences by Dhandapani, Goodman, Holmsen, and Pollack gives a way to record mutual positions of n convex planar sets [DGHP05]. 4 Around Tverberg’s theorem 4.1 Sierksma’s conjecture Conjecture 4.1 (Sierksma’s) The number of Tverberg r-partitions of a set of (r −1)(d + 1) + 1 points in Rd is at least ((r −1)!)d. 19 This question was raised by Sierksma [Sie79] and not much progress has been achieved since. The best lower bound is about the square root of the conjectured one. This is a result of [Vˇ Z93] and [Hel07]. The conjecture, if true, is sharp, as shown by the example in Figure 3 for d = 2, r = 4: the vertices of the 3 triangles plus the point in the center is a set with 10 points and 3!2 Tverberg partitions. Figure 3: 10 points with 3!2 Tverberg partitions. In Rd take analogously r −1 d-dimensional simplices with their center at the origin; their vertices together with the origin form a set of (r−1)(d+1)+1 points with (r −1)!d Tverberg partitions. There are further cases where equality holds, such as the one connected to the following problem raised by Perles. We need a deﬁnition: a Tverberg partition S1, . . . , Sr of an m-element set X ⊂Rd is of type (a1, a2, . . . , ar) if the multisets {a1, a2, . . . , ar} and {\|S1\|, \|S2\|, . . . , \|Sr\|} coincide. Problem 4.2 Suppose that a1, a2, . . . , ar is a partition of m = (r −1)(d + 1) + 1 such that 1 ≤ai ≤d + 1 for every i. Is there a conﬁguration of m points in Rd for which all of Tverberg partitions are of type (a1, a2, . . . , ar)? This problem was raised by Perles many years ago and a positive answer was given by White [Whi17]. White’s examples provide a rich family of examples for cases of equality in Sierksma’s conjecture. An even more general family of constructions for the equality cases, based on staircase convexity, is in the paper of Bukh, Loh, and Nivasch [BLN17]. A similar construction was given by P´ or [P´ or18] in connection with the so-called universal Tverberg partitions. Problem 4.3 Explore further examples of equality cases in Sierksma’s con-jecture. 20 4.2 Topological Tverberg Conjecture 4.4 (topological Tverberg conjecture) Let f be a continu-ous function from the m-dimensional simplex σm to Rd. If m ≥(d+1)(r−1) then there are r pairwise disjoint faces of σm whose images have a point in common. If f is a linear function, this conjecture reduces to Tverberg’s theo-rem. The case r = 2 was proved by Bajm´ oczy and B´ ar´ any [BB79] using the Borsuk–Ulam theorem. Moreover, for r = 2, one can replace the sim-plex by any other polytope of the same dimension. The case where r is a prime number was proved in an important paper of B´ ar´ any, Shlosman, and Sz˝ ucs [BSS81]. The prime power case was settled by ¨ Ozaydin, in an unpub-lished (yet available) paper [¨ Oza87]. For the prime power case, the proofs are quite diﬃcult and are based on computations of certain characteristic classes. In 2015 the topological Tverberg conjecture was disproved in a short note by Frick [Fri15]. This involves some early result on vanishing of topo-logical obstructions by ¨ Ozaydin, a theory developed by Mabillard and Wag-ner [MW14] extending Whitney’s trick to k-fold intersections, and a fruit-ful reduction by Gromov [Gro10], rediscovered and extended by Blagojevi´ c, Frick, and Ziegler [BFZ19]. Conjecture 4.5 Let f be a linear function from an m-dimensional polytope P to Rd. If m ≥(d + 1)(r −1), then there are r pairwise disjoint faces of P whose images have a point in common. Problem 4.6 Does the conclusion of the topological Tverberg conjecture hold if the images of the faces under f form a “good cover” (that is if all those images and all their non-empty intersections are contractible)? 4.3 Colorful Tverberg Let C1, . . . , Cd+1 be disjoint subsets of Rd, called colors, each of cardinality at least t. A (d + 1)-subset S of Rd is said to be multicolored (or rainbow) if \|S ∩Ci\| = 1 for i = 1, . . . , d + 1. Let r be an integer, and let T(r, d) denote the smallest value t such that for every collection of colors C1, . . . , Cd+1 of size at least t each there exist r disjoint multicolored sets S1, . . . , Sr such that 21 Tr i=1 conv Si ̸= ∅. The question of ﬁniteness of T(r, d) was raised in [BFL90] and proved there for the case d = 2. The general case was solved by an important theorem of ˇ Zivaljevi´ c and Vre´ cica [Vˇ Z92]. It asserts that T(r, d) ≤2r −1 if r is a prime, which implies that T(r, d) ≤4r −1 for all r and d. This theorem is one of the highlights of discrete geometry and topological combinatorics. The only known proofs of this theorem rely on topological arguments although the statement is about convex hulls, partitions, and linear algebra. The following question is a challenge for convex geometers. Problem 4.7 Find a non-topological proof of the ﬁniteness of T(r, d). B´ ar´ any and Larman [BL92] showed that T(r, 2) = r and asked the fol-lowing. Conjecture 4.8 (colorful Tverberg conjecture) T(r, d) = r. The case where r +1 is a prime was proved by Blagojevi´ c, Matschke, and Ziegler [BMZ15]. It is a neat result of Lov´ asz that appeared in [BL92] that T(2, d) = 2 for all d. Sober´ on gives an equally neat (and very diﬀerent) proof of the same result in [Sob15]. The colorful Tverberg theorem is related to a well-known problem in discrete geometry, that of halving lines and hyperplanes. Given 2n points in general position in Rd, a halving hyperplane is a hyperplane with n points on each side. Problem 4.9 What is the maximum number H(2n, d) of partitions of a set of 2n points in Rd into equal parts via halving hyperplanes? Equivalently, what is the minimum number of non-Radon partitions with parts of equal size? A well-known conjecture that is open even for d = 2 is that for a ﬁxed d, H(n, d) = nd−1+o(1). With the help of the colorful Tverberg theorem it was shown that H(n, d) = nd−ϵd, where ϵd is a positive constant depending on d. For d = 2 it is known that neC√nH(2, n) ≤O(n4/3). A matroid version of Tverberg’s theorem is the topic of [BKM17], which states the following. Assume that M is a matroid of rank d + 1. Let b(M) 22 denote the maximal number of disjoint bases in M. If f is a continuous map from the matroidal complex of M to Rd, then there exist t ≥1 4 p b(M) independent sets σ1, . . . , σt ∈M such that Tt 1 f(σi) ̸= ∅. It is not clear how good this lower estimate on t is. Conjecture 4.10 In the above theorem, p b(M) could be replaced by cb(M) for some absolute positive constant c. 5 The cascade conjecture and more When we have r < d + 2 points in Rd they have a Radon partition iﬀthey are aﬃnely dependent. Are there conditions that guarantee that the exis-tence of Tverberg partitions below the Tverberg number? In this section we will discuss the dimension of Tverberg points and the quest for conditions guaranteeing the existence of Tverberg partitions for conﬁgurations of points below the Tverberg number. 5.1 The cascade conjecture For a set A, denote by Tr(A) the set of points in Rd that belong to the convex hull of r pairwise disjoint subsets of A. We call these points Tverberg points of order r. Let ¯ tr(A) = 1 + dim Tr(A). (Note that dim ∅= −1.) Radon’s theorem can be stated as follows: if ¯ t1(A) < \|A\| then T2(A) ̸= ∅. A similar statement which is still open is: if ¯ t1(A) + ¯ t2(A) < \|A\| then T3(A) ̸= ∅. We can go one step further: if ¯ t1(A)+¯ t2(A)+¯ t3(A) < \|A\| then T4(A) ̸= ∅. These statements are special cases of Conjecture 5.1 (cascade conjecture) For every A ⊂Rd, X r≥1 ¯ tr(A) ≥\|A\|. This is a question of Kalai from 1974 [Kal95]; see also [Kal00]. The conjecture was proved for d ≤2 by Akiva Kadari (unpublished MSc thesis in Hebrew). While this conjecture is wide open we can ask for topological extensions of various kinds and for more general topological conditions for conﬁgurations of cardinality below the Tverberg number (r −1)(d + 1) + 1, that imply the existence of a Tverberg partition into r parts; see Problem 5.4. 23 5.2 Reay’s dimension conjecture The following is a 1979 question from Reay [Rea79] where general position means weak general position; that is, no d + 1 points lie in a hyperplane. Conjecture 5.2 (Reay’s conjecture) If A is a set of (d+1)(r−1)+1+k points in general position in Rd, then dim Tr(A) ≥k. In particular, Reay’s conjecture asserts that a set of (d + 1)r points in general position in Rd can be partitioned into r sets of size d + 1 such that the simplices described by these sets have an interior point in common. This is easy when the points are in very general position, for instance, when they are algebraically independent. The main diﬃculty is how to use the weak general position condition. A recent result of Frick and Sober´ on [FS20] (see Section 7.1) is perhaps relevant here. While the conclusion of the cascade conjecture seems stronger than that of Reay’s dimension conjecture, it is not known how to derive it from the cascade conjecture. 5.3 Special cases of the cascade conjecture and ex-pressing a directed graph as union of two trees A special case of the cascade conjecture asserts that given 2d+2 points in Rd, you can either partition them into two simplices whose interiors intersect, or you can ﬁnd a Tverberg partition into three parts. A reformulation based on positive hulls is: given 2d non-zero vectors in Rd such that the origin is a vertex of the cone spanned by them, it is the case that either: • We can divide the points into two sets A and B so that the cones spanned by them have a d-dimensional intersection, or • We can divide them into three sets A, B, and C so that the cones spanned by them have a non-trivial intersection. Another interesting reformulation is obtained when we dualize using the Gale transform, and this has led to the problem we consider next: a very special class of conﬁgurations arising from graphs. Start from a directed graph G with n vertices and 2n −2 edges and associate with each directed edge (i, j) the vector ej −ei. This leads to the following problem. 24 Problem 5.3 Let G be a directed graph with n vertices and 2n −2 edges. When can we divide the set of edges into two trees T1 and T2 (we disregard the orientation of edges) so that when we reverse the directions of all edges in T2 we get a strongly connected digraph? One of us (Kalai) conjectured that if G can be written as the union of two trees, the only additional obstruction is that there is a cut consisting only of two edges in reversed directions. Chudnovsky and Seymour found an additional necessary condition: there is no induced cycle v1, vk, . . . , v2k, v1 in G, such that each vertex ci is cubic, the edges of the cycle alternate in direction, and none of the vertices v1, . . . , v2k are sources or sinks of G. 5.4 Tverberg partitions of order three for conﬁgura-tions below the Tverberg number Problem 5.4 When n < 2d + 3, ﬁnd conditions for the set of Radon points and the set of Radon partitions of a set X of n points in Rd, that guarantee the existence of a Tverberg partition into three parts. The cascade conjecture asserts that if n = d + 2 + k and the dimension of Radon points is smaller than k, then there exists a Tverberg partition into three parts. While this is wide open, it would be interesting to propose a more general topological condition that suﬃces for the existence of a Tverberg partition into r parts. Conjecture 5.5 If the map from the Radon partitions of X to the Radon points of X is topologically degenerate (in some sense), then a Tverberg par-tition into three parts exists. In Problem 5.4 and Conjecture 5.5 we can relax the conclusion and can do so in various ways. For that we need a few deﬁnitions: the k-core of a ﬁnite set X in Rd is the intersection of the convex hull of all sets A ⊂X with \|X \ A\| ≤k, that is, corek X = \ {conv A : A ⊂X, \|X \ A\| ≤k}. The case k = 0 is the usual convex hull. The k-Radon core of a ﬁnite set X in Rd is the intersection of Radon points of all sets A ⊂X with \|X \ A\| ≤k; 25 this is the set of points in Rd that remain Radon points of X even after we delete k points from X in all possible ways. (Clearly, the Tverberg points of order three are in the ﬁrst Radon core, and the points in the ﬁrst Radon core are in the 2-core.) Problem 5.6 When n < 2d + 3, ﬁnd conditions for the set of Radon points and the set of Radon partitions of a set X of n points in Rd that guarantee (1) the second core of X, core2 X, is non-empty, (2) the ﬁrst Radon core of X is non-empty, (3) X admits a Reay (3,2)-partition, that is, a partition into three parts such that the convex hulls are pairwise intersecting; see Section 6.2. 5.5 Radon partitions and Radon points for conﬁgura-tions based on cubic graphs Let G be a cubic graph with n vertices {v1, v2, . . . , vn}. Associate with ev-ery edge {vi, vj} in G its characteristic vector in Rd, giving a conﬁguration Conf(G) of 3n/2 points in (n −1)-dimensional space. In [Onn01] and also in personal communication (2011), Onn observed that the existence of a Tverberg 3-partition (or even of a Reay (3,2)-partition; see Section 6.2) is equivalent to a 3-edge coloring for G, and concluded that deciding if a con-ﬁguration of 3(d + 1)/2 points in Rd (d an odd integer) admits a Tverberg partition into three part is NP-complete. The following problem is motivated by the four color theorem. Problem 5.7 (i) Study Radon partitions and Radon points for conﬁgura-tions based on cubic graphs. (ii) Find conditions for the Radon points and Radon partitions of Conf(G) that guarantee a 3-edge coloring for G. It would be interesting to ﬁnd conditions for Problems 5.4 and Conjec-ture 5.5 that would imply the 3-edge colorability of bipartite cubic graphs and, much more ambitiously, conditions that would imply the four-color the-orem, namely, the 3-edge colorability of planar cubic graphs. 26 6 More around Tverberg’s theorem 6.1 Eckhoﬀ’s partition conjecture Let X be a set endowed with an abstract closure operation X →cl(X). The only requirements of the closure operation are: (1) cl(cl(X)) = cl(X) and (2) A ⊂B implies cl(A) ⊂cl(B). Deﬁne tr(X) to be the largest size of a (multi)set in X that cannot be par-titioned into r parts whose closures have a point in common. The following conjecture is due to Eckhoﬀ[Eck00]. Conjecture 6.1 (Eckhoﬀ’s partition conjecture) For every closure op-eration, tr ≤t2 · (r −1). If X is the set of subsets of Rd and cl(A) is the convex hull operation, then Radon’s theorem asserts that t2(X) = d + 1 and Eckhoﬀ’s partition conjecture implies Tverberg’s theorem. In 2010 Eckhoﬀ’s partition conjec-ture was refuted by Boris Bukh [Buk10]. Bukh’s beautiful paper contains several important ideas and further results. We will mention one ingredient. Recall the nerve construction for moving from a family F of n convex sets to the simplicial complex that records empty and non-empty intersections for all subfamilies G of F. Bukh studied simplicial complexes whose vertex sets correspond to the power set of a set of size n: starting with n points in Rd or some abstract convexity space, consider the nerve of convex hulls of all 2n subsets of these points! In Bukh’s counterexample, tr = t2 · (r −1) + 1, which is just one larger than the conjectured bound. Perhaps tr ≤t2 · (r −1) + c for some universal constant c ≥1. There is a recent and positive development about Eckhoﬀ’s conjecture. P´ alv¨ olgyi [P´ al20] has proved that tr grows linearly in r, that is, tr ≤cr where the constant c depends only on r2. Problem 6.2 Find classes of closure operations for which tr ≤t2 · (r −1). We can ask if the inequality tr ≤t2 · (r −1) holds for Moran and Yehu-dayoﬀ’s convexity spaces considered in Section 2.5. 27 Bukh’s paper includes an interesting notion that extends the notion of nerves. Given a conﬁguration of points in the Euclidean space or in an abstract convexity space, we consider the nerve of convex hulls of all non-empty subsets of the points. This is a simplicial complex that we refer to as the B-nerve of the conﬁguration, with the additional structure that vertices are labeled by subsets, and with some additional combinatorial properties. Problem 6.3 Study properties of B-nerves of point conﬁgurations in Rd. 6.2 A conjecture by Reay For a set X ⊂Rd a Reay (r, j)-partition is a partition of X into subsets S1, S2, . . . , Sr such that Tj i=1 conv Ski ̸= ∅, for every 1 ≤k1 < · · · < kj ≤r. In other words, the convex hulls of any j sets of the partition intersect. Deﬁne R(d, r, j) as the smallest integer m such that every m-element set X ⊂Rd has a Reay (r, j)-partition. Reay [Rea79] conjectured that you cannot improve the value given by Tverberg’s theorem, namely, that Conjecture 6.4 (Reay’s conjecture) R(d, r, j) = (r −1)(d + 1) + 1. Micha A. Perles believes that Reay’s conjecture is false even for j = 2 and r = 3 for large dimensions, but with Moriah Sigron he proved [PS16] the strongest positive results in the direction of Reay’s conjecture. 6.3 Two old problems and universality Problem 6.5 (McMullen and Larman) How many points v(d) guaran-tee that for every set X of v(d) points in Rd there exists a partition into two parts X1 and X2 such that for every p ∈X, conv (X1\p) ∩conv (X2\p) ̸= ∅. This is a strong form of Radon’s theorem: the partition X1, X2 of X = X1 ∪X2 remains a Radon partition even after we delete any point from X. Similar questions can be asked about Tverberg partitions. Larman [Lar72] proved that v(d) ≤2d + 3 and this bound is sharp for d = 1, 2, 3, 4. The lower bound v(d) ≥⌈5d 3 ⌉+ 3 is a result of Ram´ ırez Alfons´ ın [RA01]. This problem is the dual form of the original question by McMullen: what is the 28 largest integer n = f(d) such that every set of n points in general position in Rd is projectively equivalent to the set of vertices of a convex polytope. A related problem is the following. Problem 6.6 How many points T(d; s, t) in Rd guarantee that they can be divided into two parts such that every union of s convex sets containing the ﬁrst part has a non-empty intersection with every union of t convex sets containing the second part. We explain next why R(d; s, t) is ﬁnite. This is a fairly general Ramsey-type argument and it gives us an opportunity to mention a few recent im-portant results. The argument has two parts: (1) Prove that T(d; s, t) is ﬁnite (with good estimates) when the points are in cyclic position (to be deﬁned shortly). (2) Use the fact that for every d and n there is f(d, n) such that among every m points in general position in Rd, m > f(d, n), one can ﬁnd n points in cyclic position. The ﬁniteness of T(d; r, s) follows (with horrible bounds) from these two ingredients by standard Ramsey-type results. It would be nice to understand the behavior of this function. Statement (2) is a kind of universality theorem. In a more precise form it says that for every d and n there is an integer f(d, n) such that the following holds. Every sequence x1, . . . , xm in Rd in general position with m ≥f(d, n) contains a subsequence y1, . . . , yn such that all simplices of this subsequence are oriented the same way. The latter condition says, in more precise form, that for every set of subscripts i1, . . . , id+1 with 1 ≤i1 < . . . < id+1 ≤n, the sign of the determinant of the (d + 1) × (d + 1)matrix yi1 yi2 · · · yid+1 1 1 · · · 1 (2) is the same (and diﬀerent from 0). Now a point set is cyclic if its elements can be ordered so that the simplices along this ordering have the same orientation. Statement (2) says that the property of being cyclic is universal because every long enough sequence of points in general position contains a cyclic subsequence of length n. Every ﬁnite sequence of points on the moment curve is cyclic. This shows that no other type of point sequence can be universal. Recently a fairly good understanding of f(d, n) has been achieved in a series of papers. 29 Theorem 6.1 f(d, n) = twrd(θ(n)). Here, twrd is the d-fold tower function. The lower bound is by Suk [Suk14] (improving earlier bounds by Conlon, Fox, Pach, Sudakov, and Suk [CFP+14]) and the upper bound comes from B´ ar´ any, Matouˇ sek, and P´ or [BMP16]. The following, somewhat vague, question emerges here naturally. Problem 6.7 Determine the universal type of n lines in R3 and in Rd. More generally, what is the universal type of n k-dimensional aﬃne ﬂats in Rd? Some preliminary results in this direction are the topic of a forthcoming paper by B´ ar´ any, Kalai, and P´ or [BKP21] We note that the order type of a sequence of points does not determine its Tverberg partitions. Problem 6.8 Develop a notion of order type based on Tverberg partitions into at most r parts, r ≥3. Here, Perles and Sigron’s work on strong general position [PS16], and P´ or’s universality theorem [P´ or18] could be relevant. 7 Carath´ eodory and weak ϵ-nets 7.1 Colorful Carath´ eodory and the Rota basis conjec-ture The following question was raised in Chow’s Polymath 12 [Cho17] dedi-cated to Rota’s basis conjecture. Consider d + 1 sets (or colors if you wish) C1, C2, · · · , Cd+1 of points in Rd. Assume that each \|Ci\| = d+1 and that the interior of each conv Ci contains the origin. Problem 7.1 (D. H. J. Polymath) Can we ﬁnd a partition of all points into d+1 rainbow parts such that the interior of the convex hulls of the parts have a point in common. (A rainbow set is a set containing one element from each Ci.) To see the connection, ﬁrst recall Rota’s basis conjecture. 30 Conjecture 7.2 (Rota’s basis conjecture) If B1, B2, . . . , Bn are disjoint bases in Rn (or even in an arbitrary matroid), then it is possible to ﬁnd n new disjoint bases C1, C2, . . . , Cn such that each Ci contains one element from every Bj. Note that Rota’s basis conjecture, can be stated (over R) as follows: Consider d+1 sets (or colors) C1, C2, · · · , Cd+1 of points in Rd. Assume that each \|Ci\| = d + 1 and that the interior of each conv Ci is nonempty. Then there exists a partition of all points into d + 1 rainbow parts such that the interior of the convex hulls of each part is non-empty. Returning to Conjecture 7.1, we note here that according to the colorful Carath´ eodory theorem there is a rainbow set whose convex hull contains the origin. Without the words “the interiors of” Problem 7.1 would be a special case of the colorful Tverberg conjecture (Section 4.3). A positive answer would be a strong variant of Reay’s conjecture (Section 5.2) on the dimension of Tverberg points, and, as explained before, also a strong form of Rota’s basis conjecture over the reals. A recent result of Frick and Sober´ on [FS20] is that a set of r(d+1) points in Rd can always be partitioned into r sets, each of size d + 1, such that the convex hulls of the parts have a point in common. This theorem is related to the uncolored case of Problem 7.1 but without the word “interior.” 7.2 The complexity of the colorful Carath´ eodory the-orem and of Tverberg partitions Problem 7.3 Consider d+1 sets C1, C2, · · · , Cd+1 of points in Rd. Assume that each \|Ci\| = d + 1 and that each conv Ci contains the origin. Is there a polynomial-time algorithm to ﬁnd a rainbow simplex containing the origin? An interesting result in this direction is due to Meunier et al. [MMSS17]. They show that the problem lies in the intersection of complexity classes PPAD and PLS. The same applies to the analogous question about Tverberg partitions: is there a polynomial-time algorithm to ﬁnd a Tverberg partition of an (r−1)(d+1)+1-element point set in Rd? There are very few geometric problems in both classes PPAD and PLS that are not known to be solvable in polynomial time. The results in [MMSS17] are the ﬁrst upper bound on the complexity of these problems. 31 7.3 Carath´ eodory-type theorem for cores Recall the deﬁnition of the k-core of a ﬁnite set X in Rd from Section 5.4. The Carath´ eodory number for the k-core is the smallest integer f(d, k) with the property that a ∈corekX (where X ⊂Rd) implies the existence of Y ⊂X such that a ∈corekY and \|Y \| ≤f(d, k). So f(d, 0) = d + 1 is just the Carath´ eodory theorem. B´ ar´ any and Perles [BP90] established the ﬁniteness of f(d, k) together with some other properties of this function, for instance, that f(d, 1) = max{2(d + 1), 1 + d + ⌊d2/4⌋}, and that f(2, k) = 3(k + 1). Several questions remain open; we mention only two of them. Problem 7.4 Determine f(d, 2) and f(3, k). 7.4 The covering number theorem Assume that X ⊂Rd is ﬁnite and \|X\| ≥d+1. A simplex of X is just conv Y where Y ⊂X and \|Y \| = d + 1. According to Carath´ eodory’s theorem every point in conv X is contained in a simplex of X; that is, conv X is covered by the simplices of X. Which point is covered maximally, and how many times is it covered? A famous result of Boros and F¨ uredi [BF84] says that in the planar case there is a point covered by 2 9 n 3 + O(n2) simplices (that is, triangles) of X, where n = \|X\|. This is a positive fraction of all triangles of X and the constant 2 9 is the best possible. In higher dimensions Tver-berg’s theorem and the colorful Carath´ eodory theorem imply (see [B´ ar82]) the following result. Theorem 7.1 (covering number) Assume X is a set of n ≥d + 1 points in Rd. Then there is a point covered by 1 (d+1)d n d+1 simplices of X. This is again a positive fraction of all simplices of X. Deﬁne bd as the supre-mum of all β > 0 such for that every set X of n ≥d + 1 points in Rd there is a point covered by β n d+1 simplices of X. So bd ≥(d+1)−d. In a remarkable paper, Gromov [Gro10] showed that bd ≥ 2d (d+1)!(d+1). Gromov’s theorem ap-plies to continuous maps from the boundary of an (n−1)-dimensional simplex to Rd. His estimate is an exponential improvement on the previous bounds. Both Gromov’s theorem and Pach’s theorem below play an important role in the emerging theory of high-dimensional expanders [FGL+12]. From the other direction Bukh, Matouˇ sek, and Nivasch [BMN11] give an example, based on the stretched grid, that shows bd ≤ (d+1)! (d+1)d+1. They conjecture that this is the right value of bd. 32 Conjecture 7.5 Show that bd = (d+1)! (d+1)d+1. More modestly, prove that bd is exponential in d. An interesting extension of the covering number theorem is the following result of Pach [Pac98]. Theorem 7.2 (Pach’s theorem) Assume that C1, . . . , Cd+1 are sets (col-ors, if you like) in Rd, each of size n. Then there is a point p ∈Rd and there are subsets Di ⊂Ci (for all i ∈[d + 1]), each of size at least c(d)n such that the convex hull of every transversal of the system D1, . . . , Dd+1 contains p. Here c(d) > 0 is a constant that depends only on d. This is a homogeneous version of the covering number theorem. It was conjectured in [BFL90], where case d = 2 was proved more generally even if the sets C1, C2, C3 need not have the same size. This raises the following question. Problem 7.6 Does Pach’s theorem remain true if the sets C1, . . . , Cd+1 have arbitrary sizes? We mention that Pach’s theorem does not have a topological extension, as shown in [BMNT18], and in [BH20] in a stonger form. 7.5 Weak ϵ-nets An important application of the covering number theorem is about weak ϵ-nets. Let ϵ > 0 be ﬁxed. Given a ﬁnite set X of n ≥d + 1 points, let C be the (ﬁnite) family of sets conv Y for all Y ⊂X with \|Y \| ≥ϵn. A set F ⊂Rd is called a weak ϵ-net for X if F ∩C ̸= ∅for every C ∈C. Theorem 7.3 (weak ϵ-net theorem) Under the above conditions, there is a weak ϵ-net F for X such that \|F\| ≤ cd ϵd+1, where cd > 0 is a constant. 33 The upper bound on the size of F is from [AK92b] and[ABFK92] and has been improved to O(ϵ−d), disregarding some logarithmic terms. The trivial lower bound on the size of F is 1 ϵ. Bukh, Matouˇ sek, and Nivasch [BMN11] give an example (based on the stretched grid or staircase convexity) where the size of the weak ϵ-net is at least of order 1 ϵ(log 1 ϵ)d−1. So the bounds on the size of a weak ϵ-net are far from each other, and the general belief is that the true behavior should be slightly superlinear in 1 ϵ. Problem 7.7 Find a better upper bound for the size of a weak ϵ-net. One remarkable improvement in this direction is a result of Rubin [Rub18] who showed that in the planar case there is always a weak ϵ-net of size of order 1 ϵ1.5+δ for any δ > 0. A more recent result of Rubin [Rub21] applies in any dimension d ≥2 and gives a weak ϵ-net of size of order 1 ϵd−1/2+δ for any δ > 0. Weak ϵ-nets can be deﬁned not only for points but for k-dimensional aﬃne ﬂats in Rd. We only state the question for lines in R3 and leave the rest of the cases to our imaginary reader. Let L be a set of n lines and C be a ﬁnite family of convex sets in R3. Assume that every C ∈C intersects an ϵ-fraction of the lines in L, that is, \|{L ∈L : C ∩L ̸= ∅}\| ≥ϵn for every C ∈C. Conjecture 7.8 (ϵ-net of lines) Under these conditions, there is a set of lines L∗whose size depends only on ϵ such that every C ∈C intersects some line in L∗. The set L∗can be thought of as a weak ϵ-net of lines for C. We will encountered this question again soon, in connection with Problem 8.1. 8 A glance at common transversals 8.1 Transversals for intersecting families A k-transversal of a family of convex sets in Rd is a k-dimensional aﬃne space that intersects every set in the family. Transversal theory deals with conditions that guarantee the existence of k-transversals. The case k = 0 is connected to Helly-type theorems, and there are some general results for 34 hyperplane transversals, namely, k = d −1, and very few general results for 0 < k < d −1 and, in particular, for line transversals in R3. The fascinating theory geometric transversals goes beyond the scope of this paper; for surveys see Goodman, Pollack, Wenger [GPW93], Wenger [Wen99], and [Hol13]. We will mention only a few problems where the conditions are in terms of the intersection pattern of the sets in the family. Problem 8.1 Assume that a family C of n convex sets in R3 satisﬁes the the property that any two sets in C intersect. Show that there is a line intersecting cn elements in C, where c > 0 is a universal constant. Partial results in this direction are given in [B´ ar21b]. Problem 8.1 is the ﬁrst, and so far most interesting, unsolved case of a series of problems of the same type. Namely, for what numbers k, r, d is it true that, given a family C of convex sets in Rd where every k tuple is intersecting, there is an r-ﬂat intersecting a positive fraction of the sets in C? Of course, the positive fraction should depend only on k, r, and d. An interesting example satisfying the conditions is when C consists of n lines in a two-dimensional plane in R3. Then, of course, every set in C is a line transversal for all sets in C. This example shows that degenerate cases are going to make the problem diﬃcult. Figure 4 is an example of ﬁve pairwise intersecting convex sets in R3 without a common line transversal. The ﬁve sets comprise three blue rectangles and two red triangles, all of whose vertices belong to two parallel planes H0 and H1. H0 H1 Figure 4: Five sets in R3 that pairwise intersect and have no line transversal. The question comes from a paper by Mart´ ınez, Rold´ an, and Rubin [MRR20] and is connected to the colorful Helly theorem. They also ask the slightly more general bipartite version of the question. 35 Problem 8.2 Assume F and G are ﬁnite families of convex sets in R3 with the property that A ∩B ̸= ∅for any two sets A ∈F and B ∈G. Show that there is a line intersecting c\|F\| elements of F or c\|G\| elements of G where c > 0 is again a universal constant. An example is two sets F and G of lines on a doubly ruled surface, which shows that degenerate cases may cause diﬃculties again. It is worth men-tioning that both questions are invariant under non-degenerate aﬃne trans-formation. We observe here that a positive answer to Conjecture 7.8 from the last section would imply that in Problem 8.1 there is a very ﬁnite set L of lines intersecting every element of C, where by “very ﬁnite” we mean that the size of L is bounded by 1000, say, or by some other absolute constant. 9 Conclusion This paper introduces the fascinating area of Helly-type theorems, and de-scribes some of its main themes and goals through a variety of open prob-lems. Often, results from convexity give a simple and strong manifestation of theorems from topology: Helly’s theorem manifests the nerve theorem from algebraic topology, and Radon’s theorem can be regarded as an early “lin-ear” appearance of the Borsuk–Ulam theorem. One of our main themes is to further explore these connections to topology. Helly-type theorems also oﬀer complex and profound combinatorial connections and applications that represent the second main theme of this paper. We note that Helly-type theorems and the interplay between convex geometry, combinatorics, and topology play an important role in the emerging theory of high-dimensional expanders. There are various related parts of this theory that we did not consider. We gave only a small taste of the theory of common transversals; we did not discuss the closely related theorems of Kirchberger and Krasnoselskiˇ i; and we did not consider the rich connections to metric geometry. For example, when you consider families of translates of a ﬁxed convex set the theory takes interesting and surprising turns, and has applications and connections, e.g., to the theory of Banach spaces. Acknowledgments. 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188034	https://www.youtube.com/watch?v=R5C9K3oedGQ	4.2 Linear Approximations and Differentials Ryan Melton 1590 subscribers 28 likes Description 3330 views Posted: 24 Jul 2018 OpenStax Calculus volume 1 Transcript: section 4.2 linear approximations and differentials we've seen how Tenjin twines allow us to approximate a function changing over time in previous sections specifically section 3.4 we dealt with this quite a bit here we continue this idea to approximate a function locally using linear functions the idea is this move sketch a small graph here if I have a function just go a generic cubic here five a generic function here and I find the tangent line let's just say right here at this point if I find the tangent line at this point if I have a value and x value that is close to that value a let's go ahead and label this a I have an x value that is close to that just call it B then the Y value the actual Y value for that function at that point f of B that's going to be very close very close to if I had used the tangent line to approximate that value there's some error given there's quite a bit of error however that might be close enough might be good enough for practical application so we consider a function f that is differentiable at a point x equals a that's the point I just really that is in blue the tangent line to the graph F at a is given by this equation that'll be the equation for this tangent line to the point a for X near a f of X my outputs are approximately approximately given by that tangent line we call this the linear approximation denoting that function l of x equals well our teens at the equation of our tangent line this function is also known as the linearization of f at x equals a so we want to use this linear approximation we're going to find this we'll use it to estimate the value of the square root of nine point one where the original function is the square root function and the original point is nine our X is very near that value our nine point one is very close to nine so this could work so what how we'll find L of X f of X is equal to f of in this case nine as our a value plus f prime of nine times X minus nine let's go ahead and find the derivative of that so f prime of X is 1 over 2 root X so then F prime of nine is going to be one-sixth square root of nine is three times two pro nine is one-sixth now F of nine there's one thing we need F of 9 which that would be the square root of 9 which is 3 so filling these things in L of X is equal to 3 plus 1/6 times X minus 9 that is the equation that fit that is the tangent line next we want to find so this is the first thing you're asked to do find the linear approximation and we could of course rewrite that in standard form or in slope intercept form now to find L of 9.1 so this would be the approximation of the square root of nine point one three plus one-sixth times nine point one minus nine which is three point zero one six repeating now to consider how close this is what the error is just want to make a note over here the square root of nine point one is approximately three point zero one six six two so the error is actually quite minimal using that this always will be true as long as our value of x is very very close to a you get too far away in your X values and your Y values change tremendously as long as it's a smooth function differentiable which implies continuous then those values will be very close find the linear approximation of f of X equals sine X and x equals PI thirds and use it to approximate sine of 62 degrees I would like to point out here that 62 degrees is fairly close to PI thirds PI thirds of 60 degrees when that conversion so that is true now I do want to use the Radian measure of this so this is going to be sine of an if we use our conversion factor of pi equals 180 degrees 31 ninetieth pi so we're trying to approximate so L of x equals F of our a value is PI thirds plus F prime of PI thirds times X minus PI thirds well f prime of X is cosine X so the F prime of PI thirds is 1/2 and of course we need F of PI thirds that'll be sine sine of PI thirds which is root 3 over 2 so that our linear approximation L of X is root 3 over 2 plus 1/2 times X minus PI thirds there is my function now to approximate this I want L of 31 ninetieths Phi which is root 3 over 2 pi minus PI thirds which is approximately 8 8 3 4 8 again for my notes over here sine of 31 PI over 90 is zero point eight eight two nine four eight which again is fairly close to the value we had that's enough however my actual approximated value 0.88 3 for 8 now for number 3 find the linear approximation of f of x equals 1 plus x quantity to the n at x equals 0 use this approximation to estimate one point zero one wanted it cute first we need to find F prime F prime of X using our power rule 9 times 1 plus X to the N minus 1 now F prime of 1 or at 0 x equals 0 there is going to be 1 to the N minus 1 that's just going to be 1 so f of 0 F prime of 0 is in now we need to find f of 0 go back over here and write this out L of x equals F of 0 plus F prime of 0 times X minus 0 now F of 0 will be 1 to the n that is 1 so this is 1 plus one plus and then F prime is n times X minus zero so that is just NX all right so to find well in this case specifically I should make sure that is stated this is one plus three X so if we want to find one point zero one cubed du L of 101.1 that seems like it's going to be too big that is 4.03 at time of zero okay well that value is going to be 4.03 now the actual value 1.01 cubed is one point zero three ah I see the issue we actually want to do is L of 0.01 because we already have the the 1 plus so in this case X is actually 0.01 that will take that down quite a bit all right now evaluating that let's be careful buzzer because I already had a 1 plus there I can't forget that so that would be 1 1 point make sure I just tapped at any one point zero 3 all right now the actual value is as I said one point zero three zero three zero one so this value is very good it's an approximation so so far we have dictated that X must be near a to formalize this we use the idea of a differential the notation dy/dx indicates the change in Y with respect to a small change in X so given a function y equals f of X that's differentiable what DX be an independent variable that can be assigned any nonzero real number and define the dependent variable dy by D y equals F prime of X DX this is called the differential form and it comes up quite a bit especially in differential equations intersection a whole field that dedicated to that now for each of these functions find dy and approximate r evaluate when x equals 3 and DX is 0.1 now by definition d y equals f prime of X DX so in this case the y equals I'm going to write this out without evaluating first two x plus two times DX now X is 3 DX is point one plus two times zero point 1 and dy is that will be 8 times 0.1 0.8 okay next we want to consider y equals cosine X so dy equals F prime of X DX F prime of X is minus sine X DX so evaluating at x equals three times point one sine of three point one let's make sure I'm in radians okay negative zero point one four one one two now we could compare this to the actual change in Y instead of the with a small change of X we can actually evaluate it at both points three and then at three point one and compare the two or we can do this now these are going to be very close if X is close to a so suppose the input X changes by a small amount here we are if X changes from a to a plus DX then the change in X is DX or Delta X and the change in Y is given by Delta y equals f of a plus DX minus f of a now based on our linear approximation the change in Y can be found let's follow the string of qualities for X near a f of X is approximately LX therefore if DX is small then F of a plus DX is approximately L of a plus DX which L of a plus DX would be evaluating here so F of a plus F prime of a times a plus DX minus a which is really going to be F of a plus F prime of a times DX and that means let me go ahead and write that here actually F of a plus F prime of a times DX you know F of a plus D X minus F of a would be then that linear approximation minus F of a which would cancel that so we'd be left with F prime of a DX so this is the actual change if we took the input of our new point minus the input of our olds point that's the actual change now I would like to point out that as X approaches a Delta Y approaches dy okay so let's go and compute both of these I'll go ahead and compute both of these so Delta Y for Delta Y we want to find f of a plus BX minus f of a in this case that is f of three point one since our DX is one or point one and our a value is three now evaluating each of these we get fifteen point eight one is fifteen so that Delta y equals zero point eight one now finding dy as we did earlier and it would be the exact same output there it would be zero point eight at the same equation same point same D X yes so look back at number four so as X approaches a these two are going to get closer and closer Delta Y is going to approach it's going to approach dy now any type of measurement and I am this is actually from the textbook large chunks of this any type of measurement is prone to a certain amount of error in many applications certain quantities are based are calculated based on measurements for example the area of a circle is calculated by measuring the radius of the circle an error in the measurement of the radius leads to an error in the computed value of the area here we examine this type of their error and study how differentials can be used to estimate that error so consider the function f with an input that's a measured quantity suppose the exact value of the measured quantity is a but the measured value is a plus DX so there's some error in our measurement of DX we say the measurement error is DX and as a result an error occurs and the calculated quantity this type of error is known as a propagated error and it's given by Delta Y now since all measurements are prone to some error we don't know the actual value so we cannot calculate the propagated error exactly however given an estimate of the accuracy the measurement we can use differentials to approximate the propagated error as I said earlier as the two approach each other there as X approaches a that points Delta Y is approximately dy so we can often use F prime of a DX so we don't often know the exact measurement we don't know the actual radius we know our estimate and so we can use Delta y is approximately dy which is f prime of a plus DX DX so in this next example we'll see how differentials can be used to estimate the error and calculating the volume of a box so if we assume the measurement of the side length is made with a certain amount of accuracy give us something to work with suppose the side length of a cube is measured to be 5 centimeters with an accuracy of 0.1 centimeters use differences to estimate the error in the computed volume of the cube well our volume because we are dealing with a cube that is X cubed which means corresponding to that DV is 3 x squared DX so we are going to estimate the computed volume okay based off of our error okay so our DX our DX is point 1 that means DX is between zero point negative zero point one and positive point one well 3 x squared DX then between two values we can use that formula 3 x squared DX now our x value is 5 so 3 x squared DX and 3x squared DX notice I have a negative point 1 there now that comes out 5 and this one will be positive at 7 point 5 so the volume of our box is going to be within seven point five centimeters cubed of way actual volume it's the idea there so an error of 0.1 could actually potentially give us an as much as 7.5 when it comes to our answer difference our variety now we want to compute the volume of this cube if the side length was 4.9 or if it was 5.1 and compute the estimated error and compare it to the actual potential error all right so the volume that were 4.9 maybe 4.9 cubed versus the volume if it were 5.1 and if we calculate those 29 cubed is 117 649 this value 132 6 5 1 and this is not Delta Delta V because we actually need to subtract the actual the actual volume so if that if it is 5 then 5 cube is 125 so our Delta V is between one seventeen point six four eight minus five five one and 132 0.65 one minus one twenty-five so in that case our Delta V the change in the actual output is slightly bigger the window of error is bigger compared to our actual item so both of these are different types of measurements different types of measurement error so if we assume that the sides are actually five but aren't or within a certain one of accuracy then we find the first answer that it's within 7.5 but if we actually assume that we were wrong and these are our two values then they're slightly different but they're very very close okay again Delta Y vs vs dy okay they will approach each other as well example seven an astronaut using a camera measure the radius of Earth as four thousand miles with an error of plus or minus 80 miles we're going to use differentials to estimate the relative and percent error of using this radius measurement to calculate the volume of Earth assuming the planet is a perfect sphere well relative error relative error is the change in a quantity divided by the actual quantity the error divided by the actual now what we can work with is we know that we are within 80 miles so the error in our radius measurement is between 80 and negative 80 now to find the volume let's begin with this volume equals 4/3 PI R cubed which means DV change in the volume it's going to be 4 PI R squared D are now our volume go ahead and figure this our volume is approximately 4,000 let's rewrite that 4/3 PI times 4,000 cubed now potentially however if we are within 80 then our DV is 4pi times 4,000 squared times TV versus 4 pi times 4,000 squared times minus DD so our error there is going to be what okay so that's a very large number so rather than actually calculate what that is there's our range on DV now our relative error is going to be our approximated values divided by our actual so for pi clip-on times 4,000 squared times a let's use that one divided by 4/3 pi 4,000 cubed and that is by 4/3 pi 4,000 cubed that should be 0.06 which means that V V over V that's the that's what it is the the error divided by the actual 6 minus 0.06 and our relative error is 6% okay that brings us to the end of this section to some applications of differentials
188035	https://www.youtube.com/watch?v=ZlotX-FxpPI	Find each angle of triangle ABC, where a=2b.cos(C) and sin(A)sin(B/2+C)=sin(C)(sin(B/2)+sin(A)). Maths Enhancer's Class 1680 subscribers 14 likes Description 327 views Posted: 1 Dec 2022 The video illustrates step-by-step how to find each angle of triangle ABC, where a=2b.cos(C) and sin(A)sin(B/2+C)=sin(C)(sin(B/2)+sin(A)). Subscribe at: MathsEnhancer #MathematicalCompetition #MathematicalOlympiad #ChallengingMathematicsProblems #MathematicsCompetition #EnhanceMathematicsProblemSolvingSkills #Mathematics ImproveMathematicsGrades Find angles of triangles Sine law Trigonometry Addition formulae Compound-angle formulae Double angle formulae Addition formula Compound-angle formula Double angle formula Law of sines Law of sine AwesomeMath Math Olympiad challenge Challenging mathematics problems Enhance mathematics problem solving skills Mathematical competition questions Mathematics competition questions Improve mathematics grades Olympics math Olympics maths Olympics mathematics Olympics math competition Math Olympiad training International Mathematical Olympiad questions and solutions International Mathematical Olympiad questions and answers How to solve Olympiad question How to solve International Mathematical Olympiad question How to solve Olympiad Mathematics Question How to prepare for Maths Olympiad How to prepare for Math Olympiad Olympiad mathematics competition Olympiad exam Olympiad exam sample paper Maths Olympiad sample question Maths Olympiad sample answer Math Olympiad 2 comments Transcript: binds each angle of triangle ABC a equals 2B cosine C PSI a PSI B over 2 plus C equals sine C sine B over 2 plus sine a one and two are given conditions a equals B cosine C plus C cosine B since a equals c d plus DB c d equals B cosine c d b equals c cosine B substitutes three in one we have 2B cosine C equals B cosine c plus c cosine B cosine C on the left hand side of the equation and that's on the right hand side of the equation cancel thus B cosine C equals c cosine B by the sine law we know that sine B over b equals sine C over C so b equals sine B over sine c times see substitutes five in four we have sine B over sine c times C cosine C equals c cosine b c on the left hand side of the equation and that's on the right hand side of the equation cancel so sign B cosine C equals sign C cosine B by rearranging the equation we have sine B cosine C minus sine C cosine b equals zero compound angle formula sine B minus C equals sine B cosine C minus cosine B sine C so we can now write PSI B minus C equals zero which means b equals c a equals Pi minus 2C substitutes six and seven into we have sine Pi minus 2C times sine C over 2 plus c equals sign c times sine C over 2 plus PSI Pi minus 2C sine Pi minus 2C is equal to sine 2 C INE C over 2 plus C is equal to sine three C over 2 PSI Pi minus two c is equal to sine two c double angle formula so into C equals 2 sine C cosine C so we can now write two sine C cosine C PSI 3C over 2 equals sine Z times PSI C over 2 plus sine to C so I see on the left hand side of the equation and that's on the right hand side of the equation cancel so we can now write 2 cosine C sine 3C over 2 equals sine C over 2 plus sine 2C sum and product formula PSI Alpha plus sine beta equals to psi Alpha plus beta over 2 cosine Alpha minus beta over 2 let's Alpha equal five over two times C beta equal C over 2 we have sine 5 over 2 times C plus sine C over 2 equals two sine three C over two cosine C substitutes 9 in 8. we have sine C over 2 plus PSI to C equals sine C over 2 plus sine 5 over 2 times C PSI C over 2 on the left hand side of the equation and that's on the right hand side of the equation cancel plus signs you see equals PSI 5 over 2 times C by rearranging the equation we have sine 5 over 2 times C minus sine to C equals zero summon product formula PSI Alpha minus sine beta equals 2 cosine Alpha plus beta over 2 sine Alpha minus beta over 2 10 can be written as 2 cosine 5 over 2 times C plus two c over two PSI 5 over 2 times C minus 2 C over 2 equals zero by simplifying the equation we have 2 cosine 9C over 4 PSI C over 4 equals zero since zero is less than C is less than pi over 2 so sine C over four doesn't equal zero thus cosine 9 C over 4 equals zero which means nine c over four equals pi over 2 by rearranging the equation and simplifying we have C equals 2 pi over nine so b equals C equals 2 pi over nine a equals Pi minus 2 C equals Pi minus 2 times 2 pi over 9 equals 5 pi over 9. thank you for watching hope you found this video helpful please subscribe to this YouTube channel for notification for new videos see you next time [Music]
188036	https://www.ixl.com/math/grade-4/multi-step-word-problems	IXL \| Multi-step word problems \| 4th grade math SKIP TO CONTENT IXL Learning Sign in Remember Sign in now Join now IXL Learning Learning Math Skills Lessons Videos Games Language arts Skills Videos Games Science Social studies Spanish Recommendations Recommendations wall Skill plans IXL plans Textbooks Test prep Awards Student awards Assessment Analytics Takeoff Inspiration Learning All Learning Math Language arts Science Social studies Spanish Recommendations Skill plans Learning Skill plans IXL plans Textbooks Test prep Awards Assessment Analytics Takeoff Inspiration Membership Sign in Math Language arts Science Social studies Spanish Recommendations Skill plans IXL plans Textbooks Test prep Awards Fourth grade N.4 Multi-step word problems EA9 Share skill Copy the link to this skill share to facebookshare to twitter Time to get in the zone! Your teacher would like you to focus on skills in . Let's pick a skill from these categories. Let's go! N.4Multi-step word problemsEA9 Share video Copy the link to this video share to facebookshare to twitter You are watching a video preview. Become a member to get full access! You've reached the end of this video preview, but the learning doesn't have to stop! Join IXL today! Become a memberSign in Incomplete answer You did not finish the question. Do you want to go back to the question? Go backSubmit Learn with an example Submit Learn with an example Learn with an example question solution Learn with an example Excellent! You got that right! Continue Learn with an example Jumping to level 1 of 1 Excellent! Now entering the Challenge Zone—are you ready? Questions answeredQuestions 0 Time elapsedTime 000002hrminsec SmartScore out of 100 IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. It tracks your skill level as you tackle progressively more difficult questions. Consistently answer questions correctly to reach excellence (90), or conquer the Challenge Zone to achieve mastery (100)! Learn more 0 You met your goal! Teacher tools Group Jam Live Classroom Leaderboards Work it out Not feeling ready yet? This can help: N.3Use strip diagrams to represent and solve multi-step word problems N.3 Use strip diagrams to represent and solve multi-step word problems - Fourth grade G8Z Company \| Membership \| Blog \| Help center \| User guides \| Tell us what you think \| Testimonials \| Careers \| Contact us \| Terms of service \| Privacy policy © 2025 IXL Learning. All rights reserved. Follow us First time here? 1 in 4 students use IXL for academic help and enrichment. Pre-K through 12th grade Sign up nowKeep exploring
188037	https://simple.wikipedia.org/wiki/Law_of_large_numbers	Contents Law of large numbers The law of large numbers, or LLN for short, is a theorem from statistics. It states that if a random process is repeatedly observed, then the average of the observed values will be stable in the long run. This means that as the number of observations increases, the average of the observed values will get closer and closer to the expected value. For example, when rolling dice, the numbers 1, 2, 3, 4, 5 and 6 are possible outcomes. They are all equally likely. The population mean (or "expected value") of the outcomes is: The following graph shows the results of an experiment of rolls of a die. In this experiment, it can be seen that the average of die rolls varies wildly at first, but as predicted by the LLN, the average stabilizes around the expected value of 3.5 as the number of observations become large. History Jacob Bernoulli first described the LLN. He said the concept was so simple that even the stupidest man instinctively knows it is true. Despite this, it took him over 20 years to develop a good mathematical proof. Once he had found it, he published the proof in Ars Conjectandi (The Art of Conjecturing) in 1713. He named this his "golden theorem". It became generally known as "Bernoulli's theorem" (not to be confused with the law in physics with the same name). In 1835, S.D. Poisson further described it under the name "la loi des grands nombres" (the law of large numbers). Thereafter, it was known under both names, but the "law of large numbers" is most frequently used. Other mathematicians also contributed to make the law better. Some of them were Chebyshev (who proved a more general version of the law for averages), Markov, Borel, Cantelli and Kolmogorov. After these studies, there are now two different forms of the law: One is called the "weak" law, and the other the "strong" law. These forms do not describe different laws. They have different ways to describe the convergence of the observed or measured probability to the actual probability. The strong form of the law implies the weak one. Related pages References
188038	https://tutors.com/lesson/types-of-polygons	Types of Polygons (Video) 17 Different Types & Examples Geometry Tutoring• 29415 Tutoring jobs 20+ tutors near you & online ready to help. Find a tutor TABLE OF CONTENTS Definition Types Regular and Irregular Convex and Concave Simple and Complex Names of Polygons TUTORS COURSES TYPES OF POLYGONS Types of Polygons Written by Malcolm McKinsey January 16, 2023 Fact-checked by Paul Mazzola Definition Types Names of Polygons What is a polygon? Polygons are some of the first shapes we learn to draw as children, and they appear all around us.Polygonscan be regular or irregular. They can be simple or complex, convex or concave. They can be the familiar shapes you see in geometry textbooks, or they can be strange shapes, like darts and bowties. The word"polygon"means"many-angled,"from Greek. To be a polygon, a flat, closed shape must use only line segments to create its sides. So a circle or any shape that has a curve is not a polygon. The three identifying properties of any polygon are that the polygon is: A two-dimensional shape Closing in a space (having an interior and exterior) Made with straight sides Get free estimates from geometry tutors near you. Search Types of polygons Let's take a look at the vast array of shapes that are polygons and go into detail. Types of Polygons A convex polygonhas no interior angle greater than 180°(it has no inward-pointing sides). A concave polygon has one interior angle greater than180°. A simple polygonencloses a single interior space (boundary) and does not have self-intersecting sides. Complex polygons have self-intersecting sides! An irregular polygondoes not have congruent sides and interior angles. A regular polygonhas congruent sides and interior angles. Various polygons Regular and irregular polygons Regular polygonshave congruent sides and interior angles. Every side is equal in length to every other side, and every interior angle is equal to all other interior angles. The number of regular polygons is limitless. Irregular polygonsdo not have congruent sides and angles. Home plate on a softball or baseball field is an irregular pentagon, because it has five sides with two 90° angles. Regular And Irregular Polygons Convex and concave polygons Aconvex polygoncloses in an interior area without looking "dented." None of its interior angles point inward. In geometry, you could have a four-sided polygon that points outward in all directions, like akite, or you could have the same four sides so two of them point inward, forming adart. The kite is convex; the dart is concave. Every interior angle of a convex polygon is less than 180°. A concave polygon has at least one angle greater than 180°. Think of a bowtie-shaped simple hexagon (6 sides). It will have two interior angles greater than 180°. Convex and Concave Polygons Simple and complex polygons Simple polygonshave no self-intersecting sides.Complex polygons, also called self-intersecting polygons, have sides that cross over each other. The classic star is a complex polygon. Most people can doodle a star on paper quickly, but few people label it apentagram, complex polygon, or self-intersecting polygon. Simple and Complex Polygons The family of complex star-shaped polygons generally share the Greek number prefix and use the suffix -gram: pentagram, hexagram, octagram, and so on. You cannot draw a complex triangle. For every polygon with four or more sides, a complex polygon can be drawn. A complex quadrilateral is that familiar bowtie shape, but it is considered to have only four sides, because one pair of opposite sides has twisted to cross each other. Just as you do not count the crossed sides as four line segments, you do not count the two angles they create as interior angles. The complex quadrilateral still only has four sides and four interior angles. Complex polygonsmay be hard to imagine unless you think of them with elastic sides. If you could lift part of the polygon up and twist it, so two sides cross one another, and then put it down flat again, you would have a complex polygon. Because you twisted two sides, you still have those two sides (they do not double in number by crossing). They also do not create new vertices where they cross. Antiparallelogram An unusual complex polygon is theantiparallelogram, which looks a bit like bird wings. An antiparallelogram (or crossed parallelogram) has the normal two pairs of congruent, opposite sides, but one pair has been crossed, forming what appears to be two touching triangles. Antiparallelogram Like any parallelogram, though, the antiparallelogram still only has four sides and four interior angles. Its diagonals, however, lie outside the shape! Names of polygons \| Polygon Shape \| Number of Sides \| --- \| \| Triangle \| 3 sides \| \| Square \| 4 sides \| \| Rectangle \| 4 sides \| \| Quadrilateral \| 4 sides \| \| Parallelogram \| 4 sides \| \| Rhombus \| 4 sides \| \| Dart \| 4 sides \| \| Kite \| 4 sides \| \| Pentagon \| 5 sides \| \| Hexagon \| 6 sides \| \| Heptagon \| 7 sides \| \| Octagon \| 8 sides \| \| Nonagon \| 9 sides \| \| Decagon \| 10 sides \| \| Dodecagon \| 12 sides \| \| Icosagon \| 20 sides \| \| Hectagon \| 100 sides \| \| n-gon \| n sides \| Is it a polygon? The interiors of all polygons can be broken up into triangles, which is a handy way to find the sum of their interior angles. Take two less than the number of sides,n, and multiply times 180°: (n- 2) x 180°. Is It a Polygon? Angles And Sides A circle is not a polygon, but a icosikaihenagon is a polygon. An icosikaihenagon is a 21-sided polygon. Most mathematicians and mathematics students would simply write "21-gon" to name it. Get free estimates from geometry tutors near you. Search Is A Circle A Polygon? No Lesson summary Polygons can be studied and classified in many different ways. You now see that polygons can be regular or irregular, convex or concave, and simple or complex. When you see an unfamiliar polygon, you can determine its properties and classify it correctly. To be a polygon, the shape must be flat, close in a space, and be made using only straight sides. Polygons with congruent sides and angles are regular; all others are irregular. Polygons with all interior angles less than 180° are convex; if a polygon has at least one interior angle greater than 180°, it is concave. Simple polygons do not cross their sides; complex polygons have self-intersecting sides. Polygons are all around you! Fact-checked and reviewed for accuracy.We adhere to stricteditorial integrity. 20+ Geometry Tutors near North Charleston, SC Get better grades with tutoring from top-rated private tutors. Local and online. View tutors Table of contents Definition Types Regular and Irregular Convex and Concave Simple and Complex Names of Polygons Find a tutor 20+ Geometry Tutors near North Charleston, SC Get better grades with tutoring from top-rated private tutors. Local and online. View tutors Related articles Regular polygons Polygon Triangle Rhombus Trapezoid Decagon Heptagon Similar polygons Nonagon Kites in geometry Related articles Regular polygons Polygon Triangle Rhombus Trapezoid Decagon Heptagon Similar polygons Nonagon Kites in geometry Find geometry tutors in your area Geometry Tutors New York Geometry Tutors Los Angeles Geometry Tutors Chicago Geometry Tutors Houston Geometry Tutors Phoenix Geometry Tutors Philadelphia Geometry Tutors San Antonio Geometry Tutors Dallas Geometry Tutors San Diego Geometry Tutors San Jose Geometry Tutors Detroit Geometry Tutors San Francisco Geometry Tutors Jacksonville Geometry Tutors Indianapolis Geometry Tutors Austin Geometry Tutors Columbus Geometry Tutors Fort Worth Geometry Tutors Charlotte Geometry Tutors Memphis Geometry Tutors Baltimore Find tutors nearby Geometry Tutors near me Math Tutors near me Algebra Tutors near me Algebra 2 Tutors near me Calculus Tutors near me Online Math Tutors near me College Algebra Tutors near me Precalculus Tutors near me College Math Tutors near me Home Tutors near me ACT Math Tutors near me AP Statistics Tutors near me AP Calculus Tutors near me Elementary Math Tutors near me Pre-Algebra Tutors near me Discrete Math Tutors near me Arithmetic Tutors near me Linear Algebra Tutors near me Middle School Math Tutors near me Statistics Tutors near me Trigonometry Tutors near me About Tutors Home Log in About Press Contact us Help Top services Math Tutors Reading Tutors English Tutors SAT Tutoring Chemistry Tutors Spanish Tutors Writing Tutors Students Tips for hiring Tutoring prices Free online courses Online tutoring Services near me Tutors Tutoring jobs Online tutoring jobs How to become a tutor Online whiteboard for teaching Tutors.com reviews © 2025 Liaison, Inc. •Terms of Use•Privacy Policy•Accessibility•Services
188039	https://www.youtube.com/watch?v=cczABrLd_uA	Monopolist optimizing price: Total revenue. \| Microeconomics \| Khan Academy Khan Academy 9090000 subscribers 508 likes Description 284768 views Posted: 25 Jan 2012 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: Starting to think about how a monopolist would rationally optimize profits Watch the next lesson: Missed the previous lesson? Microeconomics on Khan Academy: Topics covered in a traditional college level introductory microeconomics course About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academy's Microeconomics channel: Subscribe to Khan Academy: 13 comments Transcript: What I want to start thinking about in this video is, given that we do have a monopoly on something, and in this example, in this video, we're going to have a monopoly on oranges. Given that we have a monopoly on oranges and a demand curve for oranges in the market, how do we maximize our profit? And to answer that question, we're going to think about our total revenue for different quantities. And from that we'll get the marginal revenue for different quantities. And then we can compare that to our marginal cost curve. And that should give us a pretty good sense of what quantity we should produce to optimize things. So let's just figure out total revenue first. So obviously, if we produce nothing, if we produces 0 quantity, we'll have nothing to sell. Total revenue is price times quantity. Your price is 6 but your quantity is 0. So your total revenue is going to be 0 if you produce nothing. If you produce 1 unit-- and this over here is actually 1,000 pounds per day. And we'll call a unit 1,000 pounds per day. If you produce 1 unit, then your total revenue is 1 unit times $5 per pound. So it'll be $5 times, actually 1,000, so it'll be $5,000. And you can also view it as the area right over here. You have the height is price and the width is quantity. But we can plot that, 5 times 1. If you produce 1 unit, you're going to get $5,000. So this right over here is in thousands of dollars and this right over here is in thousands of pounds. Just to make sure that we're consistent with this right over here. Let's keep going. So that was this point, or when we produce 1,000 pounds, we get $5,000. If we produce 2,000 pounds, now we're talking about our price is going to be $4. Or if we could say our price is $4 we can sell 2,000 pounds, given this demand curve. And our total revenue is going to be the area of this rectangle right over here. Height is price, width is quantity. 4 times 2 is 8. So if I produce 2,000 pounds then I will get a total revenue of $8,000. So this is 7 and 1/2, 8 is going to put us something right about there. And then we can keep going. If I produce, or if the price is $3 per pound, I can sell 3,000 pounds. My total revenue is this rectangle right over here, $3 times 3 is $9,000. So if I produce 3,000 pounds, I can get a total revenue of $9,000. So right about there. And let's keep going. If I produce, or if the price, is $2 per pound, I can sell 4,000 pounds. My total revenue is $2 times 4, which is $8,000. So if I produce 4,000 pounds I can get a total revenue of $8,000. It should be even with that one right over there, just like that. And then if the price is $1 per pound I can sell 5,000 pounds. My total revenue is going to be $1 times 5, or $5,000. So it's going to be even with this here. So if I produce 5,000 units I can get $5,000 of revenue. And if the price is 0, the market will demand 6,000 pounds per day if it's free. But I'm not going to generate any revenue because I'm going to be giving it away for free. So I will not be generating any revenue in this situation. So our total revenue curve, it looks like-- and if you've taken algebra you would recognize this as a downward facing parabola-- our total revenue looks like this. It's easier for me to draw a curve with a dotted line. Our total revenue looks something like that. And you can even solve it algebraically to show that it is this downward facing parabola. The formula right over here of the demand curve, its y-intercept is 6. So if I wanted to write price as a function of quantity we have price is equal to 6 minus quantity. Or if you wanted to write in the traditional slope intercept form, or mx plus b form-- and if that doesn't make any sense you might want to review some of our algebra playlist-- you could write it as p is equal to negative q plus 6. Obviously these are the same exact thing. You have a y-intercept of six and you have a negative 1 slope. If you increase quantity by 1, you decrease price by 1. Or another way to think about it, if you decrease price by 1 you increase quantity by 1. So that's why you have a negative 1 slope. So this is price is a function of quantity. What is total revenue? Well, total revenue is equal to price times quantity. But we can write price as a function of quantity. We did it just now. This is what it is. So we can rewrite it, or we could even write it like this, we can rewrite the price part as-- so this is going to be equal to negative q plus 6 times quantity. And this is equal to total revenue. And then if you multiply this out, you get total revenue is equal to q times q is negative q squared plus 6 plus 6q. So you might recognize this. This is clearly a quadratic. Since you have a negative out front before the second degree term right over here, before the q squared, it is a downward opening parabola. So it makes complete sense. Now, I'm going to leave you there in this video. Because I'm trying to make an effort not to make my videos too long. But in the next video what we're going to think about is, what is the marginal revenue we get for each of these quantities? And just as a review, marginal revenue is equal to change in total revenue divided by change in quantity. Or another way to think about it, the marginal revenue at any one of these quantities is the slope of the line tangent to that point. And you really have to do a little bit of calculus in order to actually calculate slopes of tangent lines. But we'll approximate it with a little bit of algebra. But what we essentially want to do is figure out the slope. So if we wanted to figure out the marginal revenue when we're selling 1,000 pounds-- so exactly how much more total revenue do we get if we just barely increase, if we just started selling another millionth of a pound of oranges-- what's going to happen? And so what we do is we're essentially trying to figure out the slope of the tangent line at any point. And you can see that. Because the change in total revenue is this and change in quantity is that there. So we're trying to find the instantaneous slope at that point, or you could think of it as the slope of the tangent line. And we'll continue doing that in the next video.
188040	https://www.youtube.com/watch?v=SNbVbm5Uly8	114 Test of the Wave Propagation in 2D grid UW SBEL 424 subscribers Description 104 views Posted: 7 Jun 2019 Transcript:
188041	https://cnettleman.net/what-is-legal-description/	Tony Nettleman- June 6, 2023- Professional Land Surveyors- 0 Comments (Last updated February 2024) When it comes to real estate transactions, one term that often comes up is “legal description of property”. Whether you’re buying a real estate property, transferring ownership, or dealing with property boundaries, understanding what a legal land description is and its importance is crucial. In this blog post, we will explore the concept of a legal description and its significance in real estate. Why Legal Descriptions are Important Accurate legal descriptions are essential for a variety of reasons. Firstly, they provide clarity and certainty about the boundaries of a property, preventing any confusion or disputes over ownership or usage rights. Secondly, legal descriptions are crucial for the recording of deeds and other legal documents, ensuring that the correct property is identified and transferred. Legal descriptions are used in property surveys, zoning regulations, and tax assessments. Additionally, legal descriptions serve as the basis for: Creating maps Creating deeds Creating other legal documents related to land ownership Without precise legal descriptions, confusion and ambiguity may arise, leading to costly litigation and uncertainty regarding property rights. Thus, an accurate property description is essential for maintaining order and facilitating smooth land transactions within the legal framework. Legal Description of Property In simple terms, a legal description is a detailed and precise way of identifying a specific piece of property. It is a written representation that uniquely identifies a parcel of land, distinguishing it from all other properties. Legal descriptions are typically used in legal documents, such as deeds, contracts, and land records, to provide an accurate and unambiguous reference to a property’s location and boundaries. An exact legal description provides specific information about the boundaries, dimensions, and location of the land in a format that is legally recognized and sufficient for recording deeds and other legal documents. Legal descriptions often include details such as metes and bounds (a surveying method using distances and angles), reference to landmarks, lot numbers, subdivision names, and other pertinent information to accurately define the property’s boundaries. This property description serves as an official and legally binding way to identify the property and distinguish it from others. 3 Types of Land Descriptions A legal property description can take several forms, depending on the jurisdiction and the level of accuracy required. The most common types of legal descriptions include metes and bounds, lot and block, and government rectangular survey system. Metes and Bounds Metes and bounds is the oldest and most flexible method of legal description. It relies on physical landmarks, such as natural features (rivers, trees, etc.) or artificial monuments (stakes, markers, etc.), to define a property’s boundaries. The description starts at a designated point of beginning (POB) and follows a series of compass directions and distances, forming a closed loop that outlines the property. Lot and Block Lot and block, also known as recorded plat or subdivision, is a system commonly used in urban areas where land is divided into individual lots. In this method, the property is described by referring to a recorded plat map that shows the layout of the subdivision. Each lot is assigned a unique number or letter, making it easier to identify and locate the property. Rectangular System The government rectangular survey system, also known as the Public Land Survey System (PLSS), is a method used to describe land in many parts of the United States. Under this system, land is divided into a grid-like pattern of townships, ranges, and sections. Each township is further divided into 36 sections, each covering approximately one square mile. A legal description using the PLSS would typically include the township, range, section, and often a fractional description to indicate a smaller portion of a section. When dealing with a legal description, it is important to understand that minor errors or discrepancies can have significant consequences. Even a small mistake in the legal description can lead to problems in the future, including property boundary disputes or difficulties in selling the property. Therefore, it is advisable to consult a qualified surveyor or attorney when dealing with legal descriptions to ensure accuracy and avoid any potential issues. Schedule a Call Add Legal Description to a Property Deed When it comes to real estate transactions, ensuring accuracy and clarity in legal documents like deeds is paramount. Adding a complete legal description to a deed is a crucial step in this process, as it provides a detailed and precise identification of the property being conveyed. As mentioned above, a legal description typically includes specific details such as the property’s boundaries, dimensions, and any identifiable landmarks. This legal document serves as a definitive reference point for the property’s location and boundaries, leaving no room for ambiguity or misunderstanding of land use. There are several commonly used methods for creating legal descriptions, including metes and bounds, lot and block, and government rectangular survey systems. Each method has its own set of guidelines and requirements, but they all aim to accurately define the property in question. Incorporating a legal description into a deed not only fulfills legal requirements but also helps protect the interests of both the buyer and the seller. By clearly delineating the boundaries and characteristics of the property, potential disputes or misunderstandings can be avoided down the line. Additionally, having a precise legal description is essential for various legal and administrative purposes, such as obtaining permits, conducting surveys, and resolving property disputes. Adding a legal description to a deed is a critical step in the real estate transaction process. It provides a comprehensive and accurate depiction of the property being conveyed, ensuring clarity and certainty for all Land Description Search A legal description is a detailed and precise way of identifying a specific piece of property. It is a crucial component of real estate transactions, providing clarity, certainty, and accuracy regarding a property’s boundaries. Whether you are buying, selling, or transferring ownership, understanding the legal description is essential to ensure a smooth and legally sound transaction. When searching for property, utilizing a legal description is imperative for precision and accuracy. A legal description provides a detailed account of a property’s boundaries, enabling individuals to identify and locate the exact piece of land in question. This description typically includes information such as the property’s lot number, block number, subdivision name, metes and bounds, and/or GPS coordinates. By relying on a legal description, individuals can avoid confusion or ambiguity that might arise from using more general descriptors like street addresses or parcel numbers, which can be imprecise or subject to change. Legal descriptions are especially crucial in situations where multiple properties may share similar addresses or when properties are being divided or combined. Moreover, legal descriptions serve as a foundation for legal documents, such as deeds, contracts, and land surveys, ensuring clarity and consistency in property transactions. They are also essential for resolving disputes related to property boundaries or ownership rights. In essence, utilizing a legal description when searching for property provides a reliable framework for accurately identifying, delineating, and documenting real estate assets, facilitating smooth and transparent transactions while minimizing the risk of confusion or legal complications. Questions? Contact Dr. Nettleman or click below for an informational video about legal land descriptions. Legal Descriptions Video Share this: Click to share on Facebook (Opens in new window) Facebook Click to share on X (Opens in new window) X Please Share This: AI in Surveying May 8, 2025 interested in AI in surveying? Artificial intelligence (AI) is revolutionizing industries across the board, and land surveying is no exception. As a technical field that intersects with law, engineering, and real estate, land surveying relies heavily on data... Land Boundary Dispute: The Role of a Surveying Expert May 8, 2025 Land boundary disputes can be complex legal matters. They require precise measurements, historical analysis, and expert interpretation. In such cases, a surveying expert plays a crucial role in providing objective, technical insights to resolve conflicts. This... Water Rights in Texas: Full Guide May 8, 2025 Water is one of Texas's most vital yet contentious resources. Currently, the state is grappling with rapid population growth, extended droughts, and increasing demands from agriculture, industry, and municipalities. Because of this, water use and Texas water... 0 Comments Leave a ReplyCancel reply Recent Posts AI in Surveying Land Boundary Dispute: The Role of a Surveying Expert Water Rights in Texas: Full Guide The Difference Between Littoral and Riparian Rights Modern Land Surveying: Revolutionizing Techniques with AI Categories Case Studies Hints and Tips Professional Land Surveyors Riparian Surveying Surveyors Uncategorized This website uses cookies to improve your experience. We'll assume you're ok with this, but you can opt-out if you wish. Cookie settingsACCEPT Privacy & Cookies Policy Privacy Overview This website uses cookies to improve your experience while you navigate through the website. Out of these cookies, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. We also use third-party cookies that help us analyze and understand how you use this website. 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188042	https://www.quora.com/How-do-you-prove-that-for-all-a-b-integers-gcd-a-b-1-iff-gcd-a-b-ab-1-but-you-are-unsure-of-your-proof-for-the-converse-elementary-number-theory-proof-writing-divisibility-solution-verification-math	Something went wrong. Wait a moment and try again. GCD & LCM Solution Verification Greatest Common Divisor Proofs (mathematics) Elementary Number Theory Proofing Language 5 How do you prove that for all a,b integers gcd(a,b) =1 iff gcd (a+b, ab) =1 but you are unsure of your proof for the converse (elementary number theory, proof writing, divisibility, solution verification, math)? Rishi Bommasani PhD at Stanford Computer Science · Author has 199 answers and 399.2K answer views · 5y I interpret your question as asking for a proof of the reverse direction: gcd(a + b, ab) = 1 implies gcd(a, b) = 1 Let us show this by proving the contrapositive: gcd(a, b) > 1 implies gcd(a + b, ab) > 1 Denote gcd(a, b) by k. We have that k \| a, k \| b. Therefore, k \| a +b and k \| ab. Hence, gcd(a + b, ab) is at least k and, therefore, greater than 1. This uses very basic number theory. Related questions How do I prove if gcd (a, b) = 1 then gcd (a+b, b) = 1? How do you prove $a=1$ under given conditions (elementary number theory, solution verification, divisibility, math)? How do you prove that if (a,b) are positive integers with gcd (a,bc) = gcd(a,b) gcd(a,c) , then gcd(a,b) =1 (elementary number theory, divisibility, gcd and LCM, math)? How do you prove that $\gcd (a,ak+c) =\gcd (a,c) $ (elementary number theory, proof writing, divisibility, gcd and LCM, math)? How do you prove gcd ( n , n + 1 ) = 1 for any n ? (elementary number theory, divisibility, gcd and LCM, math) Amitabha Tripathi have more than a working knowledge of Z · Upvoted by Frank Hubeny , M.S. Mathematics, University of Illinois at Urbana-Champaign (1994) · Author has 4.7K answers and 13.9M answer views · 5y (⇒) Suppose gcd(a,b)=1. If gcd(a+b,ab)>1, there must exist a prime p such that p∣(a+b) and p∣ab. From the latter, p∣a or p∣b. From gcd(a,b)=1, p divides exactly one of a,b. But then p∤(a+b), contrary to our assumption. So there can be no common prime divisor of a+b, ab. Consequently, gcd(a+b,ab)=1. (⇐) Suppose gcd(a,b)>1. There there must exist a prime p such that p∣a and p∣b. But then p∣(a+b) and p∣ab, so that gcd(a+b,ab)>1. So gcd(a+b,ab)=1 implies gcd(a,b)=1. ■ Gustavo Sánchez M.S. in Physics & Mathematics, University of Buenos Aires (Graduated 1986) · Author has 3K answers and 2.2M answer views · 5y If gcd(a+b,ab)=1, then there exist integers p and q, and p(a+b)+q(ab)=1. This can be rewritten as pa+(p+qa)b=1, therefore gcd(a,b)=1. If gcd(a+b,ab)≠1, then there exists p, prime, and p/(a+b) and p/(ab). (p is any divisor of gcd(a+b,ab).) If p/(ab) then p/a or p/b. If p/a, as p/(a+b) then p/b. Similarly, if p/a, then p/b too. Hence p/a and p/b, therefore, the p divides gcd(a,b) and then gcd(a,b)≠1. Mohammad Afzaal Butt B.Sc in Mathematics & Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views · 5y Let gcd (a , b) = 1 and gcd (a + b, a b) =d . We need to prove that d = 1 (a+b,ab)=d⟹d\|(a+b)∧d\|ab ∵gcd(a , b) = 1d\|ab⟹either d divides a or b if d divides a and d divides a + b then d must divide b. Similarly if d divides b and d divides a + b, it must divides a. But this is a contradiction. Hence d = 1 Conversely let gcd (a + b, a b) = 1 and (a , b) = d . We need to prove d = 1 gcd (a , b) = d⟹d\|a∧d\|b⟹d\|(a+b).It also means d\|ab Let gcd (a , b) = 1 and gcd (a + b, a b) =d . We need to prove that d = 1 (a+b,ab)=d⟹d\|(a+b)∧d\|ab ∵gcd(a , b) = 1d\|ab⟹either d divides a or b if d divides a and d divides a + b then d must divide b. Similarly if d divides b and d divides a + b, it must divides a. But this is a contradiction. Hence d = 1 Conversely let gcd (a + b, a b) = 1 and (a , b) = d . We need to prove d = 1 gcd (a , b) = d⟹d\|a∧d\|b⟹d\|(a+b).It also means d\|ab.⟹gcd (a + b, ab) = d. But it is a contradiction. Hence d = 1 ⟹gcd (a , b) = 1⟺(a+b,ab)=1 Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Related questions How do you prove rove$\gcd (a_1,\ldots, a_m) \gcd (b_1,\ldots, b_n) =\gcd (\text {all products $a_ib_j$}) $ (elementary number theory, gcd and LCM, math)? How do you solve number theory problems regarding for division of numbers $1$ to $100$ (number theory, elementary number theory, graph theory, gcd and LCM, math)? How do you prove that $ar + br = \gcd (a, b) r.$ (number theory, elementary number theory, gcd and LCM, euclidean domain, math)? How do you prove that if gcd ( a , b ) = 1 then gcd ( a c , b ) = gcd ( c , b ) (elementary number theory, gcd and lcm, math)? If gcd(a,m) =gcd(b,m)=1, then how do you prove that GCD (ab, m) = 1? Assistant Bot · Sep 6 Statement. For integers a,b, gcd(a,b)=1 if and only if gcd(a+b, ab)=1. Proof. 1) If gcd(a,b)=1 then gcd(a+b,ab)=1. Let p be any prime dividing both a+b and ab. If p \| ab then p divides a or b. Case p \| a. Since p \| a and p \| (a+b), subtracting gives p \| b. Thus p divides both a and b, contradicting gcd(a,b)=1 unless no such p exists. Case p \| b is symmetrical. Therefore no prime can divide both a+b and ab, so gcd(a+b,ab)=1. (Equivalently: any common divisor d of a+b and ab divides both a and (a+b)−a=b, so d\|a and d\|b; hence d\|gcd(a,b)=1 and d=1.) 2) If gcd(a+b,ab)=1 then gcd(a,b)=1. Let d = gcd(a,b). T Statement. For integers a,b, gcd(a,b)=1 if and only if gcd(a+b, ab)=1. Proof. 1) If gcd(a,b)=1 then gcd(a+b,ab)=1. Let p be any prime dividing both a+b and ab. If p \| ab then p divides a or b. Case p \| a. Since p \| a and p \| (a+b), subtracting gives p \| b. Thus p divides both a and b, contradicting gcd(a,b)=1 unless no such p exists. Case p \| b is symmetrical. Therefore no prime can divide both a+b and ab, so gcd(a+b,ab)=1. (Equivalently: any common divisor d of a+b and ab divides both a and (a+b)−a=b, so d\|a and d\|b; hence d\|gcd(a,b)=1 and d=1.) 2) If gcd(a+b,ab)=1 then gcd(a,b)=1. Let d = gcd(a,b). Then d divides a and b, hence d divides ab and a+b. Thus d is a common divisor of a+b and ab. By hypothesis gcd(a+b,ab)=1, so d=1. Therefore gcd(a,b)=1. Conclusion. The two statements are equivalent. Tanay Karnik Mathematics is the science of what is clear by itself. · Upvoted by Robby Goetschalckx , Computer scientist for 11+ years and passionate about math since childhood. · Author has 83 answers and 352.9K answer views · 5y Related How do I prove if gcd (a, b) = 1 then gcd (a+b, b) = 1? Quite trivial to prove using the method of proof by contradiction. So, let’s assume for the sake of contradiction that there exist (a,b) such that gcd(a,b)=1 but gcd(a+b,b)≠1 This means, that there exists an integer k not equal to 1, such that k\|b and k\|a+b And thus, we can write b=kp and a+b=kq where p and q are integers. ⟹a=k(q−p)⟹k\|a This is a contradiction, since gcd(a,b)=1 but k seems to divide both a and b. Thus, it is safe to conclude that gcd(a+b,b)=1 Ram Kushwah Up and coming Most viewed writer · Author has 6.8K answers and 15.7M answer views · 5y Related Let a , b be positive integers such that a 2 + b 2 1 + a b is an integer. Let d = g c d ( a , b ) . How do you prove that a 2 + b 2 1 + a b = d 2 ? The Solution is attached here: The Solution is attached here: Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. 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Brad Ballinger Math teacher · Author has 344 answers and 350.1K answer views · 3y Related How do I prove GCD(a,b) = 1, ab\|c iff a\|c and b\|c? First, I recommend taking the time to write out the claim properly. The way you've got it written now might be taken to mean that GCD(a,b) = 1 and ab\|c iff a\|c and b\|c. That's incorrect, as shown by the example a=b=2, c=4. How would I write it? Like this: Suppose GCD(a,b) = 1. How do I prove ab\|c iff a\|c and b\|c? The point here is to clearly get the GCD assumption out of the iff. Now we continue. It's a little difficult to say how you can prove it without knowing what facts you have at your disposal, but one useful fact here is that ab=GCD(a,b)LCM(a,b). Since we are given that GCD(a,b)=1, this im First, I recommend taking the time to write out the claim properly. The way you've got it written now might be taken to mean that GCD(a,b) = 1 and ab\|c iff a\|c and b\|c. That's incorrect, as shown by the example a=b=2, c=4. How would I write it? Like this: Suppose GCD(a,b) = 1. How do I prove ab\|c iff a\|c and b\|c? The point here is to clearly get the GCD assumption out of the iff. Now we continue. It's a little difficult to say how you can prove it without knowing what facts you have at your disposal, but one useful fact here is that ab=GCD(a,b)LCM(a,b). Since we are given that GCD(a,b)=1, this implies ab=LCM(a,b). This can help with both directions of your iff, particularly if you know that every common multiple of a and b is also a multiple of their LCM. Sohel Zibara Studied at Doctor of Philosophy Degrees (Graduated 2000) · Author has 5.1K answers and 2.6M answer views · 2y Related What is the proof that gcd(a,b) =gcd (a+b, b)? Let d1 = gcd(a , b) and d2 = gcd(a + b , b) On the one hand : d1 = gcd(a , b) implies that d1 \| a and d1 \| b which means that a = d1k and b = d1k' which results in a + b = d1(k + k') hence d1 \| (a + b) and consequently d1 \| d2. On the other hand: d2 = gcd(a + b , b) implies that a + b= d2m and b = d2n which results in a = d2(m – n) hence d2 \| a and consequently d1 \| d2. Morever: d1 \| d2 and d2 \| d1 tells us that d1 = d2. Sponsored by Spokeo Is there a way to find out if someone has a dating profile? Yes. All you need to do is enter their name here to see what dating websites or apps they are on. Amitabha Tripathi have more than a working knowledge of Z · Author has 4.7K answers and 13.9M answer views · Updated 6y Related How can I prove that gcd (a, b) = gcd (ab, LCM (a, b))? Take any pair of distinct positive integers a, b with a∣b. Then gcd(a,b)=a and gcd(ab,lcm(a,b))=gcd(ab,b)=b. Therefore what you are trying to prove is not always true. Let’s attempt to find out when gcd(a,b)=gcd(ab,lcm(a,b))is true. At the outset, note that gcd(a,b) divides a and a divides both ab and lcm(a,b). Therefore, gcd(a,b)∣gcd(ab,lcm(a,b)) always holds. For prime p and positive integer n, let ep(n) denote the highest power of p dividing n. Let p be prime, and let ep(a)=α and [math]{\textbf e}_p(b)=[/math] Take any pair of distinct positive integers a, b with a∣b. Then gcd(a,b)=a and gcd(ab,lcm(a,b))=gcd(ab,b)=b. Therefore what you are trying to prove is not always true. Let’s attempt to find out when gcd(a,b)=gcd(ab,lcm(a,b))is true. At the outset, note that gcd(a,b) divides a and a divides both ab and lcm(a,b). Therefore, gcd(a,b)∣gcd(ab,lcm(a,b)) always holds. For prime p and positive integer n, let ep(n) denote the highest power of p dividing n. Let p be prime, and let ep(a)=α and ep(b)=β. Then ep(gcd(a,b))=min{α,β}, ep(lcm(a,b))=max{α,β}, and ep(ab)=α+β. Therefore, the highest power of p dividing the LHS equals the highest power of p dividing the RHS if and only if min{α,β}=min{max{α,β},α+β}=max{α,β}. This holds precisely when α=β. Hence in order for gcd(a,b)=gcd(ab,lcm(a,b)) to hold, we must have ep(a)=ep(b) for each prime p, i.e., when a=b. Conversely, both sides equal a (or b) when a=b. Remark 1. The lcm(a,b) is defined as the least positive integer in the set of common multiples of a and b. If we denote this integer by ℓ, and if m is any common multiple of a and b, then it is a fact that ℓ∣m. Sometimes, this condition of divisibility replaces the condition of least element in the definition of lcm(a,b). Since ab is a common multiple of a and b, we therefore have lcm(a,b)∣ab. Hence gcd(ab,lcm(a,b))=lcm(a,b). Remark 2. The equation lcm(a,b)=gcd(ab,lcm(a,b)) holds good for all pairs of positive integers a,b. Subramanyam Durbha Adjunct faculty,Dept of Physics and Engineering at Community College of Philadelphia · Author has 113 answers and 116.8K answer views · 5y Related How do you prove if a b ∣ ( a + b ) ( a + b + 1 ) , then gcd ( a , b ) ≤ √ a + b for positive integers a and b ? Sayan Banerjee Author has 1.3K answers and 1.3M answer views · 5y Related How do I prove if gcd (a, b) = 1 then gcd (a+b, b) = 1? It suffices to prove that for some integral M and N, M.(a+b) + N. (b) = 1. Now, it is clear from what is given that: ma + nb = 1 , for integral a and b. So, m.(a + b - b) + nb = 1, or, m.(a+b) + (n-m).b = 1. Now, if m and n are integral, so are m and n - m. Thus, it is proved that a + b and b can be written as M.(a+b) + N. (b) = 1. for integral M and N. Thus, the gcd of a+b and b is 1. Gregorio Morales Math Teacher · Author has 357 answers and 376.3K answer views · 5y Related How do I prove if gcd (a, b) = 1 then gcd (a+b, b) = 1? Remember the Euclid Algorithm? It is based on the fact that p divides x and y then p divides x−y. From this, the common divisors of x,y are the same as the common divisors of x−y,y. So, gcd(a+b,b)=gcd(a+b−b,b)=gcd(a,b)=1 Related questions How do I prove if gcd (a, b) = 1 then gcd (a+b, b) = 1? How do you prove $a=1$ under given conditions (elementary number theory, solution verification, divisibility, math)? How do you prove that if (a,b) are positive integers with gcd (a,bc) = gcd(a,b) gcd(a,c) , then gcd(a,b) =1 (elementary number theory, divisibility, gcd and LCM, math)? How do you prove that $\gcd (a,ak+c) =\gcd (a,c) $ (elementary number theory, proof writing, divisibility, gcd and LCM, math)? How do you prove gcd ( n , n + 1 ) = 1 for any n ? (elementary number theory, divisibility, gcd and LCM, math) How do you prove rove$\gcd (a_1,\ldots, a_m) \gcd (b_1,\ldots, b_n) =\gcd (\text {all products $a_ib_j$}) $ (elementary number theory, gcd and LCM, math)? How do you solve number theory problems regarding for division of numbers $1$ to $100$ (number theory, elementary number theory, graph theory, gcd and LCM, math)? How do you prove that $ar + br = \gcd (a, b) r.$ (number theory, elementary number theory, gcd and LCM, euclidean domain, math)? How do you prove that if gcd ( a , b ) = 1 then gcd ( a c , b ) = gcd ( c , b ) (elementary number theory, gcd and lcm, math)? If gcd(a,m) =gcd(b,m)=1, then how do you prove that GCD (ab, m) = 1? What is the proof that gcd(a,b) =gcd (a+b, b)? Is it true that $\gcd (p+1,PK) =1$ (elementary number theory, prime numbers, problem solving, gcd and LCM, math)? Number Theory: How do you prove that for all integers a and b , if a and b are positive and a divides b , then a ≤ b ? Gcd (a, b) =gcd (a, b-a) how do I prove it? How do you solve gcd (a,b) =1 prove gcd ((a+b), (a-b)) = 1 or 2 (number theory, divisibility, gcd and LCM, math)? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
188043	https://www.chegg.com/homework-help/questions-and-answers/mass-defect-nuclear-binding-energy-problems-8-energy-released-formation-nucleus-iron-56-78-q100819483	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: Mass Defect and Nuclear Binding Energy Problems: 8. The energy released by the formation of a nucleus of iron-56 is 7.89×10−11 J. Determine the mass defect of iron-56. Answer: 8.77×1028 kg 9. Calculate the nuclear binding energy of an atom if the mass defect is 0.000799 kg. This AI-generated tip is based on Chegg's full solution. Sign up to see more! Identify the relationship between energy and mass defect using Einstein's equation where is energy and is mass defect. Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
188044	https://www.physio-pedia.com/Burn_Wound_Assessment	Contents Editors Categories Cite 1 Burn Wound Terminology 2 Classification by Depth 2.1 Circumferential burn injury special considerations 3 Blanch Test 4 Jackson's Burn Wound Model 4.1 Burn Wound Conversion 5 Total Body Surface Area 5.1 Calculation of Percentage Burn of Total Body Surface Area 5.1.1 1. The Rule of Nine 5.1.2 2. Lund-Browder Method 5.1.3 3. Palmar Surface Method 6 References Original Editor - Carin Hunter based on the course by Diane Merwarth Top Contributors - Carin Hunter, Stacy Schiurring and Jess Bell Categories Course Pages Burns Assessment ReLAB-HS Course Page When refering to evidence in academic writing, you should always try to reference the primary (original) source. That is usually the journal article where the information was first stated. In most cases Physiopedia articles are a secondary source and so should not be used as references. Physiopedia articles are best used to find the original sources of information (see the references list at the bottom of the article). If you believe that this Physiopedia article is the primary source for the information you are refering to, you can use the button below to access a related citation statement. Cite article Burn Wound Assessment Jump to:navigation, search Original Editor - Carin Hunter based on the course by Diane Merwarth Top Contributors - Carin Hunter, Stacy Schiurring and Jess Bell Contents 1 Burn Wound Terminology 2 Classification by Depth 2.1 Circumferential burn injury special considerations 3 Blanch Test 4 Jackson's Burn Wound Model 4.1 Burn Wound Conversion 5 Total Body Surface Area 5.1 Calculation of Percentage Burn of Total Body Surface Area 5.1.1 1. The Rule of Nine 5.1.2 2. Lund-Browder Method 5.1.3 3. Palmar Surface Method 6 References Related online courses on +Physiopedia Plus Online Course: Burn Wound Assessment Wound Assessment Set the foundation for learning burn wound assessment techniques Start course 1-1.5 hours - - - - Powered by Physiopedia Course instructor Diane Merwarth Diane is a physical therapist with more than 25 years of practice in the areas of wound management and burn care. Course instructor • Online Course: Burn Wound Cleansing and Solutions Wound Cleansing and Solutions Learn the best evidence-supported methods to cleanse and prepare a burn wound for healing Start course 2-2.5 hours - - - - Powered by Physiopedia Course instructor Diane Merwarth Diane is a physical therapist with more than 25 years of practice in the areas of wound • Online Course: Management of Burn Wounds Programme Management of Burn Wounds Programme Learn advanced wound care skills to promote healing of burn wound injuries Start course 18-20 hours - - - - Powered by Physiopedia Course instructor Diane Merwarth Diane is a physical therapist with more than 25 years of practice in the areas of wound management and burn • Online Course: Assessment of Infection in Burn Injuries Assessment of Infection in Burn Injuries Save a life and optimise healing by accurately identifying infection in a burn wound injury Start course 2-2.5 hours - - - - Powered by Physiopedia Course instructor Diane Merwarth Diane is a physical therapist with more than 25 years of practice in the areas • Online Course: Non-Surgical Debridement of Burn Injuries -Surgical Debridement of Burn Injuries Explore non-invasive debridement techniques to prepare a burn wound for healing Start course 2-2.5 hours - - - - Powered by Physiopedia Course instructor Diane Merwarth Diane is a physical therapist with more than 25 years of practice in the areas of wound • ONLINE COURSE Burn Wound Assessment Presented by: Diane Merwarth ONLINE COURSE Burn Wound Cleansing and Solutions Presented by: Diane Merwarth ONLINE COURSE Management of Burn Wounds Programme Presented by: Diane Merwarth ONLINE COURSE Burn Wound Assessment Presented by: Diane Merwarth Burn Wound Terminology[edit \| edit source] Eschar: Eschar refers to the nonviable layers of skin or tissue indicating deep partial or full thickness injury. It is black, thick and leathery in appearance. This word is not synonymous with the word "scab". Scab: Dry, crusty residue accumulated on top of a wound, resulting from coagulation of blood, purulent drainage, serum or a combination of all. Pseudo-Eschar: A thick gelatinous yellow or tan film that forms with silver sulfadiazine cream combining with wound exudate. It can often be mistaken for eschar, but it can be removed with mechanical debridement. Petechiae: Pinpoint, round spots that appear on the skin as a result of bleeding. The spots can appear red, brown or purple in colour. Eschar formation over burn wound on hand Eschar formation over burn wound on elbow Scab Pseudo-eschar formation over burn wound on hand Pseudo-eschar formation over burn wound bed Petechiae on back of hand All photos used with kind permission from Diane Merwarth, PT Classification by Depth[edit \| edit source] For an overview on wound healing, and the anatomy and physiology of the skin, please read this article. \| \| \| \| \| \| --- --- \| Type \| Layers Involved \| Signs and Symptoms \| Healing Time \| Prognosis and Complications \| \| Superficial (formerly first-degree burn) \| Epidermis \| Red Dry Pain No blisters \| Re-epithelialisation takes 2-5 days \| Heals well Repeated sunburns increase the risk of skin cancer later in life \| \| Superficial Partial Thickness (formerly second-degree burn) \| Epidermis and extends into the superficial dermis \| Redness with a clear blister Blanches with pressure, but shows rapid capillary refill when released Generally moist Very painful Hair attachments are intact Wound bed pink to red \| Re-epithelialisation takes 1-2 weeks \| Low risk of infection unless patient is compromised No scarring typically Oedema is common \| \| Deep Partial Thickness (formerly deep second-degree burn) \| Extends into deep (reticular) dermis \| Appears yellow or white. Less blanching than superficial. Sluggish capillary refill indicates vascular damage Hair attachments are intact Pain is often absent at this depth but is variable Blisters are uncommon Often moist and waxy Wound bed shades of red, yellow, white \| Re-epithelialisation takes 2-5 weeks. Some require surgical closure \| Scarring, contractures (may require excision and skin grafting) Oedema Circumferential burns at risk for compartment syndrome Increased risk of infection due to depth and impaired blood flow \| \| Full Thickness (formerly third-degree burn) \| Extends through entire dermis and into the hypodermis \| Shades of brown, tan, waxy white, cherry red, sometimes with petechiae Appearance can vary from waxy white, leathery grey or charred black. Skin is dry, lacking in elasticity No blanching Not painful (nerve ending damage is common) Stiff and white/brown Initially painfree Hair attachments absent No blanch response indicates capillary destruction \| Prolonged (months) and usually requires surgical interventions to ultimately close \| Increased risk of infection due to capillary destruction Eschar, or the dead, denatured skin, is removed Results in scarring and contractures \| \| Subcutaneous (formerly fourth-degree burn) \| Destruction of dermis and hypodermis, and into underlying fat, muscle and bone \| Charred with eschar Dry No elasticity Initially painfree Hair attachments absent No blanch response indicates capillary destruction \| Does not heal on its own Requires surgery and reconstruction \| Amputation Significant functional impairment Death \| Superficial burn wound Superficial partial burn wound Superficial partial burn wound Superficial partial burn wound Deep partial burn wound Deep partial burn wound Resulting scars of a deep partial burn wound Full thickness burn wound Full thickness burn wound Full thickness burn wound Resulting scars of a full thickness burn wound Subcutaneous burn wound with exposed tendon Subcutaneous burn wound Subcutaneous burn wound All photos used with kind permission from Diane Merwarth, PT Circumferential burn injury special considerations[edit \| edit source] A circumferential burn wound is typically found around an extremity or the torso and puts the patient at a significant risk for compartment syndrome. This pattern of burn injury involves deep partial thickness, full thickness, and or subcutaneous burns. Circumferential burn injury signs and symptoms for potential compartment syndrome: Out of proportion pain with any movement distal to the circumferential injury. Diminished or lack of a pulse distal to the area of circumferential injury. Diminished or lack of capillary refill in the fingers and the toes. However, assessment for compartment syndrome can be limited if the injury prevents assessment of capillary refill due to extremity damage or amputation. A red flag sign of developing compartment syndrome is a decrease in temperature of the tissue distal to the area of circumferential injury, especially on an extremity. For patients with circumferential burn injuries around the torso: high concern for development of compartment syndrome if they experience difficulty breathing or an increase in difficulty breathing. Used with kind permission from Diane Merwarth, PT Used with kind permission from Diane Merwarth, PT If a patient is experiencing the signs and symptoms of compartment syndrome, the medical team should be immediately alerted for further assessment and intervention. A bedside or surgical escharotomy will be needed to relieve the resulting pressures of compartment syndrome. Blanch Test[edit \| edit source] The blanch test is similar to the capillary refill test. It is a bedside exam to assess blood flow to the capillaries of the skin. This can be performed over intact skin or in a wound bed itself. To perform the test: Gently but firmly compress the tissue to be tested until it turns white. Record the time taken for the area to return to the previous colour. Refill time should take 3 seconds or less. If the refill time is longer, suspect capillary damage. If there is no change in colour with applied pressure, suspect capillary destruction. The following optional video includes a demonstration of the blanch test. PAI: Your AI Clinical Assistant Powered by Physiopedia Make confident clinical decisions, cut documentation time and prioritize patient care Yes please! Jackson's Burn Wound Model[edit \| edit source] Jackson's Burn Wound Model is a model used to understand the pathophysiology of a burn would. This model divides the wound into three zones. Zone of Coagulation: (outlined in purple below) This is the central area of the injury and has experienced the greatest amount of tissue damage. It is often characterised by complete destruction of the capillaries leading to cell death. This is irreversible as there is no capillary refill. Zone of Stasis or Zone of Ischaemia: (outlined in green below) This area is adjacent to the zone of coagulation and as the name suggests, it is a zone in which the there is slowing of circulating blood due to the damage. These are areas of deep partial thickness burns, or burns of indeterminate depth. This zone can usually be saved with the correct wound care. Capillaries are often compromised by oedema due to hypovolemia and/or vasoconstrictive mediators responding to injury. It is reversible if capillary flow can be restored. Zone of Hyperaemia: (outlined in blue below) This zone is located around the edge of the previous zone and is characterised by superficial and superficial partial thickness burns and has a robust capillary refill. This is an area of increased circulation due to vasodilators, such as histamine, that are released in response to the burn injury. This tissue has a good recovery rate, as long as there are no complications such as severe sepsis or prolonged hypo-perfusion. This area will completely recover without intervention unless complications occur. Zone of coagulation outlined in purple; zone of stasis in green; zone of hyperaemia in blue Zone of coagulation outlined in purple; zone of stasis in green; zone of hyperaemia in blue Burn Wound Conversion[edit \| edit source] Burn Wound Conversion: True burn wound conversion is a deterioration of the wound due to events unrelated to the initial burn injury. This refers to the worsening of tissue damage in a burn wound which previously was expected to spontaneously heal, but instead it increases in depth to a deeper wound which may require surgical intervention. Potential Causes: Dessication Infection Oedema Example of wound prior to desiccation Example of wound after desiccation Example of wound prior to infection Example of wound after infection has set in Example of wound oedema in hand and fingers All photos used with kind permission from Diane Merwarth, PT Total Body Surface Area[edit \| edit source] Total body surface area is an important figure when applying the Parkland Burn Formula. This formula is the most widely used formula to estimate the fluid resuscitation required by a patient with a burn wound upon on hospital admission. It is usually determined within the first 24 hours of admission. When applying this formula, the first step is to calculate the percentage of body surface area (BSA) damaged. This is most commonly calculated using the "Wallace Rule of Nines". When conducting a paediatric assessment, the Lund-Browder Method is commonly used, as children have a greater percentage surface area of their head and neck compared to an adult. The formula recommends 4 millilitres per kilogram of body weight in adults (3 millilitres per kilogram in children) per percentage burn of total body surface area (%TBSA) of crystalloid solution over the first 24 hours of care. 4 mL/kg/%TBSA (3 mL/kg/%TBSA in children) = total amount of crystalloid fluid during first 24 hours The latest research indicates that while this method is still in use, the fluid levels should be constantly monitored, while assessing the urine output, to prevent over-resuscitation or under-resuscitation. Calculation of Percentage Burn of Total Body Surface Area[edit \| edit source] The Rule of Nine Lund-Browder Method Palmer Method 1. The Rule of Nine[edit \| edit source] \| \| \| --- \| \| Body Part \| Percentage for Rule of Nine \| \| Head and Neck \| 9% \| \| Entire chest \| 9% \| \| Entire abdomen \| 9% \| \| Entire back \| 18% \| \| Lower Extremity \| 18% each \| \| Upper Extremity \| 9% each \| \| Groin \| 1% \| 2. Lund-Browder Method[edit \| edit source] 3. Palmar Surface Method[edit \| edit source] The "Rule of Palm" or Palmar Surface Method can be used to estimate body surface area of a burn. This rule indicates that the patient's palm (with the exclusion of the fingers and wrist) is approximately 1% of the patient's body surface area. When a quick estimate is required, the percentage body surface area will be the number of the patient's own palm it would take to cover their injury. It is important to use the patient's palm and not the provider's palm. Related articles Introduction to Burns - Physiopedia What is a Burn? A burn is an injury to the skin or other organic tissue primarily caused by exposure to heat or other causative agents (radiation, electricity, chemicals). It is the result of energy transfer to the body. According to the World Health Organization (WHO), it is a global public health problem, causing an estimated 180,000 deaths annually. It is among the leading causes of disability in low and middle-income countries and almost two-thirds of burns occur in the WHO African and South-East Asia regions. Burn injuries occur in all genders, however females have slightly higher rates of death as compared to males. They also affect all age groups and are the fifth most common cause of non-fatal childhood injuries. Burn injuries do not only affect the skin. They can also affect tissue, organ and system networks (e.g. from smoke inhalation) and have psychological effects. For a review of skin anatomy and physiology, please read this article. For a list of wound care terminology, please see this article. Types of Burn Injuries[edit \| edit source] 1. Thermal Burns[edit \| edit source] Thermal burn injuries are caused by exposure to an external heat source or hot liquids. An external heat source can be a hot, solid object or even a cold object. Scalds are caused by something wet, such as hot water, steam from hot water or cold water. The types of thermal burns are: 1.1 Flame Burns[edit \| edit source] Flame burns are caused by exposure to an open fire. These burns are often associated with an inhalation injury and trauma. They tend to be mostly full-thickness burns. Flame burns are common in adults, but they are also associated with abuse in children, domestic violence and certain rituals. 1.2 Contact Burns[edit \| edit source] Contact burns are caused by contact with an extremely hot object or surface, commonly stoves, heaters and irons. Contact burns tend to be deep dermal or full-thickness burns. They are often seen in people with epilepsy, those who misuse addictive substances or in older people after a loss of consciousness. 1.3 Frostbite or Ice Burns[edit \| edit source] Frostbite occurs when the skin is exposed to cold, typically any temperature below -0.55C (31F), for an extended period of time. This causes the water in the cells of the skin and underlying tissue to freeze and crystalise. This crystallisation causes direct cellular injury. Indirect injury occurs when the tissue becomes ischaemic. Frostbite can affect any part of the body, but the extremities, such as the hands, feet, ears, nose and lips, are most likely to be affected. If frostbite penetrates the deeper skin layers, impacting tissue and bone, it can cause permanent damage. An ice burn (also known as a snow burn) is caused by ice or a frozen object coming into contact with the skin for an extended period of time. Prolonged exposure to freezing temperatures, snow, or high-velocity winds can increase the chance of this type of burn injury. Ice burns can also be caused in the clinical setting by applying ice or cold packs directly against the skin when treating an injury or sore muscles. 1.4 Scalds[edit \| edit source] Scalds are caused by hot liquids, such as boiling water and cooking oil coming into direct contact with the skin. Common mechanisms of injury include spilling of a hot drink or cooking oil, or being exposed to hot bath water. Scalds tend to cause superficial to superficial partial burns. Scald burns cause about 70% of burns in children. They often occur in older adults as well. This type of injury can be a sign of abuse and should be assessed accordingly. 1.5 Friction Burn[edit \| edit source] A friction burn is an abrasion injury that occurs when the skin rubs against another surface. A friction burn is not a true burn. However, because friction can generate heat, in extreme cases, a patient can present with burns to the outer layer of the skin. Common causes are rope burn, rug burn, chafing or skinning, or "road rash" from fallen cyclists or motorcycle accidents. 2. Electrical Burns[edit \| edit source] An electrical burn is an injury caused by heat, which is produced when an electrical current passes through the body. This can cause deep tissue injuries. Injury severity depends on many factors, including the pathway of the current, the resistance of the current in the tissues, and the strength and duration of the flow. AC (Alternating Current) and DC (Direct Current) are both potentially lethal, but there are differences in presentation. A DC current can cause a single strong contraction which will often expel the individual from the source, for example, touching a car battery. An AC current injury is often of a lower frequency and the patient is often unable to let go. AC injuries are more common as these are found in most home power sockets.There will be an entry and exit wound. The amount of internal damage depends on the pathway taken by the electrical current as it passes through the body. Electrical burns are often associated with cardiac arrest, ventricular fibrillation, and tetanic muscle contractions. For more information, please see AC and DC Electric Shock Effects Compared 3. Chemical Burns[edit \| edit source] Chemical burns or caustic burns are injuries that are caused by the skin coming into direct contact with a chemical agent. These can be strong acids, alkaline, or organic compounds. Chemical compounds can have different effects on human tissue depending on the following: The strength or concentration of the chemical The site of contact (eye, skin, mucous membrane) Ingestion or inhalation Skin integrity Volume of substance Duration of exposure Type of chemical acids can causes "coagulation necrosis" of the tissue coagulation necrosis is when blood flow to a certain area stops or slows, causing ischaemia and subsequent cell death alkaline burns can cause "liquefaction necrosis" liquefaction necrosis is a chemical process in which the necrotic tissue softens, becoming liquid/viscous. This can cause the burn to spread deeper than originally thought and should be monitored. 4. Radiation Burns[edit \| edit source] A radiation burn is damage due to prolonged exposure to radio frequency energy or ionising radiation. The most common type of radiation burn is sunburn caused by prolonged exposure to ultraviolet rays (UV). Radiation burns can also occur with exposure to high power radio transmitters, repeated high exposure to X-rays for medical imaging, and interventional radiology procedures or radiotherapy for the treatment of cancer. Radiation burns resulting from therapeutic radiation are rare. 5. Inhalation Burns/Injury[edit \| edit source] Inhalation injury refers to a pulmonary injury resulting from inhalation of smoke or chemical by-products of combustion. Inhalation injury results in direct cellular damage, alterations in regional blood circulation and perfusion, obstruction of the airways, and the release of pro-inflammatory cytokine and toxin release. Inhalation injuries also affect mucociliary clearance and weaken the alveolar macrophages. This injury can be split into three categories: 5.1 Heat Injury to the Upper Airway[edit \| edit source] The greatest complication of a heat injury to the upper airway is obstruction due to extensive swelling of the tongue, epiglottis, and aryepiglottic folds. Heat injuries do not commonly extend into the lower airway. Oedema/swelling can take a several hours to develop. Regular reassessment of a patient's airways is recommended as the presentation may change after the initial evaluation as fluid resuscitation commences. 5.2 Chemical Injury to the Lower Airways[edit \| edit source] Chemical combustion results in the creation of materials toxic to the respiratory tract. This may cause local irritation in the respiratory tract. Common chemical irritants created by combustion include: Smoke produces toxins that may damage both the airway epithelial cells and capillary endothelial cells which can cause acute respiratory distress syndrome Burning rubber and plastic produces sulfur dioxide, nitrogen dioxide, ammonia and chlorine, which affect the respiratory airways and alveoli Burning laminated furniture - this furniture may contain glues that release cyanide gas during combustion Burning cotton or wool produces aldehydes that are toxic to the human body For more information on chemical irritants causing acute inhalation injury: their effects and sources of exposure, please see Acute Inhalation Injury. 5.3 Systemic Toxicity due to Carbon Monoxide or Cyanide Exposure[edit \| edit source] Carbon monoxide (CO) is produced during a fire when any carbon-based product is not completely burned. When inhaled, CO binds with haemoglobin in the blood stream and reduces oxygen delivery. Diagnosis involves taking an accurate history, recording any changes in mental status, and the presence of high carboxyhaemoglobin levels. Occasionally, patients may require mechanical ventilation and treatment of shock. Common symptoms can include: Headaches Nausea and vomiting Dizziness Exhaustion Changes in mental state Chest pain Difficulty breathing Myocardial ischaemia Poisoning can lead to a brain injury with associated neurological problems. Symptoms include: Cognitive sequelae Anxiety and depression Persistent headaches, dizziness Sleep problems Motor weakness Vestibular and balance problems Gaze abnormalities Peripheral neuropathies Hearing loss Tinnitus Parkinsonian-like symptoms Cyanide is often a byproduct of burning household materials. Cyanide intoxication often occurs in conjunction with a CO inhalation injury. Cyanide intoxication lowers the lethal threshold of both cyanide and CO. Diagnosis requires an accurate history, recording changes in mental status, carboxyhaemoglobin concentrations higher than 10% and high lactate levels. For more information, please see Inhalation Injury. Burn Wound Classification[edit \| edit source] Burn injuries should be classified according to their severity, i.e. their depth and size. The characteristics of a burn injury can vary in terms of the amount of pain and colour of burn, depending on its depth. Always be on the look out for signs of inhalation burns, which are common in burns around the mouth or nose. Burns may have many complications and can cause shortness of breath, hoarseness of the voice, and stridor (noisy breathing due to airflow obstruction) or wheezing. Common symptoms are itchiness (a sign of healing), and numbness or tingling following an electrical injury. Burns can have a significant impact on an individual's mental health and this should always be taken into consideration. For more information on managing and assessing a patient's mental health following a burn injury or other trauma, please read this article. Classification by Depth[edit \| edit source] Type Layers Involved Signs and Symptoms Healing Time Prognosis and Complications Superficial (formerly first-degree burn) Epidermis Red Dry Pain No blisters Re-epithelialisation takes 2-5 days Heals well Repeated sunburns increase the risk of skin cancer later in life Superficial Partial Thickness (formerly second-degree burn) Epidermis and extends into the superficial dermis Redness with a clear blister Blanches with pressure, but shows rapid capillary refill when released Generally moist Very painful Hair attachments are intact Wound bed pink to red Re-epithelialisation takes 1-2 weeks Low risk of infection unless patient is compromised No scarring typically Oedema is common Deep Partial Thickness (formerly deep second-degree burn) Extends into deep (reticular) dermis Appears yellow or white. Less blanching than superficial. Sluggish capillary refill indicates vascular damage Hair attachments are intact Pain is often absent at this depth but is variable Blisters are uncommon Often moist and waxy Wound bed shades of red, yellow, white Re-epithelialisation takes 2-5 weeks. Some require surgical closure Scarring, contractures (may require excision and skin grafting) Oedema Circumferential burns at risk for compartment syndrome Increased risk of infection due to depth and impaired blood flow Full Thickness (formerly third-degree burn) Extends through entire dermis and into the hypodermis Shades of brown, tan, waxy white, cherry red, sometimes with petechiae Appearance can vary from waxy white, leathery grey or charred black. Skin is dry, lacking in elasticity No blanching Not painful (nerve ending damage is common) Stiff and white/brown Initially painfree Hair attachments absent No blanch response indicates capillary destruction Prolonged (months) and usually requires surgical interventions to ultimately close Increased risk of infection due to capillary destruction Eschar, or the dead, denatured skin, is removed Results in scarring and contractures Subcutaneous (formerly fourth-degree burn) Destruction of dermis and hypodermis, and into underlying fat, muscle and bone Charred with eschar Dry No elasticity Initially painfree Hair attachments absent No blanch response indicates capillary destruction Does not heal on its own Requires surgery and reconstruction Amputation Significant functional impairment Death Information for table modified from: Merwarth, D. Management of Burn Wounds. Burn Wound Assessment Course. Plus. 2022. Total Body Surface Area[edit \| edit source] Total body surface area is an important figure when applying the Parkland Burn Formula. This formula is the most widely used formula to estimate the fluid resuscitation required by a patient with a burn wound upon on hospital admission. It is usually determined within the first 24 hours of admission. When applying this formula, the first step is to calculate the percentage of body surface area (BSA) damaged. This is commonly estimated using the "Wallace Rule of Nines" in adults and the Lund-Browder Method in children. The Lund-Browder Method uses different percentages to the Wallace Rule of Nines as children have a greater percentage surface area of their head and neck compared to an adult. The Parkland Burn Formula recommends 4 milliliters per kilogram of body weight in adults (3 milliliters per kilogram in children) per percentage burn of total body surface area (%TBSA) of crystalloid solution over the first 24 hours of care. 4 mL/kg/%TBSA (3 mL/kg/%TBSA in children) = total amount of crystalloid fluid during first 24 hours The latest research has indicated that while this method is still used, fluid levels should be constantly monitored, while assessing the urine output, to prevent over-resuscitation or under-resuscitation. Calculation of Percentage Burn of Total Body Surface Area[edit \| edit source] The Rule of Nine Lund-Browder Method Palmer Method The Rule of Nine and Lund-Browder Method Percentages[edit \| edit source] When using these methods, it is important to note that there are discrepancies in the percentage assigned to each area. Initially, all areas in the Rule of Nines were assigned 9%. This has evolved over the years and there has been an attempt to break down areas to get a more accurate estimate. Always remember that the team working with the patient must agree on which method everyone will use and, once agreed upon, this should be used throughout treatment. It is also important to note that these methods give an estimate for the Fluid Resuscitation calculation. Fluid Resuscitation will need to be closely monitored and adjusted by the dietitian. Rule of Nines Lund-Browder Method Please find blank copies of Burn Wound Assessment sheets below: Burn Wound Assessment Sheet - Adult Burn Wound Assessment Sheet - Paediatric Palmar Surface Method[edit \| edit source] The "Rule of Palm" or Palmar Surface Method can be used to estimate body surface area of a burn. This rule indicates that the palm of the patient, with the exclusion of the fingers and wrist, is approximately 1% of the patient's body surface area. When a quick estimate is required, the percentage body surface area will be the number of the patient's own palm it would take to cover their injury. It is important to use the patient's palm and not the provider's palm. Jackson's Burn Wound Model[edit \| edit source] Jackson's Burn Wound Model is a model used to understand the pathophysiology of a burn would. This model divides the wound into three zones. Zone of Coagulation: This is the area in the central part of the injury. This part of the burn experiences the greatest tissue damage. Zone of Stasis or Zone of Ischaemia: This area is adjacent to the zone of coagulation and as the name suggests, it is a zone in which the there is slowing of circulating blood due to the damage. This zone can usually be saved with the correct wound care. Zone of Hyperaemia: This zone is circumferential and is characterised by superficial and superficial partial thickness burns and has a robust capillary refill. This is an area of increased circulation due to vasodilators, such as histamine, that are released in response to the burn injury. This tissue has a good recovery rate, as long as there are no complications, such as severe sepsis or prolonged hypo-perfusion. For more information please see this article: A Systematic Review on Classification Identification and Healing Process of Burn Wound Healing Summary of the Pathophysiology of Burns[edit \| edit source] Please watch the video below for a summary of the pathophysiology of burns: Burns Overview - Physiopedia Introduction A burn is an injury to the skin or other organic tissue primarily caused by exposure to heat or other causative agents (radiation, electricity, chemicals). According to WHO, it is a global public health problem, accounting for an estimated 180,000 deaths annually. It is among the leading causes of disability in low and middle-income countries and almost two-thirds occur in the WHO African and South-East Asia regions. Burns do not only affect the skin, they can have other effects on the tissue, organ and system networks such as smoke inhalation, as well as psychological effects. Burns affect all genders although females have slightly higher rates of death from burns compared to males. They also affect all age groups and are the fifth most common cause of non-fatal childhood injuries. Types of Burns[edit \| edit source] Electrical Burn[edit \| edit source] Electrical burn injury is caused by heat that is generated when the electrical energy passes through the body causing deep tissue injury. The magnitude of the injury depends on the pathway of the current, the resistance of the current flow through the tissues, the strength, and the duration of the current flow. The different types of current causes various degrees of injury. For example, an alternating current is more dangerous than a direct current and it is often associated with cardiac arrest, ventricular fibrillation, and tetanic muscle contractions. Thermal Burn[edit \| edit source] Thermal burn injuries are caused by external heat sources (hot or cold), scalds (hot liquids), as a result of energy transfer, hot solid objects, steam and cold objects. The types of thermal burns are: Scalds - Scald burns result in about 70% of burns in children. They also often occur in elderly people. The common mechanisms are spilling hot drinks or liquids or being exposed to hot bathing water. Scalds tend to cause superficial to superficial partial burns. Flame - Flame burns are often associated with inhalation injury and trauma. They comprise 50% of adult burns and tend to be mostly deep dermal or full-thickness burns. Contact Burns - These types of burns are commonly seen in people with epilepsy or those who misuse alcohol or drugs or in elderly people after a loss of consciousness. Contact burns tend to be deep dermal or full-thickness burns. They occur after contact with an extremely hot object or surface. Frostbite - Occurs when the skin is exposed to cold for a long time, causing the freezing of the skin or other underlying tissue. It is due to direct cellular injury from the crystallisation of water in tissue and indirect injury from ischemia. Chemical Burn[edit \| edit source] A chemical burn injury is caused by tissue contact with chemical agents such as strong acids, alkaline, or organic compounds. Chemical agents depending on the duration of exposure and the nature of the agent have different effects on the skin. For example, contact with acid causes coagulation necrosis of the tissue (whereby the architecture of the dead tissue can be preserved), while alkaline burns generate liquefaction necrosis (whereby the tissue is transformed into a liquid, viscous mass). Systemic absorption of some chemicals is life-threatening, and local damage can include the full thickness of skin and underlying tissues. Radiation Burn[edit \| edit source] Radiation burn is damage to the skin or other biological tissue and organs due to prolonged exposure to radiation. It is the least common burn injury and the most common type of radiation burn is the sunburn caused by prolonged exposure to Ultraviolet rays (UV). Other causes are associated with the use of ionising radiation in industry, high exposure to radiotherapy e.g. X-ray, and nuclear energy. Radiation burns are often associated with cancer due to the ability of ionising radiation to interact with and damage DNA. Classifications of Burns[edit \| edit source] Burns can be classified according to their severity, depth, and size of the burn. Classification by Depth[edit \| edit source] Superficial-thickness or first-degree burns - Superficial thickness burns are burns that affect the epidermis only and are characterised by redness, pain, dryness, and with no blisters. Mild sunburn is an example of a superficial thickness burn. Partial-thickness or second-degree burns - These burns involve the epidermis and a portion of the dermis. Partial-thickness burns are often broken down into two types, superficial partial-thickness burns and deep partial-thickness burns. Superficial partial-thickness burns - Partial-thickness burns involve the epidermis and part for the dermis layer of the skin. Superficial partial-thickness burns extend through the epidermis down into the papillary, or superficial, a layer of the dermis. The injured site become erythematous because the dermal tissue has become inflamed. When pressure is applied to the reddened area. The area will blanch, but will demonstrate rapid capillary refill upon release of the pressure. Deep partial-thickness burns- These burns extend deeper into the dermis and cause damage to the hair follicle and glandular tissue. They are painful to pressure, form blisters, are wet, waxy, or dry, and may appear ivory or pearly white. Full-thickness or third-degree burns - These burns extend through the full dermis and often affect the underlying subcutaneous tissue. Skin appearance can vary from waxy white to leathery grey to charred and black. The skin is dry and inelastic and does not blanch to pressure, it is not typically painful due to the damage to the nerve endings. The dead and the denatured skin (eschar) are removed to aid healing, and scarring is usually severe in a surgical procedure known as escharotomy, which involves incising through sections of burnt skin to release the eschar and its constrictive effects, restore distal circulation, and enable appropriate ventilation. Full-thickness burns cannot heal without surgery. If the burn is circumferential, either on the limbs or trunk, this will result in a tourniquet or splinting effect, causing defects in limb circulation. In addition, this will probably decrease respiratory muscle movement. This occurs because the damaged tissue has become inelastic due to eschar formation. As a result, it should be treated to prevent its complications, such as distal ischemia, compartment syndrome, respiratory failure, tissue necrosis, or death. Subdermal or fourth-degree burns - These involve injury to the deeper tissues, such as muscle or bone. They are often blackened and it frequently leads to loss of the burned part. Classification by Size[edit \| edit source] Burn size is determined by one of the three techniques: The Rule of Nine, The Lund-Browder Method, The Palmar Surface. The Rule of Nine- This method is also known as the Wallace Rule of Nines because it is named after Dr Alexander Wallace the surgeon who first publish the method. The Rule of Nine is used to estimate the total body surface area (TBSA) involved in burn patients and also used to estimate fluid resuscitation required by a burns patient. The body surface estimation is by assigning percentages to different body areas. Body Part Percentage Head and Neck 9% Anterior Trunk 18% Posterior Trunk 18% Lower Extremity 18% each Upper Extremity 9% each Groin 1% Lund-Browder Method - This method is used instead of the rule of nine method for assessing the total surface area affected in children. Different percentages are used because the ratio of the combined surface area of the head and neck compared to the surface area of the limbs is typically larger in children than in adults. Palmar Surface Method - The palmar surface can be used to estimate relatively small burns or large burns. But for medium size burns, it is inaccurate. The surface area of a patient’s palm including the fingers is used to calculate the TBSA. Pathophysiology of Burns[edit \| edit source] A burns injury depending on the severity of the injury can result in both local and debilitating systemic effects on all other organs and systems distant to the burn area. Local Effect[edit \| edit source] This occurs immediately after the injury and the burn wound can be divided into three zones. Zone of coagulation: This occurs at the point of maximum damage and this zone is characterised by irreversible tissue damage due to coagulation of the constituent proteins that occurs as a result of the insult. Zone of stasis or zone of ischemia: This zone lies adjacent to the zone of coagulation and it is subject to a moderate degree of damage associated with vascular leakage, elevated concentration of vasoconstrictors, and local inflammatory reactions resulting in compromised tissue perfusion. But the integrity of the tissue in this zone can be saved with proper wound care Zone of hyperemia: This is the outermost zone. It is characterised by the eased blood supply and inflammatory vasodilation. The tissue here will recover unless there is severe sepsis or prolong hypoperfusion. Systemic Response[edit \| edit source] In severe burn injury, >30% TBSA complex reaction occurs both from the burn area and in the area distant to the burn. Cytokines, chemokines and other inflammatory mediators are released in excess resulting in extensive inflammatory reactions within a few hours of injury. The initial response depending on the size of the burn injury is similar to the inflammation that is triggered after tissue destruction such as trauma or major surgery. Different factors contribute to the magnitude of the host response, they include: burn severity (percentage TBSA and burn depth), burn cause, inhalation injury, exposure to toxins, other traumatic injuries, and patient-related factors such as age, pre-existing chronic medical conditions, drug or alcohol intoxication, and timing of presentation to medical aid. This inflammatory response leads to rapid oedema formation which is caused by increased microvascular permeability, increased hydrostatic microvascular pressure, vasodilation, and increased extravascular osmotic activity. These reactions are due to the direct heat effect on the microvasculature and to the chemical mediators of inflammation. Vasodilation and increased venous permeability at the early stage of the injury are caused by the release of histamine. Also, prostaglandin is released by damage to the cell membranes which causes the release of oxygen-free radicals released from polymorphonuclear leucocytes which activate the enzymes catalyzing the hydrolysis of prostaglandin precursor. These hemodynamic changes lead to continuous loss of fluid from the blood circulation causing increased haematocrit levels and a rapid fall in plasma volume, leading to a decrease in cardiac output and hypoperfusion on the cellular level. Burn shock occurs if fluid loss is not adequately restored. Besides burn shock, the burn injury can result in other types of injury which include inhalation injury. Inhalation injury is caused by heat or inhalation of smoke or chemical products of combustion leading to various degrees of damage. Usually, it is present in conjunction with the burn and can range from a minor injury to a severe injury. Inhalation injury can be divided into three types: systemic toxicity due to products of combustion (carbon monoxide (CO) and cyanide poisoning); upper airway thermal injury; and lower (bronchi and distal) airway chemical injury. Patients can sustain all of these in a closed-space fire. CO poisoning, more accurately categorised as a systemic intoxication, is easily diagnosed from the serum carboxyhaemoglobin level determined as part of the arterial blood gas measurement at hospital admission. In addition to the effects above, a severe burn injury has an effect on different organs and systems in the body. The effects include: Effect on the Cardiovascular System[edit \| edit source] The initial response to a severe burn injury is characterised by hypovolemia and reduced venous return. This concomitantly leads to a decrease in cardiac output, increased heart rate, and peripheral resistance. In addition to the cardiac dysfunction, pulmonary resistance increases causing an increase in right and left-ventricular work-load. Effect on the Respiratory System[edit \| edit source] Following smoke inhalation, inflammatory mediators are released in the lungs leading to bronchoconstriction and adult respiratory distress syndrome. Effect on the Renal System[edit \| edit source] The renal system is affected following alterations in the cardiovascular system. Renal blood flow and glomerular filtration rate are reduced secondary to hypovolemia, diminished cardiac output, and the effects of angiotensin, vasopressin and aldosterone. These alterations are usually translated in the form of oliguria as an early sign of renal compromise. Failure to promptly and adequately manage these cases may lead to acute tubular necrosis, renal failure, and mortality. Effect on the Liver[edit \| edit source] There is substantial depletion of the hepatic proteins, alterations in serum levels of triglycerides and free fatty acids are highlighted, both of which are significantly increased secondary to a decrease in fat transporter proteins rendering the liver susceptible for fatty infiltration and hepatomegaly with resultant increased risk of sepsis and burn mortality. Effects on Gastrointestinal System/Metabolism[edit \| edit source] The basal metabolic rate increases up to three times its original rate. This coupled with splanchnic hypoperfusion, necessitates early and aggressive enteral feeding to decrease catabolism and maintain gut integrity. It causes mucosal atrophy, reduced absorptive capacity, and increased surface permeability. Effect on the Endocrine System[edit \| edit source] The stress hormones i.e. catecholamine, glucagon and cortisol among other hormones are actively involved at the onset of burns injuries. These hormones display an exponential increase in their levels; sometimes reaching 10 fold their normal values. The significance of such an upsurge resides in its influence on the cardiovascular system and the resultant fluid shifts that follow these changes. The stress hormones are thereby considered as the initiators of the hypermetabolic-catabolic and proteolytic-response. Burn Prevention[edit \| edit source] Recommendations from the World Health Organization for individuals, communities and public health officials on how to reduce burn risk. Enclose fires and limit the height of open flames in domestic environments. Promote safer cookstoves and less hazardous fuels, and educate regarding loose clothing. Apply safety regulations to housing designs and materials, and encourage home inspections. Improve the design of cookstoves, particularly about stability and prevention of access by children. Lower the temperature in hot water taps. Promote fire safety education and the use of smoke detectors, fire sprinklers, and fire-escape systems in homes. Promote the introduction of and compliance with industrial safety regulations, and the use of fire-retardant fabrics for children’s sleepwear. Avoid smoking in bed and encourage the use of child-resistant lighters. Promote legislation mandating the production of fire-safe cigarettes. Improve the treatment of epilepsy, particularly in developing countries. Encourage further development of burn-care systems, including the training of health-care providers in the appropriate triage and management of people with burns. Support the development and distribution of fire-retardant aprons to be used while cooking around an open flame or kerosene stove. Conclusion[edit \| edit source] Burns injuries have physical, socio-economic, and psychological effects especially in cases of severe burns injuries. They impact not only the affected part of the body, but also the organs and systems of the body. They require an early and prompt response to reduce the effect of an injury. Besides this, they require an interdisciplinary approach to prevent the adverse effects of the injury. Rehabilitation of Burns in Disasters and Conflicts - Physiopedia Introduction Burn injuries are complex and involved injuries that require immediate and specialised interventions. Burn injuries commonly require a prolonged rehabilitation process to return to functional independence, often with adaptation or compensatory training in the long term. These patients will require both physical and psychological support throughout their rehabilitation process. This is especially true when an injury occurs as a result of disaster or conflict. Acutely, rehabilitation will be supportive of the medical needs of the patient. However, rehabilitation plays an important role in the acute phase to prepare the patient both physically and mentally for the therapy to come. This article will focus on the physical rehabilitation of patients who have had a burn injury, but will include important wound care considerations with regards to mobility and positioning. Rehabilitation for burn injuries starts from day one of the injury, right through the period of scar maturation, and often for years after the injury, especially relevant to the prevention of contractures and in children where growth is not complete. Burns Overview[edit \| edit source] Please read the linked article for background knowledge on the anatomy and physiology of the skin. Understanding the structure of the skin is an important part of burn classification. Please read the linked article for background knowledge of wound healing. Understanding the expected timeline of wound healing and the body's response to injury is vital in patient education and differentiating normal wound healing from the signs and symptoms of infection. This article also contains links to wound assessment and wound debridement to better understand these procedures. Wound care is a specialised skill and should not be performed without the proper training. Physiotherapists can specialise in wound management with advanced training. Types of Burns[edit \| edit source] Burn injuries have several common causes which include, but are not limited to: Thermal burns: They occur due to heat sources that raise the temperature of the skin and surrounding tissues. This causes tissue cell death or charring. Heat sources can include hot metals, scalding liquids, steam, and flames. Thermal burns can also be caused by exposure to extreme colds such as a frostbite injury. Radiation burns: Radiation burns occur due to prolonged exposure to ultraviolet rays or to other sources of radiation. Radiation sources can include the sun and X-rays. Chemical burns: These happen due to strong acids, alkalies, detergents, or solvents that come into contact with the skin. Electrical burns: Electrical burns occur due to electrical current, either alternating current (AC) or direct current (DC) coming into contact with the body. These burns have increased potential for internal damage. Classification of Burns[edit \| edit source] Depth of burn classification Table.1 Depth of Burns Characteristics adapted from Lathia et al. 2020. Depth of burn Tissues destroyed Burn Appearance Pain Sensitivity Healing Time and Prognosis Superficial i.e. 1st degree burn Outer layer of epidermis Red Blistering is uncommon Slight oedema Capillary refill: affected area blanches with pressure and refills (see video example below) Painful Less than 14 days No long term scarring expected Superficial Partial Thickness (SPT) i.e. 2nd degree burn: superficial/intermediate All of epidermis Upper layers of dermis Some hair follicles and sweat and sebaceous glands destroyed Red Blisters Moist subcutaneous Oedema Capillary refill Very painful and hypersensitive 7 - 20 days May scar in rare cases Pigment changes Deep Partial Thickness (DPT) i.e. 2nd degree burn: deep Epidermal and several dermis damage Most nerve endings, hair follicles, and sweat glands destroyed Variable in colour (mottled) Wet or waxy dry Generally blisters No or slow capillary refill Eschar forms Less sensitive to pain due to destroyed nerve endings At least 21 days healing time but difficult to determine Scarring Risk of contractures May require grafting Full Thickness Burn (FTB) i.e. 3rd degree burn All skin layers damaged or destroyed; fat or bone may be visible. White, charred, dry, inelastic No blisters No pain from lost cutaneous pain receptors BUT situation is often painful for the patient. Severe scarring Risk of contractures No skin regeneration Need excision and grafting Prolonged hospitalisation Below is a video which goes into greater detail on burn classification and staging. It also includes examples of the burn's appearance. Next is a video of the capillary refill test being performed on a healthy finger. Note the blanching of the skin when pressure is removed and how long it takes for the colour to return. The Capillary Refill Test (CRT) is a rapid test used for assessing the blood flow through peripheral tissues. It is a quick test performed on the nail beds to monitor the amount of blood flow to tissues and dehydration. The CRT measures the efficacity of the vascular system of hands and feet as they are far from the heart. Burn Assessment[edit \| edit source] Total Body Surface Area (TBSA) burned: the area of the body that is affected by a burn. This is expressed in a percentage, for example if a chart reads 20% TBSA burn, this would mean 20% of the total surface area of that patient was affected by a burn. Different forms are used for adults and children For adults, a major burn involves 30% TBSA or more, children 20% TBSA or more. Location and type of burn also influence the severity and have functional implications. For example: deep burns on the hands record as a small TBSA but have a huge functional impact for that patient. The two most common recording methods of TBSA are the Rule of Nines and Lund and Browder. Examples of these two methods are shown below as reference. Example of Rule of Nines and Lund-Browder Assessments Immediate Medical Care Needs[edit \| edit source] In the acute care phase of major burns injuries, medical management will include: Fluid resuscitation Airway management Wound debridement and or surgical procedures such as escharotomy or fasciotomy Pain management Acute Rehabilitation for Burn Injury[edit \| edit source] Acute Burn Injury Mobility Precautions[edit \| edit source] Burns encountered in conflict zones or disaster areas are often combined with trauma injuries such as fractures, internal injuries, or brain and or head injuries. Appropriate mobility precautions such as extremity weightbearing status, spinal and or cervical immobilisation, or bedrest needs in the case of internal organ injuries or bleeding need to be determined with the medical team. If shrapnel is present in the burn/wound and cannot be easily removed due to the risk of further tissue damage, it may be left in place. Defer to the medical team regarding mobilisation of the affected area in these cases. Pain will develop following surgical debridements, plan accordingly with the pain medication schedule. Acute Rehabilitation[edit \| edit source] Rehabilitation should begin as soon as the patient is medically stable. Research has found that early initiation of limb positioning, mobilisation and splint use has a positive outcome in subsequent development of contractures in patients with burns. Acute rehabilitation for burn injury focuses on: Respiratory Care Oedema Management Positioning, Splinting and Pressure Relief Early Mobility Progressive Graded Exercises Maintaining Functional Independence Patient and Caregiver Education Respiratory Care[edit \| edit source] The aims of physiotherapy for respiratory care involve: airway maintenance, secretion removal, gas exchange improvement, prevention and or treatment of atelectasis, and maintenance of thoracic expansion. For deeper knowledge on these topics, please read more about inhalation injury and chest physiotherapy. Oedema Management[edit \| edit source] Oedema is a normal response to injury and an important step in wound healing. However, excessive oedema can negatively effect wound healing. Acute oedema management includes: Appropriate extremity positioning Use of muscle pump action through active AROM Facial oedema requires a patient to sit up to at least 45 degrees 24 hours/day Expect dressings used for oedema management to be firm but allowing for AROM of all joints Positioning, Splinting, Pressure Relief[edit \| edit source] Proper and correct positioning is essential for contracture prevention. Positioning for contracture prevention should be used prophylactically even when there is no sign of lost ROM upon initial assessment of affected area. Depending on the severity of the burn injury, splinting and positioning programmes may need to be continued for at least six months for optimal results. Patients with burns will typically wish to go into a position of comfort, which is generally a flexion pattern. Developing scar tissue will result in shortened muscle lengths if patients are allows to maintain these positions over time. Scar tissue after a burn can develop as quickly as over a few hours. Anti-Deformity Positioning: Shoulders: abducted to 90 degrees, horizontally adducted to 20 degrees, encourage external rotation Scapula: retracted, depressed, supinated Upper extremities: neutral rotation, forearms supinated Elbows: extended Wrist: 30-40 degrees extension, MP 45-70 degrees flexion, IP extension, thumb abducted and opposed Neck: slightly extended, no pillow Hips: slight abduction with full extension, block external rotation Bony landmarks and potential courses of pressure Patients with burns are at high risk of further skin breakdown at areas of increased pressure. Offload and provide appropriate pressure relief as needed. This is especially true following a skin graft or surgical flap. No pressure can be placed over an area following a skin graft or flap as occlusion in blood flow will make the new tissue nonviable. Splinting is typically indicated: To position extremities for contracture prevention To protect grafts or surgical flaps during the initial phase of healing According to a 2020 systematic review, "orthotic use should be considered as a treatment choice for improving ROM or reducing contracture in adults who have sustained a burn injury." This study also found that patients who used splints or orthotics had a lower incident for the need of reconstructive surgery later in their healing process and provided positive skin graft outcomes with appropriate orthotic use. Splints or orthotics can be used on any part or joint of the body. They are most commonly used for positioning of the mouth (42%), then neck (12%), hands and axillae (10% each), ankle (8%), elbow (4%), and knee (2%). Unfortunately, there is no standardised wearing time or schedule for burn contracture management. Studies in non-burn-specific literature evaluating hand and finger contracture management using dynamic splinting recommend orthosis wear at end range of motion for greater than 6 hours per day, for a minimum of 12 weeks. Please read this article for more in-depth information on splinting for burns. Early Mobility, Exercise, Functional Independence[edit \| edit source] The goal of early mobilisation and therapy interventions after a burn is to maintain functional mobility and endurance while the body heals. The degree of burn will dictate therapy intensity. The areas affected by the burn will require gentle ROM, which should be slow and smooth to decrease associated pain and inflammation with movement. Currently there is little research on early mobilisation following acute burns. A 2019 meta-analysis and systematic review of trauma patients in the ICU found that early mobilisation decreased time of mechanical ventilation, but had similar mortality and hospital length of stay compared to those patients that did not receive early mobility. However a 2020 retrospective study of trauma patients in the ICU not only states that early mobility is a safe, feasible and effective strategy to improve functional outcomes, but also that patients who received early mobility were less likely to die in both the ICU and hospital. Please read this article for more in-depth information on early mobilisation of patients in the ICU. While further research is needed on the effectiveness and risks/benefits of early mobility of patients after burns, data from similar populations suggest the benefits may outweigh no mobility intervention. Please read this article for more in-depth information on post-burn rehabilitation. Education[edit \| edit source] Due to the long healing and rehabilitation timeframe following a burn, patient and caregiver education is of vital importance for successful burn management and rehabilitation. This education and training should include: ROM and stretching exercises with special consideration given to a patient's fear of movement due to pain Contracture aetiology and prevention Functional activities and mobility to include assistive device use Positioning and pressure relief techniques Splint use and wearing schedule Infection control Anticipated healing times as a patient moves from the acute phase into the subacute phase and beyond Treatment Red Flags[edit \| edit source] Rehabilitation therapists such as physiotherapists often have the most direct contact with their patients. Therefore, it is important to monitor these patients for treatment red flags and alert the medical team as appropriate. Hypovolemic shock Infection: this is the most common cause of death of patients with burns who survive the initial injury. Compartment syndrome Inadequate pain management Familiarise yourself with the signs and symptoms of sepsis. Resources[edit \| edit source] Rehabilitation in Conflict and Disasters Field Support[edit \| edit source] Responding Internationally to Disasters: Do’s and Don’ts Early Rehabilitation in Conflict and Disasters, Humanity and Inclusion Rehabilitation in Sudden Onset Disasters, Humanity and Inclusion Rehabilitation Treatment Planning Tool for Common Conflict and Emergency Related Injuries Burns Specific Support[edit \| edit source] International Society for Burn Injuries Burn Foundations and Partnerships American Burn Association Technical Standards for Medical Teams[edit \| edit source] Minimum Technical Standards and Recommendations for Rehabilitation in Emergency Medical Teams Minimum Technical Standards and Recommendations for TBI Rehabilitation Teams in Sudden-onset Disasters Assessment of Infection in Burn Injuries - Physiopedia Introduction Burn wound injuries place critical economic burden on healthcare infrastructures worldwide. They are also associated with high mortality rates due to severe complications. Infection is the most common complication in burn wound injuries. Thus, prompt and precise diagnosis is of critical importance for this patient population to prevent detrimental consequences and optimise healing outcomes. Burn wound injuries are at an increased infection risk for multiple reasons: (1) the body's physiological response to a burn injury presents with many of the same signs and symptoms of a developing infection, (2) the burn injury can impair many body systems which limits a patient's innate ability to fight off infection. Physiological Response to Burn Injuries[edit \| edit source] Burn wound injuries have both local and system consequences. This requires the wound care professional to work closely with the interdisciplinary team to appropriately monitor and manage the patient in a holistic manner. Understanding the pathophysiological changes which occur after a burn wound injury will greatly improve the wound care professional's ability to prevent and/or manage infection. Local response to burn wounds[edit \| edit source] The local response involves the area of the burn injury and the tissue directly surrounding it. This is best described using Jackson's Burn Model which divides the wound into three zones. Jackson's Burn Model Zone of coagulation (also known as the zone of necrosis). This area sustains the most damage and suffers irreversible tissue loss. Zone of stasis. This area is found around the zone of coagulation. It demonstrates decreased tissue perfusion and is, therefore, potentially salvageable with proper care. This zone can also suffer complete tissue loss with prolonged hypotension, infection, or oedema. Zone of hyperaemia. This is the outermost zone of the injury. In this area, tissue perfusion is increased. This area will likely recover unless there is severe sepsis or prolonged hypoperfusion. For more information, please review this page. Systemic response to burn wounds[edit \| edit source] Burn wounds result in damage to multiple body systems, which can put the patient at higher risk for developing an infection. Cardiovascular function and burn shock Capillary permeability is increased, leading to the loss of intravascular proteins and fluids into the interstitial compartment, which leads to oedema Myocardial contractility is decreased, there is a decrease in cardiac output immediately post-burn These changes, when combined with fluid loss from the burn wound, result in systemic hypotension and end organ hypoperfusion Local vascular compromise occurs in deep, partial, and full-thickness injuries where the arterial supply is either damaged or destroyed as a result of the burn wound Respiratory changes Inflammatory mediators cause bronchoconstriction Acute respiratory distress syndrome (ARDS) can occur with severe burns Pulmonary function can be affected by oedema, regardless of if an inhalation injury occurred Metabolic changes Basal metabolic rate can increase to three times its original rate post-burn A systematic immunoendocrine response will occur after a large burn injury and can continue for up to 3-years post-injury. This response can cause: (1) immune incompetence, (2) sepsis, (3) increased fracture risk, (4) slowing of growth rate, (5) reduced organ function, (6) decreased wound healing, and (7) death Gastrointestinal changes Absorption is affected due to gastrointestinal mucosal atrophy and decreased intestinal blood flow A patient post-burn is in a constant hypermetabolic state as a result of their burn injury. Initially, this will boost the energy response and help fight off infection and begin the healing process. However, as the hypermetabolic state becomes a prolonged response, it will be an energy drain and it becomes more difficult for the patient to mount defences and continue the healing process. Early enteral feeding is paramount to prevent malnutrition Immunological changes The immune response becomes globally depressed following a burn injury, the severity is dependent on the size of the burn wound Burn wounds will cause both innate and adaptive immune responses Patient will have an excessive and prolonged inflammatory response Loss of the protective skin layer opens the body to the risk of infection Immune incompetence further depresses the body's ability to fight infection - sepsis is the leading cause of post-burn mortality "Burn shock results from the interplay of direct tissue injury, hypovolemia, and the release of multiple mediators of inflammation, with effects on both the microcirculation and the function of the heart and lungs." Additional Risk Factors for Burn Injury Infection[edit \| edit source] Additional risk factors of burn wound injuries for infection: An increased length of stay in the ICU Increased wait time for test results (ie wound cultures or biopsies) to identify the source of infection delays the use of targeted antibiotics or antimicrobial agents to most effectively treat infection Possibly a delay due to transportation time to a burn centre for skilled aggressive care A delay for surgery due to patient medical instability or lack of surgical facility Any burn greater than 20% total body surface area of deep partial and/or full thickness injury is a high risk for infection, regardless of the patient's underlying conditions Infection risk increases more when the patient is less than four years old or more than 55 years old Infection risk is increased for patients who are immunocompromised or who have pre-existing comorbidities that put them at risk for developing an infection The presence of eschar in the wound increases the risk of infection Signs and Symptoms of Sepsis in a Burn Wound[edit \| edit source] The detection of a burn wound infection has multiple layers of complexity. The normal response to a burn injury mimics many of the signs and symptoms of infection. Therefore, the ability to identify an infection is masked by the patient's response to the burn injury. To review covert and overt signs and symptoms of infection, please see this article. While infections are one of the most common complications following a burn wound injury, sepsis is the leading cause of death in both adult and paediatric burn patients. The diagnosis and management of sepsis in burns also has multiple layers of complexity. The diagnosis of sepsis in patients with severe burns (>20% total body surface area) is complicated by the overlap of clinical signs of the typical post-burn hypermetabolic response with those of sepsis. For a burn wound to be diagnosed as septic, the patient must exhibit at least three of the following: Table is modified from: Norbury W, et. al. Infection in burns. Surg Infect, 2016; 17(2): 250-255. Not age specific Adults Children Temperature >39°C or <36.5°C (>102.2°F or <97.7°F) Progressive tachycardia >110 beats/min (bpm) >2 SD [standard deviations] above age-specific norm (85% age-adjusted maximum heart rate) Progressive tachypnea Not ventilated: >25 bpm Minute ventilation >12 L/min when ventilated >2 SD above age-specific norm (85% age-adjusted maximum respiratory rate) Thrombocytopenia (will not apply until 3 days after initial resuscitation) <100,000/mcL <2 SD below age-specific norm Hyperglycemia (in the absence of pre-existing diabetes mellitus) Untreated plasma glucose >200 mg/dL or equivalent mM/L Insulin resistance. Examples include: >7 units of insulin/hour intravenous drip (adults) Resistance to insulin (>25% increase in insulin requirements over 24 hours) Inability to continue enteral feedings >24 hours Examples include: Abdominal distension Enteral feeding intolerance (residual >150 mL/h in children or 2× feeding rate in adults) Uncontrollable diarrhea (>2,500 mL/d for adults or >400 mL/d in children) Have a documented infection Documented one of the following: Culture-positive infection OR Pathologic tissue source identified OR Clinical response to antimicrobials Sources of Infection[edit \| edit source] Research shows that burn wounds are sterile for the first 6-12 hours after initial injury. However, contamination occurs soon there after and the proliferation of bacteria increases rapidly. Sources of contamination in a burn wound: Normal skin flora Endogenous sources, such as the lungs or the gut, or any mucosal membranes where bacteria thrive in a normal system Exogenous sources, such as the environment or cross-contamination Bacterial infections[edit \| edit source] For more in-depth information on bacterial infections, please read this article. Rough timeline of bacterial concentrations in a burn wound: Early phase of infection: gram-positive usually the first identified Staphyloccus aureus is typically the first pathogen found in burn wound infection, this includes Methicillin-resistant Staphylococcus aureus (MRSA) Staphylococcus aureus is also typically the first microbe identified in sepsis and in arterial bacterial dissemination from the wound At five days from the time of injury: gram-negative bacteria become predominant, and Pseudomonas aeruginosa is the primary bacteria identified. It is the most common pathogen found in burn wounds and is very common in the development of biofilms. Later stages: development of fungal or yeast infection independent or in combination with an already existing bacterial infection Fungal infections[edit \| edit source] For more in-depth information on fungal infections, please read this article. Fungal infections are also a major concern for patients with burn wound injuries. It can often be difficult to distinguish a fungal infection from a bacterial infection in a burn wound by appearance alone. Risk factors for developing a fungal infection in a burn wound: Total body surface area involvement >30% Long hospital length of stay Multiple comorbidities Clinical observations for a fungal infection in a burn wound: Changes in the wound appearance An unanticipated separation of eschar is a key sign of fungal infection Rapid conversion from a partial thickness burn to a deep partial or a full thickness burn wound A blackening of the burn wound tissue A persistent ongoing fever which remains unchanged with antibiotic therapy Candida albicans and Aspergillus are the most common fungal pathogens. Aspergillus can be a very serious fungal infection and requires urgent assessment and potential surgical debridement. Example of fungal infection in wound. Image used with kind permission of Diane Merwarth PT. Sample collection[edit \| edit source] Biopsy[edit \| edit source] Wound biopsies are an important diagnostic component in the management of chronic wounds to monitor for potential malignancy or infection. The tissue biopsy, especially in burn wounds, has long been recognised as the gold standard. It is the most quantitative method of collecting tissue and identifying the level of bacterial contamination or infection in a wound. Taking a wound biopsy requires sampling at the wound edge and in the wound bed. The procedure is invasive and requires skill on the part of the caregiver to collect the sample. Biopsies require extra time for results to become available but allow for accurate targeted antibiotics or antimicrobials to be used against the infection. It has been found that bacteria are not homogeneously present in the wound bed, but they are in pocketed in different areas. Therefore, multiple biopsies should be collected from a single burn wound. It is also recommended to take wound biopsies for wounds that have not responded to treatment after 2–6 weeks. Semiquantitative swab culture[edit \| edit source] There is controversy over the effectiveness of performing a swab culture and if the results will represent the infection. However, there is research which shows a correlation between bacteria identified in a swab culture and the bacteria identified from a tissue sample, specifically a biopsy. A swab culture will not give the level of quantitative results as a biopsy, but if done appropriately, the results will identify the microbes that need to be targeted with antimicrobial interventions. Benefits of performing a swab culture: Less invasive than a biopsy Can be easily performed by the bedside clinician Provides quicker results which allow more timely targeted antimicrobial interventions General Guidelines for the Treatment of Burn Wound Infections[edit \| edit source] Unfortunately, there are no standardised protocols for treatment of a burn wound infection. Topical antimicrobial agents[edit \| edit source] These treatments need to be individualised Once a local infection has been identified, begin a topical antimicrobial agents For spreading or systemic infection, begin systemic antimicrobial agents More than one broad-spectrum antibiotic may be initiated while awaiting tissue sample results Fungal infections will require topical and antifungal agents, with the potential for surgical debridement Due to the nature of rapidly forming biofilms, treatment may include biofilm-based wound management Surgical interventions[edit \| edit source] Will be needed to to remove biofilm population. The wound usually requires frequent surgical debridement. It has been found that a biofilm will start to reform within 24 hours of being disrupted, and it becomes mature in three days. Therefore, it is important to follow up with appropriate biofilm-based wound management. Fungal infections may require surgical debridement Early excision of eschar, which provides a good environment for bacterial proliferation Skin grafts to close a wound immediately and decrease exposure to new sources of infection Infection Prevention[edit \| edit source] Vigilance Close observation for overt and covert signs and symptoms of infection Closely monitoring vital signs and lab values for signs of infection Accurate and thorough documentation of wound appearance and features Frequently reassess and update care plan based on the patient's response to current interventions Manage patient's comorbidies Maintain optimal nutrition and water intake Oral intake is the most valuable and the most effective method A nasogastric tube is the second most effective method Enteral nutrition is the least favourable because it doesn't utilise the gut Manage oedema Minimising oedema by elevating the extremities Applying compression bandages to the extremities if tolerated Provide psychosocial support and education A positive mental attitude can aide in wound healing and infection prevention Include the patient in care plans Temperature control Prevent the patient from being too hot or too cold, so adjust the temperature in the patient's room or in the operating room (OR). Often hospitals are set a minimum of 29-31°C (85-88 °F) room temperature to prevent patients with a burn wound from becoming too cold. Dressing considerations: Wounds can cool down when dressings become saturated with drainage or when left exposed to air too long during a dressing change. Recommended strategies for minimising the onset of infection in medical facilities Universal precautions, proper use of personal protective equipment (PPE), and hand hygiene Patients at high risk for developing an infection should be placed in a negative pressure room with strict isolation precautions. Wound care should be done with aseptic (sterile) technique when possible, otherwise clean (non-sterile) techniques should be used. Strict aseptic technique should be used when inserting any kind of device (intravenous line, intra-arterial line, foley catheter). It is also recommended for these devices to be removed as soon as they are no longer necessary to support the patient's recovery from their burn injury. Environmental control to include: (1) a complete cleaning of the patient's room twice/day with wipe down of all high-touch surfaces, (2) terminally cleaning the room anytime the patient is not present, (3) after the patient has been discharged from the room, it is recommended to complete three days of terminal clean prior to use by another person. Surveillance cultures are recommended by many burn centres This includes nasal cultures, cultures of the wound, possibly anal cultures with the onset of vancomycin-resistant enterococcus (VRE) It is recommended that cultures be collected on admission and weekly throughout the hospital stay Resources[edit \| edit source] Optional Additional Recommended Reading: Hettiaratchy S, Dziewulski P. Pathophysiology and types of burns. Bmj. 2004 Jun 10;328(7453):1427-9. Nunez Lopez O, et. al. Predicting and managing sepsis in burn patients: current perspectives. Therap. Clin. Risk Mngmt, 2017; 17: 1107-1117 Optional Video: Please view this optional video for an overview of burn wound anatomy and classification, as well as a detailed description of burn wound pathophysiology. Burn Wound Injury Dressing Selection - Physiopedia Burn Wound Injury Standard of Care Please see this document for a growing list of wound care terminology and definitions. For a review of other steps in burn wound care, please see the following articles: Burn wound injury assessment Assessment of infection in burn wounds Burn wound injury cleansing techniques and solutions Non-surgical debridement of burn injuries Current Standard of Care[edit \| edit source] The current standard of care for large and deep burn wounds is (1) early surgical excision and (2) wound closure. Wound closure can be achieved by skin grafting or temporary biological coverage for deep partial and full thickness burns. Alternatives to Early Excision and Wound Closure[edit \| edit source] Surgical debridement followed by wound cleansing and regular dressing changes preferred option for burns of indeterminate depth, where there are areas of deep partial thickness, full thickness or superficial burn injury by performing ongoing wound care and allowing the more superficial areas to re-epithelialise, those areas that need debridement and skin grafting can be more easily defined Local wound cleansing and dressing changes indicated when surgery is not feasible or after debridement without grafting or temporary closure common for smaller burn areas, including full thickness burns standard for most superficial partial thickness burns Exceptions to Standards of Care[edit \| edit source] Surgery is not feasible due to patient status, or resource availability Grafting or skin substitutes are not appropriate due to (1) contaminated wounds, (2) native skin is too damaged or of too small an area to provide a skin graft, or (3) skin substitutes are not available Burn wounds which (1) do not undergo surgical debridement, or that (2) underwent surgical debridement without application of a skin graft or temporary cover with a skin substitute will require ongoing dressing changes throughout the course of healing. Role of Wound Dressings[edit \| edit source] In all cases where burn wounds are not grafted immediately, routine burn wound care and dressing changes are needed. The determination of dressings and frequency of interventions are based on a variety of factors that are described below. Effective burn wound dressings: absorb and manage drainage minimise the risk of burn wound conversion maintain a moist wound environment minimise peri-wound maceration prevent excessive evaporation from the wound surface that can (1) cause the wound to become desiccated and (2) result in hypothermia provide topical antimicrobial protection minimise contamination from the external environment decrease oedema protect the wound be care provider friendly (ie. easy to apply and/or remove) reduce pain during (1) removal and application of the dressings and (2) during functional activities allow movement and function Clinical Pearl: Benefits of Moist Wound Healing[edit \| edit source] Multiple studies have demonstrated that the application of moist wound dressings immediately after injury minimises the risk of burn wound conversion. The positive effects of a moist wound environment on wound healing include: increased keratin migration and re-epithelialisation increased collagen synthesis increased autolytic debridement decreased necrosis decreased pain decreased inflammation decreased scarring facilitation of cell-to-cell signaling providing a means of delivering topical treatment improved wound aesthetics after healing Determining the Dressing Care Plan[edit \| edit source] Determining Dressing Change Frequency[edit \| edit source] Based on the burn wound or patient status, and on the type of dressing: status of the burn wound (or patient): dressing changes will be daily or more than planned in the following situations: the dressing used is not antimicrobial, or the agent has short-acting antimicrobial properties verified or suspected infection significant areas of un-debrided eschar drainage is not contained by the dressing there are any other concerns it is important to monitor for conversion need to balance concern for the wound and the desire to leave the wound undisturbed and not introduce risk for additional contamination minimise risk of damage to healing tissue type of dressing or topical agent ability of the dressing to maintain a moist environment Determining Dressing and Topical Agent[edit \| edit source] Consider the following areas: depth and stage of healing of burn wound indications of infection amount of wound drainage clinical assessment of progress, or lack of progress, in wound healing ease of dressing application and removal availability of dressings and topical agents cost of topical agents and dressings dressings change as wound progresses (or doesn’t progress) Burn Wound Dressing Options[edit \| edit source] This section includes a summary of gauze and gauze-like dressings and possible solutions, creams, and ointments that can be used in the treatment of burn wound injuries. Please see this article for more information on advanced dressings for burn care. Table 1. Solutions used on dressings. Benefits Risks Saline Non-antimicrobial Non-cytotoxic Mafenide Acetate Broad-spectrum Common for the treatment of pseudomonas No antifungal coverage Monitor for metabolic acidosis Sodium Hypochlorite Broad spectrum Cytotoxic at full (Dakin’s) and half-strength Mixed reports of cytotoxicity at 0.025% concentration Hypochlorous acid Broad-spectrum Non-cytotoxic Povidone-Iodine Broad-spectrum Cytotoxic at full strength Acetic acid (various reports of strengths, 0.25% up to 5%) Broad-spectrum Used primarily for the treatment of pseudomonas Table 2. Creams and ointments used on dressings Benefits Risks Burn-specific Considerations Silver sulfadiazine Effective against gram-positive (gram +) or gram-negative (gram –) bacteria, and some yeast Transient leukocytopenia Avoid eyes and mucosal membranes Contraindications: Sulfa allergies, pregnant women, and infants <2 months of age Toxic to keratinocytes, delays re-epithelialisation Full and deep-partial thickness burns: before debridement after debridement, if graft and/or temporary coverage is not performed Mafenide acetate (sulfamyalon) Effective against gram + and gram – bacteria NOT effective against fungal infections Most effective antimicrobial agent to penetrate eschar Prolonged use, or use over large total body surface area (TBSA) may cause metabolic acidosis and respiratory complications Cytotoxic to fibroblasts and keratinocytes Delays re-epithelialisation Use can be painful, especially on more superficial burns No longer the standard of care for use with burn wounds, except for: very short-term use very small burns deep partial or full thickness burns to the ears Antibiotic ointments Triple antibiotic (eg Neosporin): typically bacitracin, neomycin and polymyxin B: effective for gram + and gram – microbes Bacitracin: effective for gram + bacteria, but NOT gram – or yeast Mupirocin: effective against gram + microbes, including MRSA Gentamicin: broad-spectrum coverage; not commonly used for burn wounds Change or discontinue the antibiotic used when: lack of effectiveness increased signs/symptoms of infection symptoms of antibiotic sensitivity or allergy Recommended primarily for superficial partial thickness burn wounds Medical grade honey Broad-spectrum activity Maintains moist environment Less toxic and less costly than silver products Antimicrobial by acidity and osmotic gradient Minimal evidence for its use with burn wounds, but research shows positive outcomes on superficial partial thickness burns Cadexomer Iodine Most effective against MSSA and MRSA Table 3. Impregnated Antimicrobial Dressings Benefits Risks Burn-specific Considerations Polyhexamethylene biguanide (PHMB)-impregnated dressing Broad-spectrum antimicrobial agent used against gram + and gram – bacteria, yeast and fungi Less cytotoxic vs other antimicrobial agents Recommended for superficial partial thickness burn wounds Bismuth (eg. Xeroform gauze) Effective against biofilm formation (inhibits polysaccharide capsule) Bacteriostatic against enteric microbes (C. diff, E. coli) Not cytotoxic Contraindicated in patients with Bismuth allergy Burn Wound Injury Areas of Special Concern[edit \| edit source] Guidelines for Blister Management[edit \| edit source] "In all of the literature that I've looked at over the past couple of decades, [blister management] hasn't changed. The approach to managing blisters is kind of a 50-50 split between that side of the issue that thinks that all blisters should be unroofed and debrided immediately and completely, and the side that thinks you should leave blisters alone and allow things to happen naturally." - Diane Merwarth, Physical Therapist, Wound Care Specialist Blisters should be deroofed (unroofed) in the following situations: During surgical cleansing and debridement When the blister itself is disrupted it has become a portal for entry of microbes there is a risk of microbes trapped under loose skin When appearance is questionable thick, cloudy or opaque fluid bloody or discoloured Blisters should be left intact in the following situations: If blisters are small and not disrupted When they are not affecting function Blisters should be drained but NOT deroofed in the following situations: Large taut blisters with clear fluid Blisters continuing to increase in size Blisters are preventing function Deroofing is the process whereby the 'roof' of the blister is removed under clean (aseptic) conditions to expose the viable tissue beneath. Intact blisters Left image: intact blister. Middle image: deroofed blister. Right image: re- epithelialization in wound bed. Left image: blister intact. Right image: deroofed blister All photos provided by and used with kind permission from Diane Merwarth, PT Resources[edit \| edit source] Clinical Resources:[edit \| edit source] Wound Antiseptics and European Guidelines for Antiseptic Application in Wound Treatment American Burn Association Guidelines for Burn Care Under Austere Conditions Optional Additional Reading:[edit \| edit source] Allorto NL. Primary management of burn injuries: Balancing best practice with pragmatism. South African Family Practice. 2020 Jan 1;62(1):1-4. References[edit \| edit source] ↑ Harish V, Li Z, Maitz PK. First aid is associated with improved outcomes in large body surface area burns. Burns. 2019 Dec 1;45(8):1743-8. ↑ Palackic A, Jay JW, Duggan RP, Branski LK, Wolf SE, Ansari N, El Ayadi A. Therapeutic Strategies to Reduce Burn Wound Conversion. Medicina. 2022 Jul;58(7):922. ↑ Bereda G. Burn Classifications with Its Treatment and Parkland Formula Fluid Resuscitation for Burn Management: Perspectives. Clinical Medicine And Health Research Journal. 2022 May 12;2(3):136-41. ↑ Mehta M, Tudor GJ. Parkland formula. 2019 ↑ Ahmed FE, Sayed AG, Gad AM, Saleh DM, Elbadawy AM. A Model for Validation of Parkland Formula for Resuscitation of Major Burn in Pediatrics. The Egyptian Journal of Plastic and Reconstructive Surgery. 2022 Apr 1;46(2):155-8. ↑ Ete G, Chaturvedi G, Barreto E, Paul M K. Effectiveness of Parkland formula in the estimation of resuscitation fluid volume in adult thermal burns. Chinese Journal of Traumatology. 2019 Apr 1;22(02):113-6. Retrieved from " Categories: Course Pages Burns Assessment ReLAB-HS Course Page Get Top Tips Tuesday and The Latest Physiopedia updates Back to top suggested results
188045	https://trustedinstitute.com/concept/pmi-rmp/quantitative-risk-analysis/expected-monetary-value-emv-analysis/	Expected Monetary Value (EMV) Analysis test - Quantitative Risk Analysis - PMI Risk Management Professional questions \| TrustEd Institute Register Back to Quantitative Risk Analysis Expected Monetary Value (EMV) Analysis 5 minutes 5 Questions Expected Monetary Value (EMV) Analysis is a quantitative risk assessment technique used to calculate the average outcome when the future includes scenarios that may or may not happen. Essentially, it involves multiplying the monetary impact of an outcome by its probability of occurrence to determin…Expected Monetary Value (EMV) Analysis is a quantitative risk assessment technique used to calculate the average outcome when the future includes scenarios that may or may not happen. Essentially, it involves multiplying the monetary impact of an outcome by its probability of occurrence to determine its EMV. In project risk management, EMV helps in quantifying risks by assigning a monetary value to both opportunities (positive risks) and threats (negative risks), thereby aiding in decision-making processes. For instance, if there's a 30% chance of incurring a $10,000 cost due to a potential risk event, the EMV of that risk is $3,000 (0.30 x $10,000). By calculating the EMV for all identified risks, project managers can sum these values to understand the overall potential impact on the project budget. This approach allows for the comparison of different risks on a common monetary scale and supports the prioritization of risk responses based on their financial significance. EMV Analysis is particularly useful when combined with decision tree analysis, where it assists in evaluating complex decisions involving multiple risks and uncertainties. It provides a clear, quantitative basis for choosing between different project options or risk response strategies by highlighting the expected financial outcomes. Moreover, EMV can be incorporated into contingency reserves, ensuring that adequate funds are allocated to address potential risks. Overall, EMV Analysis enhances the objectivity and rigor of the risk management process by translating uncertainties into actionable financial data. Read more Expected Monetary Value (EMV) Analysis Introduction to Expected Monetary Value (EMV) Analysis Expected Monetary Value (EMV) Analysis is a critical technique in quantitative risk management that helps project managers make decisions based on statistical analysis of potential outcomes. As a key component of the PMI-RMP certification, understanding EMV thoroughly is essential for effective risk assessment and response planning. Why EMV Analysis is Important EMV analysis provides a systematic approach to evaluate risk events by considering both their probability and impact in monetary terms. It enables project managers to: • Prioritize risks based on their potential financial impact • Make informed decisions between alternative project approaches • Justify contingency reserves with quantitative data • Compare different risk response strategies objectively • Communicate risk implications to stakeholders using financial metrics What is Expected Monetary Value (EMV)? EMV is a statistical concept that calculates the average outcome when the future includes scenarios that may or may not happen. It multiplies each possible outcome by the probability of that outcome occurring, then sums these products. The formula for EMV is: EMV = Probability × Impact Where: • Probability is expressed as a decimal (between 0 and 1) • Impact is the monetary value (positive for opportunities, negative for threats) How EMV Analysis Works Step 1: Identify Risk Events Identify all potential risk events that could impact your project financially. Step 2: Determine Probabilities Assign probability values to each risk event based on historical data, expert judgment, or statistical analysis. Step 3: Estimate Impact Calculate the monetary impact if each risk event occurs. Step 4: Calculate EMV Multiply the probability by the impact for each risk event. Step 5: Sum the EMVs Add all individual EMVs to get the overall project EMV. Step 6: Make Decisions Use the EMV results to select risk responses or choose between project alternatives. Decision Tree Analysis Decision trees are graphical representations that help visualize EMV calculations, especially for complex decision scenarios with multiple potential outcomes. They show: • Decision nodes (represented by squares) • Chance nodes (represented by circles) • Outcomes (represented by triangles or end points) • Probabilities and impacts of each branch By calculating the EMV at each chance node, project managers can determine the optimal decision path. EMV in Monte Carlo Analysis EMV principles are often applied in Monte Carlo simulations, which run hundreds or thousands of project scenarios to determine probability distributions of various outcomes. This provides a more comprehensive view of potential project results than single-point estimates. Practical Example of EMV Analysis Consider a construction project with the following risks: Risk 1: Weather delay • Probability: 30% (0.3) • Impact: $50,000 cost increase • EMV = 0.3 × $50,000 = $15,000 Risk 2: Material price increase • Probability: 45% (0.45) • Impact: $30,000 cost increase • EMV = 0.45 × $30,000 = $13,500 Risk 3: Early completion bonus • Probability: 20% (0.2) • Impact: $40,000 cost decrease (opportunity) • EMV = 0.2 × (-$40,000) = -$8,000 Total project EMV = $15,000 + $13,500 - $8,000 = $20,500 This $20,500 represents the expected monetary impact of risks on the project and could be used to establish contingency reserves. Limitations of EMV Analysis While valuable, EMV analysis has limitations: • Requires reliable probability and impact estimates • Assumes risk events are independent • May oversimplify complex risk scenarios • Doesn't account for risk attitudes or utility theory • Focuses on financial impacts only Exam Tips: Answering Questions on EMV Analysis Understanding Calculation Questions • Pay close attention to whether risks represent threats (positive EMV) or opportunities (negative EMV) • Double-check your probability values (must be between 0 and 1) • Practice multiplying decimals by monetary values quickly • Remember that the total EMV is the sum of all individual EMVs Interpreting EMV Results • For decision trees, the optimal choice is usually the option with the best EMV (lowest for cost, highest for profit) • If comparing project approaches, the lower EMV generally represents the less risky option • Negative total EMV may indicate a project with net positive opportunity Common Exam Pitfalls • Confusing probability percentages (ensure you convert to decimals) • Mixing up signs (threats are positive EMVs, opportunities are negative EMVs) • Misinterpreting decision tree branches and nodes • Overlooking the need to sum all EMVs for total project risk exposure Application Questions • Know how to compare alternative strategies using EMV • Understand how EMV relates to contingency reserves • Be ready to explain how EMV fits into the overall risk management process • Recognize the relationship between EMV and other quantitative risk techniques like sensitivity analysis and Monte Carlo simulation Conceptual Questions • Be prepared to explain the strengths and weaknesses of EMV analysis • Understand when EMV is most appropriate to use • Know how EMV contributes to objective decision-making • Recognize EMV's role in justifying risk responses Remember that EMV analysis is just one tool in the quantitative risk analysis toolkit. The PMI-RMP exam may ask you to identify when EMV is the most appropriate technique compared to other methods. Test mode: Exam (Timed) Practice (With explanations) Start practice test PMI-RMP - Expected Monetary Value (EMV) Analysis Example Questions Test your knowledge of Expected Monetary Value (EMV) Analysis Question 1 In EMV analysis, which statement best describes a key characteristic of opportunity risks with positive impacts? They are computed by dividing the probability of success by the expected financial return They are determined by subtracting negative impacts from the total project budget baseline They are calculated by multiplying probability by positive monetary gains They are calculated by adding probability percentages to potential monetary benefits Show Explanation Correct Answer: They are calculated by multiplying probability by positive monetary gains In EMV (Expected Monetary Value) analysis, opportunity risks with positive impacts are calculated by multiplying the probability of the opportunity occurring by the potential positive monetary value (gain) that would result if the opportunity materializes. This is the fundamental calculation method in EMV analysis for positive risks/opportunities: EMV = Probability × Impact(positive monetary value) The other answers are incorrect because: - Adding probability percentages to monetary benefits is mathematically invalid and would give meaningless results - Subtracting negative impacts from the budget baseline does not factor in probabilities and isn't related to opportunity calculation - Dividing probability by expected return would produce an invalid metric and reverses the proper mathematical relationship Question 2 In Expected Monetary Value (EMV) analysis, a risk event has a 30% probability of occurrence and would result in a $50,000 loss. What is the EMV for this risk? -$15,000 -$25,000, after applying organizational risk tolerance factors -$35,000, as the risk impact needs to be adjusted for market conditions -$7,500, which accounts for risk appetite threshold adjustment Show Explanation Correct Answer: -$15,000 The Expected Monetary Value (EMV) calculation is straightforward: EMV = Probability × Impact EMV = 0.30 × (-$50,000) EMV = -$15,000 This is a simple probability calculation that multiplies the likelihood of the risk occurring (30% or 0.30) by the monetary impact if it does occur (-$50,000). The negative sign indicates a loss. The other answers are incorrect because they introduce factors that are not part of EMV calculation. EMV is purely mathematical and does not consider market conditions, risk appetite thresholds, or organizational risk tolerance factors. These may be important in other risk analysis contexts, but they are not part of basic EMV calculations. Question 3 When applying EMV analysis to multiple risks, which mathematical approach is most appropriate for comparing different risk mitigation strategies? Calculate the EMV for each option and choose the lowest total EMV Average the EMVs across all risks and select the median value as the optimal solution Calculate the EMV for each option and select the strategy with the highest probability of success Sum all potential losses and multiply by the combined probability of occurrence Show Explanation Correct Answer: Calculate the EMV for each option and choose the lowest total EMV In EMV (Expected Monetary Value) analysis for multiple risks, calculating the EMV for each option and selecting the lowest total EMV is the most appropriate mathematical approach. Here's why: The correct approach involves: 1. Calculating individual EMVs for each risk in each strategy 2. Summing up all EMVs within each strategy 3. Comparing the total EMVs across strategies 4. Selecting the strategy with the lowest total EMV This method is optimal because: - It considers both probability and impact for each risk - It allows for comprehensive comparison of different strategies - Lower EMV indicates less expected monetary loss - It aligns with risk minimization objectives The other approaches are incorrect because: - Selecting based on highest probability of success does not account for impact magnitude - Simply multiplying combined probabilities loses individual risk characteristics - Using average or median values can mask significant risk variations and lead to incorrect strategy selection Go Premium PMI Risk Management Professional Preparation Package (2025) 3223 Superior-grade PMI Risk Management Professional practice questions. Accelerated Mastery: Deep dive into critical topics to fast-track your mastery. Unlock Effortless PMI-RMP preparation: 5 full exams. 100% Satisfaction Guaranteed: Full refund with no questions if unsatisfied. 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188046	https://ojs.sabinet.co.za/index.php/sapj/article/download/810/416/5731	S Afr Pharm J 2024 Vol 91 No 4 26 REVIEW Introduction Gout is a common form of inflammatory arthritis that occurs due to a buildup of uric acid in the body over time. 1,2 Since the body cannot easily dissolve and excrete high uric acid levels via urine, the uric acid starts to crystallise and form sharp crystals known as tophi in the joints, usually in the joint of the big toe. 3,4 The tophi initially cause no pain; however, they can become painful over time and may result in damage to the bone and soft tissue, leading to misshapen joints. 2 Although the big toe is more commonly affected, other joints affected by gout are the knees, ankles, feet, hands, wrists, and elbows. 1,5 The presence of high levels of uric acid in these joints causes severe pain and inflammation. 2Gout can affect anyone, but it is more prevalent in men, and women usually develop it after menopause. 4 The condition typically begins in middle age, but if it starts at a younger age, the symptoms are usually more severe. 2 Gout is a progressive disease that can go through several stages. 3In the first stage, known as hyperuricaemia, elevated urate levels in the blood lead to the formation of crystals in the joints, as shown in Figure 1. 3 Typically, there are no symptoms during this stage. 2 The second stage is characterised by gout flares, which involve periodic attacks of intense joint pain and swelling. 3 Intercritical gout, the third stage, is the period between gout attacks when there are no symptoms. 2 The final stage, chronic gout, involves the accumulation of tophi in the joints, skin, or other parts of the body. 1 Depending on their location, tophi can cause permanent damage to the joints and other internal organs, increasing the risk of developing other conditions or complications, especially related to the heart and kidneys. 2,3 Comorbidities that may increase the prevalence of gout include: 3• Hypertension (high blood pressure) • Chronic kidney disease • Obesity • Diabetes • Nephrolithiasis (kidney stones) • Myocardial Infarction (heart attack) • Congestive heart failure • Sleep apnoea • Depression The diagnosis of gout is not always straightforward, and a differential diagnosis may be necessary. Other diseases can Crystals of pain: navigating gout and its management SD Vambe, CV Mchavi, E Hough, E Bronkhorst School of Pharmacy, Sefako Makgatho Health Sciences University, South Africa Corresponding author, email: elmien.bronkhorst@smu.ac.za Abstract Gout is a form of inflammatory arthritis, caused by the buildup of uric acid crystals in the joints, especially the big toe. If left untreated these tophi, or crystals can become extremely painful, and over time may result in damage to bone and soft tissue. It is important to get a correct diagnosis on gout and to differentiate with other diseases like septic arthritis, rheumatoid arthritis and even stress fractures. Non-pharmacological treatment and prevention strategies include sufficient rest and adequate dietary and lifestyle modifications. The management of gout distinguishes between treatment for acute gout symptoms and the prevention of a gout attack or the lowering of uric acid in the serum. Urate-lowering therapy, like allopurinol and febuxostat, lowers blood urate levels, can prevent gout flare-ups and diminishes tophi over time. Treatment with one or more potent anti-inflammatory medication is necessary for the management of acute flares. Four categories of medicine are available for treatment of acute symptoms of pain and inflammation. They include nonsteroidal anti-inflammatory medicine, corticosteroids, colchicine, and anti-IL-1β biologics. Efficacy between these agents is similar, thus focus should be on minimising individual risks. People with a tendency to develop gout must limit their consumption of red meat, fish, shellfish and alcohol, particularly those that have additional purines such as beer, wine and whiskey. Keywords: gout, urate-lowering therapy, allopurinol, colchicine © Authors Figure 1: Stages of gout progression 3S Afr Pharm J 2024 Vol 91 No 4 27 REVIEW present similarly to gout and cause a misdiagnosis as shown in Figure 2. 6 Pseudogout Pseudogout, formerly known as calcium pyrophosphate deposition disease or CPPD, is now commonly referred to as pseudogout due to its similarity to gout. 7 Both gout and pseudogout cause sudden joint pain, swelling, and redness, which makes them difficult to differentiate. 8 It is the type of crystals formed in the two conditions that differ. 7 For gout, it is uric acid, while in pseudogout, it is crystallised calcium pyrophosphate (CPP). 6 Infected joint (septic arthritis) Both gout and an infected joint can cause fever and an increase in white blood cells. 4 However, the presence of an offending microorganism in the fluid taken from the affected joint indicates septic arthritis, as it is an infection, unlike gout. 6 Treatment of septic arthritis is directed at eliminating the offending bacteria. Bacterial skin infection (cellulitis) Both gout and cellulitis can cause inflammation and pain in the lower leg. 7 The difference is that in gout there is an accumulation of uric acid crystals in a joint, while cellulitis is a bacterial infection in the deep layer of the skin. 6 A blood culture can be used to differentiate the two conditions. Stress fracture Gout is often mistaken for injuries to the toes caused by dropping heavy items on the toes or jamming the big toe against a hard surface. Stress fractures can occur without the individual being aware and are frequently confused with gout. 6 An X-ray can assist with identifying the cause of the pain if a stress fracture is suspected. Rheumatoid arthritis In individuals with polyarticular gout, which affects several joints, gout is often mistaken for rheumatoid arthritis. 6 The key distinction is that gout typically starts by affecting one or a few joints, while rheumatoid arthritis tends to involve multiple, larger joints symmetrically and can affect many organs in the body. 4Blood tests, such as anti-CCP, C-reactive protein, erythrocyte sedimentation rate, and rheumatoid factor, can help doctors distinguish between gout and rheumatoid arthritis. 6 Psoriatic arthritis As with rheumatoid arthritis, psoriatic arthritis (PsA) can cause swelling around the fingers or toes, which may resemble gout tophi. 6 However, with PsA there is no buildup of uric acid crystals in the joints. 4,6 Pathophysiology and clinical presentations Table I: Pathophysiology of gout Aspect Details Pathophysiology Gout is characterised by elevated serum uric acid levels (hyperuricaemia), typically exceeding 6.8 mg/dL. 9 Uric acid crystal formation As blood uric acid levels increase, urate crystals form. Clinical presentation Kidney Stones: Formation of uric acid crystals can lead to kidney stones Tophi: Deposits of urate crystals in joints and tissues can form tophi (chalky nodules). Gouty Arthritis: Urate crystal deposition in joints can cause episodes of gouty arthritis, characterised by sudden and severe joint pain. 10 Hyperuricaemia Hyperuricaemia is characterised by elevated levels of uric acid in the bloodstream, typically exceeding 6 mg/dL in women and 7 mg/dL in men. 11 Uric acid is produced during the breakdown of purines in the body as shown in Figure 3. 12,13 Research has additionally demonstrated a correlation between elevated uric acid levels and various other health conditions, such as kidney disease, heart disease, hypertension, diabetes, non-alcoholic fatty liver disease, and metabolic syndrome. 14,15,16 Hyperuricaemia causes cardiovascular disease and chronic kidney disease by prompting abnormal growth of vascular smooth muscle cells and impaired endothelial function, which triggers inflammation. 17 Inflammatory response Hyperuricaemia gradually progresses and promotes the formation of monosodium urate (MSU) crystals, triggered by various factors such as dehydration, alcohol, hypertension, thereby causing inflammation in the joints. 18,19 Inflammatory cytokines, particularly IL-1β, are the key mediators of gouty inflammation. 20 The NLRP3 inflammasome is the major pathway by which MSU crystals trigger the cellular inflammatory response as shown in Figure 4. 21 Delivery of ingested MSU crystals to the inflammasome in phagocytes subsequently triggers intracellular assembly of the cytosolic NALP3 (cryopyrin) inflammasome protein complex. 21 The MSU crystals cause the inflammasome assembly, which in turn causes caspase-1 activation, phagocyte maturation, and the production of IL-1β. 20 Figure 2: Diseases that mimic gout 6 S Afr Pharm J 2024 Vol 91 No 4 28 REVIEW Acute gout attacks Acute gout attacks start suddenly and escalate quickly, with joint pain usually reaching its peak within 24 hours of onset. These attacks often begin to improve within 5–12 days even without treatment, although full recovery may take longer for some individuals. 22 Chronic gout Chronic gout develops due to ongoing inflammation that follows repeated gout attacks. It is characterised by persistent synovitis (inflammation of the synovial membrane), erosion of bone, damage to cartilage, and the formation of tophi (deposits of uric acid crystals) in tissue. 23 Causes and risk factors The causes of gout typically involve multiple factors, such as genetic predisposition, existing medical conditions, and dietary habits. 4 In uncommon instances, a single genetic anomaly may lead to gout, often linked with other health issues. Regardless of the specific cause, elevated levels of uric acid in the blood can lead to clinical symptoms of gout in susceptible individuals. 25 Risk factors associated with gout and high uric acid levels include advancing age, male gender, obesity, a diet rich in purines, alcohol consumption, and genetic susceptibility. Medications such as diuretics, low-dose aspirin, ethambutol, pyrazinamide, and cyclosporine are known to potentially raise uric acid levels and contribute to the development of gout. 19 Foods that can increase uric acid levels and contribute to gout include animal products such as seafood (like shrimp and lobster), organ meats (such as liver and kidney), and red meats (like mutton and beef ). Additionally, beverages such as alcohol, sweetened drinks, sodas, and those containing high-fructose corn syrup may also play a role in the development of this condition. 25 Signs and symptoms Gout attacks are intensely painful and typically occur suddenly, often overnight. Symptoms in the affected joints may include severe pain, redness or discoloration, stiffness, swelling, tenderness (even to light touch, such as from a bedsheet), and a sensation of warmth or intense heat in the joint. 4 Triggers of symptoms Factors that can trigger gout flares include consuming foods high in purines and taking medications such as furosemide. Environmental factors such as exposure to lead, particulate matter, temperature changes, and physiological stress have also been identified as triggers for gout flares. 26 Onset of symptoms Gout episodes often last a week or two, however, the patient may not exhibit any gout symptoms in between attacks. Nevertheless, Figure 4: Inflammatory response in gout 21 MSU crystals Recognition of MSU Phagocytosis of MSU Inflammasome activation IL-1β IL-1RI IL-1RAcP Signal Transduction IL-1β release Monocyte Endothelium Pro-inflammatory mediators release Neurophil recruitment AMP AMPD APRT HGPRT1 HGPRT1 5'-nucleotidase 5'-nucleotidase Xanthine oxidase Xanthine oxidase Guanine deaminase Uricase (Not present in humans 5'-nucleotidase 5'-nucleotidase PRPP PNP PNP PNP PRPP PRPP Adenosine Inosine Xanthosine Guanosine Guanine Xanthine Uric acid Allantoic acid Hypoxanthine Adenine IMP XMP GMP Figure 3: Uric acid synthesis and purine metabolism in gout 13 S Afr Pharm J 2024 Vol 91 No 4 29 REVIEW some flares continue longer than others and may result in more severe symptoms. 4 Treatment Non-pharmacological treatment and prevention strategies The management of gout involves non-pharmacological measures as adjuncts to managing acute gout attacks. 27 These measures include: 1. Sufficient rest 2. Topical ice application 3. Reduce the intake of sugar-sweetened soft drinks 4. Dietary and lifestyle modifications It is recommended that people with gout limit their consumption of red meat, fish, shellfish, and alcohol, particularly those that have additional purines such as beer, lager, and whiskey. 27 It has been widely held that diet can reduce the chance of developing gout; in particular, consuming fewer alcoholic beverages and foods high in purines, as these are linked to elevated blood urate levels. 28 Asummary of foods one should eat and avoid when they have gout is shown in Figure 5. 29 Because of its uricosuric effects, which are more pronounced at greater dosages, increasing vitamin C intake above 500 mg/day reduces the incidence of gout, whereas the intake of soy protein, non-soy legumes, and fresh fruit (> 2 portions/day) is negatively correlated with the incidence of gout. 27 Therefore, these dietary and lifestyle modifications can be suggested as supplementary to ULT. 27 Pharmacological treatment Urate-lowering therapy (ULT), lowers blood urate levels, stops gout flare-ups, and diminishes tophi over time. ULT comprises of xanthine oxidase inhibitors (allopurinol and febuxostat), uricosuric low-fat dairy Whole grain bread cherries dark berries vegetables brown rice tuna liver kidneys beer salmon Figure 5: Foods to eat and avoid with gout 29 Figure 6: Management of gout: An algorithm 28 S Afr Pharm J 2024 Vol 91 No 4 30 REVIEW Table II: Treatment of acute gout flares 28 Dosing Duration of treatment NSAIDs: Naproxen 500 mg twice daily 3–5 days Celecoxib 400 mg twice daily 3–5 days Indomethacin 50 mg three times a day 5 days Ibuprofen 800 mg three times a day 5 days Etoricoxib 120 mg daily 8 days Corticosteroids: Prednisone 0.5 mg/kg or 40 mg daily 2–5 days Colchicine 0.5–1 mg, followed by 0.5 mg two hours later. Maximum of 6 mg daily. 3 days Anti-IL-1β biologics: Anakinra 100 mg daily 3–5 days Canakinumab 150 mg SC 1 injection (t =26 days) Table III: Treatment of established and acute gout 31,32 Allopurinol Colchicine Houdé Dosing Prophylactic treatment of gout and hyperuricaemia :• 50 mg, 12 hourly; Increase dose as required up to 200–400 mg. • Treatment of hyperuricaemia : Initial: 200 mg, 8 hourly. Maintenance: 300–400 mg daily. Acute attacks of gout :• Initial: 0.5–1 mg immediately, followed by 0.5 mg every 2 hours until pain relief is obtained or until vomiting or diarrhoea occurs. • Maximum: 6 mg for a minimum of 3 days, but preferably 7 days, should elapse between courses of gout treatment with colchicine. Drug interactions • Warfarin: increased risk of bleeding and bruising • Azathioprine: increased risk of bone marrow toxicity • Theophylline: increases effects by slowing drug metabolism. • Enalapril (ACE-I): increased risk for anaphylaxis(rash) and Stevens-Johnson syndrome. • Quinidine: increase the effect of colchicine by affecting elimination • Itraconazole: increased effects with fatal side effects in kidney and hepatic dysfunction. • Verapamil, ketoconazole, clarithromycin, erythromycin, atazanavir, ritonavir, cyclosporine: increase effects. • Digoxin and statins increase the risk of toxicity of the other, rhabdomyolysis including fatality. Contraindications • Hypersensitivity to allopurinol or to any of the excipients • Severe renal disorder • Severe hepatic disorder • An acute gout attack • Patients who have exhibited serious adverse effects from the medicine • In children, except those with malignancy • Pregnancy • Lactation • Hypersensitivity to colchicine or any of its excipients • Patients undergoing haemodialysis • Severe renal impairment (CrCl < 10ml/min) • Severe hepatic impairment • Blood disorders - Myelosuppression - Leukopenia - Granulocytopenia - Thrombocytopenia - Aplastic anaemia • Coadministration with P-glycoprotein inhibitors such as ciclosporin, verapamil, or quinidine in patients with renal or hepatic impairment • Coadministration with strong CYP3A4 inhibitors such as ritonavir, atazanavir, indinavir, clarithromycin, telithromycin, itraconazole, or ketoconazole in patients with renal or hepatic impairment • Pregnancy and lactation Adverse effects • Stevens-Johnson syndrome (rash) • Angioedema • Thrombocytopenia • Agitation • Ammonia-like breath odour • Bleeding gums • Joint or muscle pain • Bloody or black, tarry stools • Cloudy urine • Peripheral neuritis • Neuropathy • Rhabdomyolysis • Hepatic impairment • Rash • Alopecia • Bone marrow depression with agranulocytosis • Aplastic anaemia • Thrombocytopenia • Burning, “crawling”, or tingling feeling in the skin • Muscle weakness • Numbness in the fingers or toes (usually mild) S Afr Pharm J 2024 Vol 91 No 4 32 REVIEW agents (probenecid and lesinurad), and uricases (rasburicase and pegloticase). ULT is normally started several weeks after the resolution of a gout flare, as it is believed that commencing during a flare will exacerbate the current flare. When ULT is started the flare becomes worse, which leads to patients discontinuing treatment. 30 Treatment of acute gout flares Treatment with one or more potent anti-inflammatory medications is necessary for the management of acute flares. There are four categories accessible: nonsteroidal anti-inflammatory drugs (NSAIDs), corticosteroids, colchicine, and anti-IL-1β biologics. Since they are all efficient, selecting one should focus on minimising individual risks. 28 Treatment of established gout Urate-lowering therapy (ULT) Guidelines published by the Rheumatologic Society support the idea that urate-lowering is essential for the treatment of established gout. 28 Allopurinol is the oldest available and most used xanthine oxidase inhibitor. 28 Flare prophylaxis during urate lowering Patients with gout frequently had more flare-ups during the early stages of urate reduction, most likely because of crystals being released from dissolving urate collections. 28 When feasible, prophylactic anti-inflammatory medication should be supplied to patients in addition to the initial ULT. Low doses of colchicine (0.6 mg once or twice daily) are frequently used, even if other anti-inflammatories may be appropriate for treating acute flares. This may be because colchicine is most likely to be tolerated during the three to nine months following the initial ULT, which is necessary to lower the risk of gout flares below pre-treatment levels. 28 Conclusion Gout is a common form of inflammatory arthritis that occurs due to a buildup of uric acid, forming urate crystals in the joints over a long period. The management involves non-pharmacological measures to prevent flareups, that include sufficient rest, reduced intake of sugar-sweetened drinks and general dietary and lifestyle modifications. ULT is essential for the treatment of established gout, with the aim to reduce flare-ups. Acute symptoms can be treated with colchicine, or in severe cases with NSAIDs in combination with glucocorticoids like prednisone. References Dalbeth N, Gosling AL, Gaffo A, Abhishek A. Gout. The Lancet. 2021;397(10287):1843-1855. 2. National Institute of Arthritis and Musculoskeletal and Skin Diseases. Gout [Internet]. 2020. Available from: 3. Donvito T. The 4 stages of Gout progression (and how to stop Gout from getting worse) [Internet]. CreakyJoints. 2019. Available from: treatment-and-wcare/medications/gout-stages-progression/. Accessed 26 Jul 2024. 4. Cleveland Clinic. Gout: Symptoms, causes, treatments [Internet]. Cleveland Clinic. 2023. Available from: 5. Mayo Clinic. Gout - Symptoms and causes [Internet]. Mayo Clinic. Mayo Foundation for Medical Education and Research. 2022. Available from: 6. Brody B. 6 Diseases that can mimic Gout (and delay your diagnosis) [Internet]. Creaky-Joints. 2019. Available from: gout-misdiagnosis/. 7. Mayo Clinic. Pseudogout - Symptoms and causes [Internet]. Mayo Clinic. 2018. Available from: syc-20376983. 8. DeVries C. 4 ways Gout and pseudogout are different [Internet]. Arthritis-health. 2021. Available from: 9. Dalbeth N, Choi HK, Joosten LAB, et al. Gout. Nature Reviews Disease Primers [Internet]. 2019;5(1). 10. Kanwal A, Bajwa MA, Samreen T, et al. A comprehensive review on gout: the epidemiologi-cal trends, pathophysiology, clinical presentation, diagnosis and treatment. Journal of the Pakistan Medical Association. 2020;71(4):1-12. 11. George C, Minter DA. Hyperuricemia [Internet]. Nih.gov. StatPearls Publishing; 2019. Avail-able from: 12. Breshears MA, Confer AW. The urinary system. Pathologic Basis of Veterinary Disease [Inter-net]. 2017;617-681.e1. 13. Maharajan D. Uric acid metabolism and Gout [Internet]. SlideShare. 2019. Available from: 14. Xiong Q, Liu J, Xu Y. Effects of uric acid on Diabetes Mellitus and its chronic complications. International Journal of Endocrinology. 2019;2019:1-8. 15. 16. Shahin L, Patel KM, Heydari MK, Kesselman MM. Hyperuricemia and cardiovascular risk. Cureus. 2021;13(5):e14855. 17. National Institute of Arthritis and Musculoskeletal and Skin Diseases. Gout [Internet]. 2020. Available from: 18. Kim JH, Kwon MJ, Choi HG, et al. The association between hyperuricemia and car-diovascular disease history: A cross-sectional study using KoGES HEXA data. Medicine. 2022;101(51):e32338. 19. Chittoor G, Voruganti VS. Hyperuricemia and Gout. Principles of Nutrigenetics and Nutrig-enomics. 2020;389-94. 20. Zhao J, Wei K, Jiang P, et al. Inflammatory response to regulated cell death in Gout and its functional implications. Frontiers in Immunology. 2022;13. fimmu.2022.888306. 21. So AK, Martinon F. Inflammation in gout: mechanisms and therapeutic targets. Nature Reviews Rheumatology. 2017;13(11):639-47. 22. Blevins HM, Xu Y, Biby S, Zhang S. The NLRP3 inflammasome pathway: A review of mecha-nisms and inhibitors for the treatment of inflammatory diseases. Frontiers in Aging Neuro-science [Internet]. 2022;14. 23. Coburn BW, Mikuls TR. Treatment options for acute Gout. Federal Practitioner [Internet]. 2016;33(1):35-40. 24. Ragab G, Elshahaly M, Bardin T. Gout: An old disease in new perspective. Journal of Ad-vanced Research [Internet]. 2017;8(5):495-511. 25. Cleveland Clinic. Gout: Symptoms, causes, treatments [Internet]. Cleveland Clinic. 2023. Available from: 26. Fenando A, Widrich J, Rednam M, Gujarathi R. Gout [Internet]. PubMed. Treasure Island (FL): StatPearls Publishing; 2022. Available from: NBK546606/. 27. Helget LN, Mikuls TR. Environmental triggers of hyperuricemia and Gout. Rheumat-ic Disease Clinics of North America. 2022;48(4):891-906. rdc.2022.06.009. 28. Abhishek A, Doherty M. Education and non-pharmacological approaches for gout. Rheu-matology [Internet]. 2018;57(suppl_1):i51-8. kex421. 29. Pillinger MH, Mandell BF. Therapeutic approaches in the treatment of gout. Seminars in Arthritis and Rheumatism. 2020;50(3):S24-30. 30. Poulson B. Foods to avoid with Gout: Seafood, yeast, red meat, and more [Internet]. Very-well Health. 2023. Available from: 31. Evidence reviews for timing of urate-lowering therapy in relation to a flare in people with gout: Gout: diagnosis and management: Evidence review F [Internet]. PubMed. London: National Institute for Health and Care Excellence (NICE); 2022. Available from: https:// www.ncbi.nlm.nih.gov/books/NBK583523/#:~:text=Long%2Dterm%20management%20 of%20gout. 32. Allopurinol (Oral Route) side effects - Mayo Clinic [Internet]. www.mayoclinic.org. Avail-able from: 33. Colchicine (Oral Route) side effects - Mayo Clinic [Internet]. www.mayoclinic.org. Available from: drg-20067653?p=1#:~:text=Stop%20taking%20this%20medicine%20as.
188047	https://en.wikipedia.org/wiki/Lapse_rate	Published Time: Thu, 04 Sep 2025 01:17:58 GMT Lapse rate - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Contents move to sidebar hide (Top) 1 Environmental lapse rate 2 Cause 3 Convection and adiabatic expansion 4 Mathematics of the adiabatic lapse rateToggle Mathematics of the adiabatic lapse rate subsection 4.1 Dry adiabatic lapse rate 4.2 Moist adiabatic lapse rate 5 Effect on weather 6 Impact on the greenhouse effect 7 Lapse rate in an isolated column of gas 8 See also 9 Notes 10 References 11 Further reading 12 External links [x] Toggle the table of contents Lapse rate [x] 16 languages العربية Deutsch Español فارسی 한국어 Hrvatski Bahasa Indonesia Italiano Nederlands 日本語 Norsk nynorsk Srpskohrvatski / српскохрватски ไทย Українська 粵語中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance move to sidebar hide From Wikipedia, the free encyclopedia Vertical rate of change of temperature in atmosphere Higher Czarny Staw pod Rysami lake (elevation 1,583 metres (5,194 ft)) is still frozen as the lower Morskie Oko lake has already almost melted (elevation 1,395 metres (4,577 ft)). Photo from Polish side of the Tatra mountains, May 2019. The lapse rate is the rate at which an atmospheric variable, normally temperature in Earth's atmosphere, falls with altitude.Lapse rate arises from the word lapse (in its "becoming less" sense, not its "interruption" sense). In dry air, the adiabatic lapse rate (i.e., decrease in temperature of a parcel of air that rises in the atmosphere without exchanging energy with surrounding air) is 9.8 °C/km (5.4 °F per 1,000 ft). The saturated adiabatic lapse rate (SALR), or moist adiabatic lapse rate (MALR), is the decrease in temperature of a parcel of water-saturated air that rises in the atmosphere. It varies with the temperature and pressure of the parcel and is often in the range 3.6 to 9.2 °C/km (2 to 5 °F/1000 ft), as obtained from the International Civil Aviation Organization (ICAO). The environmental lapse rate is the decrease in temperature of air with altitude for a specific time and place (see below). It can be highly variable between circumstances. Lapse rate corresponds to the vertical component of the spatial gradient of temperature. Although this concept is most often applied to the Earth's troposphere, it can be extended to any gravitationally supported parcel of gas. Environmental lapse rate [edit] A formal definition from the Glossary of Meteorology is: The decrease of an atmospheric variable with height, the variable being temperature unless otherwise specified. Typically, the lapse rate is the negative of the rate of temperature change with altitude change: Γ=−d T d z{\displaystyle \Gamma =-{\frac {\mathrm {d} T}{\mathrm {d} z}}} where Γ{\displaystyle \Gamma } (sometimes L{\displaystyle L}) is the lapse rate given in units of temperature divided by units of altitude, T is temperature, and z is altitude.[a] The environmental lapse rate (ELR), is the actual rate of decrease of temperature with altitude in the atmosphere at a given time and location. As an average, the International Civil Aviation Organization (ICAO) defines an international standard atmosphere (ISA) with a temperature lapse rate of 6.50°C/km7 from sea level to 11 km (36,090 ft or 6.8 mi). From 11 km up to 20 km (65,620 ft or 12.4 mi), the constant temperature is −56.5°C(−69.7°F), which is the lowest assumed temperature in the ISA. The standard atmosphere contains no moisture. Unlike the idealized ISA, the temperature of the actual atmosphere does not always fall at a uniform rate with height. For example, there can be an inversion layer in which the temperature increases with altitude. Cause [edit] The temperature profile of the atmosphere is a result of the interaction between radiative heating from sunlight, cooling to space via thermal radiation, and upward heat transport via natural convection (which carries hot air and latent heat upward). Above the tropopause, convection does not occur and all cooling is radiative. Within the troposphere, the lapse rate is essentially the consequence of a balance between (a) radiative cooling of the air, which by itself would lead to a high lapse rate; and (b) convection, which is activated when the lapse rate exceeds a critical value; convection stabilizes the environmental lapse rate. Sunlight hits the surface of the earth (land and sea) and heats them. The warm surface heats the air above it. In addition, nearly a third of absorbed sunlight is absorbed within the atmosphere, heating the atmosphere directly. Thermal conduction helps transfer heat from the surface to the air; this conduction occurs within the few millimeters of air closest to the surface. However, above that thin interface layer, thermal conduction plays a negligible role in transferring heat within the atmosphere; this is because the thermal conductivity of air is very low.: 387 The air is radiatively cooled by greenhouse gases (water vapor, carbon dioxide, etc.) and clouds emitting longwave thermal radiation to space. If radiation were the only way to transfer energy within the atmosphere, then the lapse rate near the surface would be roughly 40°C/km and the greenhouse effect of gases in the atmosphere would keep the ground at roughly 333 K (60°C; 140°F).: 59–60 However, when air gets hot or humid, its density decreases. Thus, air which has been heated by the surface tends to rise and carry internal energy upward, especially if the air has been moistened by evaporation from water surfaces. This is the process of convection. Vertical convective motion stops when a parcel of air at a given altitude has the same density as the other air at the same elevation. Convection carries hot, moist air upward and cold, dry air downward, with a net effect of transferring heat upward. This makes the air below cooler than it would otherwise be and the air above warmer. Because convection is available to transfer heat within the atmosphere, the lapse rate in the troposphere is reduced to around 6.5°C/km and the greenhouse effect is reduced to a point where Earth has its observed surface temperature of around 288 K (15°C; 59°F). Convection and adiabatic expansion [edit] Emagram diagram showing variation of dry adiabats (bold lines) and moist adiabats (dash lines) according to pressure and temperature As convection causes parcels of air to rise or fall, there is little heat transfer between those parcels and the surrounding air. Air has low thermal conductivity, and the bodies of air involved are very large; so transfer of heat by conduction is negligibly small. Also, intra-atmospheric radiative heat transfer is relatively slow and so is negligible for moving air. Thus, when air ascends or descends, there is little exchange of heat with the surrounding air. A process in which no heat is exchanged with the environment is referred to as an adiabatic process. Air expands as it moves upward, and contracts as it moves downward. The expansion of rising air parcels, and the contraction of descending air parcels, are adiabatic processes, to a good approximation. When a parcel of air expands, it pushes on the air around it, doing thermodynamic work. Since the upward-moving and expanding parcel does work but gains no heat, it loses internal energy so that its temperature decreases. Downward-moving and contracting air has work done on it, so it gains internal energy and its temperature increases. Adiabatic processes for air have a characteristic temperature-pressure curve. As air circulates vertically, the air takes on that characteristic gradient, called the adiabatic lapse rate. When the air contains little water, this lapse rate is known as the dry adiabatic lapse rate: the rate of temperature decrease is 9.8°C/km (5.4°F per 1,000 ft) (3.0°C/1,000 ft). The reverse occurs for a sinking parcel of air. When the environmental lapse rate is less than the adiabatic lapse rate the atmosphere is stable and convection will not occur.: 63 The environmental lapse is forced towards the adiabatic lapse rate whenever air is convecting vertically. Only the troposphere (up to approximately 12 kilometres (39,000 ft) of altitude) in the Earth's atmosphere undergoes convection: the stratosphere does not generally convect. However, some exceptionally energetic convection processes, such as volcanic eruption columns and overshooting tops associated with severe supercell thunderstorms, may locally and temporarily inject convection through the tropopause and into the stratosphere. Energy transport in the atmosphere is more complex than the interaction between radiation and dry convection. The water cycle (including evaporation, condensation, precipitation) transports latent heat and affects atmospheric humidity levels, significantly influencing the temperature profile, as described below. Mathematics of the adiabatic lapse rate [edit] Simplified graph of atmospheric lapse rate near sea level The following calculations derive the temperature as a function of altitude for a packet of air which is ascending or descending without exchanging heat with its environment. Dry adiabatic lapse rate [edit] Thermodynamics defines an adiabatic process as: P d V=−V d P γ{\displaystyle P\,\mathrm {d} V=-{\frac {V\,\mathrm {d} P}{\gamma }}} the first law of thermodynamics can be written as m c v d T−V d P γ=0{\displaystyle mc_{\text{v}}\,\mathrm {d} T-{\frac {V\,\mathrm {d} P}{\gamma }}=0} Also, since the density ρ=m/V{\displaystyle \rho =m/V} and γ=c p/c v{\displaystyle \gamma =c_{\text{p}}/c_{\text{v}}}, we can show that: ρ c p d T−d P=0{\displaystyle \rho c_{\text{p}}\,\mathrm {d} T-\mathrm {d} P=0} where c p{\displaystyle c_{\text{p}}} is the specific heat at constant pressure. Assuming an atmosphere in hydrostatic equilibrium: d P=−ρ g d z{\displaystyle \mathrm {d} P=-\rho g\,\mathrm {d} z} where g is the standard gravity. Combining these two equations to eliminate the pressure, one arrives at the result for the dry adiabatic lapse rate (DALR), Γ d=−d T d z=g c p=9.8∘C/km{\displaystyle \Gamma {\text{d}}=-{\frac {\mathrm {d} T}{\mathrm {d} z}}={\frac {g}{c{\text{p}}}}=9.8\ ^{\circ }{\text{C}}/{\text{km}}} The DALR (Γ d{\displaystyle \Gamma _{\text{d}}}) is the temperature gradient experienced in an ascending or descending packet of air that is not saturated with water vapor, i.e., with less than 100% relative humidity. Moist adiabatic lapse rate [edit] The presence of water within the atmosphere (usually the troposphere) complicates the process of convection. Water vapor contains latent heat of vaporization. As a parcel of air rises and cools, it eventually becomes saturated; that is, the vapor pressure of water in equilibrium with liquid water has decreased (as temperature has decreased) to the point where it is equal to the actual vapor pressure of water. With further decrease in temperature the water vapor in excess of the equilibrium amount condenses, forming cloud, and releasing heat (latent heat of condensation). Before saturation, the rising air follows the dry adiabatic lapse rate. After saturation, the rising air follows the moist (or wet) adiabatic lapse rate. The release of latent heat is an important source of energy in the development of thunderstorms. While the dry adiabatic lapse rate is a constant 9.8°C/km (5.4°F per 1,000 ft, 3°C/1,000 ft), the moist adiabatic lapse rate varies strongly with temperature. A typical value is around 5°C/km, (9°F/km, 2.7°F/1,000 ft, 1.5°C/1,000 ft). The formula for the saturated adiabatic lapse rate (SALR) or moist adiabatic lapse rate (MALR) is given by: Γ w=g(1+H v r R sd T)(c pd+H v 2 r R sw T 2){\displaystyle \Gamma {\text{w}}=g\,{\frac {\left(1+{\dfrac {H{\text{v}}\,r}{R_{\text{sd}}\,T}}\right)}{\left(c_{\text{pd}}+{\dfrac {H_{\text{v}}^{2}\,r}{R_{\text{sw}}\,T^{2}}}\right)}}} where: Γ w{\displaystyle \Gamma {\text{w}}},wet adiabatic lapse rate, K/m g{\displaystyle g},Earth's gravitational acceleration = 9.8067 m/s 2 H v{\displaystyle H{v}},heat of vaporization of water = 2 501 000 J/kg R sd{\displaystyle R_{\text{sd}}},specific gas constant of dry air = 287 J/kg·K R sw{\displaystyle R_{\text{sw}}},specific gas constant of water vapour = 461.5 J/kg·K ϵ=R sd R sw{\displaystyle \epsilon ={\frac {R_{\text{sd}}}{R_{\text{sw}}}}},the dimensionless ratio of the specific gas constant of dry air to the specific gas constant for water vapour = 0.622 e{\displaystyle e},the water vapour pressure of the saturated air r=ϵ e p−e{\displaystyle r={\frac {\epsilon e}{p-e}}},the mixing ratio of the mass of water vapour to the mass of dry air p{\displaystyle p},the pressure of the saturated air T{\displaystyle T},temperature of the saturated air, K c pd{\displaystyle c_{\text{pd}}},the specific heat of dry air at constant pressure, = 1003.5 J/kg·K The SALR or MALR (Γ w{\displaystyle \Gamma _{\text{w}}}) is the temperature gradient experienced in an ascending or descending packet of air that is saturated with water vapor, i.e., with 100% relative humidity. Effect on weather [edit] This section relies largely or entirely on a single source. Relevant discussion may be found on the talk page. Please help improve this article by introducing citations to additional sources. Find sources:"Lapse rate"–news·newspapers·books·scholar·JSTOR(March 2022) The latent heat of vaporization adds energy to clouds and storms. The varying environmental lapse rates throughout the Earth's atmosphere are of critical importance in meteorology, particularly within the troposphere. They are used to determine if the parcel of rising air will rise high enough for its water to condense to form clouds, and, having formed clouds, whether the air will continue to rise and form bigger shower clouds, and whether these clouds will get even bigger and form cumulonimbus clouds (thunder clouds). As unsaturated air rises, its temperature drops at the dry adiabatic rate. The dew point also drops (as a result of decreasing air pressure) but much more slowly, typically about 2°C per 1,000 m. If unsaturated air rises far enough, eventually its temperature will reach its dew point, and condensation will begin to form. This altitude is known as the lifting condensation level (LCL) when mechanical lift is present and the convective condensation level (CCL) when mechanical lift is absent, in which case, the parcel must be heated from below to its convective temperature. The cloud base will be somewhere within the layer bounded by these parameters. The difference between the dry adiabatic lapse rate and the rate at which the dew point drops is around 4.5°C per 1,000 m. Given a difference in temperature and dew point readings on the ground, one can easily find the LCL by multiplying the difference by 125 m/°C. If the environmental lapse rate is less than the moist adiabatic lapse rate, the air is absolutely stable — rising air will cool faster than the surrounding air and lose buoyancy. This often happens in the early morning, when the air near the ground has cooled overnight. Cloud formation in stable air is unlikely. If the environmental lapse rate is between the moist and dry adiabatic lapse rates, the air is conditionally unstable — an unsaturated parcel of air does not have sufficient buoyancy to rise to the LCL or CCL, and it is stable to weak vertical displacements in either direction. If the parcel is saturated it is unstable and will rise to the LCL or CCL, and either be halted due to an inversion layer of convective inhibition, or if lifting continues, deep, moist convection (DMC) may ensue, as a parcel rises to the level of free convection (LFC), after which it enters the free convective layer (FCL) and usually rises to the equilibrium level (EL). If the environmental lapse rate is larger than the dry adiabatic lapse rate, it has a superadiabatic lapse rate, the air is absolutely unstable — a parcel of air will gain buoyancy as it rises both below and above the lifting condensation level or convective condensation level. This often happens in the afternoon mainly over land masses. In these conditions, the likelihood of cumulus clouds, showers or even thunderstorms is increased. Meteorologists use radiosondes to measure the environmental lapse rate and compare it to the predicted adiabatic lapse rate to forecast the likelihood that air will rise. Charts of the environmental lapse rate are known as thermodynamic diagrams, examples of which include Skew-T log-P diagrams and tephigrams. (See also Thermals). The difference in moist adiabatic lapse rate and the dry rate is the cause of foehn wind phenomenon (also known as "Chinook winds" in parts of North America). The phenomenon exists because warm moist air rises through orographic lifting up and over the top of a mountain range or large mountain. The temperature decreases with the dry adiabatic lapse rate, until it hits the dew point, where water vapor in the air begins to condense. Above that altitude, the adiabatic lapse rate decreases to the moist adiabatic lapse rate as the air continues to rise. Condensation is also commonly followed by precipitation on the top and windward sides of the mountain. As the air descends on the leeward side, it is warmed by adiabatic compression at the dry adiabatic lapse rate. Thus, the foehn wind at a certain altitude is warmer than the corresponding altitude on the windward side of the mountain range. In addition, because the air has lost much of its original water vapor content, the descending air creates an arid region on the leeward side of the mountain. Impact on the greenhouse effect [edit] If the environmental lapse rate was zero, so that the atmosphere was the same temperature at all elevations, then there would be no greenhouse effect. This doesn't mean the lapse rate and the greenhouse effect are the same thing, just that the lapse rate is a prerequisite for the greenhouse effect. The presence of greenhouse gases on a planet causes radiative cooling of the air, which leads to the formation of a non-zero lapse rate. So, the presence of greenhouse gases leads to there being a greenhouse effect at a global level. However, this need not be the case at a localized level. The localized greenhouse effect is stronger in locations where the lapse rate is stronger. In Antarctica, thermal inversions in the atmosphere (so that air at higher altitudes is warmer) sometimes cause the localized greenhouse effect to become negative (signifying enhanced radiative cooling to space instead of inhibited radiative cooling as is the case for a positive greenhouse effect). Lapse rate in an isolated column of gas [edit] A question has sometimes arisen as to whether a temperature gradient will arise in a column of still air in a gravitational field without external energy flows. This issue was addressed by James Clerk Maxwell, who established in 1868 that if any temperature gradient forms, then that temperature gradient must be universal (i.e., the gradient must be same for all materials) or the second law of thermodynamics would be violated. Maxwell also concluded that the universal result must be one in which the temperature is uniform, i.e., the lapse rate is zero. Santiago and Visser (2019) confirm the correctness of Maxwell's conclusion (zero lapse rate) provided relativistic effects are neglected. When relativity is taken into account, gravity gives rise to an extremely small lapse rate, the Tolman gradient (derived by R. C. Tolman in 1930). At Earth's surface, the Tolman gradient would be about Γ t=T s×(10−16{\displaystyle \Gamma {t}=T{s}\times (10^{-16}}m−1){\displaystyle ^{-1})}, where T s{\displaystyle T_{s}} is the temperature of the gas at the elevation of Earth's surface. Santiago and Visser remark that "gravity is the only force capable of creating temperature gradients in thermal equilibrium states without violating the laws of thermodynamics" and "the existence of Tolman's temperature gradient is not at all controversial (at least not within the general relativity community)." See also [edit] Adiabatic process Atmospheric thermodynamics Fluid dynamics Foehn wind Lapse rate climate feedback Scale height Notes [edit] ^Note: Γ{\displaystyle \Gamma } and γ{\displaystyle \gamma } are both used in this article but with very distinct meanings. References [edit] ^Jacobson, Mark Zachary (2005). Fundamentals of Atmospheric Modeling (2nd ed.). Cambridge University Press. ISBN978-0-521-83970-9. ^Ahrens, C. Donald (2006). Meteorology Today (8th ed.). Brooks/Cole Publishing. ISBN978-0-495-01162-0. ^Todd S. Glickman (June 2000). Glossary of Meteorology (2nd ed.). American Meteorological Society, Boston. ISBN978-1-878220-34-9.(Glossary of Meteorology)Archived 2024-07-26 at the Wayback Machine ^Salomons, Erik M. (2001). Computational Atmospheric Acoustics (1st ed.). Kluwer Academic Publishers. ISBN978-1-4020-0390-5. ^Stull, Roland B. (2001). An Introduction to Boundary Layer Meteorology (1st ed.). Kluwer Academic Publishers. ISBN978-90-277-2769-5. ^Daidzic, Nihad E. (2019). "On Atmospheric Lapse Rates". International Journal of Aviation, Aeronautics, and Aerospace. 6 (4). doi:10.15394/ijaaa.2019.1374. ^Manual of the ICAO Standard Atmosphere (extended to 80 kilometres (262 500 feet)) (Third ed.). International Civil Aviation Organization. 1993. ISBN978-92-9194-004-2. Doc 7488-CD. ^ abManabe, Syukuro; Strickler, Robert F. (1964). "Thermal Equilibrium of the Atmosphere with a Convective Adjustment". Journal of the Atmospheric Sciences. 21 (4): 361–385. Bibcode:1964JAtS...21..361M. doi:10.1175/1520-0469(1964)021<0361:TEOTAW>2.0.CO;2. Retrieved 1 September 2024. ^"What is Earth's Energy Budget? Five Questions with a Guy Who Knows". NASA. 10 April 2017. Retrieved 1 September 2024. ^"Conduction". Center for Science Education. Retrieved 1 September 2024. ^Wallace, John M.; Hobbs, Peter V. (2006). Atmospheric Science (2 ed.). Elsevier. ISBN9780080499536. ^Hartmann, Dennis L.; Dygert, Brittany D.; Blossey, Peter N.; Fu, Qiang; Sokol, Adam B. (2022). "The Vertical Profile of Radiative Cooling and Lapse Rate in a Warming Climate". Journal of Climate. 35 (19): 6253–6265. Bibcode:2022JCli...35.2653H. doi:10.1175/JCLI-D-21-0861.1. Retrieved 1 September 2024. ^ abRichard M. Goody; James C.G. Walker (1972). "Atmospheric Temperatures"(PDF). Atmospheres. Prentice-Hall. Archived from the original(PDF) on 2016-06-03. ^Williams, Jack. "Understanding Air Density and its Effects". USAToday.com. Retrieved 1 September 2024. ^"Is humid air heavier than dry air?". howstuffworks. Retrieved 1 September 2024. ^Danielson, EW; Levin, J; Abrams, E (2002). Meteorology. McGraw Hill Higher Education. ISBN9780072420722. ^"The stratosphere: overview". UCAR. Retrieved 2016-05-02. ^Landau and Lifshitz, Fluid Mechanics, Pergamon, 1979 ^Kittel; Kroemer (1980). "6". Thermal Physics. W. H. Freeman. p.179. ISBN978-0-7167-1088-2. problem 11 ^"Dry Adiabatic Lapse Rate". tpub.com. Archived from the original on 2016-06-03. Retrieved 2016-05-02. ^Minder, JR; Mote, PW; Lundquist, JD (2010). "Surface temperature lapse rates over complex terrain: Lessons from the Cascade Mountains". J. Geophys. Res. 115 (D14): D14122. Bibcode:2010JGRD..11514122M. doi:10.1029/2009JD013493. ^"Saturation adiabatic lapse rate". Glossary. American Meteorological Society. ^"Mixing ratio". Glossary. American Meteorological Society. ^Whiteman, C. David (2000). Mountain Meteorology: Fundamentals and Applications. Oxford University Press. ISBN978-0-19-513271-7. ^Thomas, Gary E.; Stamnes, Knut (1999). Radiative Transfer in the Atmosphere and Ocean. Cambridge University Press. ISBN0-521-40124-0. ^Schmithüsen, Holger; Notholt, Justus; König-Langlo, Gert; Lemke, Peter; Jung, Thomas (16 December 2015). "How increasing CO 2 leads to an increased negative greenhouse effect in Antarctica". Geophysical Research Letters. 42 (23). doi:10.1002/2015GL066749. ISSN0094-8276. S2CID131351000. ^Sejas, S.A.; Taylor, P. C.; Cai, M. (2018). "Unmasking the negative greenhouse effect over the Antarctic Plateau". npj Clim Atmos Sci. 1 (17): 17. Bibcode:2018npCAS...1...17S. doi:10.1038/s41612-018-0031-y. PMC7580794. PMID33102742. ^Maxwell, J Clerk (1868). "XXII. On the dynamical theory of gases". The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science. 35 (236): 185–217. ^Santiago, Jessica; Visser, Matt (2019). "Tolman temperature gradients in a gravitational field". European Journal of Physics. 40 (25604): 025604. arXiv:1803.04106. Bibcode:2019EJPh...40b5604S. doi:10.1088/1361-6404/aaff1c. ^Tolman, R. C. (1930). "On the weight of heat and thermal equilibrium in general relativity". Phys. Rev. 35 (8): 904. Bibcode:1930PhRv...35..904T. doi:10.1103/PhysRev.35.904. Further reading [edit] Beychok, Milton R. (2005). Fundamentals Of Stack Gas Dispersion (4th ed.). author-published. ISBN978-0-9644588-0-2.www.air-dispersion.com R. R. Rogers and M. K. Yau (1989). Short Course in Cloud Physics (3rd ed.). Butterworth-Heinemann. ISBN978-0-7506-3215-7. External links [edit] Definition, equations and tables of lapse rate from the Planetary Data system. National Science Digital Library glossary: Lapse Rate Environmental lapse rate Absolute stable air An introduction to lapse rate calculation from first principles from U. 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188048	https://cheenta.com/calculate-geometric-mean/	Book Store Success Story Whatsapp Support How to Calculate Geometric Mean \| Learn the Concept Join Trial or Access Free Resources Let's learn how to calculate the geometric mean. This is a concept video useful for Mathematics Olympiad and ISI and CMI Entrance. Watch and Learn: Read and Learn: What is the Geometric mean of two numbers a and b & how to calculate it? Suppose a and b are positive numbers then their geometric mean is defined as square root of a times b. This is the formula of geometric mean. We will construct the geometric mean of a and b geometrically. So, here are the steps: Let's start by drawing the diameter of a circle. Suppose the two endpoints are X and Y of the diameter. Choose any point M on this particular diameter Now, let's construct a perpendicular at M which hits the circle at N. M divides the diameter into two parts XM and YM. Suppose, the length of XM and YM be a and b respectively. Now, we want to show that the length of MN is the square root of a times b, i.e., the geometric mean of a and b. Now, join XN and YN, and then look at the triangles YMN and XMN and try to show that these two triangles are actually similar. Now, see the following proof: Since, we know that similar triangles have proportional sides. So, we can say, Hence, this is proved. Hope you liked it! Some Useful Links: AM-GM Inequality - Math Olympiad Concepts Video Triangular Number Sequence - Explanation with Application Our Math Olympiad Program Share: 🠄 Cheenta Toppers of the Month - January ISI Entrance 2020 Problems and Solutions - B.Stat... 🠆 More Posts Series and Trigonometry \| ISI B.Stat Entrance 2009 Gaps in Permutation \| TOMATO Objective Problem 145 Number Theory \| PRMO 2019 \| Problem 3 Geometry and Trigonometry \| PRMO 2019 \| Problem 11 1 2 3 … 51 Next » Leave a Reply Cancel reply This site uses Akismet to reduce spam. Learn how your comment data is processed.
188049	https://www.oed.com/dictionary/salvage_v	Advanced search Oxford English Dictionary The historical English dictionary An unsurpassed guide for researchers in any discipline to the meaning, history, and usage of over 500,000 words and phrases across the English-speaking world. Find out more about OED Understanding entries Glossaries, abbreviations, pronunciation guides, frequency, symbols, and more Explore resources Personal account Change display settings, save searches and purchase subscriptions Account features Getting started Videos and guides about how to use the new OED website Read our guides Recently added hoaching kottu roti busy bee private bag PMS-ing shwmae bedrid chiollagh digestivo Texas gate hyperpop bombil jibaro camera roll Uglish bowlered Word of the day aequoreal adjective Marine, oceanic. Recently updated Hobbist Boran tsarian brechan typing shurt hobbet tone arm shipload duma hubbly shipward tsar flaily hobbled beat-up ## Word stories Read our collection of word stories detailing the etymology and semantic development of a wide range of words, including ‘dungarees’, ‘codswallop’, and ‘witch’. ## Word lists Access our word lists and commentaries on an array of fascinating topics, from film-based coinages to Tex-Mex terms. ## World Englishes Explore our World Englishes hub and access our resources on the varieties of English spoken throughout the world by people of diverse cultural backgrounds. ## History of English Here you can find a series of commentaries on the History of English, charting the history of the English language from Old English to the present day. Sign in Personal account Access or purchase personal subscriptions Get our newsletter Save searches Set display preferences Sign in Register Institutional access Sign in through your institution Sign in with library card Sign in with username / password Recommend to your librarian Institutional account management Sign in as administrator on Oxford Academic Word of the Day Sign up to receive the Oxford English Dictionary Word of the Day email every day. Our Privacy Policy sets out how Oxford University Press handles your personal information, and your rights to object to your personal information being used for marketing to you or being processed as part of our business activities. We will only use your personal information for providing you with this service.
188050	https://jennvadnais.com/2016/02/20/desmos-dilations-the-vertical-classroom/	Desmos, Dilations & The Vertical Classroom \| Communicating Mathematically Communicating Mathematically Jenn Vadnais (@RilesBlue) Skip to content Home 3 Act Lessons About Jenn Posts Genius Hour ← Desmos, Complementary Angles &SMP#3 Collaborative Group Roles – How I Became a Fan → Desmos, Dilations & The Vertical Classroom Posted onFebruary 20, 2016byjgvadnais This is a story about a Desmos Activity Builder that flopped The changes that transformed it into a successful activity 1st time teaching the lesson: Mrs. Paine’s room I’m excited to test out theDesmos Activity Builder on Dilationswith students. The lesson started off good, students were engaged and then the activity fell flat. In between classes, I’d go back into the activity builder and make adjustments. None of the little adjustments worked. Since I was only scheduled to be in Mrs.Paine’s room for 1 day, the opportunity to clarify misconceptions didn’t exist. I felt a bit defeated. 2nd time teaching the lesson: Mr. Peterson’s room I’m grateful for the opportunity to redeem myself and the activity. I reworked the activity’s organization. Again, it started off good, then fell flat. UGH! During the last class of the day, a student began submitting off topic responses. When this issue was addressed, he shouted, “I hate computers. Math is NOT supposed to be done on computers!” My initial reaction was, “Did he just dis Desmos? What.is.he.thinking?” Teaching rule #37:Don’t take student comments personally (especially when said in frustration) but do listen to them. Seth had a point. The lesson was too computer heavy. Desmos was not the right medium for the given line of questioning. Thankful for the follow up day, I completely changed the next day’s lesson and emailed Mr. P the plan. It was time to combine the Activity Builder with his vertical classroom set up. In the new format, Desmos would play a supporting role to the work completed on the vertical white boards. The Desmos/Vertical Classroom format was a success. Students were engaged, their attitudes were positive and learning increased 200% from the previous day. And Seth – He was a happy and active participant. Here’s how it flowed … Desmos Dilations Activity Builder Combined w/ Vertical White Boards. Slides 1- 6: These slides help students understand how to move the triangles and measure an angle. Vocabulary such as, pre-image, image, dilation, similar was introduced. Day 1 Mr. Peterson: On the first day, It took roughly 15 minutes to discuss the information contained in slides 1 – 6. Since the rest of the lesson flopped, I restructured the activity builder. The problem slides were scrapped Advertisement Slide 8 Starting off day 2, we separated students into groups and positioned them by a vertical white board. One team member moved the black dots to create a unique triangle. The team picked a value for k that’s greater than 1 and recorded it on their white board. At first, k was labeled d for dilation. But since the dilation represents the constant of proportionality, I changed the d to k within the activity builder. That changed occurred after the pictures were taken. Teams were required to recreate the pre-image and image/dilation on their white board. Slide 9: Label all parts of the diagram. This required students to Read the left side of slide 8 Understand the left side of slide 8 Label appropriately Use the measuring tool to find the angles measures. Left side of slide 8 Slide 10: Write 3 ratios in the form Dilated side/Pre-image side. Then simplify. Advertisement Slide 11:This slide connects their data to the direct variation equation, y = kx. Fill out the x/y charts based on your ratios. Use the red line to help you determine the equation that represents your data (see activity builder) Enter the equation that represents your data. Write the equation on your white board. Extension of Slide 11/Slide 12:Turn the equation into a sentence Slide 13 – 17: Repeat the process outlined in slides 8 – 12 but with k representing a number less than 1. Students were independent throughout these slides allowing Mr. Peterson and me to spend a considerable amount of time checking for understanding. Closing Thoughts One of the many aspects of a teacher’s job is to create and guide students through lessons. If the initial lesson plan isn’t making the desired impact, teachers hold the power to alter the plan. Teachers reflect and fine-tune throughout the day, therefore building their capacity to quickly assess student learning and change course when necessary. The bottom line regarding the Dilations Activity Builder:All the students (in both classes) began to lose interest around the same slide, regardless of the little adjustments made in between classes. Advertisement Why?A couple of slides required students to analyze their pre-image and dilation and then submit responses. The slides were packed with information. Too much information for a single student to analyze. Students were not making the desired mathematical connections. They needed to work with the information, make it tangible– draw, label and discuss. Once the investigation involved physically creating and labeling diagrams on the vertical white boards, not only did students easily worked through the questions but also their understanding of dilations increased. Could students have drawn and labeled the images on paper?Yes, but I don’t believe it’s the best medium. The freedom to draw on a big space where one can stand back to scan all the pieces is undeniably more appealing to students. They worked with Desmos and the vertical white boards fluidly. With this format, I observed students’ independence shine. They were owning their learning. Other posts in the Vertical Classroom series The Vertical Classroom Project Begins The Vertical Classroom: Seeing is Believing A link to all my Desmos related posts & Desmos Activities All Things Desmos Advertisement Share this: Click to email a link to a friend (Opens in new window)Email Click to share on X (Opens in new window)X Click to share on Facebook (Opens in new window)Facebook Click to share on Pinterest (Opens in new window)1 Pinterest 1 Click to share on LinkedIn (Opens in new window)LinkedIn Click to print (Opens in new window)Print Click to share on Tumblr (Opens in new window)Tumblr Click to share on Pocket (Opens in new window)Pocket Click to share on WhatsApp (Opens in new window)WhatsApp More Click to share on Reddit (Opens in new window)Reddit Like Loading... Related The Zeroth Power: Desmos &VNPS Inspiration Negative exponents can be a confusing topic for many students. Due to time constraints, I've focused more on the rules in lieu of introducing negative exponents conceptually. As a result, students become confused on how the negative exponent effects the base. I wanted to change this outcome. One night,… April 5, 2016 In "desmos" Rethinking Classroom Design: Collaborative Stations, VNPS &Desmos Twitter Math Camp '15 sent me down the vertical classroom path (Click here to read post). Throughout the 2015-2016 school year, I promoted vertical non-permanent surfaces (VNPS) whenever possible. Lucky for me, Joe, a 7th/8th grade teacher, was interested and additional white (shower) boards were installed in his room. As I observed his… August 6, 2016 In "desmos" The Vertical Classroom Project Begins In This Article The inspiration for the Vertical Classroom Project My first VCP lesson General and specific observations Questions, concerns and ponderables The next step Materials The Inspiration By the end of Alex Overwijk's vertical classroom session at TMC15, I had decided on my #1TMCthing and my goal for the… August 27, 2015 In "math" About jgvadnais Instructional Technology Coach. Desmos Fellow. Google Level 1 Certified. SoCal transplant. New Englander at heart. Lover of yoga, dogs, green smoothies and coffee creamer View all posts by jgvadnais → This entry was posted in desmos, Math Education, proportional reasoning, Uncategorized, vertical classroom, vnps, white boards and tagged desmos, dilations, Direct variation, education, math, pre-image, proportional Reasoning, Vertical classroom, VNPS, y=kx. Bookmark the permalink. ← Desmos, Complementary Angles &SMP#3 Collaborative Group Roles – How I Became a Fan → 4 Responses to Desmos, Dilations & The Vertical Classroom Laura Jenkins says: February 20, 2016 at 6:47 pm I love this blog. I’m also considering being instructional coach. Do you have a Twitter? LikeLike Reply jgvadnaissays: February 20, 2016 at 6:58 pm Thank you for reading! My Twitter handle is @RilesBlue. I love being an instructional coach – very rewarding! 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188051	https://whereseric.com/faq/2004-eric-clapton-chrisites-guitar-auction-complete/	Home News Tour Dates Upcoming Concerts Eric Clapton Tour Archive Photos Discography EC FAQ EC Official Magazine About All Issues Subscription 2004 Eric Clapton Chrisite’s Guitar Auction: Complete Sale Details On 24 June 1999, Christie’s New York auctioned off a selection of Eric Clapton’s guitars which raised over $5 million for the Crossroads Centre in Antigua. On that sale’s fifth anniversary, 24 June 2004, Christie’s presented the sequel by selling the cream of Eric Clapton’s collection along with instruments donated by his musician friends. The 88 lots in the 2004 sale garnered a record-breaking $7,438,624 million (including buyer’s premium of 19.5% on sales up to $100,000 plus 12% of any amount in excess of $100,000. The proceeds (approximately $6.4 million) benefit the Crossroads Centre Foundation. The foundation funds Crossroads Centre at Antigua, a drug and alcohol residential rehabilitation facility, founded by Eric Clapton in 1997. The guitar which earned the most interest was Eric Clapton’s legendary Fender Stratocaster, Blackie. It sold for a staggering $959,500 (including buyer’s premium). This sale held the distinction of the most money paid at auction for a guitar of the rock era but was surpassed in later years. Lot 41 – Eric’s beloved Cherry Red ES335-TDC came in second selling for $847,500 (including buyer’s premium). Rounding out the top 3 was Lot 19 – the 1939 Martin 000-42 which Clapton played on MTV Unplugged. It garnered a staggering $791,500 (including buyer’s premium) List of the lots sold and their final selling price (which includes the buyer’s premium): Lot #1 – C.F. Martin & Company Style 0-18 (1898) $35,850 (Pre-Sale Estimate: $3,000 – $5,000) Lot #2 – C.F. Martin & Company Style 0-18 (1944) $23,900 (Pre-Sale Estimate: $3,000 – $5,000) Lot #3 – C.F. Martin & Company Style 0-16NY (1961) $19,120 (Pre-Sale Estimate: $2,000 – $3,000) Lot #4 – C.F. Martin & Company Style 0-21 (1925) $15,535 (Pre-Sale Estimate: $4,000 -$6,000) Lot #5 – C.F. Martin & Company Style 0-27 (1915) $35,850 (Pre-Sale Estimate: $5,000 – $7,000) Lot #6 – C.F. Martin & Company Style 00-21 (1966) $89,625 (Pre-Sale Estimate: $20,000 – $30,000) A gift from George Harrison to Eric Clapton Lot # 7 – C.F. Martin & Company Style 00-17 (1948) $20,315 (Pre-Sale Estimate: $2,500 – $3,500) Lot #8 – C.F. Martin & Company Style 0-45 (1915) $31,070 (Pre-Sale Estimate: $20,000 – $30,000) Lot #9 – C.F. Martin & Company Style 00-45 (1929) $71,700 (Pre-Sale Estimate: $30,000 – $50,000) Lot #10 – C.F. Martin & Company Style F-7/28 (‘35-‘73) $44,215 (Pre-Sale Estimate: $10,000 – $15,000) The “Andy Guitar” – converted from arch-top to flat-top – and played extensively by Andy Fairweather Low during his tenure in Eric Clapton’s band. Lot #11 – C.F. Martin & Company Style J-40BK (1994) $21,510 (Pre-Sale Estimate: $2,000 – $3,000) Lot #12 – C.F. Martin & Company Style J-40-12 Custom (1994) $45,410 (Pre-Sale Estimate: $4,000 – $6,000) Used by Eric Clapton during the recording of “From The Cradle” Lot #13 C.F. Martin & Company Sytle D-28 (1948) $50,190 (Pre-Sale Estimate: $10,000 – $15,000) Used by Eric Clapton at the Music For Montserrat benefit concert in 1997 Lot #14 – David Russell Young Dreadnought (1978) $31,070 (Pre-Sale Estimate: $3,000 – $5,000) Lot #15 – Dana Bourgeois Model D-150 (circa 1996) $3,107 (Pre-Sale Estimate: $1,000 – $2,000) Donated by Dan Tyminski Lot #16 – C.F. Martin & Company Style 000-28/45 Conversion (1966) $186,700 (Pre-Sale Estimate $20,000 – $30,000) “The Longworth” Lot #17 – C.F. Martin & Company Style 000-28EC (2002) $28,680 (Pre-Sale Estimate: $4,000 – $6,000) Purchased by Eric Clapton for use at Concert For George Rehearsals Lot #18 – C.F. Martin & Company Style 000-42ECB (2000) $71,700 (Pre-Sale Estimate: $15,000 – $20,000) Used on a number of recordings by Eric Clapton, including his “Reptile” Album Lot #19 – C.F. Martin & Company Style 000-42 (1939) $791,500 (Pre-Sale Estimate: $60,000 – $80,000) Main instrument used by Eric Clapton at the filming of MTV Unplugged in January 1992. This sale set a WORLD AUCTION RECORD FOR A MARTIN GUITAR at the time. Lot #20 – C.F. Martin & Company Style 000-42 (1939) $113,525 (Pre-Sale Estimate: $35,000 – $55,000) Home and studio guitar Lot #21 – William B. Tilton Style Grade 2 (circa 1865) $26,290 (Pre-Sale Estimate: $2,000 – $3,000) “The Ligvoder Strut” Lot #22 – Dobro by The Regal Company (late 1930s) $62,140 (Pre-Sale Estimate: $3,000 – $5,000 Used by Eric Clapton at The White House and on the Pilgrim Tour of Japan 1999 Lot #23 – Dobro Model R-5 Duolian (1989) $14,340 (Pre-Sale Estimate: $1,000 – $1,500) Hawaiin motif Lot #24 – Tacoma “Papoose” (1998) $2,032 (Pre-Sale Estimate: $1,000 – $2,000) Donated by Eric Johnson Lot #25 – Jose Ramirez III Classical Guitar (1992) $28,680 (Pre-Sale Estimate: $7,000 – $9,000) Used by Eric Clapton on “Pilgrim” and other 1990s recordings Lot #26 – Fernandez Gerundino Flamenco Guitar (1976) $16,730 (Pre-Sale Estimate: $6,000 – $8,000) Purchased by Eric Clapton for use at Concert For George Rehearsals Lot #27 – Robert S. Ruck Classical Guitar (1975) $7,768 (Pre-Sale Estimate: $3,000 – $5,000) Lot #28 – Guitarras Alhambra / Contemporary Flamenco Guitar $14,340 (Pre-Sale Estimate: $800 – $1,200) Lot #29 – Juan Alvarez Classical Guitar (1977) $253,900 (Pre-Sale Estimate: $10,000 – $15,000) Donated by Giorgio Armani, used by Eric Clapton in filming the “Tears In Heaven” video Lot #30 – Coppertone / Brian Knight Acoustic Steel String Guitar $10,158 (Pre-Sale Estimate: $1,000 – $1,500) Back is copper sheeting and engraved with the figure of a knight Lot #31 – Maurice Dupont Model MD-30 (1995) $7,170 (Pre-Sale Estimate: $3,000 – $5,000) Lot #32 – George Lowden Model 038 (circa 1996) $41,825 (Pre-Sale Estimate: $5,000 – $7,000) Used by Eric Clapton at the 39th Annual Grammy Awards Ceremony NYC Lot #33 – Anthony C. Zemaitis 12-String Guitar (1969) $253,900 (Pre-Sale Estimate: $30,000 – $50,000) “Ivan The Terrible” and co-designed by Eric Clapton Lot #34 – Gibson L-5 (1929) $27,485 (Pre-Sale Estimate: $10,000 – $15,000) Lot #35 – Gibson ’34 L-5 Custom Shop (1997) $23,900 (Pre-Sale Estimate: $6,000 – $8,000) Used by Eric Clapton when recording “Pilgrim” Lot #36 – Gibson L-75 (1935) $9,560 (Pre-Sale Estimate: $2,000 – $3,000) Lot #37 – Gibson L-5 Premier (circa 1948) $15,535 (Pre-Sale Estimate: $12,000 – $18,000) Studio guitar used primarily by Andy Fairweather Low who was in Eric Claptons band from 1992 – 2003. Lot #38 – Gibson L-5-P / CES Conversion (1948) $101,575 (Pre-Sale Estimate: $10,000 – $15,000) Lot #39 – Gibson L-5-CES (1956) $65,725 (Pre-Sale Estimate: $15,000 -$20,000 Used as a spare guitar for Eric Clapton’s main Gibson L-5 Lot #40 – Gibson ES-350T (1956) $65,725 (Pre-Sale Estimate: $8,000 – $12,000) Used by Eric Clapton on the Nothing But The Blues Tour Lot #41 – Gibson ES-335 TDC $847,500 (Pre-Sale Estimate: $60,000 – $80,000) The “Cherry Red” Gibson, owned and used extensively by Eric Clapton from 1964. WORLD AUCTION RECORD FOR A GIBSON GUITAR / THIRD HIGHEST PRICE PAID FOR A GUITAR at the time. Lot #42 – Gibson ES335 DOT/CH $5,378 (Pre-Sale Estimate: $2,000 – $3,000) Donated by Otis Rush Lot #43 – Gibson Custom Shop ES-336 (1996) $8,365 (Pre-Sale Estimate: $2,000 – $3,000) Lot #44 – Gibson BB King Lucille Model (2003) $23,900 (Pre-Sale Estimate: $2,000 – $3,000) Donated by B.B. King and Gibson Guitars Lot #45 0 Gibson Custom Shop Les Paul (circa 2001) $52,580 (Pre-Sale Estimate: $4,000 – $6,000) Played by Eric Clapton at the Rose Garden Arena, Portland, Oregon in 2001 Lot #46 – Gibson Les Paul Studio (1992) $23,900 (Pre-Sale Estimate: $1,000 – $2,000) Donated by Brian May, signed by artists at Freddie Mercury Tribute Concert at Wembley Stadium in 1992 Lot #47 – Gibson SG Pete Townshend Signature Model Prototype (2000) $38,240 (Pre-Sale Estimate: $3,000 – $5,000) Donated by Pete Townshend, Number 7 of 10 Lot #48 – Gibson Firebird V Re-Issue (1995) $8,963 (Pre-Sale Estimate: $1,000 – $2,000) Donated by Sonny Landreth Lot #49 – Gibson Chet Atkins CE (1998) $35,850 (Pre-Sale Estimate: $2,000 – $3,000) Used by Eric Clapton during the 2003 Japan Tour for “Can’t Find My Way Home” Lot #50 – Ibanez Pat Metheny PM-100NT $7,170 (Pre-Sale Estimate: $2,000 – $3,000) Donated by Pat Metheny Lot #51 – Terry C. McInturff Model TCM Glory Standard Custom (1998) $12,548 (Pre-Sale Estimate: $4,000 – $6,000) Lot #52 – Gretsch Burgundy Duo Jet (1962) $20,315 (Pre-Sale Estimate: $4,000 – $6,000) Lot #53 – Valco-Supro Model 1540 Bermuda (1963) $8,365 (Pre-Sale Estimate: $1,000 – $2,000) Donated by Hubert Sumlin Lot #54 – Ovation Preacher 12-String Model 1285 (1975) $31,070 (Pre-Sale Estimate: $3,000 – $5,000) Donated by Roger Waters Lot #55 – Ibanez JEM 7D (circa 2002) $5,378 (Pre-Sale Estimate: $1,000 – $2,000) Donated by Steve Vai Lot #56 – Ibanez JS 1000 (circa 2003) $16,370 (Pre-Sale Estimate: $1,000 – $2,000) Donated by Joe Satriani Lot #57 – Ernie Ball – Music Man Albert Lee Model (1997) $11,950 (Pre-Sale Estimate: $1,000 – $2,000 Donated by Albert Lee, his main stage guitar for the last 6 years Lot #58 – Schecter in the style of a Fender Stratocaster (circa 1980) $50,190 (Pre-Sale Estimate: $4,000 – $6,000 Donated by Mark Knopfler Lot #59 Tom Anderson Guitarworks – Tom Anderson Classic (1991) $16,370 (Pre-Sale Estimate: $3,000 – $5,000) Donated by Carlos Santana Lot #60 – Valley Art Guitar Classic Pro Model (circa 1984) $7,768 (Pre-Sale Estimate: $1,000 – $2,000 Donated by Larry Carlton Lot #61 – Robin Octave Double Neck (circa 1981 – 1982) $45,410 (Pre-Sale Estimate: $4,000 – $6,000) Donated by Jimmie Vaughan Lot #62 – Crossroads – Hard Rock CafÃ© EC Autographed Poster $3,107 (Pre-Sale Estimate: $400 – $600) Lot #63 – Crossroads – Hard Rock CafÃ© EC Autographed Poster $4,183 (Pre-Sale Estimate: $400 – $600) Lot #64 – Gianni Versace Stage Suit (two-piece, purple) $4,541 (Pre-Sale Estimate: $800 – $1,200) Worn onstage by Eric Clapton in the 1990s Lot #65 – Gianni Versace Stage Suit (three-piece, ivory) $28,680 (Pre-Sale Estimate: $2,000 – $3,000 Worn onstage by Eric Clapton in the 1990s Lot #66 – Gianni Versace Stage Suit (two-piece, black and cream) $19,120 (Pre-Sale Estimate: $800 – $1,200) Worn onstage by Eric Clapton in the 1990s Lot #67 – Gianni Versace / Fender Guitar Strap (circa early 1990s) $10,158 (Pre-Sale Estimate: $400 – $600 Intricate design of black and clear rhinestones Lot #68 – Gianni Versace / Ernie Ball Guitar Strap (circa early 1990s)$10,158 (Pre-Sale Estimate: $400 – $600 Stripes of black and clear rhinestones Lot #69 – Fender Stratocaster – Master Built Production Sample (circa 1996) $186,700 (Pre-Sale Estimate: $8,000 – $12,000) Black finish – one of Eric Clapton’s main stage guitars between 1998 – 1999 Lot #70 – Fender Electric Mandolin (1957) $11,950 (Pre-Sale Estimate: $4,000 – $6,000) Lot #71 – Fender Twin Amp Model 5F8-4 (circa 1958) $23,900 (Pre-Sale Estimate: $2,000 – $3,000) Lot #72 – Fender Jazzmaster (1958) $15,535 (Pre-Sale Estimate: $5,000 – $7,000) Lot #73 – Fender Composite Stratocaster (circa 1959 and 1960s) $41,825 (Pre-Sale Estimate: $3,000 – $4,000) Rosewood fingerboard Lot #74 – Fender Telecaster Lefty with A B-Bender (2000) $10,158 (Pre-Sale Estimate: $1,000 – $2,000) Donated by Doyle Bramhall II Lot #75 – Fender Telecaster (1967) $50,190 (Pre-Sale Estimate: $3,000 – $4,000) “The Virginia Telecaster” Lot #76 – Fender Stratocaster (1979) $16,730 (Pre-Sale Estimate: $2,000 – $3,000) Donated by J.J. Cale Lot #77 – Fender Stratocaster “The Strat” (circa 1980) $21,510 (Pre-Sale Estimate: $2,000 -$3,000) Donated by Jimmy Page Lot #78 – Fender Stratocaster Custom Built by J.W. Black (1993) $22,705 (Pre-Sale Estimate: $4,000 – $6,000) Aluminum body (chambered) in mirror finish – prototype Lot #79 – Fender Stratocaster 50th Anniversary Issue Master Built (1996) $455,500 (Pre-Sale Estimate: $8,000 – $12,000) Eric Clapton Signature Model, 23 carat gold leaf finish. It was Eric Clapton’s main stage guitar during 1997. Lot #80 – Fender Stratocaster Custom Shop (1994) $13,145 (Pre-Sale Estimate: $3,000 – $5,000) Olympic White with numerous signatures – presentation guitar from T.J. Martell Foundation and MTV Lot #81 – Fender Stratocaster Eric Clapton Signature Model (1997) $16,730 (Pre-Sale Estimate: $4,000 – $6,000) Body has Volkswagen Sound Foundation logo accompanied by 1998 tour memorabilia created by VW Lot #82 – Fender Stratocaster (circa 1985) $16,730 (Pre-Sale Estimate: $1,000 – $2,000) Donated by Robert Cray – used extensively on stage and in the film Hail Hail Rock N Roll Lot #83 – Fender Composite Stratocaster (circa 1965 and later) $623,500 (Pre-Sale Estimate: $15,000 – $20,000) “Lenny” – Donated by Jimmie Vaughan from the Estate of Stevie Ray Vaughan – used by SRV throughout his career. This is the only personal guitar of SRV to be released by the Estate into the public domain. SECOND HIGHEST PRICE FOR A FENDER STRATOCASTER at the time. Lot #84 – Fender Custom Shop Stratocaster Stevie Ray Vaughan Relic (2004) $41,825 (Pre-Sale Estimate: $6,000 – $8,000) Donated by Jimmie Vaughan and Fender – replica of SRV’s “Number One” Lot #85 – Fender Custom Shop Stratocaster (2000) $220,300 (Pre-Sale Estimate: $7,000 – $9,000) Iridescent purple / blue finish – used on stage by EC during 2001 Lot #86 – Fender Custom Shop Stratocaster (1988) $231,500 (Pre-Sale Estimate: $10,000 – $15,000) Eric Clapton Signature Model – black. Eric Clapton’s main stage guitar between 1990 and 1993 Lot #87 – Fender Stratocaster Master Built Crash Concept Model (2004) $321,100 (Pre-Sale Estimate: $8,000 – $12,000) Third guitar painted by grafitti artist, Crash, for Eric Clapton. Used from 15 March 2004 up until a concert the night before the auction in Albany, New York. Lot #88 – Fender Composite Stratocaster (circa 1956 and 1957) $959,500 (Pre-Sale Estimate: $100,000 – $150,000) “Blackie” – The guitar most famously associated with Eric Clapton. He has referred to this guitar as being “part of him.” Blackie was practically his sole stage and studio guitar from late 1970 to 1985. Occasionally used in the 1990s. WORLD AUCTION RECORD FOR A GUITAR at the time. Where’s Eric! Find us on Facebook Visit Our FB Page
188052	https://taoanalysis.wordpress.com/2020/04/26/exercise-4-3-1/	Tao Analysis Solutions Exercise 4.3.1 Exercise statement Prove Proposition 4.3.3. Proposition 4.3.3 (Basic properties of absolute value and distance). Let be rational numbers. (a) (Non-degeneracy of absolute value) We have . Also, if and only if is . (b) (Triangle inequality for absolute value) We have . (c) We have the inequalities if and only if . In particular, we have . (d) (Multiplicativity of absolute value) We have . In particular, . (e) (Non-degeneracy of distance) We have . Also, if and only if . (f) (Symmetry of distance) . (g) (Triangle inequality for distance) . Hints How to think about the exercise I want to prove a few supplementary facts here that will be useful in the proofs below. I think proving these will make some of the proofs below conceptually cleaner. It’s totally possible to do this exercise without proving any of these supplementary facts. In two of the proofs for the exercise (parts (b) and (d)) I prove a square version first, and use it to conclude the non-square version. I wish I had an explanation of why this works, but I don’t. It’s a simple and standard trick, but I don’t know why it should work. Fact 1. Let be a rational number. Then . Proof. If is positive or zero, this is true by definition of absolute value. If is negative, then we have . Fact 2. Let be rational numbers. If and is non-negative, then . Proof. We have several cases. If and is positive, then this follows from Proposition 4.2.9(e). If , then . If , then so . Fact 3. Let be non-negative rational numbers. If , then . Proof. Suppose first that or . If then . Thus so cannot be positive or negative. (For if then by Proposition 4.2.9(e) we could multiply by on both sides to obtain , a contradiction. Similarly for .) So by trichotomy we see that . Next, if , then so we have . In either case, . Now suppose both and are positive. Suppose for sake of contradiction that . Then by order trichotomy we have two cases, or . If then , a contradiction. And if then , which again is a contradiction. Thus we are left to conclude that . Fact 4. Let be non-negative rational numbers. If , then . Proof. We prove the contrapositive. Suppose . We want to show that . Since , we must have and . By Fact 2, we have . If then by Fact 3 we would have , which contradicts the fact that . Thus we must have and , i.e. . Finally, a psychological note: I found this to be the hardest post to write so far. I wrote parts of it one day, then gave up, then wrote some other parts on a second day, then put it down again, and came back to it on a third day and didn’t do much, put it down, and finally came back to finish it on a fourth day. For some reason, it felt very annoying to type everything up. I don’t think I got this feeling when working through the book myself, originally. Model solution (a) By order trichotomy, we have three cases: , , and . If then is a positive number, so . If then . If then is negative, so is positive, hence is positive which means we have . Therefore, . Now suppose . If were positive or negative, the absolute value would be a positive number. So by trichotomy, we see that must be zero. Conversely, if then by definition of absolute value we have . (b) There are several ways to prove this. Here are two: (c) Suppose . If then so we have . If then we have . Using and a slight modification of Exercise 4.2.6, we have . Now suppose . If then . This means , so . Thus we have . If then . Thus we have . This means is positive, so is negative. So we have , which means . To show that , take . Then we have so by the above result we must have . (d) Here are two ways to prove this: To show , take . (e) By part (a), . By part (a), we have if and only if . But we know that and that iff . Thus if and only if . (f) By part (d), . (g) By part (b) using and as inputs, we have . But and so the result follows. Share this: Related Post navigation 8 thoughts on “Exercise 4.3.1” I believe you made a typo (understandably) for Fact 2. xz <= yz. If it really was xz <= xy, i’d have a counter-example x=3, y=4, z=8. LikeLike Thanks, fixed! LikeLike Leave a comment Cancel reply Δ Pages Archives Categories Blog Stats Subscribe to Blog via Email Enter your email address to subscribe to this blog and receive notifications of new posts by email. Email Address: Subscribe RSS
188053	https://pubmed.ncbi.nlm.nih.gov/35561121/	Diagnostic accuracy of transvaginal ultrasound for detection of endometriosis using International Deep Endometriosis Analysis (IDEA) approach: prospective international pilot study - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Diagnostic accuracy of transvaginal ultrasound for detection of endometriosis using International Deep Endometriosis Analysis (IDEA) approach: prospective international pilot study M Leonardi12,C Uzuner1,W Mestdagh3,C Lu4,S Guerriero5,M Zajicek6,A Dueckelmann7,F Filippi8,F Buonomo9,M A Pascual10,A Stepniewska11,M Ceccaroni11,T Van den Bosch1213,D Timmerman1213,G Hudelist14,G Condous1 Affiliations Expand Affiliations 1 Acute Gynaecology, Early Pregnancy and Advanced Endosurgery Unit, Nepean Hospital, Sydney Medical School Nepean, University of Sydney, Sydney, NSW, Australia. 2 Department of Obstetrics and Gynecology, McMaster University, Hamilton, ON, Canada. 3 KU Leuven, Department of Electrical Engineering (ESAT), STADIUS Center for Dynamical Systems, Signal Processing and Data Analytics, Leuven, Belgium. 4 Computer Science, Aberystwyth University, Aberystwyth, UK. 5 Department of Obstetrics and Gynecology, University of Cagliari, Cagliari, Italy. 6 Department of Obstetrics and Gynecology, affiliated with the Sackler School of Medicine at Tel Aviv University, The Chaim Sheba Medical Center, Tel Hashomer, Ramat Gan, Israel. 7 Department of Gynecology, Charité University Hospital, Berlin, Germany. 8 Centro Procreazione Medicalmente Assistita, Fondazione IRCCS Ca' Granda Ospedale Maggiore Policlinico di Milano, Milan, Italy. 9 Institute for Maternal and Child Health IRCCS Burlo Garofolo, Trieste, Italy. 10 Department of Obstetrics, Gynecology and Reproduction, Hospital Universitari Dexeus, Barcelona, Spain. 11 Department of Obstetrics and Gynecology, Gynecology Oncology and Minimally Invasive Pelvic Surgery, International School of Surgical Anatomy (ISSA), IRCCS Ospedale Sacro Cuore - Don Calabria Negrar, Verona, Italy. 12 Department of Obstetrics and Gynecology, University Hospitals Leuven, Leuven, Belgium. 13 Department of Development and Regeneration, KU Leuven, Leuven, Belgium. 14 Department of Gynaecology, Hospital St John of God, Vienna, Austria. PMID: 35561121 DOI: 10.1002/uog.24936 Item in Clipboard Multicenter Study Diagnostic accuracy of transvaginal ultrasound for detection of endometriosis using International Deep Endometriosis Analysis (IDEA) approach: prospective international pilot study M Leonardi et al. Ultrasound Obstet Gynecol.2022 Sep. Show details Display options Display options Format Ultrasound Obstet Gynecol Actions Search in PubMed Search in NLM Catalog Add to Search . 2022 Sep;60(3):404-413. doi: 10.1002/uog.24936. Authors M Leonardi12,C Uzuner1,W Mestdagh3,C Lu4,S Guerriero5,M Zajicek6,A Dueckelmann7,F Filippi8,F Buonomo9,M A Pascual10,A Stepniewska11,M Ceccaroni11,T Van den Bosch1213,D Timmerman1213,G Hudelist14,G Condous1 Affiliations 1 Acute Gynaecology, Early Pregnancy and Advanced Endosurgery Unit, Nepean Hospital, Sydney Medical School Nepean, University of Sydney, Sydney, NSW, Australia. 2 Department of Obstetrics and Gynecology, McMaster University, Hamilton, ON, Canada. 3 KU Leuven, Department of Electrical Engineering (ESAT), STADIUS Center for Dynamical Systems, Signal Processing and Data Analytics, Leuven, Belgium. 4 Computer Science, Aberystwyth University, Aberystwyth, UK. 5 Department of Obstetrics and Gynecology, University of Cagliari, Cagliari, Italy. 6 Department of Obstetrics and Gynecology, affiliated with the Sackler School of Medicine at Tel Aviv University, The Chaim Sheba Medical Center, Tel Hashomer, Ramat Gan, Israel. 7 Department of Gynecology, Charité University Hospital, Berlin, Germany. 8 Centro Procreazione Medicalmente Assistita, Fondazione IRCCS Ca' Granda Ospedale Maggiore Policlinico di Milano, Milan, Italy. 9 Institute for Maternal and Child Health IRCCS Burlo Garofolo, Trieste, Italy. 10 Department of Obstetrics, Gynecology and Reproduction, Hospital Universitari Dexeus, Barcelona, Spain. 11 Department of Obstetrics and Gynecology, Gynecology Oncology and Minimally Invasive Pelvic Surgery, International School of Surgical Anatomy (ISSA), IRCCS Ospedale Sacro Cuore - Don Calabria Negrar, Verona, Italy. 12 Department of Obstetrics and Gynecology, University Hospitals Leuven, Leuven, Belgium. 13 Department of Development and Regeneration, KU Leuven, Leuven, Belgium. 14 Department of Gynaecology, Hospital St John of God, Vienna, Austria. PMID: 35561121 DOI: 10.1002/uog.24936 Item in Clipboard Cite Display options Display options Format Abstract Objective: To evaluate the diagnostic accuracy of transvaginal ultrasound (TVS) in predicting deep endometriosis (DE) following the International Deep Endometriosis Analysis (IDEA) consensus methodology. Methods: This was an international multicenter prospective diagnostic accuracy study involving eight centers across six countries (August 2018-November 2019). Consecutive participants with endometriosis suspected based on clinical symptoms or historical diagnosis of endometriosis were included. The index test was TVS performed preoperatively in accordance with the IDEA consensus statement. At each center, the index test was interpreted by a single sonologist. Reference standards were: (1) direct visualization of endometriosis at laparoscopy, as determined by a non-blinded surgeon with expertise in endometriosis surgery; and (2) histological assessment of biopsied/excised tissue. Surgery was performed within 12 months following the index TVS. Accuracy, sensitivity, specificity, positive and negative predictive values (PPV and NPV) and positive and negative likelihood ratios (LR+ and LR-) of TVS in the diagnosis of DE were calculated. Results: Included in the study were 273 participants with complete clinical, TVS, laparoscopic and histological data. Of these, based on histology, 256 (93.8%) were confirmed to have endometriosis, including superficial endometriosis, and 190 (69.6%) were confirmed to have DE. Based on surgical visualization, 207/273 (75.8%) patients had DE. For DE overall, the diagnostic performance of TVS based on surgical visualization as the reference standard was as follows: accuracy, 86.1%; sensitivity, 88.4%; specificity, 78.8%; PPV, 92.9%; NPV, 68.4%; LR+, 4.17; LR-, 0.15, and the diagnostic performance of TVS based on histology as the reference standard was as follows: accuracy, 85.9%; sensitivity, 89.8%; specificity, 75.9%; PPV, 90.4%; NPV, 74.6%; LR+, 3.72; LR-, 0.13. Conclusions: Using the IDEA consensus methodology provides strong diagnostic accuracy for TVS assessment of DE. We found a higher TVS detection rate of DE overall than that reported by the most recent meta-analysis on the topic (sensitivity, 79%), albeit with a lower specificity. © 2022 International Society of Ultrasound in Obstetrics and Gynecology. Keywords: diagnostic accuracy; endometriosis; pelvic pain; sliding sign; transvaginal ultrasound. © 2022 International Society of Ultrasound in Obstetrics and Gynecology. PubMed Disclaimer Similar articles Prospective diagnostic test accuracy of transvaginal ultrasound posterior approach for uterosacral ligament and torus uterinus deep endometriosis.Freger SM, Turnbull V, McGowan K, Leonardi M.Freger SM, et al.Ultrasound Obstet Gynecol. 2024 Feb;63(2):263-270. doi: 10.1002/uog.27492.Ultrasound Obstet Gynecol. 2024.PMID: 37725753 Transvaginal ultrasound for diagnosis of deep endometriosis involving uterosacral ligaments, torus uterinus and posterior vaginal fornix: prospective study.Ros C, de Guirior C, Mension E, Rius M, Valdés-Bango M, Tortajada M, Matas I, Martínez-Zamora MÁ, Gracia M, Carmona F.Ros C, et al.Ultrasound Obstet Gynecol. 2021 Dec;58(6):926-932. doi: 10.1002/uog.23696. Epub 2021 Nov 9.Ultrasound Obstet Gynecol. 2021.PMID: 34090310 To determine the optimal ultrasonographic screening method for rectal/rectosigmoid deep endometriosis: Ultrasound "sliding sign," transvaginal ultrasound direct visualization or both?Reid S, Espada M, Lu C, Condous G.Reid S, et al.Acta Obstet Gynecol Scand. 2018 Nov;97(11):1287-1292. doi: 10.1111/aogs.13425. Epub 2018 Aug 9.Acta Obstet Gynecol Scand. 2018.PMID: 30007066 Diagnostic accuracy of transvaginal sonography for detecting parametrial involvement in women with deep endometriosis: systematic review and meta-analysis.Guerriero S, Martinez L, Gomez I, Pascual MA, Ajossa S, Pagliuca M, Alcázar JL.Guerriero S, et al.Ultrasound Obstet Gynecol. 2021 Nov;58(5):669-676. doi: 10.1002/uog.23754.Ultrasound Obstet Gynecol. 2021.PMID: 34358386 Free PMC article. Optimal imaging modality for detection of rectosigmoid deep endometriosis: systematic review and meta-analysis.Gerges B, Li W, Leonardi M, Mol BW, Condous G.Gerges B, et al.Ultrasound Obstet Gynecol. 2021 Aug;58(2):190-200. doi: 10.1002/uog.23148. Epub 2021 Jul 12.Ultrasound Obstet Gynecol. 2021.PMID: 33038269 See all similar articles Cited by The bidirectional relationship between endometriosis and microbiome.Uzuner C, Mak J, El-Assaad F, Condous G.Uzuner C, et al.Front Endocrinol (Lausanne). 2023 Mar 7;14:1110824. doi: 10.3389/fendo.2023.1110824. eCollection 2023.Front Endocrinol (Lausanne). 2023.PMID: 36960395 Free PMC article.Review. Transvaginal Ultrasound in the Diagnosis and Assessment of Endometriosis-An Overview: How, Why, and When.Daniilidis A, Grigoriadis G, Dalakoura D, D'Alterio MN, Angioni S, Roman H.Daniilidis A, et al.Diagnostics (Basel). 2022 Nov 23;12(12):2912. doi: 10.3390/diagnostics12122912.Diagnostics (Basel). 2022.PMID: 36552919 Free PMC article.Review. Non-invasive imaging techniques for diagnosis of pelvic deep endometriosis and endometriosis classification systems: an International Consensus Statement†,‡.Condous G, Gerges B, Thomassin-Naggara I, Becker C, Tomassetti C, Krentel H, van Herendael BJ, Malzoni M, Abrao MS, Saridogan E, Keckstein J, Hudelist G; Intersociety Consensus Group.Condous G, et al.Facts Views Vis Obgyn. 2024 Jun;16(2):127-144. doi: 10.52054/FVVO.16.2.012.Facts Views Vis Obgyn. 2024.PMID: 38807551 Free PMC article. Transvaginal Ultrasound for the Diagnosis of Endometriosis: Current Practices and Barriers in Australian Sonographers.Yang X, Deslandes A, Cross T, Childs J.Yang X, et al.Australas J Ultrasound Med. 2025 May 22;28(2):e70003. doi: 10.1002/ajum.70003. eCollection 2025 May.Australas J Ultrasound Med. 2025.PMID: 40415951 Non-invasive imaging techniques for diagnosis of pelvic deep endometriosis and endometriosis classification systems: an International Consensus Statement.Condous G, Gerges B, Thomassin-Naggara I, Becker C, Tomassetti C, Krentel H, van Herendael BJ, Malzoni M, Abrao MS, Saridogan E, Keckstein J, Hudelist G; Intersociety Consensus Group §.Condous G, et al.Hum Reprod Open. 2024 May 29;2024(3):hoae029. doi: 10.1093/hropen/hoae029. eCollection 2024.Hum Reprod Open. 2024.PMID: 38812884 Free PMC article. See all "Cited by" articles References REFERENCES Guerriero S, Condous G, van den Bosch T, Valentin L, Leone FP, Van Schoubroeck D, Exacoustos C, Installe AJ, Martins WP, Abrao MS, Hudelist G, Bazot M, Alcazar JL, Goncalves MO, Pascual MA, Ajossa S, Savelli L, Dunham R, Reid S, Menakaya U, Bourne T, Ferrero S, Leon M, Bignardi T, Holland T, Jurkovic D, Benacerraf B, Osuga Y, Somigliana E, Timmerman D. Systematic approach to sonographic evaluation of the pelvis in women with suspected endometriosis, including terms, definitions and measurements: a consensus opinion from the International Deep Endometriosis Analysis (IDEA) group. Ultrasound Obstet Gynecol 2016; 48: 318-332. Nisenblat V, Bossuyt PM, Farquhar C, Johnson N, Hull ML. Imaging modalities for the non-invasive diagnosis of endometriosis. Cochrane Database Syst Rev 2016; 2: CD009591. Gerges B, Li W, Leonardi M, Mol BW, Condous G. Optimal imaging modality for detection of rectosigmoid deep endometriosis: systematic review and meta-analysis. Ultrasound Obstet Gynecol 2021; 58: 190-200. Indrielle-Kelly T, Fruhauf F, Fanta M, Burgetova A, Lavu D, Dundr P, Cibula D, Fischerova D. Application of International Deep Endometriosis Analysis (IDEA) group consensus in preoperative ultrasound and magnetic resonance imaging of deep pelvic endometriosis. Ultrasound Obstet Gynecol 2020; 56: 115-116. Indrielle-Kelly T, Fruhauf F, Fanta M, Burgetova A, Lavu D, Dundr P, Cibula D, Fischerova D. Diagnostic Accuracy of Ultrasound and MRI in the Mapping of Deep Pelvic Endometriosis Using the International Deep Endometriosis Analysis (IDEA) Consensus. Biomed Res Int 2020; 2020: 3583989. Show all 21 references Publication types Multicenter Study Actions Search in PubMed Search in MeSH Add to Search MeSH terms Endometriosis / diagnostic imaging Actions Search in PubMed Search in MeSH Add to Search Endometriosis / pathology Actions Search in PubMed Search in MeSH Add to Search Endometriosis / surgery Actions Search in PubMed Search in MeSH Add to Search Female Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Pilot Projects Actions Search in PubMed Search in MeSH Add to Search Pregnancy Actions Search in PubMed Search in MeSH Add to Search Prospective Studies Actions Search in PubMed Search in MeSH Add to Search Sensitivity and Specificity Actions Search in PubMed Search in MeSH Add to Search Ultrasonography / methods Actions Search in PubMed Search in MeSH Add to Search Vagina / diagnostic imaging Actions Search in PubMed Search in MeSH Add to Search Vagina / pathology Actions Search in PubMed Search in MeSH Add to Search Related information MedGen [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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188054	http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polabs.html	\| \| \| \| --- \| Polarization by Absorption A number of crystalline materials absorb more light in one incident plane than another, so that light progressing through the material become more and more polarized as they proceed. This anisotropy in absorption is called dichroism. There are several naturally occurring dichroic materials, and the commercial material polaroid also polarizes by selective absorption. \| \| \| Crossed polarizers \| \| IndexPolarization concepts \| \| \| \| \| --- \| \| HyperPhysics Light and Vision \| R Nave \| \| Go Back \| \| \| \| \| \| \| --- --- \| Dichroic Materials Materials which have different absorption for perpendicular incident planes for light are said to be dichroic. The mineral tourmaline is the best known of natural materials. Tourmaline refers to a class of boron silicates. A tourmaline crystal has a unique optic axis, and any electric field vector which is perpendicular to that axis is strongly absorbed. \| \| \| Polaroid is strongly dichroic and therefore an effective polarizer. If the transmission axes of ideal polarizers are perpendicular, no light is transmitted. The light tranmitted at other angles follows the Law of Malus. \| \| \| \| --- \| \| Liquid crystal display \| Crossed polarizers \| \| IndexPolarization concepts \| \| \| \| \| --- \| \| HyperPhysics Light and Vision \| R Nave \| \| Go Back \| \| \| \| \| --- \| Polaroid Material Polaroid is the trade name for the most commonly used dichroic material. It selectively absorbs light from one plane, typically transmitting less than 1% through a sheet of polaroid. It may transmit more than 80% of light in the perpendicular plane. The word "polaroid" usually refers to polaroid H-sheet, which is a sheet of iodine-impregnated polyvinyl alcohol. A sheet of polyvinyl alcohol is heated and stretched in one direction while softened, which has the effect of aligning the long polymeric molecules in the direction of stretch. When dipped in iodine, the iodine atoms attach themselves to the aligned chains. The iodine atoms provide electrons which can move easily along the aligned chains, but not perpendicular to them. Light waves with electric fields parallel to these chains are strongly absorbed because of the dissipative effects of the electron motion in the chains. The direction perpendicular to the polyvinyl alcohol chains is the "pass" direction since the electrons cannot move freely to absorb energy. \| \| \| Polaroid sunglasses \| \| IndexPolarization concepts \| \| \| \| \| --- \| \| HyperPhysics Light and Vision \| R Nave \| \| Go Back \| \| \| \| \| --- \| Polaroid Sunglasses The polaroid material used in sunglasses makes use of dichroism, or selective absorption, to achieve polarization. \| \| \| Polarization by reflection \| \| IndexPolarization concepts \| \| \| \| \| --- \| \| HyperPhysics Light and Vision \| R Nave \| \| Go Back \|
188055	https://proofwiki.org/wiki/Quotient_of_Complex_Conjugates	Quotient of Complex Conjugates From ProofWiki Jump to navigation Jump to search Theorem Let $z_1, z_2 \in \C$ be complex numbers. Let $\overline z$ be the complex conjugate of the complex number $z$. Then: : $\overline {\paren {\dfrac {z_1} {z_2} } } = \dfrac {\paren {\overline {z_1} } } {\paren {\overline {z_2} } }$ for $z_2 \ne 0$. Proof Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$, where $x_1, y_1, x_2, y_2 \in \R$. Then: \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- \| \| \| \| \| \| (\ds \overline {\paren {\frac {z_1} {z_2} } }) \| (=) \| \| \| \| (\ds \overline {\paren {\frac {x_1 x_2 + y_1 y_2} { {x_2}^2 + {y_2}^2} + i \frac {x_2 y_1 - x_1 y_2} { {x_2}^2 + {y_2}^2} } }) \| \| \| Division of Complex Numbers \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \frac {x_1 x_2 + y_1 y_2} { {x_2}^2 + {y_2}^2} - i \frac {x_2 y_1 - x_1 y_2} { {x_2}^2 + {y_2}^2}) \| \| \| Definition of Complex Conjugate \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \frac {x_1 x_2 + \paren {-y_1} \paren {-y_2} } { {x_2}^2 + \paren {-y_2}^2} + i \frac {x_2 \paren {-y_1} - x_1 \paren {-y_2} } { {x_2}^2 + \paren {-y_2}^2}) \| \| \| \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \frac {x_1 - i y_1} {x_2 - i y_2}) \| \| \| Division of Complex Numbers \| \| \| \| \| \| \| \| (\ds ) \| (=) \| \| \| \| (\ds \frac {\paren {\overline {z_1} } } {\paren {\overline {z_2} } }) \| \| \| Definition of Complex Conjugate \| \| $\blacksquare$ Sources 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $56 \ \text{(a)}$ 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): conjugate (of a complex number) 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): conjugate (of a complex number) Retrieved from " Categories: Proven Results Complex Conjugates Navigation menu Search
188056	https://www.rxkinetics.com/antibiotic_pk_pd.html	PK/PD Approach to Antibiotic Therapy Review ### A PK/PD Approach to Antibiotic Therapy #### Introduction Pharmacokinetics (PK) is concerned with the time course of antimicrobial concentrations in the body, while pharmacodyamics (PD) is concerned with the relationship between those concentrations and the antimicrobial effect. Antibiotic dosing regimens have traditionally been determined by PK parameters only. However, PD plays an equal, if not more important, role. In this age of increasing antimicrobial resistance, PD becomes even more important because these parameters may be used to design dosing regimens which counteract or prevent resistance. Discussion The primary measure of antibiotic activity is the minimum inhibitory concentration (MIC). The MIC is the lowest concentration of an antibiotic that completely inhibits the growth of a microorganism in vitro. While the MIC is a good indicator of the potency of an antibiotic, it indicates nothing about the time course of antimicrobial activity. PK parameters quantify the serum level time course of an antibiotic. The three pharmacokinetic parameters that are most important for evaluating antibiotic efficacy are the peak serum level (Cmax), the trough level (Cmin), and the Area Under the serum concentration time Curve (AUC). While these parameters quantify the serum level time course, they do not describe the killing activity of an antibiotic. Integrating the PK parameters with the MIC gives us three PK/PD parameters which quantify the activity of an antibiotic: the Peak/MIC ratio, the T>MIC, and the 24h-AUC/MIC ratio. The Peak/MIC ratio is simply the Cpmax divided by the MIC. The T>MIC (time above MIC) is the percentage of a dosage interval in which the serum level exceeds the MIC. The 24h-AUC/MIC ratio is determined by dividing the 24-hour-AUC by the MIC. #### Antimicrobial Patterns The three pharmacodyamic properties of antibiotics that best describe killing activity are time-dependence, concentration-dependence, and persistent effects. The rate of killing is determined by either the length of time necessary to kill (time-dependent), or the effect of increasing concentrations (concentration-dependent). Persistent effects include the Post-Antibiotic Effect (PAE). PAE is the persistant suppression of bacterial growth following antibiotic exposure. Using these parameters, antibiotics can be divided into 3 categories: Pattern of ActivityAntibioticsGoal of TherapyPK/PD Parameter Type I Concentration-dependent killing and Prolonged persistent effects Aminoglycosides Daptomycin Fluoroquinolones Ketolides Maximize concentrations Peak/MIC Type II Time-dependent killing and Minimal persistent effects Carbapenems Cephalosporins Erythromycin Linezolid Penicillins Maximize duration of exposure T>MIC Type III Time-dependent killing and Moderate to prolonged persistent effects.Azithromycin Clindamycin Oxazolidinones Tetracyclines Vancomycin Maximize amount of drug 24h-AUC/MIC For Type I antibiotics (AG's, fluoroquinolones, daptomycin and the ketolides), the ideal dosing regimen would maximize concentration, because the higher the concentration, the more extensive and the faster is the degree of killing. Therefore, the Peak/MIC ratio is the important predictors of antibiotic efficacy. For aminoglycosides, it is best to have a Peak/MIC ratio of at least 8-10 to prevent resistence. Type II antibiotics (beta-lactams, clindamycin, erythromcyin, and linezolid) demonstrate the complete opposite properties. The ideal dosing regimen for these antibiotics maximizes the duration of exposure. The T>MIC is the parameter that best correlates with efficacy. For beta-lactams and erythromycin, maximum killing is seen when the time above MIC is at least 70% of the dosing interval. Type III antibiotics (vancomycin, tetracyclines, azithromycin, and the dalfopristin-quinupristin combination) have mixed properties, they have time-dependent killing and moderate persistent effects. The ideal dosing regimen for these antibiotics maximizes the amount of drug received. Therefore, the 24h-AUC/MIC ratio is the parameter that correlates with efficacy. For vancomycin, a 24h-AUC/MIC ratio of at least 400 is necessary for MRSA. #### Outcome studies Aminoglycoside Pharmacodynamics in Vivo Initial serum peak levelDiedSurvived < 5mcg/ml21%79% >= 5mcg/ml2%98% Moore et al, J Infect Dis 149: 443, 1984 Aminoglycoside Pharmacodynamics in vivo Moore et al, J Infect Dis 155: 93, 1987 Vancomycin Outcome vs 24h-AUC/MIC ratio An early study by Hyatt et al, found: 24h-AUC/MIC ratioSatisfactoryUnsatisfactory < 125 4 (50%)4 > 125 71 (97%)2 A 2012 study by Brown et al found that patients with an AUC24/MIC ratio of less than 211 had a greater that 4-fold increase in attributable mortality than patients who received vancomycin doses that achieved an AUC24/MIC ratio of greater than 211. A 2016 meta-analysis by Men et al, demonstrated that achieving a high 24-hr AUC/MIC of vancomycin significantly decreases mortality rates by 53% and rates of infection treatment failure by 61%, with 400 being a reasonable target. Fluoroquinolone Pharmacodynamics vs S. pneumoniae 24h-AUC/MIC ratioMicrobiological Response < 33.7(64%) > 33.7(100%) Ambrose et al, Antimicrob Agents Chemo 10: 2793, 2001 Pharmacodynamics of Beta-Lactams and Macrolides in Otitis Media Craig et al, Ped Infect Dis 15: 255, 1996 #### Conclusion PK dosing has shown us that one dose is not appropriate for all patients. Pharmacodynamics shows us that one target level is not appropriate for all patients. We need to evalaute both the serum level data and the MIC, taking into consideration the PD properties of the drug. Numerous outcome studies have shown that class-appropriate PK/PD parameters are excellent predictors of antibiotic efficacy. #### References 1. Rodvold KA. Pharmacodynamics of antiinfective therapy: taking what we know to the patient's bedside. Pharmacotherapy. 2001 Nov;21(11 Pt 2):319S-330S. [ PubMed ] 2. Gunderson BW, Ross GH, Ibrahim KH, Rotschafer JC. What do we really know about antibiotic pharmacodynamics? Pharmacotherapy. 2001 Nov;21(11 Pt 2):302S-318S. [ PubMed ] 3. Nicolau DP. Optimizing outcomes with antimicrobial therapy through pharmacodynamic profiling. J Infect Chemother. 2003 Dec;9(4):292-6. [ PubMed ] 4. Craig Wm. The Role of Pharmacodynamics in Effective Treatment of Community Acquired Pathogens. Advanced Studies in Medicine 2002;2(4):126-134. 5. Li RC, Zhu ZY. The integration of four major determinants of antibiotic action: bactericidal activity, postantibiotic effect, susceptibility, and pharmacokinetics. J Chemother. 2002 Dec;14(6):579-83. [ PubMed ] 6. Frimodt-Moller N. How predictive is PK/PD for antibacterial agents? Int J Antimicrob Agents. 2002 Apr;19(4):333-9. [ PubMed ] 7. Schentag JJ. Pharmacokinetic and pharmacodynamic surrogate markers: studies with fluoroquinolones in patients. Am J Health Syst Pharm. 1999 Nov 15;56(22 Suppl 3):S21-4. [ PubMed ] 8. Wright DH, Brown GH, Peterson ML, Rotschafer JC. Application of fluoroquinolone pharmacodynamics. J Antimicrob Chemother. 2000 Nov;46(5):669-83. [ PubMed ] 9. Van Bambeke F, Tulkens PM. Macrolides: pharmacokinetics and pharmacodynamics. Int J Antimicrob Agents. 2001;18 Suppl 1:S17-23. [ PubMed ] 10. Turnidge JD. The pharmacodynamics of beta-lactams. Clin Infect Dis. 1998 Jul;27(1):10-22. [ PubMed ] 11. Ambrose PG, et al. Pharmacodynamics of fluoroquinolones against Streptococcus pneumoniae. Antimicrob Agents Chemother. 2001 Oct;45(10):2793-7. [ PubMed ] 12. Moore RD, Smith CR, Lietman PS. The association of aminoglycoside plasma levels with mortality in patients with gram-negative bacteremia. J Infect Dis. 1984 Mar;149(3):443-8. [ PubMed ] 13. Moore RD, Lietman PS, Smith CR. Clinical response to aminoglycoside therapy: importance of the ratio of peak concentration to minimal inhibitory concentration. J Infect Dis. 1987 Jan;155(1):93-9. [ PubMed ] 14. Hyatt JM, McKinnon PS, Zimmer GS, Schentag JJ. The importance of pharmacokinetic pharmacodynamic surrogate markers to outcome. Clin Pharmacokinet. 1995 Feb;28(2):143-60. [ PubMed ] 15. Brown J, Brown K, Forrest A. Vancomycin AUC24/MIC Ratio in Patients with Complicated Bacteremia and Infective Endocarditis Due to Methicillin-Resistant Staphylococcus aureus and Its Association with Attributable Mortality during Hospitalization. Antimicrobial Agents and Chemotherapy. 2012;56(2):634-638. [ PubMed ] 16. Men P, Li H-B, Zhai S-D, Zhao R-S (2016) Association between the AUC0-24/MIC Ratio of Vancomycin and Its Clinical Effectiveness: A Systematic Review and Meta-Analysis. PLoS ONE 11(1): e0146224. [ PubMed ] Click here to return to previous page www.rxkinetics.com ©Copyright 1984 - 2025. All rights reserved. RxKinetics, Plattsburg, MO 64477
188057	https://www.khanacademy.org/math/geometry/hs-geo-congruence/hs-geo-working-with-triangles/v/equilateral-and-isosceles-example-problems	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
188058	https://www.youtube.com/watch?v=A_ESfuN1Pkg	Introduction to complex numbers \| Imaginary and complex numbers \| Algebra II \| Khan Academy Khan Academy 9090000 subscribers 1948 likes Description 335598 views Posted: 18 Dec 2013 Practice this lesson yourself on KhanAcademy.org right now: Watch the next lesson: Missed the previous lesson? Algebra II on Khan Academy: Your studies in algebra 1 have built a solid foundation from which you can explore linear equations, inequalities, and functions. In algebra 2 we build upon that foundation and not only extend our knowledge of algebra 1, but slowly become capable of tackling the BIG questions of the universe. We'll again touch on systems of equations, inequalities, and functions...but we'll also address exponential and logarithmic functions, logarithms, imaginary and complex numbers, conic sections, and matrices. Don't let these big words intimidate you. We're on this journey with you! About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academy’s Algebra II channel: Subscribe to Khan Academy: 37 comments Transcript: Now that we know a little bit about the imaginary unit i, let's see if we can simplify more involved expressions, like this one right over here. 2 plus 3i plus 7i squared plus 5i to the third power plus 9i to the fourth power. And I encourage you to pause the video right now and try to simplify this on your own. So as you can see here, we have various powers of i. You could view this as i to the first power. We have i squared here. And we already know that i squared is defined to be negative 1. Then we have i to the third power. I to the third power would just be i times this, or negative i. And we already reviewed this when we first introduced the imaginary unit, i, but I'll do it again. i to the fourth power is just going to be i times this, which is the same thing as negative 1 times i. That's i to the third power times i again. i times i is negative 1. So that's negative 1 times negative 1, which is equal to 1 again. So we can rewrite this whole thing as 2 plus 3i. 7i squared is going to be the same thing, so i squared is negative 1. So this is the same thing as 7 times negative 1. So that's just going to be minus 7. And then we have 5i to the third power. i to the third power is negative i. So this could be rewritten as negative i. So this term right over here we could write as minus 5i, or negative 5i, depending on how you want to think about it. And then finally, i to the fourth power is just 1. So this is just equal to 1. So this whole term just simplifies to 9. So how could we simplify this more? Well we have several terms that are not imaginary, that they are real numbers. For example, we have this 2 is a real number. Negative 7 is a real number. And 9 is a real number. So we could just add those up. So 2 plus negative 7 would be negative 5. Negative 5 plus 9 would be 4. So the real numbers add up to 4. And now we have these imaginary numbers. So 3 times i minus 5 times i. So if you have 3 of something and then I were to subtract 5 of that same something from it, now you're going to have negative 2 of that something. Or another way of thinking about it is the coefficients. 3 minus 5 is negative 2. So three i's minus five i's, that's going to give you negative 2i. Now you might say, well, can we simplify this any further? Well no, you really can't. This right over here is a real number. 4 is a number that we've been dealing with throughout our mathematical careers. And negative 2i, that's an imaginary number. And so what we really consider this is this 4 minus 2i, we can now consider this entire expression to really be a number. So this is a number that has a real part and an imaginary part. And numbers like this we call complex numbers. It is a complex number. Why is it complex? Well, it has a real part and an imaginary part. And you might say, well, gee, can't any real number be considered a complex number? For example, if I have the real number 3, can't I just write the real number 3 as 3 plus 0i? And you would be correct. Any real number is a complex number. You could view this right over here as a complex number. And actually, the real numbers are a subset of the complex numbers. Likewise, imaginary numbers are a subset of the complex numbers. For example, you could rewrite i as a real part-- 0 is a real number-- 0 plus i. So the imaginaries are a subset of complex numbers. Real numbers are a subset of complex numbers. And then complex numbers also have all of the sums and differences, or all of the numbers that have both real and imaginary parts.
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188060	https://www.onemathematicalcat.org/Math/Precalculus_obj/parabolasPrecalc.htm	Parabolas: Definition, Reflectors/Collectors, Derivation of Equations (Click for cat book) Parabolas were introduced in the Algebra II curriculum. Links to these earlier lessons are given below, together with concepts that you must know from each section. Be sure to check your understanding by doing some of each problem type in these earlier exercises. These concepts are then further explored in this current lesson. Parabolas You must know the: Definition of a parabola as the set of points in a plane that are the same distance from a fixed point and a fixed line Terminology: focus (the fixed point); directrix (the fixed line); vertex Reflecting property of parabolas Collecting property of parabolas Equations of Simple Parabolas Know how to derive the equation ‘$\,x^2 = 4py\,$’ using the definition. The equation $\,x^2 = 4py\,$ is one of the two standard forms for a parabola. The other standard form, $\,y^2 = 4px\,,$ is derived on this page (below). The parabola described by $\,x^2 = 4py\,$ is a function of $\,x\,$; it can be equivalently written as $\,y = \frac{1}{4p}x^2\,.$ Know how to use graphical transformations to describe all parabolas with directrix parallel to the $x$-axis. Know the significance of the parameter $\,p\,$ in the equations for a parabola. Be able to graph the equation $\,x^2 = 4py\,$ (and shifted versions). In particular, be able to: Determine if the parabola is concave up or concave down Find the coordinates of the focus Find the equation of the directrix Plot an additional point that gives a sense of the ‘width’ of the parabola Play With Parabolas You can play with parabolas below: Move the focus: How does the shape change as the focus moves closer to the directrix? How does the shape change as the focus moves farther away from the directrix? Move the directrix: When the directrix is parallel to the $y$-axis, which way does the parabola open? When the directrix is parallel to the $x$-axis, which way does the parabola open? Below is some information that wasn't covered in the Algebra II curriculum. The exercises on this page address only this new information. However, it is assumed that you've mastered the exercises from the two earlier parabola lessons (Parabolas and Equations of Simple Parabolas). Axis of Symmetry for a Parabola Recall that a parabola is determined by two pieces of information: a line (the directrix) a point not on the line (the focus) The axis of symmetry for a parabola is the line through the focus that is perpendicular to the directrix. If the parabola is ‘folded’ along this line, then ‘half’ the parabola perfectly coincides with the other half. All conic sections have (at least one) axis of symmetry. For the standard forms of conics, the axes of symmetry are always the $x$-axis and/or the $y$-axis. For the standard form $\,x^2 = 4py\,$ of a parabola, the axis of symmetry is the $y$-axis. For the standard form $\,y^2 = 4px\,$ of a parabola (derived next), the axis of symmetry is the $x$-axis. Deriving the Standard Form $\,y^2 = 4px\,$ For a Parabola Place a parabola with its vertex at the origin, as shown below. If we put the focus on the $x$-axis, then the directrix will be parallel to the $y$-axis. Why? The axis of symmetry passes through the vertex (the origin) and the focus, hence is the $x$-axis. The directrix is perpendicular to the axis of symmetry. Or, if we put the directrix parallel to the $y$-axis, then the focus will be on the $x$-axis. Why? The axis of symmetry is perpendicular to the directrix and contains the vertex (the origin), hence is the $x$-axis. The focus always lies on the axis of symmetry. In either case, let $\,p \ne 0\,$ denote the $x$-value of the focus. Thus, the focus has coordinates $\,(p,0)\,.$ Although the sketch below shows the situation where $\,p\gt 0\,,$ the following derivation also holds for $\,p \lt 0\,.$ Since the vertex is a point on the parabola, the definition of parabola dictates that it must be the same distance from the focus and the directrix. Thus, the directrix must cross the $x$-axis at $\,-p\,$; indeed, every $x$-value on the directrix equals $\,-p\,.$ Let $\,(x,y)\,$ denote a typical point on the parabola. The distance from $\,(x,y)\,$ to the focus $\,(p,0)\,$ is found using the distance formula: $$ \begin{align} &\sqrt{(x-p)^2 + (y-0)^2 }\cr &\qquad = \sqrt{(x-p)^2 + y^2} \tag{1} \end{align} $$ To find the distance from $\,(x,y)\,$ to the directrix, first drop a perpendicular from $\,(x,y)\,$ to the directrix. This perpendicular intersects the directrix at $\,(-p,y)\,.$ The distance from $\,(x,y)\,$ to the directrix is therefore the distance from $\,(x,y)\,$ to $\,(-p,y)\,$: $$ \begin{align} &\sqrt{(x-(-p))^2 + (y-y)^2}\cr &\qquad = \sqrt{(x+p)^2} \tag{2} \end{align} $$ From the definition of parabola, distances $\,(1)\,$ and $\,(2)\,$ must be equal: $$ \cssId{s77}{\sqrt{(x-p)^2 + y^2} = \sqrt{(x+p)^2}} $$ This equation simplifies considerably, as follows: Squaring both sides: $$\cssId{s80}{(x-p)^2 + y^2 = (x+p)^2}$$ Multiplying out: $$\cssId{s82}{x^2 - 2px + p^2 + y^2 = x^2 + 2px + p^2}$$ Subtracting $\,x^2 + p^2\,$ from both sides: $$\cssId{s84}{y^2 - 2px = 2px}$$ Adding $\,2px\,$ to both sides: $$ \cssId{s86}{y^2 = 4px} $$ This is the second standard form for a parabola, where the axis of symmetry is the $x$-axis. The most critical thing to notice is the coefficient of $\,x\,,$ since it holds the key to locating the focus of the parabola . As an example, consider the equation $\,y^2 = 10x\,.$ Comparing with $\,y^2 = 4px\,,$ we see that $\,10 = 4p\,,$ or $\,p = \frac{10}4 = \frac 52\,.$ Thus, $\,y^2 = 10x\,$ graphs as a parabola with vertex at the origin and focus $\,(\frac 52,0)\,.$ Standard Forms for Parabolas In summary, we have: \| \| \| --- \| \| Standard Form of Parabola \| $x^2 = 4py$ \| \| Axis of Symmetry \| $y$-axis \| \| Vertex \| $(0,0)$ \| \| Focus \| $(0,p)$ \| \| Directrix \| $y = -p$ \| \| \| \| --- \| \| Standard Form of Parabola \| $y^2 = 4px$ \| \| Axis of Symmetry \| $x$-axis \| \| Vertex \| $(0,0)$ \| \| Focus \| $(p,0)$ \| \| Directrix \| $x = -p$ \| Sketching Parabolas You do not need to memorize lots of things to sketch parabolas in standard form! The key thing to memorize is that the important coefficient is $\,4p\,,$ where the size (absolute value) of $\,p\,$ gives the distance from the focus to the origin. You'll be letting the equation tell you the proper shape, as indicated next: Recognize Form Recognize that you have an equation of the form $\,x^2 = 4py\,$ or $\,y^2 = 4px\,.$ If needed, find the discriminant to jog your memory: $\,B^2 - 4AC = 0\,$ indicates a parabola. There's only one interesting coefficient in these equations ($\,4p\,$), so it is referred to as ‘the coefficient’ below. Isolate Squared Term Isolate the squared term. That is, get the squared term (with a coefficient of $\,1\,$) all by itself on one side of the equation. Determine Shape (Opening Up/Down/Left/Right) Since a perfect square is always nonnegative, the other side of the equation must also be nonnegative. Depending on the equation, this will force one of the following four situations: All the $x$-values positive (parabola opens to the right) All the $x$-values negative (parabola opens to the left) All the $y$-values positive (parabola is concave up) All the $y$-values negative (parabola is concave down) The examples below illustrate how this works. Use this information to (lightly) sketch the proper shape. Remember that the vertex is the origin. Plot the Focus The focus is always inside the parabola. To find its distance from the origin, throw away any minus signs (as needed) on the coefficient and divide by $\,4\,.$ Why? The coefficient is $\,4p\,,$ and $\,\|p\|\,$ gives the distance from the focus to the origin. Or, you can set the coefficient to $\,4p\,$ and solve for $\,p\,$ (which will include the correct sign). Draw in the Directrix Sketch the directrix. It is the same distance from the origin as the focus, on the opposite side. Plot Easy Points to Get the Proper ‘Width’ There are always three easy points to get on a parabola, once you know the focus and the directrix. Refer to the sketch below. The vertex is always halfway between the focus and directrix. In the standard forms, the vertex is the origin. Use two fingers to mark the distance from the focus to the directrix (shown in yellow). Keeping one finger on the focus, sweep around ($\,90^\circ\,$ in each direction) to get the two ‘easy points’. Why are they points on the parabola? Their distances to the focus and directrix both equal the radius of the circle shown (see the red segments). These two easy points give a sense of the ‘width’ of the parabola. Three easy points on a parabola: the vertex; two points to give a sense of the ‘width’ Example 1 Graph: $\,y^2 = 3x$ Solution The equation $\,y^2 = 3x\,$ is in the form $\,y^2 = 4px\,.$ The $\,y^2\,$ term is already isolated, with a coefficient of $\,1\,.$ $$ \begin{align} &y^2\ge 0\cr &\quad \implies 3x\ge 0\cr &\quad \implies x\ge 0 \end{align} $$ Thus, the $x$-value of every point on the parabola is nonnegative; the parabola opens to the right. $$ \begin{align} &4p = 3\cr &\quad \implies p = \frac 34\cr &\quad \implies \text{focus } = (\frac 34,0) \end{align} $$ Equation of directrix: $\,x = -\frac 34\,$ Two Easy Points Distance from focus to directrix: $\,2(\frac 34) = \frac 32\,$ Coordinates of easy points: $\,(\frac 34,\frac 32)\,$ and $\,(\frac 34,-\frac 32)$ It's confidence-boosting to check that these points satisfy the equation $\,y^2 = 3x\,$: $$\begin{gather} \cssId{s175}{\bigl(\pm \frac 32\bigr)^2\ \ \overset{?}{=}\ \ 3\bigl(\frac 34\bigr)}\cr\cr \cssId{s176}{\frac 94 = \frac 94}\quad \cssId{s177}{\text{Yep!}} \end{gather} $$ Example 2 Graph: $\,2x^2 + 5y = 0$ Solution The equation $\,2x^2 + 5y = 0\,$ has only $\,x^2\,$ and $\,y\,$ terms. It can be put in the form $\,x^2 = 4py\,.$ Isolate the square term, and get a coefficient of $\,1\,$: $$ \begin{gather} \cssId{s184}{2x^2 + 5y = 0}\cr\cr \cssId{s185}{2x^2 = -5y}\cr\cr \cssId{s186}{x^2 = -\frac 52y} \end{gather} $$ $$ \begin{align} &x^2\ge 0\cr &\quad \implies -\frac 52y\ge 0\cr &\quad \implies y\le 0 \end{align} $$ Thus, the $y$-value of every point on the parabola is negative (or zero); the parabola opens down. $$ \begin{align} &4p = -\frac 52\cr &\quad \implies p = -\frac 58\cr &\quad \implies \text{focus } = (0,-\frac 58) \end{align} $$ Equation of directrix: $\,y = \frac 58$ Two Easy Points Distance from focus to directrix: $\,2(\frac 58) = \frac 54$ Coordinates of easy points: $\,(\frac 54,-\frac 58)\,$ and $\,(-\frac 54,-\frac 58)$ It's confidence-boosting to check that these points satisfy the equation $\,2x^2 + 5y = 0\,$: $$ \begin{gather} \cssId{s199}{2\bigl(\pm \frac 54\bigr)^2 + 5(-\frac 58)\ \ \overset{?}{=}\ \ 0}\cr\cr \cssId{s200}{2\bigl(\frac {25}{16}\bigr) - \frac{25}{8}\ \ \ \overset{?}{=}\ \ 0}\cr\cr \cssId{s201}{0 = 0}\quad \cssId{s202}{\text{Yep!}} \end{gather} $$ Concept Practice
188061	https://www.sensorsone.com/sphere-surface-area-to-diameter-calculator/	Surface Area to Diameter of Sphere Calculator Skip to primary navigation Skip to main content Skip to primary sidebar SensorsONE Products Contact Surface Area to Diameter of Sphere Calculator \| Calculate Diameter of Sphere \| \| Surface Area (A) [?] \| \| \| \| \| Diameter (ø) [?] \| \| \| \| Click save settings to reload page with unique web page address for bookmarking and sharing the current tool settings, or click flip tool to reverse the tool function with current settings ✕ clear settings Switch tool with the current settings for diameter, and calculate sphere volume, circle area, circle circumference or radius instead Sorry, a graphic could not be displayed here, because your browser does not support HTML5 Canvas. Related Tools Calculate surface area of sphere from diameter Sphere volume to diameter calculator Convert an area to different units Length and distance converter Circle area to diameter calculator Circle circumference to diameter calculator Circle radius to diameter calculator Contents Toggle User Guide Formula Symbols Surface Area of Sphere Diameter of Sphere User Guide This tool will calculate the diameter of a sphere from the surface area, and will convert different measurement units for surface area and diameter. Formula The formula used to calculate the sphere diameter is: ø = √(A / π) Symbols ø = Sphere diameter A = Sphere surface area π= Pi = 3.14159… Surface Area of Sphere Enter the surface area of a sphere. The sphere surface area represents the total area of the outer surface of the sphere if it was to be laid out flat as a two-dimensional shape, e.g. a true scale map of the world is a 2D scaled representation of the surface area of the world. Diameter of Sphere This is the diameter of the sphere which corresponds to the volume specified. The diameter of a sphere is the length of a straight line drawn between two points on the surface of the sphere, where the line also passes through the centre of the sphere. Primary Sidebar Tools Unit Converters Pressure Liquid Level Temperature & Heat Electrical & Electromagnetic Flow Force, Load & Torque Mass & Weight Dimensions Density Time & Frequency Speed & Acceleration Energy & Power Sensors & Instrumentation Shapes Movement & Motion Travel & Transportation Pricing & Quantities Chemical Quantities Health & Fitness Tool Feedback Please provide feedback on using this Surface Area to Diameter of Sphere Calculator. ©2025 SensorsONE Ltd, all rights reserved Home Products Tools Legal Contact By clicking "OK" or continuing to use this website you are providing your consent to use cookies as described in our Privacy Policy
188062	https://en.wikipedia.org/wiki/Pauling%27s_rules	Jump to content Pauling's rules Deutsch Español فارسی Français 한국어 Bahasa Indonesia Русский Türkçe Українська 中文 Edit links From Wikipedia, the free encyclopedia Sorry to interrupt, but our fundraiser won't last long. This Wednesday, we ask you to join the 2% of readers who give. If everyone reading this right now gave just $2.75, we'd hit our goal quickly. $2.75 is all we ask. September 3: Knowledge is human. We're sorry we've asked you a few times recently, but it's Wednesday, September 3—please don't wait until tomorrow to help. We're happy you consult Wikipedia often. If just 2% of our most loyal readers gave $2.75 today, we'd reach our goal quickly. Most readers donate because Wikipedia is useful, others because they realize knowledge needs humans. If you agree, please give. Any contribution helps, whether it's $2.75 one time or monthly. 25 years ago Wikipedia was a dream. A dream built piece by piece by people, not machines. Now, with 65 million articles and 260,000 volunteers across the world, Wikipedia is proof that knowledge is human—a place of free, collaborative, and accessible knowledge. Your donation isn't just supporting a website; it's investing in the world's largest collaborative project of human intelligence—crafted by humans, for humans. Please join the 2% of readers who give what they can to help keep Wikipedia strong and growing. Thank you. Rules to predict ionic compounds' crystal structures This article is about Pauling's rules on crystal structures. For Pauling's rules on oxoacid strengths, see Acid dissociation constant. Pauling's rules are five rules published by Linus Pauling in 1929 for predicting and rationalizing the crystal structures of ionic compounds. First rule: the radius ratio rule [edit] Main article: Cation-anion radius ratio For typical ionic solids, the cations are smaller than the anions, and each cation is surrounded by coordinated anions which form a polyhedron. The sum of the ionic radii determines the cation-anion distance, while the cation-anion radius ratio (or ) determines the coordination number (C.N.) of the cation, as well as the shape of the coordinated polyhedron of anions.: 524 For the coordination numbers and corresponding polyhedra in the table below, Pauling mathematically derived the minimum radius ratio for which the cation is in contact with the given number of anions (considering the ions as rigid spheres). If the cation is smaller, it will not be in contact with the anions which results in instability leading to a lower coordination number. Polyhedron and minimum radius ratio for each coordination number \| C.N. \| Polyhedron \| Radius ratio \| \| 3 \| triangular \| 0.155 \| \| 4 \| tetrahedron \| 0.225 \| \| 6 \| octahedron \| 0.414 \| \| 7 \| capped octahedron \| 0.592 \| \| 8 \| square antiprism (anticube) \| 0.645 \| \| 8 \| cube \| 0.732 \| \| 9 \| triaugmented triangular prism \| 0.732 \| \| 12 \| cuboctahedron \| 1.00 \| The three diagrams at right correspond to octahedral coordination with a coordination number of six: four anions in the plane of the diagrams, and two (not shown) above and below this plane. The central diagram shows the minimal radius ratio. The cation and any two anions form a right triangle, with , or . Then . Similar geometrical proofs yield the minimum radius ratios for the highly symmetrical cases C.N. = 3, 4 and 8. For C.N. = 6 and a radius ratio greater than the minimum, the crystal is more stable since the cation is still in contact with six anions, but the anions are further from each other so that their mutual repulsion is reduced. An octahedron may then form with a radius ratio greater than or equal to 0.414, but as the ratio rises above 0.732, a cubic geometry becomes more stable. This explains why Na+ in NaCl with a radius ratio of 0.55 has octahedral coordination, whereas Cs+ in CsCl with a radius ratio of 0.93 has cubic coordination. If the radius ratio is less than the minimum, two anions will tend to depart and the remaining four will rearrange into a tetrahedral geometry where they are all in contact with the cation. The radius ratio rules are a first approximation which have some success in predicting coordination numbers, but many exceptions do exist. In a set of over 5000 oxides, only 66% of coordination environments agree with Pauling's first rule. Oxides formed with alkali or alkali-earth metal cations that contain multiple cation coordinations are common deviations from this rule. Second rule: the electrostatic valence rule [edit] For a given cation, Pauling defined the electrostatic bond strength to each coordinated anion as , where z is the cation charge and ν is the cation coordination number. A stable ionic structure is arranged to preserve local electroneutrality, so that the sum of the strengths of the electrostatic bonds to an anion equals the charge on that anion. where is the anion charge and the summation is over the adjacent cations. For simple solids, the are equal for all cations coordinated to a given anion, so that the anion coordination number is the anion charge divided by each electrostatic bond strength. Some examples are given in the table. Cations with oxide O2− ion \| Cation \| Radius ratio \| Cation C.N. \| Electrostatic bond strength \| Anion C.N. \| \| Li+ \| 0.34 \| 4 \| 0.25 \| 8 \| \| Mg2+ \| 0.47 \| 6 \| 0.33 \| 6 \| \| Sc3+ \| 0.60 \| 6 \| 0.5 \| 4 \| Pauling showed that this rule is useful in limiting the possible structures to consider for more complex crystals such as the aluminosilicate mineral orthoclase, KAlSi3O8, with three different cations. However, from data analysis of oxides from the Inorganic Crystal Structure Database (ICSD), the result showed that only 20% of all oxygen atoms matched with the prediction from second rule (using a cutoff of 0.01). Third rule: sharing of polyhedron corners, edges and faces [edit] The sharing of edges and particularly faces by two anion polyhedra decreases the stability of an ionic structure. Sharing of corners does not decrease stability as much, so (for example) octahedra may share corners with one another.: 559 The decrease in stability is due to the fact that sharing edges and faces places cations in closer proximity to each other, so that cation-cation electrostatic repulsion is increased. The effect is largest for cations with high charge and low C.N. (especially when r+/r- approaches the lower limit of the polyhedral stability). Generally, smaller elements fulfill the rule better. As one example, Pauling considered the three mineral forms of titanium dioxide, each with a coordination number of 6 for the Ti4+ cations. The most stable (and most abundant) form is rutile, in which the coordination octahedra are arranged so that each one shares only two edges (and no faces) with adjoining octahedra. The other two, less stable, forms are brookite and anatase, in which each octahedron shares three and four edges respectively with adjoining octahedra.: 559 Fourth rule: crystals containing different cations [edit] In a crystal containing different cations, those of high valency and small coordination number tend not to share polyhedron elements with one another.: 561 This rule tends to increase the distance between highly charged cations, so as to reduce the electrostatic repulsion between them. One of Pauling's examples is olivine, M2SiO4, where M is a mixture of Mg2+ at some sites and Fe2+ at others. The structure contains distinct SiO4 tetrahedra which do not share any oxygens (at corners, edges or faces) with each other. The lower-valence Mg2+ and Fe2+ cations are surrounded by polyhedra which do share oxygens. Fifth rule: the rule of parsimony [edit] The number of essentially different kinds of constituents in a crystal tends to be small. The repeating units will tend to be identical because each atom in the structure is most stable in a specific environment. There may be two or three types of polyhedra, such as tetrahedra or octahedra, but there will not be many different types. Limitation [edit] In a study of 5000 oxides, only 13% of them satisfy all of the last 4 rules, indicating limited universality of Pauling's rules. See also [edit] Goldschmidt tolerance factor Octet rule References [edit] ^ Pauling, Linus (1929). "The principles determining the structure of complex ionic crystals". J. Am. Chem. Soc. 51 (4): 1010–1026. Bibcode:1929JAChS..51.1010P. doi:10.1021/ja01379a006. ^ Jump up to: a b c d e f g h Pauling, Linus (1960). The nature of the chemical bond and the structure of molecules and crystals; an introduction to modern structural chemistry (3rd ed.). Ithaca (NY): Cornell University Press. pp. 543–562. ISBN 0-8014-0333-2. {{cite book}}: ISBN / Date incompatibility (help) ^ Jump up to: a b Housecroft, Catherine E.; Sharpe, Alan G. (2005). Inorganic chemistry (2nd ed.). Upper Saddle River, NJ: Pearson Prentice Hall. p. 145. ISBN 9780130399137. ^ Toofan, Jahansooz (February 1994). "A Simple Expression between Critical Radius Ratio and Coordination Number". Journal of Chemical Education. 71 (2): 147. Bibcode:1994JChEd..71..147T. doi:10.1021/ed071p147. Following the erratum, equations should read and , (where bond angle) ^ Petrucci, Ralph H.; Harwood, William S.; Herring, F. Geoffrey (2002). General chemistry: principles and modern applications (8th ed.). Upper Saddle River, NJ: Prentice Hall. p. 518. ISBN 0-13-014329-4. ^ Jump up to: a b c d George, Janine; Waroquiers, David; Di Stefano, Davide; Petretto, Guido; Rignanese, Gian-Marco; Hautier, Geoffroy (2020-05-04). "The Limited Predictive Power of the Pauling Rules". Angewandte Chemie International Edition. 59 (19): 7569–7575. doi:10.1002/anie.202000829. ISSN 1433-7851. PMC 7217010. PMID 32065708. \| Linus Pauling \| \| --- \| \| Publications \| The Nature of the Chemical Bond (1939 book) Sickle Cell Anemia, a Molecular Disease (1949 paper) Vitamin C and the Common Cold (1970 book) How to Live Longer and Feel Better (1986) \| \| Concepts \| Valence bond theory + Orbital hybridisation + Resonance Pauling Electronegativity Scale Pauling's rules \| \| Founded \| Linus Pauling Institute International League of Humanists \| \| Related \| Ava Helen Pauling (wife) Molecular medicine Vitamin C megadosage Intravenous ascorbic acid Linus Pauling Award Pauling Field 4674 Pauling \| Retrieved from " Categories: Molecular geometry Crystals Atomic radius Coordination chemistry Empirical laws Eponymous chemical rules Hidden categories: CS1 errors: ISBN date Articles with short description Short description matches Wikidata
188063	https://study.com/skill/learn/how-to-write-the-equation-of-a-circle-centered-at-the-origin-given-the-radius-explanation.html	How to Write the Equation of a Circle Centered at the Origin Given the Radius \| Geometry \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up How to Write the Equation of a Circle Centered at the Origin Given the Radius Geometry Skills Practice Click for sound 2:43 You must c C reate an account to continue watching Register to access this and thousands of other videos Are you a student or a teacher? I am a student I am a teacher Try Study.com, risk-free As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it risk-free It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time. Already registered? Log in here for access Back What teachers are saying about Study.com Try it risk-free for 30 days Already registered? Log in here for access 00:04 How to write the… 01:12 How to write the… 02:00 How to write the… Jump to a specific example Speed Normal 0.5x Normal 1.25x 1.5x 1.75x 2x Speed Marissa Maniaci, Kathryn Boddie Instructors Marissa Maniaci Marissa has bachelor's and master's degrees in Biomedical Engineering from the University of Michigan, and a master's in Secondary Education from Mercyhurst University. She has previously taught 7th grade science, and is currently a high school Biology teacher. She has several years of tutoring experience, particularly in math and science, and she loves getting the chance to interact with students one-on-one to expand their math and science knowledge. View bio Kathryn Boddie Kathryn has taught high school or university mathematics for over 10 years. She has a Ph.D. in Applied Mathematics from the University of Wisconsin-Milwaukee, an M.S. in Mathematics from Florida State University, and a B.S. in Mathematics from the University of Wisconsin-Madison. View bio Example SolutionsPractice Questions How to Write the Equation of a Circle Centered at the Origin Given the Radius Step 1: Since the center of the circle is at the origin, (0, 0), simplify the general form of the equation of the circle to remove h and k. Step 2: Identify the radius of the circle that was given in the problem and calculate the value of the squared radius. Step 3: Write the equation for the given circle. How to Write the Equation of a Circle Centered at the Origin Given the Radius Vocabulary and General Formula Center of a circle: The center of the circle exists at a location that is equally distanced from all points that lie along the circle. The center is a point with x and y-coordinates, given in the format (x, y). Radius of a circle: The radius of a circle is the distance from the center of the circle to any and all points that lie along the circle. The radius of a circle is a positive numerical value. General equation for a circle:(x−h)2+(y−k)2=R 2, where (h,k) are the coordinates of the center of the circle, and R is the radius of the circle. Now we will look at two examples of how to write the equation of a circle centered at the origin if given the radius of the circle. How to Write the Equation of a Circle Centered at the Origin Given the Radius Example 1 Write the equation of a circle centered at the origin with a radius of 7. Step 1: Since the center of the circle is at the origin, (0, 0), simplify the general form of the equation of the circle to remove h and k. The general form for the equation of a circle is (x−h)2+(y−k)2=R 2, but since our circle is centered at the origin we have both h = 0 and k = 0. This allows us to simplify the equation of the circle to x 2+y 2=R 2. Step 2: Identify the radius of the circle that was given in the problem and calculate the value of the squared radius. The radius of the circle was given as 7, so R = 7. We need to calculate 7 2, which is 49. R 2=7 2=49 Step 3: Write the equation for the given circle. We can now replace the R 2 in our equation with 49, and we have the final equation for our circle as shown below. x 2+y 2=49 How to Write the Equation of a Circle Centered at the Origin Given the Radius Example 2 Write the equation of a circle centered at the origin with a radius of 3. Step 1: Since the center of the circle is at the origin, (0, 0), simplify the general form of the equation of the circle to remove h and k. Since the center of the circle is again at the origin, we can skip to the following simplified version of the general form of the equation for a circle. x 2+y 2=R 2. Step 2: Identify the radius of the circle that was given in the problem and calculate the value of the squared radius. The radius, R, was given as 3. 3 2=9 so we will use R 2=9 in our equation. Step 3: Write the equation for the given circle. The equation for the circle centered at the origin with a radius of 3 is x 2+y 2=49. Get access to thousands of practice questions and explanations! 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188064	https://www.youtube.com/watch?v=HOqzGN42Akk	Converting Unit-Step Functions to Piecewise Functions, Quiz 5 Problem 1 Adam Glesser 4140 subscribers 10 likes Description 1602 views Posted: 7 Oct 2021 We convert a combination of unit-step functions into a piecewise defined function and we graph the resulting function. Transcript: hello everybody this is our set of video solutions to quiz 5 this is the fall 2021 semester at Cal State Fullerton it's math 106 which is differential equations and integral calculus in this very first problem from the quiz uh well the first three problems in fact are all going to be exploring the connections between unit step functions and PE wise defined functions as well as their graphs so in this one we're given a uh unit step function or rather a combination of unit step functions plus some extra junk uh and we want to write it as a piecewise defined function and then we want to graph it all right uh okay fine we can do that so let's see first thing uh it might actually be helpful to graph it uh as we go and then writing the piece wise shouldn't be so bad so I'm going to you like break the rules of the directions right they can't stop me uh and and I'm going to try to graph this thing right off the bat so the first thing that this says is that the function is going to begin at two meaning from negative Infinity on until some point particularly one we're just going to be the constant function two uh at one something is going to happen so I'm going to mark this down there's one and then at three something else is going to happen so there could be two and then maybe here's three so at one something is going to happen but before one it's just going to be at height two so let's see there's maybe a two right there and so I just get a nice flat constant line all the way up until one and then something is going to change oops so I'm going to put a nice Circle and I'll use a little arrow on the left here to denote it going off to negative Infinity okay now what happens when we get to time equals uh one well here we're going to be turning off the unit step function at one meaning we're going to be going down right that's that minus tells you right we're going to be going down and by how much well by three so if I go down by three I go one two right so this was a one zero oh I'm going to be at 1 okay and I'm going to keep going in a straight line until I get to three how do I know three well that's the next time there's a switch all right so I go there I put a little open Dot cuz something is going to happen now what going to happen well this u3 tells me at three we're going to apply the unit step function so it's going to go up because of this plus and how much it's going to go up by five so if we're at Nega - 1 now we need to go all the way up to three and there's a four so we jump up here to four and then we head off to Infinity because there's no more switches okay so there's actually our graph and so writing down f as a piecewise function should be pretty routine at this point so first from Infinity all the way up to one we get a constant value of two okay then from 1 to three so we're going to include one but not include three we're down here at -1 and then finally when T is at least three oops when T is at least three and then heading off to Infinity we get a value of four
188065	https://www.reddit.com/r/HomeworkHelp/comments/sb1vry/university_calc_domain_of_fx_tan_x/	[University Calc] Domain of f(x)= tan x? : r/HomeworkHelp Skip to main content[University Calc] Domain of f(x)= tan x? : r/HomeworkHelp Open menu Open navigationGo to Reddit Home r/HomeworkHelp A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to HomeworkHelp r/HomeworkHelp•4 yr. ago [deleted] [University Calc] Domain of f(x)= tan x? I’m currently doing a review of high school trig for my uni calc class and I can’t remember how domain of tan works. My professor says it’s: x ≠ π/2 + kπ How did he get to here? I know it involves values of the unit circle but I’m confused! Read more Share Related Answers Section Related Answers domain of tan x function range of tan x function domain of inverse tan function discontinuity points of tangent function derivative of tan x New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of January 23, 2022 Reddit reReddit: Top posts of January 2022 Reddit reReddit: Top posts of 2022 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
188066	https://oeis.org/A264963	A264963 - OEIS login The OEIS is supported by the many generous donors to the OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A264963 Number of down-up parking functions of length n. 1 1, 1, 1, 3, 17, 104, 915, 8618, 104436, 1337282, 20709209, 336013996, 6343919118, 124736306407, 2780356513594, 64249797198125, 1651884203936474, 43874277964032394, 1278413274487999471, 38372024627757454128, 1249821733374560346851, 41835183404896657899658 (list; graph; refs; listen; history; text; internal format) OFFSET 0,4 LINKS Table of n, a(n) for n=0..21. R. Stanley, Parking Functions, 2011 FORMULA a(2n) = A260694(2n). EXAMPLE a(3) = 3: 212, 213, 312. a(4) = 17: 2121, 2131, 2132, 2141, 2142, 2143, 3121, 3131, 3132, 3141, 3142, 3231, 3241, 4121, 4131, 4132, 4231. a(5) = 104: 21212, 21213, 21214, 21215, ..., 52314, 52412, 52413, 53412. CROSSREFS Cf. A000111, A260694. Sequence in context: A163064A020069A020024 A036551A162479A194780 Adjacent sequences: A264960A264961A264962 A264964A264965A264966 KEYWORD nonn AUTHOR Alois P. Heinz, Nov 29 2015 STATUS approved LookupWelcomeWikiRegisterMusicPlot 2DemosIndexWebCamContributeFormatStyle SheetTransformsSuperseekerRecents The OEIS Community Maintained by The OEIS Foundation Inc. Last modified September 29 03:55 EDT 2025. Contains 388827 sequences. License Agreements, Terms of Use, Privacy Policy
188067	https://www.khanacademy.org/math/algebra-home/alg-series-and-induction/alg-finite-arithmetic-series/v/alternate-proof-to-induction-for-integer-sum	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
188068	https://www.ck12.org/flexi/geometry/equations-of-circles/what-is-the-equation-of-a-circle-with-center-at-(h-k)-and-radius-r/	Flexi answers - What is the equation of a circle with center at (h, k) and radius r? \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects Geometry Equations of Circles Question What is the equation of a circle with center at (h, k) and radius r? Flexi Says: The equation of a circle with center at point (h,k) and radius r is given by: (x−h)2+(y−k)2=r 2 Analogy / Example Try Asking: A radio signal can transmit messages up to a distance of 5km. If the radio signal's origin is located at a point whose coordinates are (-2, 7), what is the equation of the circle that defines the boundary up to which messages can be transmitted?What is the equation of a circle in the xy plane?Find the center of the circle whose equation is: Unexpected text node: ' x^{2}+y^{2}-10x+20y+89=0 '? How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy
188069	https://openstax.org/books/calculus-volume-2/pages/preface	Preface - Calculus Volume 2 \| OpenStax This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising purposes. Privacy Notice Customize Reject All Accept All Customize Consent Preferences We use cookies to help you navigate efficiently and perform certain functions. You will find detailed information about all cookies under each consent category below. The cookies that are categorized as "Necessary" are stored on your browser as they are essential for enabling the basic functionalities of the site. ...Show more For more information on how Google's third-party cookies operate and handle your data, see:Google Privacy Policy Necessary Always Active Necessary cookies are required to enable the basic features of this site, such as providing secure log-in or adjusting your consent preferences. 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Reject All Save My Preferences Accept All Skip to ContentGo to accessibility pageKeyboard shortcuts menu Log in Calculus Volume 2 Preface Calculus Volume 2Preface Contents Contents Highlights Table of contents Preface 1 Integration 2 Applications of Integration 3 Techniques of Integration 4 Introduction to Differential Equations 5 Sequences and Series 6 Power Series 7 Parametric Equations and Polar Coordinates A \| Table of Integrals B \| Table of Derivatives C \| Review of Pre-Calculus Answer Key Index Search for key terms or text. Close Welcome to Calculus Volume 2, an OpenStax resource. This textbook was written to increase student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost. About OpenStax OpenStax is a nonprofit based at Rice University, and it’s our mission to improve student access to education. 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About Calculus Volume 2 Calculus is designed for the typical two- or three-semester general calculus course, incorporating innovative features to enhance student learning. The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them. Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency. Volume 2 covers integration, differential equations, sequences and series, and parametric equations and polar coordinates. Coverage and Scope Our Calculus Volume 2 textbook adheres to the scope and sequence of most general calculus courses nationwide. We have worked to make calculus interesting and accessible to students while maintaining the mathematical rigor inherent in the subject. With this objective in mind, the content of the three volumes of Calculus have been developed and arranged to provide a logical progression from fundamental to more advanced concepts, building upon what students have already learned and emphasizing connections between topics and between theory and applications. The goal of each section is to enable students not just to recognize concepts, but work with them in ways that will be useful in later courses and future careers. The organization and pedagogical features were developed and vetted with feedback from mathematics educators dedicated to the project. Volume 1 Chapter 1: Functions and Graphs Chapter 2: Limits Chapter 3: Derivatives Chapter 4: Applications of Derivatives Chapter 5: Integration Chapter 6: Applications of Integration Volume 2 Chapter 1: Integration Chapter 2: Applications of Integration Chapter 3: Techniques of Integration Chapter 4: Introduction to Differential Equations Chapter 5: Sequences and Series Chapter 6: Power Series Chapter 7: Parametric Equations and Polar Coordinates Volume 3 Chapter 1: Parametric Equations and Polar Coordinates Chapter 2: Vectors in Space Chapter 3: Vector-Valued Functions Chapter 4: Differentiation of Functions of Several Variables Chapter 5: Multiple Integration Chapter 6: Vector Calculus Chapter 7: Second-Order Differential Equations Pedagogical Foundation Throughout Calculus Volume 2 you will find examples and exercises that present classical ideas and techniques as well as modern applications and methods. Derivations and explanations are based on years of classroom experience on the part of long-time calculus professors, striving for a balance of clarity and rigor that has proven successful with their students. Motivational applications cover important topics in probability, biology, ecology, business, and economics, as well as areas of physics, chemistry, engineering, and computer science. Student Projects in each chapter give students opportunities to explore interesting sidelights in pure and applied mathematics, from showing that the number e is irrational, to calculating the center of mass of the Grand Canyon Skywalk or the terminal speed of a skydiver. Chapter Opening Applications pose problems that are solved later in the chapter, using the ideas covered in that chapter. Problems include the hydraulic force against the Hoover Dam, and the comparison of the relative intensity of two earthquakes. Definitions, Rules, and Theorems are highlighted throughout the text, including over 60 Proofs of theorems. Assessments That Reinforce Key Concepts In-chapter Examples walk students through problems by posing a question, stepping out a solution, and then asking students to practice the skill with a “Check Your Learning” component. The book also includes assessments at the end of each chapter so students can apply what they’ve learned through practice problems. Many exercises are marked with a [T] to indicate they are suitable for solution by technology, including calculators or Computer Algebra Systems (CAS). Answers for selected exercises are available in the Answer Key at the back of the book. Early or Late Transcendentals Calculus Volume 2 is designed to accommodate both Early and Late Transcendental approaches to calculus. Exponential and logarithmic functions are presented in Chapter 2. Integration of these functions is covered in Chapters 1 for instructors who want to include them with other types of functions. These discussions, however, are in separate sections that can be skipped for instructors who prefer to wait until the integral definitions are given before teaching the calculus derivations of exponentials and logarithms. Comprehensive Art Program Our art program is designed to enhance students’ understanding of concepts through clear and effective illustrations, diagrams, and photographs. Answers to Questions in the Book Answers to Examples are provided just below the question in the book. All Checkpoint answers are provided in the Answer Key. Odd-numbered Exercises and Review Exercises questions are provided to students in the Answer Key as well as the Student Solution Guide on the Student Resources page. Even-numbered answers are provided only to instructors in the Instructor Answer Guide via the Instructor Resources page. Additional Resources Student and Instructor Resources We’ve compiled additional resources for both students and instructors, including Getting Started Guides, an instructor solution manual, and PowerPoint slides. Instructor resources require a verified instructor account, which can be requested on your openstax.org log-in. Take advantage of these resources to supplement your OpenStax book. Partner Resources OpenStax Partners are our allies in the mission to make high-quality learning materials affordable and accessible to students and instructors everywhere. Their tools integrate seamlessly with our OpenStax titles at a low cost. To access the partner resources for your text, visit your book page on openstax.org. About The Authors Senior Contributing Authors Gilbert Strang, Massachusetts Institute of Technology Dr. Strang received his PhD from UCLA in 1959 and has been teaching mathematics at MIT ever since. His Calculus online textbook is one of eleven that he has published and is the basis from which our final product has been derived and updated for today’s student. Strang is a decorated mathematician and past Rhodes Scholar at Oxford University. Edwin “Jed” Herman, University of Wisconsin-Stevens Point Dr. Herman earned a BS in Mathematics from Harvey Mudd College in 1985, an MA in Mathematics from UCLA in 1987, and a PhD in Mathematics from the University of Oregon in 1997. He is currently a Professor at the University of Wisconsin-Stevens Point. He has more than 20 years of experience teaching college mathematics, is a student research mentor, is experienced in course development/design, and is also an avid board game designer and player. Contributing Authors Catherine Abbott, Keuka College Nicoleta Virginia Bila, Fayetteville State University Sheri J. 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188070	https://math.stackexchange.com/questions/210720/inverse-function-of-a-polynomial	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Inverse function of a polynomial Ask Question Asked Modified 11 months ago Viewed 104k times $\begingroup$ What is the inverse function of $f(x) = x^5 + 2x^3 + x - 1?$ I have no idea how to find the inverse of a polynomial, so I would greatly appreciate it if someone could show me the steps to solving this problem. Thank you in advance! polynomials inverse Share edited Oct 10, 2012 at 21:44 Ross Millikan 384k2828 gold badges264264 silver badges472472 bronze badges asked Oct 10, 2012 at 21:37 Jaden M.Jaden M. 40722 gold badges55 silver badges77 bronze badges $\endgroup$ 5 2 $\begingroup$ Even Mathematica can't find inverse function, but you can be confident - inverse function does exist $\endgroup$ Norbert – Norbert 2012-10-10 21:42:10 +00:00 Commented Oct 10, 2012 at 21:42 11 $\begingroup$ Your polynomial is increasing, and its range is all reals, so there is an inverse. Finding a pleasant expression for the inverse is another matter. But one can find information about the derivative of the inverse without knowing a formula. $\endgroup$ André Nicolas – André Nicolas 2012-10-10 22:05:10 +00:00 Commented Oct 10, 2012 at 22:05 $\begingroup$ I think, it's very hard. Is it homework? Or where is this problem from? $\endgroup$ Berci – Berci 2012-10-10 22:21:38 +00:00 Commented Oct 10, 2012 at 22:21 $\begingroup$ This question has been asked before, albeit in greater detail. Most likely we're having trouble answering it because we're missing this information: math.stackexchange.com/questions/85226/… $\endgroup$ Gyu Eun Lee – Gyu Eun Lee 2012-12-15 03:42:21 +00:00 Commented Dec 15, 2012 at 3:42 $\begingroup$ @Gyu Eun Lee That has the same equation, so maybe is why Jaden M. asked this question, but the question you linked looks like a homework problem that asks for the inverse of particular numbers and the derivative of the inverse at a particular number, but specifically avoids the much harder problem of finding an explicit formula for the inverse, which is what this question asks. $\endgroup$ Mr. Nichan – Mr. Nichan 2024-03-03 18:26:50 +00:00 Commented Mar 3, 2024 at 18:26 Add a comment \| 6 Answers 6 Reset to default 29 $\begingroup$ This is an experimental way of working out the inverse. We can treat the polynomial like an expansion \begin{equation} f(x) = -1 + x + 0x^2 + 2x^3 + 0x^4 + x^5 + 0x^6 + 0x^7 + \cdots \end{equation} then we can perform a Series Reversion on this to give the inverse series (as an infinite expansion) \begin{equation} f^{-1}(x) = (1+x) -2(1+x)^3 +11(1+x)^5-80(1+x)^7+665(1+x)^9-\cdots \end{equation} at a glance this doesn't seem to helpful, but if we search the coefficients in the OEIS we seem to get a hit! It would appear (conjecture) that the general coefficient is then \begin{equation} a(n) = \binom{5n+1}{n}\frac{(-1)^n}{2n+1} \end{equation} and we could write that \begin{equation} f^{-1}(x) = \sum_{n=0}^\infty \binom{5n+1}{n}\frac{(-1)^n}{2n+1}(x+1)^{2n+1} \end{equation} if we evaluate that in Mathematica, it gives \begin{equation} f^{-1}(x)=(1+x)\;_4F_3\left(\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{6}{5}\bigg\|\frac{3}{4},\frac{5}{4},\frac{6}{4}\bigg\|-\frac{5^5}{4^4}(1+x)^2 \right) \end{equation} a generalised hypergeometric function. If we plot the composition of these two i.e $f(f^{-1}(x))$ or $f^{-1}(f(x))$ the plot seems to indicate \begin{equation} f(f^{-1}(x))=f^{-1}(f(x))=x \end{equation} of course this is not a proof, and evaluating the inverse function may become numerically unstable if $x$ becomes too large. Share edited Jan 3, 2017 at 17:55 answered Jan 3, 2017 at 17:44 Benedict W. J. IrwinBenedict W. J. Irwin 4,4871919 silver badges4545 bronze badges $\endgroup$ 1 4 $\begingroup$ I don't understand why this isn't the accepted answer. This is brilliant and helped me check the stability of some IIR filters for nonuniformly spaced samples that I have been building. $\endgroup$ The Dude – The Dude 2022-05-26 17:03:02 +00:00 Commented May 26, 2022 at 17:03 Add a comment \| 11 $\begingroup$ Generally, you say $y=$ your polynomial and solve for $x$. Fifth degree polynomials are generally not solvable. The general approach for a quadratic would be essentially the quadratic formula. Given $y=ax^2+bx+c$, you find $x=\frac {-b \pm \sqrt{b^2-4a(c-y)}}{2a}$. You need to pick one sign to get a function. Share edited Nov 23, 2018 at 15:07 answered Oct 10, 2012 at 21:47 Ross MillikanRoss Millikan 384k2828 gold badges264264 silver badges472472 bronze badges $\endgroup$ 3 $\begingroup$ Thank you! I will try it again. $\endgroup$ Jaden M. – Jaden M. 2012-10-10 23:47:32 +00:00 Commented Oct 10, 2012 at 23:47 $\begingroup$ The solution that is expected to be x = f(y) does not contain y. Substitute c by c-y ? $\endgroup$ h22 – h22 2018-11-23 08:02:55 +00:00 Commented Nov 23, 2018 at 8:02 $\begingroup$ @h22: yes, that is correct. Thanks $\endgroup$ Ross Millikan – Ross Millikan 2018-11-23 15:07:55 +00:00 Commented Nov 23, 2018 at 15:07 Add a comment \| 5 $\begingroup$ As others indicated, there is no algebraic formula for the inverse function $f^{-1}$. The inverse functions exists (since $f$ is increasing), but there are serious algebraic obstructions to solving $y=x^5 + 2x^3 + x - 1$ for $x$. But we can find particular values of $f^{-1}$ and of its derivative. For example, to find $f^{-1}(3)$ we would just have to note that $f(1)=3$. Therefore, $f^{-1}(3)=1$. (The number $3$ is lucky here; if asked about $f^{-1}(4)$, one would need a numerical method, e.g., a calculator.) Also, the inverse function theorem can be used to find the derivative of $f^{-1}$ at $3$: $$ (f^{-1})'(3) = \frac{1}{f'(1)} = \frac{1}{5+6+1} = \frac{1}{12} $$ Related question. Share edited Apr 13, 2017 at 12:19 community wiki 2 revsuser147263 $\endgroup$ Add a comment \| 3 $\begingroup$ You need to solve the equation $$x^5 + 2x^3 + x - 1=y$$ for $x$. Unfortunately, such quintic equations are known to have no closed-form solution in general, and this one does not escape the rule. Anyway, there is a little backdoor, as a quintic can be (after painful computation) reduced to the form known as Bring Quintic Form $$x^5-x-a=0.$$ Under this particular form, the solutions of $x$ in terms of $a$ are called the Bring radicals of $a$. So if you accept this special univariate function in your toolbox, then you can invert the quintic polynomials. The cases of linear, quadratic, cubic and quartic polynomials can be solved with the usual functions, with increasing difficulty. Share answered Feb 23, 2021 at 20:18 user65203user65203 $\endgroup$ Add a comment \| 0 $\begingroup$ The original question seems to have come from a homework problem which does NOT require explicitly finding the inverse of a polynomial; however, since this page is one of the first things that pop up when you search for how to find the inverse of a polynomial function, and a general method for this WAS asked for, this general question is what I will discuss. Also, I realize that, until I discuss numerical methods at the bottom, I'm just repeating what others have said with more explanation. Solving a polynomial $y = p(x) = a_{0} + a_{1}x + a_{2}x^2 + ... + a_{n}x^n$ for x is equivalent to solving the equation $p(x) - y = 0$ i.e., $a_{n}x^n + a_{n-1}x^n-1 + ... + a_{1}x + (a_0 - y) = 0$ for x, i.e., to getting the zeros of the polynomial on the left as a function of y. (That is, for each value of y, we treat y as a constant and find the zeros of the polynomial, so in the general case of polynomials on the complex numbers, this is a function from complex numbers (any value of y) to sets of complex numbers (the up to n possible complex x values that satisfy this equation for that value of y). Using map notation and set-builder notation: $X: \mathbb{C} \rightarrow \wp(\mathbb{C}): X(y) = {x \in \mathbb{C} \| p(x) - y = 0}$ The specific case of a polynomial that is invertible on real numbers (and therefore of odd number order, incidentally) corresponds to a case where, for any real number y, X(y) has exactly one element which is a real number. $p^{-1}$ is then the function that maps each real number y to that one real number element of X(y). In addition to this multi-function interpretation using sets, any function can be made invertible by changing the domain and range, though the domain and range will be switched in the inverse. What all this means is that you can definitely find an equation (albeit a nasty one) for the inverses of polynomials of degree up to 4, but some (I think "almost all") polynomials of degree 5 or above probably don't have an inverse that can be written in terms of addition, subtraction, multiplication, division, and rational number exponents (which can be created from radicals and multiplication). This latter result I'm guessing is a corollary to the Abel-Ruffini theorem or some variation on it, which says there is no general equation for the roots of polynomials of degree 5 or above using only the arithmetic operations described above. You should look at it yourself for more details because I'm not familiar with the details. For degree 0 polynomials (constant functions): $y=a_{0}$ x is irrelevent, so: $X(a_{0}) = \mathbb{C}$, and $\forall y \in \mathbb{C}, y a_{0} \implies X(y) = \emptyset$ For degree 1 polynomials (linear/affine equations): $y=a_{0} + a_{1}x$ the most basic algebra shows that $X(y) = { \frac{y - a_{0}}{a_{1}} }$ For degree 2 polynomials (quadratic equations): $y=a_{0} + a_{1}x + a_{2}x^2$ one can use the quadratic formula to find that $X(y) = \left{ \frac{-a_{1} - \sqrt{a_{1}^2 - 4a_{2}(a_{0} - y)}}{2a_{2}}, \frac{ -a_{1} + \sqrt{a_{1}^2 - 4a_{2}(a_{0} - y)} }{2a_{2}} \right} = \frac{-a_{1} \pm \sqrt{a_{1}^2 - 4a_{2}(a_{0} - y)}}{2a_{2}} = \frac{-a_{1} + \sqrt{a_{1}^2 - 4a_{2}(a_{0} - y)}}{2a_{2}}$ That last expression $\left(\frac{-a_{1} + \sqrt{a_{1}^2 - 4a_{2}(a_{0} - y)}}{2a_{2}}\right)$ is equal to the others if we adopt the convention that an nth root of a complex number is actually the SET of all complex numbers whose nth power is what is under the radical, and that putting a set into an algebraic equations gives its image under the function represented by that expression, or, if that image would be a set of sets, the set of elements of elements of that set of sets, so, if $S = \sqrt{a_{1}^2 - 4a_{2}(a_{0} - y)} = \left{ -\sqrt{a_{1}^2 - 4a_{2}(a_{0} - y)}, +\sqrt{a_{1}^2 - 4a_{2}(a_{0} - y)} \right}$, then $\frac{-a_{1} + S}{2a_{2}} = { \frac{-a_{1} + s}{2a_{2}} \| s \in S }$, and, for example, if T is some set, then $\sqrt{\frac{f(y) + T}{g(y)}} = { z \in \mathbb{C} \| \exists t \in T$ such that $z^3 = \frac{f(y) + t}{g(y)}}$, recalling that y is treated as an already known complex number, since these sets are values of X(y) for some y. This is formally useful to simplify the notation of the cubic and quartic cases, but perhaps you can understand it better with intuition than with my explanation. For degree 3 polynomials (cubic equations): $y = a_{0} + a_{1}x + a_{2}x^2 + a_{3}x^3$, so $0 = (a_{0} - y) + a_{1}x + a_{2}x^2 + a_{3}x^3$ One could use the much more complicated cubic equation. This is known for often giving answers in strange forms that are hard to simplify, however, sometimes it's not that bad, especially if you do it by steps and $Î_{1}^2(y) - 4Î_{0}^3$ is always positive in the region of interest: $X(y) = -\frac{1}{3a_{3}}\left(a_{2} + \sqrt{\frac{(2a_{2}^3 - 9a_{3}a_{2}a_{1} + 27a_{3}^2 (a_{0} - y)) \pm \sqrt{(2a_{2}^3 - 9a_{3}a_{2}a_{1} + 27a_{3}^2 (a_{0} - y))^2 - 4(a_{2}^2 - 3a_{3}a_{1})^3}}{2}} + \frac{a_{2}^2 - 3a_{3}a_{1}}{\sqrt{\frac{(2a_{2}^3 - 9a_{3}a_{2}a_{1} + 27a_{3}^2 (a_{0} - y)) \pm \sqrt{(2a_{2}^3 - 9a_{3}a_{2}a_{1} + 27a_{3}^2 (a_{0} - y))^2 - 4(a_{2}^2 - 3a_{3}a_{1})^3}}{2}}}\right)$ I got this from the following way of breaking it down Wikipedia used (where I have substituded my coefficients in place of their "a","b","c", and "d"). Obviously you could check other sources, or even just Wikipedia in case I copied it wrong. $Î_{0} = a_{2}^2 - 3a_{3}a_{1}$ $Î_{1}(y) = 2a_{2}^3 - 9a_{3}a_{2}a_{1} + 27a_{3}^2 (a_{0} - y)$ $C(y) = \sqrt{\frac{Î_{1}(y) \pm \sqrt{Î_{1}^2(y) - 4Î_{0}^3}}{2}} = \sqrt{\frac{(2a_{2}^3 - 9a_{3}a_{2}a_{1} + 27a_{3}^2 (a_{0} - y)) \pm \sqrt{(2a_{2}^3 - 9a_{3}a_{2}a_{1} + 27a_{3}^2 (a_{0} - y))^2 - 4(a_{2}^2 - 3a_{3}a_{1})^3}}{2}}$ $X(y) = -\frac{1}{3a_{3}}(a_{2} + C(y) + \frac{Î_{0}}{C(y)}) = -\frac{1}{3a_{3}} \left( a_{2} + \sqrt{\frac{Î_{1}(y) \pm \sqrt{Î_{1}^2(y) - 4Î_{0}^3}}{2}} + \frac{Î_{0}}{\sqrt{\frac{Î_{1}(y) \pm \sqrt{Î_{1}^2(y) - 4Î_{0}^3}}{2}}} \right)$ There are other ways of solving cubic equations as well. For example, there is one using trigonometric functions that I think is supposed to be neater (or at least use only real numbers for real number roots) in cases the cubic equation using radicals can't be simplified to and a+bi form. Wikipedia writes it for a "depressed" cubic equation ($t^3+pt+q=0$), which can be formed from any cubic equation using a change of variable: Finding the inverse of a cubic equation is necessary to eliminate the parameter from the curves generated by cubic splines (e.g., cubic Bezier splines), which is the reason I looked this question up. There is also a quartic formula for solving general degree 4 polynomials (quartic equations) and this is even worse in terms of how complicated it is, but I don't think any new concepts will be revealed by writing it out, so I won't today. (Maybe I will add it later.) You can look it up yourself if you want to use it. As I mentioned, the cubic and quartic formulas give hard-to-simplify equations for these roots (and thus for inverses of cubic and quartic equations), and there are no such equations for higher degree polynomials (at least not using just the basic arithmetic functions previously described). Also, in practice, even nth roots are just approximated using numerical methods (usually by a computer nowadays). Thus, for practical purposes, it might be better to explore numerical methods: First of all, note that plugging in any value of $x$ gives you a value of $y$. Thus, you know one of the elements of $X(y)$ for that particular value of $y$, and, if $y=f(x)$ where you know $f $is invertible, then you know $f^{-1}(y) = x$. If you just want a reasonably smooth graph of $f^{-1}$ or $X$ (in the non-invertable case) in a particular (x,y) region (so you don't care about (x,y) pairs where either x or y, and most notably x, is out of the region of interest), then you can do something like the following: Evaluate y=f(x) for a regular grid of x values (like using x=(0.00, 0.01, 0.02, 0.03, ... 0.98, 0.99, 1.00) to find f(x) = (f(0.00), f(0.01), ..., f(1.00)), using MatLab-like notation for a graph with x values from 0 to 1). Then, if two adjacent y values are far apart, then you can check more x values between those two y values until you achieve the desired density of y values. (Since polynomials are continuous, you will always eventually reach the desired density by repeating this.) To approximate the value of the inverse at a particular point (i.e., find x for a particular y), we could round to the nearest y value in grid like that found above if we had one. However, a more standard method is to use "Newton's Method", which is to use the following recurrence relation: $x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}$ Since $f$ is polynomial, $f'$ is easy to find. (If it wasn't we would replace it with a numerical approximation, making this the "Secant Method".) For polynomials in general, Newton's method does not always converge to a zero of the polynomial (though it always does if your initial guess (called $x_{0}$ or $x_{1}$) is close enough to the zero), and even if it does, it can only converge to one zero, which depends on your initial guess. However, in the special case of inverting a polynomial which is invertible on real numbers, we ALMOST have the conditions to ensure it always converges. (The one problem is that the derivative could equal zero at an isolated point, even if the polynomial is invertible.) Formally: If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a function whose derivative is continuous on (all of) $\mathbb{R}$ and is either always positive or always negative, and $\forall n \in \mathbb{Z^+}, x_{n}$ is a function $\mathbb{R} \rightarrow \mathbb{R}$ such that $\forall y \in \mathbb{R}, x_{1}(y) \in \mathbb{R}$, and $\forall y \in \mathbb{R}, \forall n \in \mathbb{Z^+}, x_{n+1}(y) = x_{n}(y) - \frac{f(x_{n}(y)) - y}{f'(x_{n}(y))}$, then $\forall y \in \mathbb{R}, f^{-1}(y) = \lim \limits_{n \to \infty} x_n(y)$ Normally, you don't try to explicitly calculate this limit, which I suspect would just amount to trying to find $f^{-1}(y)$ analytically in a way that makes this recursion irrelevant, but rather we just have a computer evaluate $x_{n}(y)$ for successive values of n until adjacent values get close enough together. Proving what maximum possible error a particular stopping condition for Newton's Algorithm gives is a numerical analysis topic I don't have the source to discuss right now, but is probably mentioned in any introductory text on the subject. Another method of finding zeros specifically in the case where you know a range where exactly one zero is in is the bisection method, which is basically a binary search (if you know any computer science). Similar to this and to the secant method is the false position, or regula falsi, method. Notation that might need explanation (just guessing based on thinking about the expertise of people who might ask or look up this question): If $f$ is a function, then $f^2(y) = (f(y))^2$, e.g., $Î_{1}^2(y) = (Î_{1}(y))^2$. Cubes of functions are written the same way; however, $f^{-1}(y)$ refers to the inverse function of f evaluated at y, so $f^{-1}(y) (f(y))^{-1}$ in most cases. $\forall$ is "for all" or "for any" $\exists$ is "there exists at least one" $\exists!$ is "there exists one and only one" (used below) $\implies$ means logically implies/entails $x \in S$ means x is an element of the set S. $\mathbb{Z^+}$ is positive integers $\mathbb{R}$ is real numbers $\mathbb{C}$ is complex numbers I use $f: A \to B: f(x) = ...$ to mean f is a "function" or "map" or "mapping" from the set A (it's domain) to the set B (it's codomain), i.e., $\forall x \in A, \exists! y \in B$ such that $f(x) = y$, and, specifically, $\forall x \in A, f(x) = ...$ $\wp(S)$ is the set of all subsets of a set S, including S itself and the empty set ($\emptyset$). This is called the power set of S, and the range of of X is actually a much smaller set than this, but it's okay for the codomain of the function to be larger than the function's actual range ("range" = the values that are actually f(x) for some x in it's domain). That just means the function is not "surjective"/"onto". ${x \in S \| P(x)}$ is the subset of S containing only those elements such that the predicate P is true for that element. This is called "set-builder notation". E.g., the range of the function $f: A \to B$ is the set ${y \in B \| \exists x \in A$ such that y = f(x)$}$. $f'$ is the derivative of f, i.e., a function related to f which gives the rate of change of f(x) as x changes, i.e., the slope of y=f(x) at a the point (x,y). Based on the homework problem I think this question may have come from, I'm sure the original asker knows this one already. It is worth noting, though, that $f'$ is a specific function from $\mathbb{R} \to \mathbb{R}$ the rate of change of the f(z) as WHATEVER z is changes, proportional to the change in z, formally, $f'(z) = \lim \limits_{h \to 0}\frac{f(z+h) - f(z)}{h}$ so $f'(x_{n}(y)) = \frac{df(x_{n}(y))}{dx_{n}(y)}$ NOT $\frac{df(x_{n}(y))}{dy}$ or anything else. (I just used another notation for derivatives, alluding to their being ratios of infinitessimals or limits of ratios, as described in the formal definition I gave.) Share edited Mar 7, 2024 at 2:47 answered Mar 3, 2024 at 20:50 Mr. NichanMr. Nichan 26322 silver badges88 bronze badges $\endgroup$ 1 $\begingroup$ I always write answers that are too long, because I get absorbed when I'm answering the question, but then I don't want to spend just as long going through all the stuff I already bothered to write deciding which parts are worth deleting to improve readability and declutter the page. $\endgroup$ Mr. Nichan – Mr. Nichan 2024-03-03 22:37:26 +00:00 Commented Mar 3, 2024 at 22:37 Add a comment \| 0 $\begingroup$ We can solve this algebraically if we introduce non-commutative objects $\mathbf{a}$ and $\mathbf{b}$ such that $$(\mathbf{a}x+ \mathbf{b})^5=x^5+2x^3+x-1.$$ Thus gives $\mathbf{a}^5=1, \mathbf b^5=-1, \mathbf a^4\mathbf b + \mathbf a^3 \mathbf b \mathbf a+\cdots =0,$ etc.; thus $$x=-\mathbf a^{-1}\mathbf b.$$ Then construct the corresponding context and hope $\mathbf a $ is not a zero divisor. One such possibility is that they are matrices, though there are six equations and therefore the matrices must be $3\times 3$ or larger. So thats just a general algebraic way. One can even leave it as general as the restriction for non-comutativity. Share answered Oct 20, 2024 at 20:41 Alexander ConradAlexander Conrad 1,80788 silver badges1414 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions polynomials inverse See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked Find a function such that $f^{-1}=f'$ 5 Inverse function of a polynomial and its derivative 5 Functions with "ugly" inverses 3 Does every power series have an inverse? 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188071	https://www.bigideasmath.com/external/state-resources/pdfs/NC_math2_05_04.pdf	Copyright © Big Ideas Learning, LLC All rights reserved. 5.4 Proving Statements about Segments and Angles For use with Exploration 5.4 Name ________ Date _ Essential Question How can you prove a mathematical statement? A proof is a logical argument that uses deductive reasoning to show that a statement is true. Work with a partner. Four steps of a proof are shown. Write the reasons for each statement. Given AC AB AB = + Prove AB BC = Work with a partner. Six steps of a proof are shown. Complete the statements that correspond to each reason. Given 1 3 m m ∠ = ∠ Prove m EBA m CBD ∠ = ∠ 1 EXPLORATION: Writing Reasons in a Proof 2 EXPLORATION: Writing Steps in a Proof C A B C A B 2 3 1 D E STATEMENTS REASONS 1. AC AB AB = + 1. Given 2. AB BC AC + = 2. _______ 3. AB AB AB BC + = + 3. ______ 4. AB BC = 4. _____ 163 Copyright © Big Ideas Learning, LLC All rights reserved. 5.4 Proving Statements about Segments and Angles (continued) Name ________ Date _ Communicate Your Answer 3. How can you prove a mathematical statement? 4. Use the given information and the figure to write a proof for the statement. Given B is the midpoint of . AC C is the midpoint of . BD Prove AB CD = 2 EXPLORATION: Writing Steps in a Proof (continued) STATEMENTS REASONS 1. ____ 1. Given 2. 2 3 m EBA m m ∠ = ∠ + ∠ 2. Angle Addition Postulate 3. 2 1 m EBA m m ∠ = ∠ + ∠ 3. Substitution Property of Equality 4. __ m EBA ∠ = 4. Commutative Property of Addition 5. 1 2 ___ m m ∠+ ∠ = 5. Angle Addition Postulate 6. ____ 6. Transitive Property of Equality C A B D 164 Copyright © Big Ideas Learning, LLC All rights reserved. 5.4 For use after Lesson 5.4 Name ________ Date _ In your own words, write the meaning of each vocabulary term. proof two-column proof Core Concepts Reflexive, Symmetric, and Transitive Properties of Equality Real Numbers Segment Lengths Angle Measures Reflexive Property a a = AB AB = m A m A ∠ = ∠ Symmetric Property If , a b = then If , AB CD = then If , m A m B ∠ = ∠ then . b a = . CD AB = . m B m A ∠ = ∠ Transitive Property If a b = and If AB CD = and If m A m B ∠ = ∠ and , b c = then , CD EF = then , m B m C ∠ = ∠ then . a c = . AB EF = . m A m C ∠ = ∠ Notes: Theorems Properties of Segment Congruence Segment congruence is reflexive, symmetric, and transitive. Reflexive For any segment AB, . AB AB ≅ Symmetric If , AB CD ≅ then . CD AB ≅ Transitive If AB CD ≅ and , CD EF ≅ then . AB EF ≅ Copyright © Big Ideas Learning, LLC All rights reserved. 5.4 For use after Lesson 5.4 Name ________ Date _ In your own words, write the meaning of each vocabulary term. proof two-column proof Core Concepts Reflexive, Symmetric, and Transitive Properties of Equality Real Numbers Segment Lengths Angle Measures Reflexive Property a a = AB AB = m A m A ∠ = ∠ Symmetric Property If , a b = then If , AB CD = then If , m A m B ∠ = ∠ then . b a = . CD AB = . m B m A ∠ = ∠ Transitive Property If a b = and If AB CD = and If m A m B ∠ = ∠ and , b c = then , CD EF = then , m B m C ∠ = ∠ then . a c = . AB EF = . m A m C ∠ = ∠ Notes: Theorems Properties of Segment Congruence Segment congruence is reflexive, symmetric, and transitive. Reflexive For any segment AB, . AB AB ≅ Symmetric If , AB CD ≅ then . CD AB ≅ Transitive If AB CD ≅ and , CD EF ≅ then . AB EF ≅ 165 Copyright © Big Ideas Learning, LLC All rights reserved. In your own words, write the meaning of each vocabulary term. proof two-column proof Core Concepts Reflexive, Symmetric, and Transitive Properties of Equality Real Numbers Segment Lengths Angle Measures Reflexive Property a a = AB AB = m A m A ∠ = ∠ Symmetric Property If , a b = then If , AB CD = then If , m A m B ∠ = ∠ then . b a = . CD AB = . m B m A ∠ = ∠ Transitive Property If a b = and If AB CD = and If m A m B ∠ = ∠ and , b c = then , CD EF = then , m B m C ∠ = ∠ then . a c = . AB EF = . m A m C ∠ = ∠ Notes: Theorems Properties of Segment Congruence Segment congruence is reflexive, symmetric, and transitive. Reflexive For any segment AB, . AB AB ≅ Symmetric If , AB CD ≅ then . CD AB ≅ Transitive If AB CD ≅ and , CD EF ≅ then . AB EF ≅ 2.4 Notetaking with Vocabulary (continued) Notetaking with Vocabulary (continued) 5.4 Name ________ Date ___ Properties of Angle Congruence Angle congruence is reflexive, symmetric, and transitive. Reflexive For any angle A, . A A ∠ ≅∠ Symmetric If , A B ∠ ≅∠ then . B A ∠ ≅∠ Transitive If A B ∠ ≅∠ and , B C ∠ ≅∠ then . A C ∠ ≅∠ Notes: 165 Copyright © Big Ideas Learning, LLC All rights reserved. proof two-column proof Core Concepts Reflexive, Symmetric, and Transitive Properties of Equality Real Numbers Segment Lengths Angle Measures Reflexive Property a a = AB AB = m A m A ∠ = ∠ Symmetric Property If , a b = then If , AB CD = then If , m A m B ∠ = ∠ then . b a = . CD AB = . m B m A ∠ = ∠ Transitive Property If a b = and If AB CD = and If m A m B ∠ = ∠ and , b c = then , CD EF = then , m B m C ∠ = ∠ then . a c = . AB EF = . m A m C ∠ = ∠ Notes: Theorems Properties of Segment Congruence Segment congruence is reflexive, symmetric, and transitive. Reflexive For any segment AB, . AB AB ≅ Symmetric If , AB CD ≅ then . CD AB ≅ Transitive If AB CD ≅ and , CD EF ≅ then . AB EF ≅ 165 Copyright © Big Ideas Learning, LLC All rights reserved. In your own words, write the meaning of each vocabulary term. proof two-column proof Core Concepts Reflexive, Symmetric, and Transitive Properties of Equality Real Numbers Segment Lengths Angle Measures Reflexive Property a a = AB AB = m A m A ∠ = ∠ Symmetric Property If , a b = then If , AB CD = then If , m A m B ∠ = ∠ then . b a = . CD AB = . m B m A ∠ = ∠ Transitive Property If a b = and If AB CD = and If m A m B ∠ = ∠ and , b c = then , CD EF = then , m B m C ∠ = ∠ then . a c = . AB EF = . m A m C ∠ = ∠ Notes: Theorems Properties of Segment Congruence Segment congruence is reflexive, symmetric, and transitive. Reflexive For any segment AB, . AB AB ≅ Symmetric If , AB CD ≅ then . CD AB ≅ Transitive If AB CD ≅ and , CD EF ≅ then . AB EF ≅ 165 Practice Copyright © Big Ideas Learning, LLC All rights reserved. 2.4 5.4 Name ________ Date _ Properties of Angle Congruence Angle congruence is reflexive, symmetric, and transitive. Reflexive For any angle A, . A A ∠ ≅∠ Symmetric If , A B ∠ ≅∠ then . B A ∠ ≅∠ Transitive If A B ∠ ≅∠ and , B C ∠ ≅∠ then . A C ∠ ≅∠ Notes: Writing a Two-Column Proof In a proof, you make one statement at a time until you reach the conclusion. Because you make statements based on facts, you are using deductive reasoning. Usually the first statement-and-reason pair you write is given information. Proof of the Symmetric Property of Angle Congruence Given 1 2 ∠ ≅∠ Prove 2 1 ∠ ≅∠ m m m m ∠ ≅∠ ∠ = ∠ STATEMENTS REASONS 1. 1. 2. 2. 3. 3. 2 1 Copy or draw diagrams and label given information to help develop proofs. Do not mark or label the information in the Prove statement on the diagram. statements based on facts on conclusions Copyright © Big Ideas Learning, LLC All rights reserved. 2.4 5.4 Name ________ Date ______ Properties of Angle Congruence Angle congruence is reflexive, symmetric, and transitive. Reflexive For any angle A, . A A ∠ ≅∠ Symmetric If , A B ∠ ≅∠ then . B A ∠ ≅∠ Transitive If A B ∠ ≅∠ and , B C ∠ ≅∠ then . A C ∠ ≅∠ Notes: Writing a Two-Column Proof In a proof, you make one statement at a time until you reach the conclusion. Because you make statements based on facts, you are using deductive reasoning. Usually the first statement-and-reason pair you write is given information. Proof of the Symmetric Property of Angle Congruence Given 1 2 ∠ ≅∠ Prove 2 1 ∠ ≅∠ m m m m ∠ ≅∠ ∠ = ∠ STATEMENTS REASONS 1. 1. 2. 2. 3. 3. 2 1 Copy or draw diagrams and label given information to help develop proofs. Do not mark or label the information in the Prove statement on the diagram. statements based on facts on conclusions All rights reserved. Notes: Writing a Two-Column Proof In a proof, you make one statement at a time until you reach the conclusion. Because you make statements based on facts, you are using deductive reasoning. Usually the first statement-and-reason pair you write is given information. Proof of the Symmetric Property of Angle Congruence Given 1 2 ∠ ≅∠ Prove 2 1 ∠ ≅∠ 1 2 Given 1 2 Definition of congruent angles 2 1 Symmetric Property of Equality 2 1 Definition of congruent angles m m m m ∠ ≅∠     ∠ = ∠     ∠ = ∠     ∠ ≅∠   STATEMENTS REASONS 1. 1. 2. 2. 3. 3. 4. 4. Notes: 2 1 Copy or draw diagrams and label given information to help develop proofs. Do not mark or label the information in the Prove statement on the diagram. statements based on facts that you know or on conclusions from deductive reasoning The number of statements will vary. Remember to give a reason for the last statement. definitions, postulates, or proven theorems that allow you to state the corresponding statement Chapter 9 Given PQ = RS S P Q R STATEMENTS REASONS 1. PQ = RS 1. Given 2. PQ + QR = RS + QR 2. Addition Property of Equality 3. PQ + QR = PR 3. Segment Addition Postulate 4. RS + QR = QS 4. Segment Addition Postulate 5. PR = QS 5. Transitive Property of Equality 26. Given ∠1 is a complement of ∠2. ∠2 ≅ ∠3 1 2 Prove ∠1 is a complement of ∠3. of the segment with endpoints A and B. 3. AB = AB 3. Reflexive Property of Worked-Out Examples Example #1 Copy and complete the proof. 166 Practice (continued) Prove PR = QS Given PQ = RS S P Q R STATEMENTS REASONS 1. PQ = RS 1. 2. PQ + QR = RS + QR 2. 3. 3. Segment Addition Postulate 4. RS + QR = QS 4. Segment Addition Postulate 5. PR = QS 5. Prove PR = QS Copyright © Big Ideas Learning, LLC All rights reserved. 5.4 Name ________ Date _ Extra Practice In Exercises 1 and 2, complete the proof. 1. Given and AB CD bisect each other at point M and . BM CM ≅ Prove AB AM DM = + 2. Given is a complement of . AEB BEC ∠ ∠ Prove 90 m AED ∠ = ° STATEMENTS REASONS 1. AEB ∠ is a complement of . BEC ∠ 1. Given 2. ___ 2. Definition of complementary angles 3. m AEC m AEB m BEC ∠ = ∠ + ∠ 3. ______ 4. 90 m AEC ∠ = ° 4. _____ 5. 180 m AED m AEC ∠ + ∠ = ° 5. Definition of supplementary angles 6. ___ 6. Substitution Property of Equality 7. 90 m AED ∠ = ° 7. ______ A D E B C STATEMENTS REASONS 1. BM CM ≅ 1. Given 2. CM DM ≅ 2. _____ 3. BM DM ≅ 3. _______ 4. BM DM = 4. ______ 5. ___ 5. Segment Addition Postulate (Post. 1.2) 6. AB AM DM = + 6. ______ A D M B C k Practice A 442 Integrated Mathematics I Copyright © Big Idea Worked-Out Solutions All 4. ∠EFG and ∠GFH are supplementary. 4. Defi nition of linear pair 5. m∠EFG + m∠GFH = 180° 5. Defi nition of supplementary angles 6. m∠EFG + m∠ = 180° Equality 7. ∠EFG and ∠GHF are supplementary. 7. Defi nition of supplementary angles 38. Given — AB ≅ — FG ⃖⃗ BF bisects — AC and — DG . A C B F D G Prove — BC ≅ — DF STATEMENTS REASONS 1. — AB ≅ — FG 1. Given 2. ⃖⃗ BF bisects — AC and — DG . 2. Given 3. — BC ≅ — AB , — FG ≅ — DF 3. Defi nition of segment bisector 4. — BC ≅ — FG 4. Transitive Property of Equality 5. — BC ≅ — DF 5. Transitive Property of Segment Congruence segments 5. RM = RS + SM 5. Segment A Postulate 6. CD = CF + FD 6. Segment A Postulate 7. RS + SM = CD 7. Substitutio of Equalit 8. RM = CD 8. Substitutio of Equalit 9. — RM ≅ — CD 9. Defi nition segments 40. Sample answer: Refl exive: Employee 1 wor same number of hours as Employee 1. Symme Employee 4 worked the same number of hour Employee 5, then Employee 5 worked the sam of hours as Employee 4. Transitive: If Emplo worked the same number of hours as Employ Employee 4 worked the same number of hou Employee 5, then Employee 2 worked the sam of hours as Employee 5. 41. The Symmetric Property of Equality is illustra 42. Sample answer: Refl exive: I earned the same points as myself on my favorite video game. T because a quantity is equal to itself. Symmetri the same score as Tyeesha on our math quiz, th the same score as John. This is Symmetric bec two quantities are equal to each other. Transiti has the same number of pets as Ella, and Ella h number of pets as Brady, then Dominic has the of pets as Brady. This is transitive because the that two quantities are equal is because they ar a third quantity. 43. The triangle is an equiangular (or equilateral) the Transitive Property of Angle Congruence ∠1 ≅ ∠2 and ∠2 ≅ ∠3, you know that ∠1 all three angles are congruent, the triangle is (It is also equilateral and acute.) Copyright © Big Ideas Learning, LLC All rights reserved. 5.4 Name _________ Date _ Extra Practice In Exercises 1 and 2, complete the proof. 1. Given and AB CD bisect each other at point M and . BM CM ≅ Prove AB AM DM = + 2. Given is a complement of . AEB BEC ∠ ∠ Prove 90 m AED ∠ = ° STATEMENTS REASONS 1. AEB ∠ is a complement of . BEC ∠ 1. Given 2. ___ 2. Definition of complementary angles 3. m AEC m AEB m BEC ∠ = ∠ + ∠ 3. _______ 4. 90 m AEC ∠ = ° 4. ______ 5. 180 m AED m AEC ∠ + ∠ = ° 5. Definition of supplementary angles 6. ___ 6. Substitution Property of Equality 7. 90 m AED ∠ = ° 7. ______ A D E B C STATEMENTS REASONS 1. BM CM ≅ 1. Given 2. CM DM ≅ 2. _____ 3. BM DM ≅ 3. _____ 4. BM DM = 4. ______ 5. ___ 5. Segment Addition Postulate (Post. 1.2) 6. AB AM DM = + 6. _______ A D M B C k Practice A 5.4 Notetaking with Vocabulary (continued) Name ________ Date _ Extra Practice In Exercises 1 and 2, complete the proof. 1. Given and AB CD bisect each other at point M and . BM CM ≅ Prove AB AM DM = + 2. Given is a complement of . AEB BEC ∠ ∠ STATEMENTS REASONS 1. BM CM ≅ 1. Given 2. CM DM ≅ 2. _______ 3. BM DM ≅ 3. ______ 4. BM DM = 4. _____ 5. ___ 5. Segment Addition Postulate (Post. 1.2) 6. AB AM DM = + 6. ______ A D M B C k Practice A Example #2 Write a two-column proof. 167 (continued) Practice Copyright © Big Ideas Learning, LLC All rights reserved. 5.4 Name _________ Date _ In Exercises 3 and 4, name the property that the statement illustrates. 3. If and , then . RST TSU TSU VWX RST VWX ∠ ≅∠ ∠ ≅∠ ∠ ≅∠ 4. If , then . GH JK JK GH ≅ ≅ 5. Write a two-column proof. Given M is the midpoint of . RT Prove MT RS SM = + STATEMENTS REASONS R S M T Copyright © Big Ideas Learning, LLC All rights reserved. 5.4 Name ________ Date _ In Exercises 3 and 4, name the property that the statement illustrates. 3. If and , then . RST TSU TSU VWX RST VWX ∠ ≅∠ ∠ ≅∠ ∠ ≅∠ 4. If , then . GH JK JK GH ≅ ≅ 5. Write a two-column proof. Given M is the midpoint of . RT Prove MT RS SM = + STATEMENTS REASONS R S M T 5.4 Notetaking with Vocabulary (continued) Name ________ Date ______ In Exercises 3 and 4, name the property that the statement illustrates. 3. If and , then . RST TSU TSU VWX RST VWX ∠ ≅∠ ∠ ≅∠ ∠ ≅∠ 4. If , then . GH JK JK GH ≅ ≅ 5. Write a two-column proof. Given M is the midpoint of . RT Prove MT RS SM = + STATEMENTS REASONS R S M T 167 Copyright © Big Ideas Learning, LLC All rights reserved. 2. Given is a complement of . AEB BEC ∠ ∠ Prove 90 m AED ∠ = ° STATEMENTS REASONS 1. AEB ∠ is a complement of . BEC ∠ 1. Given 2. ___ 2. Definition of complementary angles 3. m AEC m AEB m BEC ∠ = ∠ + ∠ 3. _____ 4. 90 m AEC ∠ = ° 4. ______ 5. 180 m AED m AEC ∠ + ∠ = ° 5. Definition of supplementary angles 6. ___ 6. Substitution Property of Equality 7. 90 m AED ∠ = ° 7. _______ A D E B C 5. ___ 5. Segment Addition Postulate (Post. 1.2) 6. AB AM DM = + 6. _____ 168 Practice (continued) Copyright © Big Ideas Learning, LLC All rights reserved. 9.4 Practice B Name ________ Date ______ In Exercises 1 and 2, name the properties of equality that the statement illustrates. 1. If , then 2 6 2 6. x y x y = − = − 2. If and 42 , then 10 52 . m A m B m B m A ∠ = ∠ ∠ = ° ∠ + = ° In Exercises 3 and 4, write a two-column proof for the property. 3. Symmetric Property of Segment Congruence 4. Transitive Property of Angle Congruence In Exercises 5–7, write a two-column proof. 5. Given bisects , bisects , and bisects . Prove . E AI BC AE FH EI AD EG ≅ 6. Given 28 and 118 . Prove . m KMN m PTS JMK STR ∠ = ° ∠ = ° ∠ ≅∠ 7. Given . Prove . ADC BDE ADE BDC ∠ ≅∠ ∠ ≅∠ A B D H G C E F I N S P Q R T K J L M 28° 118° C D E B A Practice B 169
188072	https://www.nagwa.com/en/videos/965136250287/	Lesson Video: Euler’s Formula for Trigonometric Identities Mathematics In this video, we will learn how to use Euler’s formula to prove trigonometric identities like double angle and half angle. Video Transcript In this video, we’ll explore the derivation of a number of trigonometric identities using Euler’s formula. There’s a very good chance you will have already worked with some of these identities extensively but perhaps aren’t quite sure where they come from. So in this lesson, we’ll see how Euler’s formula links to the double angle formulae, additional multiple angle formulae, and the product to sum formulae. We begin by recalling Euler’s formula, sometimes called Euler’s relation. This states that for a real number 𝜃, 𝑒 to the 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃. Here, 𝑖 is the imaginary unit denoted as the solution to the equation 𝑥 squared equals negative one and 𝜃 must be a real number given in radians. As we can see, the formula provides a powerful connection between complex analysis and trigonometry. But it also has many applications in physics, engineering, and quantum mechanics. In our first example, we’ll see how to derive a well-used trigonometric identity by considering the properties of the exponential function and Euler’s formula. 1) Use Euler’s formula to express 𝑒 to the negative 𝑖𝜃 in terms of sine and cosine. 2) Given that 𝑒 to the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 equals one, what trigonometric identity can be derived by expanding the exponential in terms of trigonometric functions? For part one, we’ll begin by rewriting 𝑒 to the negative 𝑖𝜃. It’s the same as 𝑒 to the 𝑖 times negative 𝜃. We can now apply Euler’s formula. Since Euler’s formula says that 𝑒 to the 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃, we can see that 𝑒 to the 𝑖 negative 𝜃 is equal to cos of negative 𝜃 plus 𝑖 sin of negative 𝜃. And then, we recall the properties of the cosine and sine functions. Cos is an even function. So cos of negative 𝜃 is equal to cos of 𝜃. Sin, however, is an odd function. So sin of negative 𝜃 is the same as negative sin 𝜃. And we can therefore rewrite our expression. And we see that 𝑒 to the negative 𝑖𝜃 is the same as cos 𝜃 minus 𝑖 sin 𝜃. Now, let’s consider part two of this question. We’re going to use the answer we got from part one. When we do, we can see that 𝑒 to the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 is the same as cos 𝜃 plus 𝑖 sin 𝜃 times cos 𝜃 minus 𝑖 sin 𝜃. Let’s distribute these parentheses, perhaps noticing that this is an expression factored using the difference of two squares. cos 𝜃 times cos 𝜃 is cos squared 𝜃. cos 𝜃 times negative 𝑖 sin 𝜃 is negative 𝑖 cos 𝜃 sin 𝜃. We then get plus 𝑖 sin 𝜃 cos 𝜃 and 𝑖 sin 𝜃 times negative 𝑖 sin 𝜃 is negative 𝑖 squared sin squared 𝜃. We see then negative 𝑖 cos 𝜃 sin 𝜃 plus 𝑖 cos 𝜃 sin 𝜃 is zero. And of course, we know that 𝑖 squared is equal to negative one. So we can simplify this somewhat. And we see that 𝑒 to the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 is cos squared 𝜃 plus sin squared 𝜃. We were told however that 𝑒 to the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 was equal to one. So you can see that we’ve derived the formula sin squared 𝜃 plus cos squared 𝜃 equals one. This is a fairly succinct derivation of the trigonometric identity sin squared 𝜃 plus cos squared 𝜃 equals one. We can perform a similar process to help us derive the double angle formulae. Let’s see what that might look like. Use Euler’s formula two drive a formula for cos two 𝜃 and sin two 𝜃 in terms of sin 𝜃 and cos 𝜃. There are actually two methods we could use to derive the formulae for cos two 𝜃 and sin two 𝜃. The first is to consider this expression; it’s 𝑒 to the 𝑖𝜃 plus 𝜙. We know that this must be the same as 𝑒 to the 𝑖𝜃 times 𝑒 to the 𝑖𝜙. We’re going to apply Euler’s formula to both parts of this equation. On the left-hand side, we can see that 𝑒 to the 𝑖𝜃 plus 𝜙 is equal to cos 𝜃 plus 𝜙 plus 𝑖 sin of 𝜃 plus 𝜙. And on the right, we have cos 𝜃 plus 𝑖 sin 𝜃 times cos 𝜙 plus 𝑖 sin 𝜙. We’re going to distribute the parentheses on the right-hand side. And when we do, we get the given expression. Remember though 𝑖 squared is equal to negative one. And we can simplify and we get cos 𝜃 cos 𝜙 minus sin 𝜃 sin 𝜙 plus 𝑖 cos 𝜃 sin 𝜙 plus 𝑖 cos 𝜙 sin 𝜃. Our next step is to equate the real and imaginary parts of the equation. On the left-hand side, the real part is cos 𝜃 plus 𝜙 and on the right is cos 𝜃 cos 𝜙 minus sin 𝜃 sin 𝜙. And so, we see that cos 𝜃 plus 𝜙 is equal to cos 𝜃 cos 𝜙 minus sin 𝜃 sin 𝜙. Next, we equate the imaginary parts. On the left-hand side, we have sin 𝜃 plus 𝜙. And on the right, we have cos 𝜃 sin 𝜙 plus cos 𝜙 sin 𝜃. And we can see then that sin 𝜃 plus 𝜙 is equal to cos 𝜃 sin 𝜙 plus cos 𝜙 sin 𝜃. Now, these two formulae are useful in their own right. But what we could actually do is replace 𝜙 with 𝜃 and we get the double angle formulae. In the first one, we get cos two 𝜃 equals cos squared 𝜃 minus sin squared 𝜃. And with our second identity, we get sin two 𝜃 equals two cos 𝜃 sin 𝜃. And there is an alternative approach we could have used. This time, we could have gone straight to the double angle formulae by choosing the expression 𝑒 to the two 𝑖𝜃 and then writing that as 𝑒 to the 𝑖𝜃 squared. This time, when we apply Euler’s formula, on the left-hand side, we get cos two 𝜃 plus 𝑖 sin two 𝜃. And on the right-hand side, we get cos 𝜃 plus 𝑖 sin 𝜃 all squared. Then, distributing these parentheses, we see that the right-hand side becomes cos squared 𝜃 plus two 𝑖 cos 𝜃 sin 𝜃 plus 𝑖 squared sin squared 𝜃. And once again, 𝑖 squared is equal to negative one. So we can rewrite the right-hand side as cos squared 𝜃 minus sin squared 𝜃 plus two 𝑖 cos 𝜃 sin 𝜃. This time when we equate the real parts, we see that cos two 𝜃 equals cos squared 𝜃 minus sin squared 𝜃. And when we equate the imaginary parts, we see that sin two 𝜃 is equal to two cos 𝜃 sin 𝜃. Now, you’ve probably noticed there isn’t a huge amount of difference in these two methods. The latter is slightly more succinct. However, the former has the benefit of deriving those extra identities for cosine and sine. It’s also useful to know that we can incorporate the binomial theorem to derive multiple angle formulae in sine and cosine. The binomial theorem says that for integer values of 𝑛, we can write 𝑎 plus 𝑏 to the power of 𝑛 as 𝑎 to the power of 𝑛 plus 𝑛 choose one 𝑎 to the power of 𝑛 minus one 𝑏. And we continue this pattern with descending powers of 𝑎 and ascending powers of 𝑏 all the way through to 𝑏 to the power of 𝑛. Our next example is going to use the binomial theorem to help us evaluate multiple angles in terms of powers of trigonometric functions. 1) Use Euler’s formula to derive a formula for cos of four 𝜃 in terms of cos 𝜃. 2) Use Euler’s formula to drive a formula for sin of four 𝜃 in terms of cos 𝜃 and sin 𝜃. For part one, we’ll use the properties of the exponential function. And we’ll write 𝑒 to the four 𝑖𝜃 as 𝑒 to the 𝑖𝜃 to the power of four. And now, we can use Euler’s formula. And we write the left-hand side as cos four 𝜃 plus 𝑖 sin four 𝜃. And on the right-hand side, we can say that this is equal to cos 𝜃 plus 𝑖 sin 𝜃 all to the power of four. Now, we’re going to apply the binomial theorem to distribute cos 𝜃 plus 𝑖 sin 𝜃 to the power of four. In our equation, 𝑎 is equal to cos of 𝜃, 𝑏 is equal to 𝑖 sin of 𝜃, and 𝑛 is the power; it’s four. And this means we can say that cos 𝜃 plus 𝑖 sin 𝜃 to the power of four is the same as cos 𝜃 to the power of four plus four choose one cos cubed 𝜃 times 𝑖 sin 𝜃 and so on. We know that four choose one is four, four choose two is six, and four choose three is also four. We also know that 𝑖 squared is negative one, 𝑖 cubed is negative 𝑖, and 𝑖 to the power of four is one. And we can further rewrite our equation as shown. Now, we’re going to equate the real parts of this equation. And that will give us a formula for cos of four 𝜃 in terms of cos 𝜃 and sin 𝜃. Let’s clear some space. The real part on the left-hand side is cos four 𝜃. And then on the right-hand side, we have cos 𝜃 to the power of four. We’ve got negative cos squared 𝜃 sin squared 𝜃. And we’ve got sin 𝜃 to the power of four. So we equate these. But we’re not quite finished. We were asked to derive a formula for cos four 𝜃 in terms of cos 𝜃 only. So here, we use the identity cos squared 𝜃 plus sin squared 𝜃 is equal to one. And we rearrange this. And we say that well, that means that sin squared 𝜃 must be equal to one minus cos squared 𝜃. And we can rewrite this as cos 𝜃 to the power of four plus six cos squared 𝜃 times one minus cos squared 𝜃 plus one minus cos squared 𝜃 squared. We distribute the parentheses. And our final step is to collect like terms. And we see that we’ve derived the formula for cos of four 𝜃 in terms of cos 𝜃. cos four 𝜃 is equal to eight cos 𝜃 to the power of four minus eight cos squared 𝜃 plus one. For part two, we can repeat this process equating the imaginary parts. They are sin of four 𝜃 on the left. And then on the right, we have four cos cubed 𝜃 sin 𝜃, negative four cos 𝜃 sin cubed 𝜃. And we see this sin four 𝜃 must be equal to four cos cubed 𝜃 sin 𝜃 minus four cos 𝜃 sin cubed 𝜃. And we could — if we so wish — factor four cos 𝜃 sin 𝜃. And we’ll be left with four cos 𝜃 sin 𝜃 times cos squared 𝜃 minus sin squared 𝜃. And last, we’ve been asked to derive a formula for sin four 𝜃 in terms of cos 𝜃 and sin 𝜃. You might now see a link between sin four 𝜃 and the double angle formulae. Now, an interesting application of Euler’s formula is that we can use it to derive an expression for sin 𝜃 and cos 𝜃 in terms of 𝑒 to the 𝑖𝜃. We’ve already seen that we can write 𝑒 to the negative 𝑖𝜃 as cos 𝜃 minus 𝑖 sin 𝜃. And since 𝑒 to the 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃, we can say that the sum of 𝑒 to the 𝑖𝜃 and 𝑒 to the negative 𝑖𝜃 is cos 𝜃 plus 𝑖 sin 𝜃 plus cos 𝜃 minus 𝑖 sin 𝜃. Well, this expression on the right-hand side simplifies to two cos 𝜃. And we can make cos 𝜃 the subject by dividing through by two. And we see we have an expression for cos 𝜃 in terms of powers of 𝑒 to the 𝑖𝜃. It’s a half 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃. Similarly, we can find the difference. And we get cos 𝜃 plus 𝑖 sin 𝜃 minus cos 𝜃 minus 𝑖 sin 𝜃. This simplifies to two 𝑖 sin 𝜃. This time we divide through by two 𝑖. And we can see that sin 𝜃 is equal to one over two 𝑖 times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃. At these two formulae have plenty of applications in their own right. But for the purposes of this video, we’ll look at one final example. And we’ll see how they can be used to derive further trigonometric identities. Use Euler’s formula to express sin cubed 𝜃 cos squared 𝜃 in the form 𝑎 sin 𝜃 plus 𝑏 sin three 𝜃 plus 𝑐 sin five 𝜃, where 𝑎, 𝑏, and 𝑐 are constants to be found. Hence, find the solutions of sin five 𝜃 minus sin three 𝜃 equals zero in the interval 𝜃 is greater than or equal to zero and less than 𝜋. Give your answer in exact form. We begin by recalling the fact that sin 𝜃 is equal to one over two 𝑖 times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃. And cos 𝜃 is equal to a half times 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃. This means we can find the product of sin cubed 𝜃 and cos squared 𝜃. We can write it as one over two 𝑖 times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃 cubed times a half times 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 squared. One over two 𝑖 cubed is negative one over eight 𝑖. And a half squared is one-quarter. So we can rewrite our expression a little bit further. We find the product of negative one over eight 𝑖 and a quarter. And we get negative one over 32𝑖. And we can rewrite the rest of our expression as shown. We’re now going to use the binomial theorem to expand each of the sets of parentheses. The first part becomes 𝑒 to three 𝑖𝜃 plus three choose one 𝑒 to the two 𝑖𝜃 times negative 𝑒 to the negative 𝑖𝜃 and so on. And this simplifies to 𝑒 to the three 𝑖𝜃 minus three 𝑒 to the 𝑖𝜃 plus three 𝑒 to the negative 𝑖𝜃 minus 𝑒 to the negative three 𝑖𝜃. Let’s repeat this process for 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 squared. When we do, we get 𝑒 to the two 𝑖𝜃 plus two 𝑒 to the zero which is just two plus 𝑒 to the negative two 𝑖𝜃. We’re going to need to find the product of these two expressions. We’ll need to do that really carefully. We’ll need to ensure that each term in the first expression is multiplied by each term in the second expression. And we can write sin cubed 𝜃 cos squared 𝜃 as shown. Now, there’s quite a lot going on here. So you might wish to pause the video and double-check your answer against mine. We’re going to gather the corresponding powers of 𝑒 together. We’ll gather 𝑒 to the five 𝑖𝜃 and 𝑒 to the negative five 𝑖𝜃. We’ll collect 𝑒 to the plus and minus three 𝑖𝜃. And we’ll gather 𝑒 the 𝑖𝜃 and 𝑒 to the negative 𝑖𝜃. Let’s neaten things up somewhat. We end up with negative one over 32𝑖 times 𝑒 to the five 𝑖𝜃 minus 𝑒 to the negative five 𝑖𝜃 minus 𝑒 to the three 𝑖𝜃 minus 𝑒 to the negative three 𝑖𝜃 minus two times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃. And now, you might be able to spot why we chose to do this. We can now go back to the given formulae. Let’s clear some space for the next step. We kind of unfactorize a little. And we can rewrite sin cubed 𝜃 cos squared 𝜃 as shown. And we can therefore replace 𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 with sin 𝜃 and so on. And we can see that sin cubed 𝜃 cos squared 𝜃 is equal to a 16th times two sin 𝜃 plus sin three 𝜃 minus sin five 𝜃. Since 𝑎, 𝑏, and 𝑐 are constants to be found, we can say that 𝑎, the coefficient of sin 𝜃, is one-eighth. 𝑏, the coefficient of sin three 𝜃, is a 16th. And 𝑐, the coefficient sin five 𝜃, is negative one 16th. Let’s now consider part two of this question. We begin by using our answer to part one and multiplying both sides by 16. We then subtract two sin 𝜃 from both sides and multiply through by negative one. And we can now see that we’ve got an equation in sin five 𝜃 minus sin three 𝜃. We’re told that sin five 𝜃 minus sin three 𝜃 is equal to zero. So we let two sin 𝜃 minus 16 sin cubed 𝜃 cos squared 𝜃 be equal to zero. And then, we factor by two sin 𝜃. Since the product of these two terms is equal to zero, this means that either of these terms must be equal to zero. So either two sin 𝜃 is equal to zero and dividing by two, we could see that sin 𝜃 is equal to zero or one minus eight sin squared 𝜃 cos squared 𝜃 is equal to zero. Given the interval 𝜃 is greater than or equal to zero and less than 𝜋, we can see that one of our solutions is when 𝜃 is equal to zero. We’re going to rewrite our other equations somewhat. We know that sin two 𝜃 is equal to two sin 𝜃 cos 𝜃. Squaring this, we get sin squared two 𝜃 equals four sin squared 𝜃 cos squared 𝜃. And that’s in turn means that our equation is one minus two sin squared two 𝜃 equals zero. Rearranging to make sin two 𝜃 the subject, we see that sin two 𝜃 is equal to plus or minus one over root two. Starting with the positive square root for 𝜃 in the interval given, we know that sin two 𝜃 is equal to one over root two when 𝜃 is equal to 𝜋 by eight or three 𝜋 by eight. Similarly, we can solve for the negative square root. And we get five 𝜋 by eight and seven 𝜋 by eight. And there are therefore five solutions to the equation sin five 𝜃 minus sin three 𝜃 equals zero in the interval 𝜃 is greater than or equal to zero and less than 𝜋. They are zero 𝜋 by eight, three 𝜋 by eight, five 𝜋 by eight, and seven 𝜋 by eight. In this video, we’ve seen that we can use Euler’s formula in conjunction with the properties of the exponential functions. And we can derive many trigonometric identities such as the Pythagorean identity and multiple angle formulae. We also saw the we can use the theorem to express sine and cosine in terms of the complex exponential function as shown. We’ve also seen that we can use the identities derived from Euler’s formula to help us simplify expressions. And these in turn can help us to solve trigonometric equations. 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188073	https://www.youtube.com/watch?v=7ye5SnqsNJo	How changing a value affects the mean and median larryschmidt 21700 subscribers 20 likes Description 5069 views Posted: 4 Jan 2018 How to calculate how changing a value affects the mean and median 4 comments Transcript: in this problem we are being given a list of numbers of data set here in this case their salaries for different employees at a business and we're being asked to tell what happens to the mean the average when you change one of those values and also what happens to the median so one of the things they've done for us that's pretty nice is they've given us this list of salaries these are weekly salaries and they've ordered them from smallest to largest so if we wanted to know what the median is finding that middle value is not going to be too tough we've got an even number though so these two are the middle two values if it was an odd number of data points we could have one single middle value so the median let's just go ahead and calculate that what you would do is take 773 and 718 take the average of those two so I'll add them together and we'll divide by 2 and I get seven hundred and seventy eight point five so that's our median right now our average or mean well that takes a little more calculating you're gonna have to add up all these values and then divide by 10 which is the number of values what I've done is actually I've typed this into a spreadsheet because I find it easier to do this stuff on a spreadsheet so let me show you here's that set of values and then you can calculate your average by using some formulas I've clicked into this empty cell here I'm going to hit the equal sign and then I'm going to type in s um and then a parenthesis and then select over all these lines and then in parenthesis that gives us the sum of all of those then I hit the slash key that means divide and then 10 so that's the sum of all those divided by 10 so when I hit enter I should get the average seven hundred and eleven point four so that's our mean so let's go back here so in this problem what they're saying is let's take this 10:36 and change it to 786 I guess in the real world this means you you fire or somebody quits who earns the most and then you hire a new person who earns a little bit less so this seven this 1036 goes away and the 786 goes in its place so we want to know what happens to the mean well let's go ahead and try that on our spreadsheet our mean right now is seven hundred and eleven point four so let's actually write that down seven hundred and eleven point four and then we'll take that ten thirty six and put 786 in place of it so right here we're gonna take that and type in 786 now let's see what happens to the mean oh it goes down to six eighty six point four so let's write that down so this is it goes to six eighty six point four and if you want to find out well definitely decreases doesn't stay the same so seven eleven point four I'm just pulling out on calculator now minus six eighty six point four it decreases by twenty five dollars alright so when we took a higher value out and replaced it with a lower value it put the average of the mean down and that makes sense that should happen any value here is going to affect the mean how about the medium does the median change well to figure out the median here we need to put this new item in order on this list so 786 would fit in right here so 786 goes there now the two middle values oh well the two middle values are still the same so this event new value is a little bit above our median so it actually doesn't change the two middle values and that means the median here stays the same so that's a little bit of work with how changing a value in a data set will change the mean and the median
188074	https://www.quora.com/What-does-in-retrospect-mean-and-how-can-it-be-used-in-a-sentence	What does “in retrospect” mean and how can it be used in a sentence? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In What does “in retrospect” mean and how can it be used in a sentence? All related (44) Sort Recommended Aditya Goutam U.G in English (language)&Digital Marketing, University of Rajasthan (Graduated 2019) · Author has 52 answers and 1.5M answer views ·5y Originally Answered: What is the actual meaning of the word "retrospect"? Can you give an example of it? · noun contemplation of the past; a survey of past time, events, etc. verb (used without object) to look back in thought; refer back (often followed by to): to retrospect to a period in one's youth. verb (used with object) to look back upon; contemplate. synonyms:- hindsight recollection reconsideration reexamination remembering remembrance reminiscence retrospection review revision survey source: Upvote · 9 2 Promoted by Grammarly Grammarly Knows English ·Updated 2y What are the different errors and mistakes in English? Grammatical mistakes can be easy to make. But fear not! Our team at Grammarly has compiled a handy list of common grammatical errors to help make your writing accurate, clear, and professional. Ambiguous (“Squinting”) modifiers Incorrect: Listening to loud music slowly gives me a headache. Correct: When I listen to loud music, I slowly develop a headache. A squinting modifier is a misplaced modifier that, because of its location in a sentence, could modify either the phrase that precedes it or the one that follows it. In the example sentence, is the subject listening to music slowly or slowly gett Continue Reading Grammatical mistakes can be easy to make. But fear not! Our team at Grammarly has compiled a handy list of common grammatical errors to help make your writing accurate, clear, and professional. Ambiguous (“Squinting”) modifiers Incorrect: Listening to loud music slowly gives me a headache. Correct: When I listen to loud music, I slowly develop a headache. A squinting modifier is a misplaced modifier that, because of its location in a sentence, could modify either the phrase that precedes it or the one that follows it. In the example sentence, is the subject listening to music slowly or slowly getting a headache? To correct a squinting modifier, move its position in the sentence to clarify to the reader which word you intend to modify. Misuse of lie/lay Incorrect: He was laying on the couch. Correct: He was lying on the couch. If you plan to place or put an object somewhere, such as a plate on a table, you should use “lay.” If you intend to stretch out on a bed for a nap, you should use “lie.” The verb “lie” is an intransitive verb, which means it does not need an object. The transitive verb “lay” requires an object. It may take some getting used to this “lay” or “lie” business; after all, misuse of these verbs is common. But if you remember to lay down your fork before you’re full, then you won’t have to lie down later from overeating. Comma splices Incorrect: He was very hungry, he ate a whole pizza. Correct: He was very hungry. He ate a whole pizza. He was very hungry, so he ate a whole pizza. To splice means to connect or join. When a writer joins two independent sentences with a comma instead of separating them with a period or coordinating conjunction, that’s a comma splice. The comma has its jobs to do, but connecting two independent sentences isn’t one of them. Besides, the period gets testy when his sister, the comma, steals his thunder. Periods have their jobs, and so do commas, but never the twain shall meet—unless it’s in the form of a semicolon. Semicolons can also take the place of a coordinating conjunction, such as “and,” “but,” or “so,” among others. Run-on sentences Incorrect: Lila enjoyed the bouquet of tulips John gave her on prom night however she prefers roses. Correct: Lila enjoyed the bouquet of tulips John gave her on prom night; however, she prefers roses. Run-on sentences, also known as fused sentences, occur when two complete sentences are squashed together without using a coordinating conjunction or proper punctuation, such as a period or a semicolon. Run-on sentences can be short or long. A long sentence isn’t necessarily a run-on sentence. To avoid run-on sentences, see if there is more than one idea communicated by two or more independent clauses. In our examples, there are two complete sentences: Example: Lily enjoyed the bouquet of tulips John gave her on prom night. Example: She prefers roses. Both sentences are complete ideas by themselves; therefore, use a semicolon or a period to indicate that they are separate independent clauses. Using “could of” instead of “could have” Incorrect: Sam could of received an A on his essay, but he made too many grammatical mistakes. Correct: Sam could have received an A on his essay, but he made too many grammatical mistakes. “Could have” is always correct; “could of” never is. Writers probably make this grammar gaffe because, when we speak, the contraction “could’ve” sounds an awful lot like “could of.” Tautologies Incorrect: Jack made a water pail with his own hands for Jill. Correct: Jack made a water pail for Jill. Tautologies express the same thing twice with different words. In our example, the word “made” implies that Jack used his own two hands to create the pail. The prepositional phrase “with his own hands” creates a redundancy. Once you know what they are, it’s fun to discover tautologies: dilapidated ruins, close proximity, added bonus, large crowd...The list goes on and on! After reading through this list of common grammatical mistakes, you might be wondering how to remember all these rules as you write. The free Grammarly for Windows and Mac is here to help. It provides a second set of eyes on your writing in real-time, so you can avoid everyday grammar and spelling errors. Moreover, Grammarly Premium offers features that evaluate conciseness and readability as well as vocabulary enhancement suggestions and genre-specific writing style checks. These tools can help you identify when you are making these common writing mistakes so you can proactively learn and improve your writing. Upvote · 1.5K 1.5K 999 520 99 55 Related questions More answers below What does the word "retrospect" mean? Can you use it in a sentence? How is the word "retrospect" used in a sentence? What does obvious in retrospect mean? What are the ways to use retrospect in a sentence? What does "inborn" mean? Can you use it in a sentence? Michael Ruby a genealogy programmer · Author has 563 answers and 3M answer views ·9y Originally Answered: What does it mean when someone says in retrospect? · "Retrospect" is the consideration of past times, from Latin retro 'backwards' + spectum 'looked at'. If I made a decision that I later regret, after I have seen the negative consequences of the decision, I will say "In retrospect (that is, looking backward), I should not have done that." Upvote · 9 1 Frank Dauenhauer Former Technical Writer & Editor of Company Publications at Eastman Kodak Products and Services (1960–1991) · Author has 45.4K answers and 277.4M answer views ·1y What does “in retrospect” mean, and how can it be used in a sentence? “In retrospect” means “when looking backward in one’s memory to an earlier time.” Examples: “In retrospect, I now see that I should have provided better direction during that session.” “In retrospect, he realized he should have taken Michelle’s comment as a hint of stormy waters ahead.” “Looking back at my junior years, most of the things that I felt were a big deal at the time now seem trivial in retrospect.” “The Ashton years were, in retrospect, The Royal Ballet's golden age.” “It all seemed quite funny at the time. Not funny i Continue Reading What does “in retrospect” mean, and how can it be used in a sentence? “In retrospect” means “when looking backward in one’s memory to an earlier time.” Examples: “In retrospect, I now see that I should have provided better direction during that session.” “In retrospect, he realized he should have taken Michelle’s comment as a hint of stormy waters ahead.” “Looking back at my junior years, most of the things that I felt were a big deal at the time now seem trivial in retrospect.” “The Ashton years were, in retrospect, The Royal Ballet's golden age.” “It all seemed quite funny at the time. Not funny in retrospect, though.” Upvote · 9 5 9 1 Jim Davis Native speaker (American), English teacher, writing coach, editor. · Author has 14.1K answers and 10.1M answer views ·9y Originally Answered: What are the ways to use retrospect in a sentence? · It means a look into the past, so you can use it any where you look back in time, as in, "In retrospect, I might have done things a lot differently." "Retrospective" describes a short history, often a biography, as in, "The museum featured a retrospective collection featuring the early paintings of Picasso." Upvote · 9 1 Related questions More answers below What does "alternate" mean? Can you use it in a sentence? How would you use the word 'cattywampus' in a sentence? What does transitory mean? Can you use it in a sentence? What does incipient mean? Can you use it in a sentence? What does "imprudent" mean? Can you use it in a sentence? Rasheed Ahmed Masters in M A English&Education (academic discipline), Regional Institute of Education, Mysore (RIE Mysore) (Graduated 1975) · Author has 1.8K answers and 1.1M answer views ·1y it means to look back into the Under Construction retrospect most of the movies were of a high standard and timeless classics Upvote · Sponsored by Complex Law Attention UK Drivers: Check PCP refund eligibility now. If you took out car finance between 2007–2021, use Complex Law to check your refund eligibility. Learn More 999 232 Dr Zubair Ahmad BUMS.PGDHHM Medical Director and Integrated Medicine Physician. · Author has 160 answers and 132.5K answer views ·5y Originally Answered: What does the word "reconsider" mean? Can you use it in a sentence? · It's a verb. Reconsider means to consider something again. Sentence:”Let's reconsider the fact that he killed the man”, said the lawyer to his client. Upvote · Lori Jordan A.A. in Writing&Psychology, Three Rivers College ·2y Originally Answered: How do you use retrospectively in a sentence? · You could either want to use retrospectively to mean with consideration to past events, such as “The researchers retrospectively reviewed 6 years of files.” Or to mean with effect from a day in the past, such as “The late fees will apply retrospectively from January 1.” Upvote · Promoted by Betterbuck Anthony Madden Writer for Betterbuck ·Updated Jul 15 What are the weirdest mistakes people make on the internet right now? Here are some 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 privac Continue Reading Here are some 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. Thinking you can only access your home equity by selling If you’re over 55, own your home, and need cash - equity release could help you tap into your home’s value without the need to sell. It’s like a mortgage that lets you access equity, either by making repayments while you stay in your home, or by letting the loan (+interest) get paid off when you pass away or move into care. Use this free calculator if you want to find out how much money you could access. You can take the money as a lump sum (usually £10k+), smaller amounts when you need it (a “drawdown”), or a mix of both. It’s important to know that the earlier you take out equity release, the more interest can build up over time. If you want a quote from a major UK provider, answer a few questions here. 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). Upvote · 999 247 99 25 9 4 Manfred Kramer Works at Information Technology · Author has 3.4K answers and 13.7M answer views ·9y Originally Answered: What exactly does the word "retrospective" mean? · Retrospective means looking back at events or things that have happened or were produced in the past. In comes from the latin words "retrospectare" and is itself a composite of the words "retro", meaning "back" or "backwards" and "spectare" meaning "to look, to view". Upvote · 9 2 Jim Pruitt Senior Editor at VMEdu, Inc., educator, writer · Author has 1.5K answers and 2.7M answer views ·9y Originally Answered: What exactly does the word "retrospective" mean? · One of the interesting things about retrospective is that it is used both as an adjective and as a noun without changing its spelling. As an adjective it means that the thing being described looks back or offers the opportunity for others to look back at prior conditions, actions, or characteristics. As a noun it refers to any situation or event that is purposed to have its participants look back at prior conditions, actions, or characteristics--usually of a chosen topic. In the Scrum project management/product development framework, there are two meetings set apart as retrospectives for team Continue Reading One of the interesting things about retrospective is that it is used both as an adjective and as a noun without changing its spelling. As an adjective it means that the thing being described looks back or offers the opportunity for others to look back at prior conditions, actions, or characteristics. As a noun it refers to any situation or event that is purposed to have its participants look back at prior conditions, actions, or characteristics--usually of a chosen topic. In the Scrum project management/product development framework, there are two meetings set apart as retrospectives for team members and stakeholders: the Retrospect Sprint and Retrospect Project meetings. Upvote · Promoted by ASOWorld Abigail buy installs for app ·9mo Where can we buy app installs? Good morning ladies and gentlemen, There a lot of of websites for you to buy app installs like ASOWorld, Appranktop, and so on. I want to introduce the benefits of getting more app installs. 65% of app downloads are achieved through searching. 83% of people are only focus on TOP 10 results, when downloading an app. We can boost your app to #TOP on any keyword you would like. Upvote · 99 19 Matt Egan Professional curmudgion · Author has 13.6K answers and 17M answer views ·2y Originally Answered: How do you use retrospectively in a sentence? · Retrospectively, I cannot remember ever being asked to use this word in a sentence. Upvote · 9 3 Berwan Dekker Knows English ·5y Originally Answered: What are the ways to use retrospect in a sentence? · You can say 'in retrospect' as looking back at an event. As in, 'in retrospect, I should not have done that.’ Upvote · Roger Silverthorn Only sixty years experience - still learning ! · Author has 190 answers and 581.6K answer views ·9y Originally Answered: What are the ways to use retrospect in a sentence? · 'In retrospect, the idea of giving the loaded shotgun to the small boy was not a good one.' Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 9 1 Cris Smith I love to play with words. · Author has 978 answers and 2.7M answer views ·9y Originally Answered: What exactly does the word "retrospective" mean? · Retrospective means looking back at things that happened in the past. His autobiography was a retrospective of his life during the 1970s. Upvote · Related questions What does the word "retrospect" mean? Can you use it in a sentence? How is the word "retrospect" used in a sentence? What does obvious in retrospect mean? What are the ways to use retrospect in a sentence? What does "inborn" mean? Can you use it in a sentence? What does "alternate" mean? Can you use it in a sentence? How would you use the word 'cattywampus' in a sentence? What does transitory mean? Can you use it in a sentence? What does incipient mean? Can you use it in a sentence? What does "imprudent" mean? Can you use it in a sentence? What does "inadvisable" mean? Can you use it in a sentence? What does "intermediate" mean? Can you use it in a sentence? What does deluge mean? Can you use it in a sentence? What does countervail mean? Can you use it in a sentence? What does recreancy mean? Can you use it in a sentence? Related questions What does the word "retrospect" mean? Can you use it in a sentence? How is the word "retrospect" used in a sentence? What does obvious in retrospect mean? What are the ways to use retrospect in a sentence? What does "inborn" mean? Can you use it in a sentence? What does "alternate" mean? 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188075	https://brainly.ph/question/6806363	What is the speed of sound in air of 26°C? - Brainly.ph Skip to main content Ask Question Log in Join for free For parents For teachers Honor code Brainly App annabelanggot3 11.11.2020 Science Senior High School answered What is the speed of sound in air of 26°C? 1 See answer See what the community says and unlock a badge. 0:03 / 0:15 Read More Answer 7 people found it helpful ztikboyz724 ztikboyz724 Ambitious 7 answers 4.8K people helped 5 Answer: The speed of sound in air is approximately 331.5 m/s at 0 °C or around 1200 km per hour. The speed of sound through air is approximately 343 m/s at normal room temperature, which is at 20 °C. The speed of sound through air is 346 m/s at 25 °C. Explanation: Explore all similar answers Thanks 7 rating answer section Answer rating 3.8 (12 votes) For the Moduler's the answer are 346.6 m/s 348.4 m/s 353.8 m/s Step 1 multiply 0.6 and 26 (0.6 × 26 = 15.6) Step 2 Add 15.6 to 331 (15.6 + 331) Step 3 The answer is 346.6 m/s Step 1 Multiply 0.6 and 29 (0.6 × 29 = 17.4) Step 2 Add 17.4 to 331 (17.4 + 331) Step 3 The answer is 348.4 m/s Step 1 multiply 0.6 and 38 (0.6 + 38 = 22.8) Step 2 Add 22.8 and 331 (22.8 + 331) Step 3 The answer is 353.8 tanks i mean correct tanls wrong thank you❤❤❤❤❤ See more comments Advertisement Still have questions? Find more answers Ask your question New questions in Science the part of the plant that grows underground and absorb water can you drawing a cobalt with 27 shell what is the states of matter? explain each. Sponges and cnidarians are said to be sessile or stationary what does this means? What would happen if our body did not have bones and muscles working together?3-5 sentences PreviousNext Advertisement Ask your question Free help with homework Why join Brainly? ask questions about your assignment get answers with explanations find similar questions I want a free account Company Careers Advertise with us Terms of Use Copyright Policy Privacy Policy Cookie Preferences Help Signup Help Center Safety Center Responsible Disclosure Agreement Get the Brainly App ⬈(opens in a new tab)⬈(opens in a new tab) Brainly.ph We're in the know (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)
188076	https://www.youtube.com/watch?v=QaubWEa9fW0	Sum of all five digit numbers formed using digits 1,3,5,7,9 without repetition - Simple Explanation Thakar Training Centre - Mechanical Engineering 242 subscribers 29 likes Description 1366 views Posted: 27 Jun 2023 Sum of all five digit numbers formed using digits 1,3,5,7,9 without repetition is explained in this video. This is one of the challenging questions asked in GATE Mechanical Engineering Exam. The question was asked for 2 marks in GATE 2014 (ME Paper, Set 4). Such type of questions are also asked in other entrance exams like JEE. The solution has been explained in a very simple manner so that all students can understand it clearly and will be able to solve such type of questions easily. gatepreparation permutations_and_combinations jeemaths 13 comments Transcript: नंबर इस फॉर्म यूजिंग डी डिजिट्स 13579 विदाउट रिपीटिंग अन्य ऑफ डेम व्हाट इसे डी सम ऑफ जो सच पॉसिबल 5 डिजिटल नंबर्स इक्वेशन थोड़ा चालीसिंग है तो हमें यहां पर पूछा गया है की व्हाट बिल बी डी सम ऑफ जो पॉसिबल 5 डिजिटल नंबर्स जो हम लोग डिजिट्स 135 और 9 उसे करके फॉर्म कर सकते ठीक है और कंडीशन दे दिए है की हम लोग डिजिट्स को रिपीट नहीं कर सकते तो अगर हम सिर्फ ये देखेगी कितने नंबर्स पॉसिबल है ऐसे अगर वही चेक करेंगे तो हम लोग चेक कर सकते हैं कितने नंबर्स पॉसिबल है तो हमारे पास 5 डिजिट्स है 5 डिजिटल नंबर और वे हैव तू उसे फाइव डिजिट्स विदाउट रिपीटिंग अन्य ऑफ दे तो पहले डिजिटल सिलेक्ट करने के लिए मेरे पास 5s होंगे राइट क्योंकि यह प्लेस ऑक्युपी कर सकते हैं ठीक है ना अपर इन अन्य ऑफ दिस प्लेस नो बिकॉज़ यू हैव तू फॉर्म डी नंबर्स विदाउट रिपीटिंग डी डिजिट्स तो फायर अभी यहां पे नहीं ए सकते सो नो फाइव आउट ऑफ कंसीडरेशन वे आर लेफ्ट विथ डिजिट्स 1379 तो अन्य ऑफ डेम कैन कम इन दिस प्लेस सो यू हैव इन विच वे कैन ऑक्युपी सो सिमिलरली अब यहां पे सपोज डिजिटल थ्री आगे ऑक्यूपाइड दिस प्लेस सो थ्री के नोट अपर इन अन्य ऑफ दिस प्लेस इसे नो डिजिट्स 3 और 5 आर आउट ऑफ कंसीडरेशन 179 सो वन सेकंड बिल हैव थ्री वे इन विच प्लेस कैन बी एक्वायर राइट और सिमिलरली नेक्स्ट स्पेस सो एवरी टाइम पर्टिकुलर प्लेस वन ली डिजिटल तो हर बार एक डिजिटल फिक्स करेंगे तो हमारे पास एक डिजिटल कम होते जाता है सो वे हैव डी टोटल नंबर्स पॉसिबल टोटल नंबर्स पॉसिबल आर नथिंग बट बाय पॉसिबल विच कैन बी फाउंड विथ डी हेल्प ऑफ डिजिट्स 13579 विदाउट रिपीटिंग अन्य ऑफ डी ठीक है तो 125 डिजिटल नंबर्स होंगे और हमको यहां पर पूछा गया है व्हाट इसे डी सम ऑफ जो सर्च नंबर्स ठीक है व्हाट इसे डी सम ऑफ नंबर 120 नंबर्स लिख के उनको एड नहीं कर सकते हैं पॉसिबल सो 3 फैक्टोरियल विच इसे सिक्स सो सिक्स नंबर तो अगर सेंडिंग ऑर्डर में लिखेंगे दिल भी 567 576 57 6756 और 765 ओके तो इन नंबर्स का टोटल मिलता है ना तो आईटी इसे नोट पॉसिबल तू डू आईटी इन दिस वे करेक्ट नहीं कर सकते ओके नेक्स्ट 600 + 50 + 7 600 + 70 + 5 700 + 50 + 600 प्लस फाइव और यूनिट्स जो राइट और हमको एड करना है तो मुझे क्या करना है मुझे इन डिजिट्स के टर्म में एक एक्सप्रेशन लिखना है ठीक है तू टाइम्स + 10 रहेगा से थिंक है सिक्स का देखो बार ए रहा है और सिक्स दोबारा निकलेंगे तो हमारे पास ए जाएगा 2 6 100 + 10 + 1 डेट इस नथिंग एन 2 6 1 = अंडरस्टूड ये भी आपको समझ जाएगा की सेवन का जो कंट्रीब्यूशन है वो ये रहने वाला है 2 7 11 वही बिकॉज़ 7 इन 100 प्लेस इस कमिंग टिस प्ले स्टोर नंबर टोटल समय नथिंग बट तू इन सिक्स इन 11 प्लस 7 ठीक है तो मैंने पहले बताया कैलकुलेट करोगे तो यह जब हम लोग ने ऐसे नंबर को लिख दिया तो हम लोग को सम लिखने के लिए एक और मेथड मिल गया है तो उसके लिए क्या रहेगा ठीक है ओके तो अभी जब हमने डिलीट आएगा तो उसकी वैल्यू फाइंड वन फाइव रहेगी अलराइट तो जो ये वाला एक्सप्रेशन है इसमें 111 होता अगर वो 4 डिजिटल नंबर होता ठीक है 4 डिजिटल नंबर प्लस 10 प्लस वन तो यह आने वाले टाइम 111 और इसके आगे एक नंबर रहेगा की ये जो नंबर ए रहा है हमको 111 यहां पे ए रहा है वन वन वन तो थ्री डिजिटल के लिए हमको मिला 11 4 डिजिटल के लिए मेरा वन वन तू क्लीयरली आई थिंक यू कैन रिलाइज समथिंग लाइक वन वन वन रहेगा है तो एक एक्सप्रेशन आने वाला यहां पर विच इस सम ऑफ एन डिजिट्स सम ऑफ एन डिजिट्स अभी यहां पर जो आने वाला है 2s कमिंग पिक्चर कितनी बार कोई नंबर किसी पर्टिकुलर बेस पर रहेगा वो कैसे पता चलेगा ठीक है यूजिंग डिजिट्स 5678 तो ऐसे कितने नंबर्स पॉसिबल होंगे जिसमें 5 हंड्रेड बिकॉज़ तो जैसे यहां पर जब मैं थ्री डिजिटल नंबर बना रहा था तो 500 स्पेस में दो बार था तो इसलिए 500 तू टाइम्स काउंट हुआ सिमिलरली 700 दो बार अकाउंट होगा अभी जब मैं 4 डिजिटल नंबर 500 6 टाइम्स काउंट होगा अरे फॉलोइंग आईटी सिमिलरली जब मैं सिक्स को 100 पीस पे रखूंगा तो 600 भी छह बार काउंट होगा 7 होगा 800 भी छह बार काउंट होगा तो आप चेक कर सकते हो की सारे डिजिट्स पे 67810 और यूनिट स्टेज ऑक्युपी करेंगे तब हमारे पास हर एक प्लेस के लिए छह नंबर बनेंगे इसका मतलब 1000 यूनिट प्लेस में तब छह नंबर बनेंगे और जब आएगा टोटल कंट्रीब्यूशन फाइंड आउट करूंगा मैं उसको ऐसे लिख सकता हूं टोटल कंट्रीब्यूशन ऑफ फाइव बिल बी 6 5 1000 + 100 + 10 + 1 तो यह सिक्स क्या था कितनी बार कोई पर्टिकुलर डेट कोई पर्टिकुलर प्लेस ऑक्युपी करता है उतनी बार उसका कंट्रीब्यूशन आएगा मल्टीप्लाई वैल्यू राइट बट कितनी बार आता है डेट इसे एन - 1 पार्ट कंसर्ट करना है मतलब मुझे 500 कितनी बार कंसीडर करना है 600 कितनी बार कंसीडर करना है या फिर 50 कितनी बार कंसीडर करना है वो मुझे पता चला है एन - 1 फैक्टोरियल से आई होप यू हैव फॉलोज दिस तू जब मैं टोटल सब निकलेगा और इन डिजिटल नंबर फॉर्म यूजिंग टोटल संख्या सो पर फैक्टर 24 25 दिस टर्स आउट तू बी 600 1111 सो दिस सिंपली इस गोइंग तू बी सिक्स सिक्स फाइव टाइम्स और दें फॉलोज में 20 सो 5 टाइम्स 6 फॉलोज में 2 0 शुड बी डी आंसर वे हैव डेट इन ऑप्शन बी पी टाइम्स सिक्स फॉलोज में 20 सो डी आंसर बिल बी ऑप्शन बी एन सम ऑफ जो सच पॉसिबल फाइव डिजिटल नंबर्स इस गोइंग तू बी फाइव टाइम्स सिक्स फॉलोज बाय 20 सोमवार आई होप एवरीबॉडी इस फॉलोज थे मेथड इन विच वे कैन अचीव दत सम ठीक है तो मेथड क्या है वह आपको सबको समझ में ए गया तो हम तब वहां पर यह सारे नंबर्स लिख नहीं सकते राइट वे कैन नोट राइट डॉ जो डी पॉसिबल फाइव डिजिटल नंबर्स तो जब हम लोग यहां पर देखा वे हैव 120 नंबर सो आईटी इसे नोट पॉसिबल तू राइट 120 नंबर्स तो वे डू एग्जांपल ऑफ थ्री डिजिटल नंबर्स डेट कैन बी फाउंड विथ हेल्प ऑफ सम डिजिट्स 56 लिए हम लोग एग्जांपल के लिए तो उनका सम हमलोग जब नंबर्स लिख के निकलते हैं तो हमको इजीली मिल जाता है बट वे वांट एन एक्सप्रेशन जिससे हम लोग डिजिट्स को उसे करके सब निकाल सकते हैं लॉन्ग हैंड में लिखा मतलब 567 मेरे डैनी 500 + 60 + 7 और वैसे ही हम लोग अपने सारे नंबर्स लिखे फिर हर एक डिजिटल का कंट्रीब्यूशन देखा हम लोग ने तो फाइव के लिए 500 आता है और फाइव आता है तो उसका जो कंट्रीब्यूशन है तो हम लोग ने इस तरह लिखा सिमिलरली सिक्स के लिए ऐसे लिखा जा सकता है और जब हम लोग सम करते हैं तो टोटल है जिसमें हम लोग डिजिट्स को उसे कर सकते हैं हम लोग टोटल सम निकाल सकते एग्जांपल और विंडोज एक्सप्रेशन और फिर जो हमारा क्वेश्चन था उसको हम लोग में इजीली सॉल्व कर दिया मैंने बोला था आपको की दिस इसे एप्लीकेबल व्हेन यू आर लुकिंग आते कैसे ऑफ विदाउट रिपीटेशन रिपीट नहीं हो सकते तभी ये एक्सप्रेशन वैलिड है इसका आंसर इस ऑप्शन बी
188077	http://www.mclements.net/Mike/mrc-BikePhysics.html	Basic Motorcycling Physics Basic Motorcycling Physics Here is a free body diagram of the forces on a motorcycle while leaned in a turn. This is a rear view of the bike as it turns to the right. The big dot is the bike's center of gravity. It shows how the lean angle is determined by the radius of the turn and the speed of the bike. There are two opposing torques, both of which share a "hinge" which is where the bike's tire meets the road. The counterclockwise torque is the "high side" torque. Its force is proportional to V squared over R. Its moment arm is proportional to the cosine of the lean angle. The clockwise torque is the "low side" torque. Its force is proportional to G (the gravitational constant). Its moment arm is proportional to the sine of the lean angle. M (V V) / R cos(Theta) = M G sin(Theta) Divide both sides by M (it cancels out) and you get: (V V) / R cos(Theta) = G sin(Theta) There are other factors like Mass, height of the center of gravity, etc. But they are not shown here because they cancel each other out and thus do not affect the equations. That doesn't mean they don't affect handling. They can affect handling tremendously - obviously, a GSXR-750 handles better than a Harley Road King. But they don't change the lean angle for a given speed and turn. Of course, G is just a constant too. There are only 3 variables here: V: the speed of the bike R: the radius of the turn Theta: the lean angle As the bike moves through the turn, both torques are equal.
188078	https://ibalmaths.com/index.php/ibdp-math-hl-2/binomial-theorem/ibdp-practice-questions-binomial-theorem/	Binomial Theorem – Practice Questions – IBDP Math HL/SL Skip to the content IBDP Math HL/SL Analysis & Approaches Menu IBDP Math HL Applications of differentiation Binomial Theorem Bivariate Statistics Circular measure Complex Numbers Continuous random variables Definite Integrals Differential Equations Discrete random variables Exponents and Logarithms Functions Introduction to differential calculus Kinematics Limits Introduction to Integration Maclaurin Series Mathematical Induction Non right-angled triangle trigonometry Permutations & Combinations Polynomials Previous Year Question Paper Properties of curves Quadratics Reasoning and proof Rules of differentiation Sequence and Series Techniques for Integration Transformation Trigonometric Functions Trigonometric identities and equations Maths Exploration (IA) ideas Paper 3 examples (HL) About Author Yoga Pre-Diploma Quizzes Quiz 1 Quiz 2 Quiz 3 Quiz 4 Quiz 5 Menu Close Menu IBDP Math HLShow sub menu Applications of differentiation Binomial Theorem Bivariate Statistics Circular measure Complex Numbers Continuous random variables Definite Integrals Differential Equations Discrete random variables Exponents and Logarithms Functions Introduction to differential calculus Introduction to Integration Kinematics Limits Maclaurin Series Mathematical Induction Non right-angled triangle trigonometry Permutations & Combinations Polynomials Properties of curves Quadratics Reasoning and proof Rules of differentiation Sequence and Series Techniques for Integration Transformation Trigonometric Functions Trigonometric identities and equations Maths Exploration (IA) ideas ibalasia Maths Books Yoga Pre-Diploma QuizzesShow sub menu Quiz 1 Quiz 2 Quiz 3 Quiz 4 Quiz 5 About Author IBDP Math HLShow sub menu Applications of differentiation Binomial Theorem Bivariate Statistics Circular measure Complex Numbers Continuous random variables Definite Integrals Differential Equations Discrete random variables Exponents and Logarithms Functions Introduction to differential calculus Introduction to Integration Kinematics Limits Maclaurin Series Mathematical Induction Non right-angled triangle trigonometry Permutations & Combinations Polynomials Properties of curves Quadratics Reasoning and proof Rules of differentiation Sequence and Series Techniques for Integration Transformation Trigonometric Functions Trigonometric identities and equations Maths Exploration (IA) ideas ibalasia Maths Books Yoga Pre-Diploma QuizzesShow sub menu Quiz 1 Quiz 2 Quiz 3 Quiz 4 Quiz 5 About Author Bals Mathematics Online Notes Home IBDP Math HL Binomial Theorem Binomial Theorem – Practice Questions Binomial Theorem – Practice Questions 10372 May 28, 2020 In the expansion of (a–3 b)n a–3 b n , the sum of 9 th and 10 th term is zero. Find the value of a b a b in terms of n n . If the coefficient of 4 th, 5 th and 6 th terms in the expansion of(1+x)n 1+x n are in arithmetic sequence, then find the value(s) of n n . If the last term in the expansion of(3−−√–1 3√)n 3–1 3 n is –log 2 81 2 3 4/–log 2 81 2 3 4 , find the value of n n . Find the remainder when 2 1003 2 1003 is divided by 7. (Non Calculator question) If x n–y n 2–√=(1–2–√)n x n–y n 2=1–2 n, then show that a.x n 2–2 y n 2=(–1)n x n 2–2 y n 2=–1 n (i) Find the first three terms in the expansion, in ascending powers of x x , of (1–2 x)5 1–2 x 5 . [2 marks] (ii) Given that the coefficient of x 2 x 2 in the expansion of (1+a x+2 x 2)(1–2 x)5 1+a x+2 x 2 1–2 x 5 is 12, find the value of the constant a a . [ 3 marks] Solution a) Use the binomial theorem to expand(a+b√)4 a+b 4 . b) Hence, deduce an expression in terms of a a and b b for (a+b√)4+(a–b√)4 a+b 4+a–b 4 . Solution Q8. Write down and simplify the general term in the binomial expansion of (2 x 2–d x 3)7 2 x 2–d x 3 7 , where d d is a constant. (b) Given that the coefficient of 1 x 1 x is −70 000, find the value of d d . Solution Sections Applications of differentiation Binomial Theorem Bivariate Statistics Circular measure Complex Numbers Continuous random variables Definite Integrals Differential Equations Discrete random variables Exponents and Logarithms Functions IBDP Maths – Exploration (IA) ideas Introduction to Integration Introduction to differential calculus Kinematics Limits Maclaurin Series Mathematical Induction Non right-angled triangle trigonometry Permutations & Combinations Polynomials Properties of curves Quadratics Reasoning and proof Sequence and Series Techniques for Integration Transformation Trigonometric Functions Trigonometric identities and equations Related Articles Notes – Binomial Theorem © 2025 IBDP Math HL/SL Powered by WordPress To the top ↑ Up ↑ Translate » Powered by Translate Original text Rate this translation Your feedback will be used to help improve Google Translate
188079	https://calculat.io/en/number/percent-of/10--75	Send You can also email us on infocalculat.io 10 percent of 75 What is % of Calculate What's 10 percent of 75? (Seven) 7.5 is 10% of 75 10 percent of 75 calculation explanation In order to calculate 10% of 75 let's write it as fractional equation. We have 75 = 100% and X = 10%. So our fraction will look like: Now we can solve our fraction by writing it as an equation: X = (75 × 10) ÷ 100 X = 750 ÷ 100 X = 7.5 Therefore, 10% of 75 is 7.5 Another way to solve our problem is to find the value of 1% of the number and then multiply it by the number of percent (10). To find 1% of a number 75 you need to divide it by 100: X = (75 ÷ 100) × 10 X = 0.75 × 10 X = 7.5 So we got the same result again: 7.5 Related Calculations See Also Percentage table \| Percent \| Value \| --- \| \| 1% of 75 \| 0.75 \| \| 2% of 75 \| 1.5 \| \| 3% of 75 \| 2.25 \| \| 4% of 75 \| 3 \| \| 5% of 75 \| 3.75 \| \| 6% of 75 \| 4.5 \| \| 7% of 75 \| 5.25 \| \| 8% of 75 \| 6 \| \| 9% of 75 \| 6.75 \| \| 10% of 75 \| 7.5 \| \| 11% of 75 \| 8.25 \| \| 12% of 75 \| 9 \| \| 13% of 75 \| 9.75 \| \| 14% of 75 \| 10.5 \| \| 15% of 75 \| 11.25 \| \| 16% of 75 \| 12 \| \| 17% of 75 \| 12.75 \| \| 18% of 75 \| 13.5 \| \| 19% of 75 \| 14.25 \| \| 20% of 75 \| 15 \| \| 21% of 75 \| 15.75 \| \| 22% of 75 \| 16.5 \| \| 23% of 75 \| 17.25 \| \| 24% of 75 \| 18 \| \| 25% of 75 \| 18.75 \| \| 26% of 75 \| 19.5 \| \| 27% of 75 \| 20.25 \| \| 28% of 75 \| 21 \| \| 29% of 75 \| 21.75 \| \| 30% of 75 \| 22.5 \| About "Percent of Number" Calculator "Percent of Number" Calculator What is % of Calculate FAQ What's 10 percent of 75? See Also
188080	https://www.facebook.com/groups/neildgtyson/posts/2621221154791477/	Neil deGrasse Tyson \| Water can boil and freeze at the same time \| Facebook Log In Log In Forgot Account? Neil deGrasse Tyson · Join Nathen Robertson · March 21, 2020 · Water can boil and freeze at the same time. Seriously, it's called the 'triple point', and it occurs when the temperature and pressure is just right for the three phases (gas, liquid, and solid) of a substance to coexist in thermodynamic equilibrium. All reactions: 1.3K 260 comments 261 shares Like Comment Share Most relevant Jerry Bryson My grandfather said once he tossed a pan of boiling water out the back door. It froze before it hit the ground and when he tried to pick it up, the ice was still hot 5y 6 Nancy Cole-Auguste Neither is God under any obligation to make sense to OUR little tiny brains. The evidence is awesome but The Essence is Unknowable 5y 39 View all 125 replies Michael Daris Kind of like time having three states at once: the future is now and the present is past...just like that, all in a snap. 5y See more on Facebook See more on Facebook Email or phone number Password Log In Forgot password? or Create new account
188081	https://www.moneymuseum.com/en/archive/cowry-shell-as-global-currency--25?&slbox=true	Cypraea moneta, a global money unit: fourthousand years of history of the Cowry Shell, c. 2000 BC to 1960 AD back Long before the first man on earth thought about striking a coin, cowries were used for payment. And at least in 1960 AD in some parts of the world cowries were still accepted as money. Come along with us on our journey through the world of money. Today we are stopping in... stop, where are we supposed to stop? The currency we are talking about today was widely used in Asia, Africa and Oceania, and some evidence suggests that it was also known in Russia and North America. Okay, let’s take a trip around the world and through time. As our currency was used for a good 4,000 years, from the end of the 3rd century BC to the second half of the 20th century AD. Yes, that used to be a currency and an extremely successful one too. The porcelain-like shell of the cowry circulated around the globe longer than any other currency in the history of money. The oldest written evidence of cowries being used as money was found in China. A ritual bronze vessel from around the turn of the 1st century BC, for instance, bears the inscription: “This precious vessel was made for Yuan, Count of Chu, who paid 14 cowry double strings for it” – that is 140 cowries. A lot of money at the time. We learn about the relative value of shell money from another inscription from the year 925 BC: “Citizen Ju Bai received from Qiu Wei one jade tablet worth 80 double strings of cowry for six fields of his lands.” So, 800 cowries were worth a jade tablet or six fields. No wonder the Chinese soon started to make cowry imitations from bone or bronze. This is how the cowry became the model for one of the oldest Chinese currencies, the so-called ant nose money, which circulated in the Southern Chinese Kingdom Chu around the time between the 3rd and the 5th centuries BC. Evidence of the immense importance of the cowry in the history of Chinese coinage is the fact that our modern day symbol for money has its origin in the stylized image of a cowry. The shell of the cowry, also known as “porcellana” in Old Italian, was highly coveted. Her Latin name – Cypraea moneta – indicates that the shell served as money for centuries. The species was native to tropical and subtropical coral reefs from the Red Sea in the West to Northern Australia, Japan, Hawaii and the Galapagos Islands. So it shouldn’t come as a surprise that cowries were a means of payment in so many different parts of the world. This is for instance a jetak, a plaited string decorated with cowries and other precious objects which was used by the Dani people in New Guinea to determine and to pay for the exact price of a pig. This cowry string is originally from Africa. Cowries were exported to Africa since the 14th century, first by Arabic, then by English and even by German merchants. This cowry imitation from stone, too, has its origin in Africa. Today, the coins and banknotes of several states still refer to this ancient currency as this banknote from the Maldives shows. By the way, even cowries weren’t safe from the danger of inflation. Probably the biggest devaluation of the porcelain-like shell was caused by the trade company Adolf Jakob Hertz & Söhne in Hamburg. The company’s fast ships transported cheap cowries from Africa’s East Coast to its West Coast. The business boomed. The profit margin was 400 per cent. This went well for a while because the Bornu Empire in Central Sudan introduced the cowry currency roughly at the same time. But when more cowries kept being imported to West Africa although eventually demand was declining, inflation hit. Oh yes, and from time to time people still tell the curious tale of Dutch explorers in the 1960s who couldn’t continue their expedition in the middle of New Guinea because they ran out of cowries and the carriers refused to accept a currency other than that. Thank you for listening. And you can find more podcasts about coins and money on the Sunflower Foundation Web page.
188082	https://math.stackexchange.com/questions/412563/determine-if-vectors-are-linearly-independent	Skip to main content Determine if vectors are linearly independent Ask Question Asked Modified 1 year, 4 months ago Viewed 589k times This question shows research effort; it is useful and clear 56 Save this question. Show activity on this post. Determine if the following set of vectors is linearly independent: ⎡⎣⎢220⎤⎦⎥,⎡⎣⎢1−11⎤⎦⎥,⎡⎣⎢42−2⎤⎦⎥ I've done the following system of equations, and I think I did it right... It's been such a long time since I did this sort of thing... Assume the following: a⎡⎣⎢220⎤⎦⎥+b⎡⎣⎢1−11⎤⎦⎥+c⎡⎣⎢42−2⎤⎦⎥=⎡⎣⎢000⎤⎦⎥ Determine if a=b=c=0: 2a+b+4c2a−b+2cb−2c=0=0=0(1)(2)(3) Subtract (2) from (1): b+cb−2c=0=0(4)(5) Substitute (5) into (4), we get c=0. So now what do I do with this fact? I'm tempted to say that only c=0, and a and b can be something else, but I don't trust that my intuition is right. linear-algebra vectors Share CC BY-SA 4.0 Follow this question to receive notifications edited Sep 3, 2018 at 14:20 amWhy 211k197197 gold badges282282 silver badges503503 bronze badges asked Jun 6, 2013 at 1:39 MirranaMirrana 9,3193636 gold badges8787 silver badges127127 bronze badges 5 4 If c=0 then you must have b=0 and then you must have a=0. Hence they are linearly independent. – copper.hat Commented Jun 6, 2013 at 1:42 1 From c=0 and b−2c=0 you can conclude? And then what about a? You were doing fine. The same thing, with less writing, can be done using row reduction. – André Nicolas Commented Jun 6, 2013 at 1:42 substitute c=0 back into (4) or (5) to show that b=0 and then both b=0 and c=0 into (1) or (2) to show that a=0. By definition they are then linearly independent. – Tpofofn Commented Jun 6, 2013 at 1:43 @AndréNicolas I'm only starting to learn about matrices now... this was taught in this method so I presume this is how I have to do it on the assignment. – Mirrana Commented Jun 6, 2013 at 1:50 Yes, at the beginning it makes sense do do things directly from the definition. – André Nicolas Commented Jun 6, 2013 at 1:51 Add a comment \| 4 Answers 4 Reset to default This answer is useful 52 Save this answer. Show activity on this post. You just stopped too early: Since you have 3 varibles with 3 equations, you can simply obtain a,b,c by substituting c=0 back into the two equations: From equation (3), c=0⟹b=0. With b=0,c=0 substituted into equation (1) or (2), b=c=0⟹a=0. So in the end, since a⎡⎣⎢220⎤⎦⎥+b⎡⎣⎢1−11⎤⎦⎥+c⎡⎣⎢42−2⎤⎦⎥=⎡⎣⎢000⎤⎦⎥⟹a=b=c=0, the vectors are linearly independent, based on the definition(shown below). The list of vectors is said to be linearly independent if the only c1,...,cn solving the equation 0=c1v1+...+cnvn are c1=c2=...=cn=0. You could have, similarly, constructed a 3×3 matrix M with the three given vectors as its columns, and computed the determinant of M. Why would this help? Because we know that if detM≠0, the given vectors are linearly independent. (However, this method applies only when the number of vectors is equal to the dimension of the Euclidean space.) M=⎡⎣⎢2201−1142−2⎤⎦⎥ detM=12≠0⟹linear independence of the columns. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Mar 28, 2020 at 19:32 answered Jun 6, 2013 at 1:47 amWhyamWhy 211k197197 gold badges282282 silver badges503503 bronze badges 0 Add a comment \| This answer is useful 12 Save this answer. Show activity on this post. you can take the vectors to form a matrix and check its determinant. If the determinant is non zero, then the vectors are linearly independent. Otherwise, they are linearly dependent. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Jun 6, 2013 at 1:50 WallaceWallace 12122 bronze badges Add a comment \| This answer is useful 5 Save this answer. Show activity on this post. just as simple,make these three vectors to be a matrix,as follows: ``` 2 2 0 1 -1 1 4 2 -2 ``` and then change it to its row-echelon form,you can get the rank of this matrix. its rank is 3,so the three vectors are linearly independent. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Oct 23, 2017 at 2:42 PeakerPeaker 6111 silver badge22 bronze badges 0 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Firstly, you are to arrange the vectors in a matrix form the reduce them to a row-reduced echelon form. (If the last row becomes all zeros then it is linearly dependent, but if the last row isn't all zeros then it is linearly independent). Let's get to it now. Arranging the vectors in matrix form we have : ⎡⎣⎢2142−1201−2⎤⎦⎥ After the first reduction we get: ⎡⎣⎢20022200−2⎤⎦⎥ Then, after the third reduction which is the last we have: ⎡⎣⎢20022000−2⎤⎦⎥ Since the last row isn't equal to all zeros after the reduction then the vectors are linearly independent. You're welcome. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Mar 31, 2024 at 11:35 ShivCK 11566 bronze badges answered Jan 15, 2020 at 21:40 Adewole OlumideAdewole Olumide 11 2 1 Please use mathjax formatting for the matrices. – ViktorStein Commented Jan 15, 2020 at 23:22 This is not always correct. See youtu.be/SOzO9EcQdQc for an example where the last row is all 0s, but the vectors are linearly independent. – Gigi Bayte 2 Commented Feb 8, 2020 at 17:36 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions linear-algebra vectors See similar questions with these tags. Featured on Meta Upcoming initiatives on Stack Overflow and across the Stack Exchange network... Community help needed to clean up goo.gl links (by August 25) Linked 1 Contradictory result when testing Linear independence using Gaussian elimination Related 1 Determine variable vector compontent so vectors are independent 2 How does this reduced matrix indicate that the vectors are linearly independent? 0 checking if matrix columns are linearly independent 3 Determine if the set of vectors are linearly independent or linearly dependent 1 Are the vectors linearly dependent or independent? 0 Check if three row vectors are linearly dependent or independent 1 Calculating the reduced row echelon form to find the maximum set of linearly independent vectors. Hot Network Questions How can I explain 15 years no work history on my resume as a NEET? 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188083	https://www.youtube.com/watch?v=FiFwxoxOmNo	ImageJ Analysis: Length Measurement, Area Measurement and Thresholding SMS TechEdu 2810 subscribers 4329 likes Description 633896 views Posted: 3 Oct 2017 In this ImageJ tutorial basic analysis of any image like length and area measurement are demonstrated both by manual and thresholding process. Instructor: Santanu Mandal Like \| Comment \| Share \| Subscribe 162 comments Transcript: Hello friends welcome to our technical tutorial by Team SMS In this tutorial I will demonstrate how to use ImageJ software to analyze any distance or area by thresholding or manually. ImageJ is an open source image processing program, which is highly extensible and With its thousands of plugins it can perform a wide variety of tasks and that's why it is a very useful tool for the scientific community. It is a Java based program, that is developed at National Institutes of Health Before starting the demonstration one thing to mention here is that The first software that is opened here is a basic ImageJ, that is downloaded from National Institutes of Health website Which contains plugins that is sufficient to do basic analysis of any image But if you want to go to That is sufficient to do basic analysis of any image, but if you want to go to sophisticated one or you want to get more plugins then you can choose the Fiji ImageJ that is also a version of ImageJ, and it is also a open source software. Which contains lot of plugins comparing to basic ImageJ in this case I will be demonstrating the basic analysis like distance between particles and area that's why the basic Software is preferred one because it is lighter to use Let's use the basic one only. This is the image located in my desktop I want to analysis. Let's open it You cam simply drag the image to the software or you can go to file then open it by selecting from where it is located Here the image comes It is an image of Nickel particles in silica Matrix. where the white particles are representing as Nikel particles If you observe carefully you will see it is written RGB at the top of the image Before doing any analysis we need to convert it into 8-Bit mode Go to Image---> type and then click on 8-Bit. Now it is converted to 8 bit image. But, before going to analysis the system does not know about the scale of the image that's why the image need to be calibrated or the scale should be set. How to set the scale Select the magnifying glass option here and click on the scale bar Left click will zoom it in and right click zoom it out Zoom it in till it is sufficient and then select the straight line. Draw a straight line over the scale bar When you will try to draw a straight line you will observe it very difficult to draw a perfect straight line. Hence to draw it as a perfect one you can press the shift key in the keyboard and then draw it You will find that now it is moving in 45 in 45 degree angle irrespective of the cursor or mouse location Once The straight line is drawn go to analyze an then set scale The dialog box appears. It's showing the pixel 101.25 that is the selected one. Known distance it is showing as 0,but actually is 300nm Let's write it as 300 and unit of lenght is nm Now our image is calibrated and ready for any analysis First I will show how to measure the inter particle distance Let's zoom out little bit. and then select the straight line and draw straight line in between the Centers of two particles and Then click on analyse and then measure Dialog box with result appears showing the length as 160 nanometer Here you see click on "Edit" then "Fill" If you observe one straight line with grey colour appeared here We need to do it multiple times to get an average of the interparticle Distance Let's do it another time Once we have done it; the first reading is already came, for the next reading we can use the shortcut key to save time like ctrl M to measure the second line appears with the length at 161 and Control+F to fill in Why we are filling it because when we will be doing it for multiple times to get an average value of the length there will be a chance that we will be taking a particular distance twice We don't want to select a particular distance twice that why we are filling it To make it marked Let's do it for some more times Draw straight line , then Control M and Control F third line appears. Do it for another one Control M Control F. Once you will feel that you have taken sufficient data Either Click on "Result" and then click on "Summarize" you will get the mean standard deviation value here 138 is a mean value and standard deviation is 43 or else you can do is that by simply saving as Excel file Excel file appears here where that length value is showing These value you can utilize to measure the mean or average or standard deviation. Thus we can measure any distance using ImageJ Now I will show how to measure any area Let's open the image once again We need to convert it to a 8-bit image; it is converted Lts zoom it little bit and then and then select the area that you want to analyze Draw a straight line above the scale bar by pressing Shift key Go to analysis. Set scale. Known distance is 300nm when it is caliberated. WE can analysis area. So there are two methods by which we can measure the area. One is using thresholding and another is manually. First I will describe the thresholding Before we start the thresholding We need to select the Specific area of this image that we want to analyze To select it click on rectangular tab here Area and then select the area that we want to analyze Why we are doing it here!! Because All the area of the image is not perfect; like here you see it is written (B) 300 nanometer as it is a software it will consider this values also when it will calculate the area That's why the particular area, need to be selected. Once that you selected go to image and duplicate it click ok Selected area comes here separately Now we will do thresholding. Go to image then adjust and then threshold See red one here, and then if you observe When I am scrolling the power Red colour is coming more and more the red colour is representing the selected area I want to select the particle that I need to analyse area white particles need to recovered wholly the red mark. Let's scroll it till all the particles are covered by red mark and then apply it you see here, it is converted into Specific image where the background is eliminated and all the particles are visible Now close it and now go to analyze and then analyze particle Here one important thing to mention is that you want to include this Particles that is located at the edges or not That depend on you if you want to exclude those particles click on this This box else leave it Click this box also to display results, clear result and add to manager Another thing is there is that "Size", "Circularity" and "Show". There are a lot of options are available we can do nothing outlines "Bare Outlines" etc. We will select only "outlines" here Circularity: circularity actually a range It is showing the type of the circle one means full circle and zero means it is not a circle at all I want in this case to select all the (types) images So the range would be 0 to 1 let it be like that. Here it is showing the size value Is also like a range It can be started from 0 to infinity So let's try By using 0 to infinity and click on OK You will see that it's selected 84 particles here And some perticals are excluded which are located at the boundary Let's do one more time to make it clear Go to "Analysis" analyze particle this time I will put it as 2000 you will see that lot of particles inside also is excluded that is happening due to the Size that we have taken as 2000. Because it is setting a limit above which only Selecting the particles so it's depend on you, Which particles you want to select. in this case this Selection is looking good so or we can decrease it little bit more little bit more So that we can get Some more particles, go to "Analyze Particle" and then I can change it to 1000 click OK Here it is showing some reasonable one with having 65 particles That area is written here, you can directly get the mean and standard dev value by clicking on the "Results" and then "Summarize" Mean value is this one 128.51 nanometer Square standard deviation 1217 nm cell and You can also save it as excel file. You see "test 2" appears here Inside the file all the area value is given of all 65 particles You can select these values and Draw a histogram also You can get the mean value using the Excel functions or you can calculate the standard deviation You can calculate radius and diameter of the particles also and you can get the histogram of diameter or radius of the particles also You can draw the histogram of the area Using origin Paste the values in the origin and then go to plot and then Statistic-->histogram you will get a nice histogram of the values with the frequency values in Y axis and particle area in the x-axis That you can do using Origin or using Excel Till now I have explained how to use thresholding To measure area of the particles If you look carefully, then you will observe that in these case The example I have taken where the contrast is very well clear Contrast in the scence particle boundary or the grain boundary is very well distinguished and therefore thresholding is easier But sometimes it happens that this boundary is not clearly visible and in that case We need to do it by manually how to do it manually!!! That I will so now is first select the portion that do you want to Measure then duplicate it This is the portion is selected. Now zoom it in And then select the freehand selection, like we have selected straight line in earlier case freehand selection this time Let's draw around the particle And area and then click on the "Analysis" tab and "Measure". It's showing with 5117nm square area Of this particle.Go to edit and then fill it so that the chance to select it another time will be eliminated when we will be doing it for multiple times To get the average value for multiple grains/particles. Let's do it another time now Once it is Done the first data Is appeared here Now you can use the shortcut key Control M You see The value comes here and Ctrl F to fill the selected particle. Do it for another time Control M and Control F Control M and control F. Do it for multiple times like 50-60 times that seems to be good one to take an average of the particle, and it's depends on the distribution of the particles also If the distribution of the particles is not that much scattered then it will not be required to take a lot of data, but If the distribution is scattered very much, then it will be required to take a lot of values like hundred, two hundred also need to be taken And Obviously it will be time-consuming process So I have demonstrated both the thresholding and manually Process to measure the area If you observe the tabs here under the "image" and the "process" there are a lot of options like smooth, sharpen and noise, add noise, Binary, FFT, Inverse FFT, Bandpass filter. Lots of filters are available The purpose of these functions is to convert any image, that is not of Good contrast; to change it into a threshold-able image In our coming tutorials I will be demonstrating is functions the threshold is automatically taking all the taking care of all the particles one at a time Thank you for watching our tutorial Like Comment and Share and Subscribe our channel for our upcoming tutorials Thank you!!!!!!!!
188084	https://jbioleng.biomedcentral.com/articles/10.1186/s13036-024-00460-1	Typesetting math: 100% Skip to main content Journal of Biological Engineering Download PDF Research Open access Published: Dynamic control of the plasmid copy number maintained without antibiotics in Escherichia coli Geunyung Park1, Jina Yang2 & Sang Woo Seo1,3,4,5,6 Journal of Biological Engineering volume 18, Article number: 71 (2024) Cite this article 2052 Accesses 1 Citations Metrics details Abstract Background Manipulating the gene expression is the key strategy to optimize the metabolic flux. Not only transcription, translation, and post-translation level control, but also the dynamic plasmid copy number (PCN) control has been studied. The dynamic PCN control systems that have been developed to date are based on the understanding of origin replication mechanisms, which limits their application to specific origins of replication and requires the use of antibiotics for plasmid maintenance. In this study, we developed a dynamic PCN control system for Escherichia coli that is maintained without antibiotics. This is achieved by regulating the transcription level of the translation initiation factor IF-1 (infA), an essential gene encoded on the plasmid, while deleting it from the plasmid-bearing host cell. Results When validated using GFP as a reporter protein, our system demonstrated a 22-fold dynamic range in PCN within the CloDF13 origin. The system was employed to determine the optimal copy number of the plasmid carrying the cad gene, which converts an intermediate of the tricarboxylic acid cycle (TCA cycle) to itaconic acid. By optimizing the PCN, we could achieve an itaconic acid titer of 3 g/L, which is 5.3-fold higher than the control strain. Conclusions Our system offers a strategy to identify the optimal expression level of genes that have a competitive relationship with metabolic pathways crucial for the growth of the host organism. This approach can potentially be applied to other bacterial hosts by substituting the sensing module or the essential gene. Background Overexpressing the genes in the synthesis pathway is a common strategy to obtain a high titer of a target metabolite. However, each metabolite producing gene has its optimal expression level for the best production. Overexpression of genes does not always guarantee the expected result. For example, the overexpression of genes could lead to growth inhibition or toxicity in the host bacteria [1, 2]. Therefore, the optimization of gene expression levels in metabolic engineering is important, and this has been conducted at transcriptional, translational, or even post-translational levels. At the transcriptional level, gene expression can be modulated by replacing a promoter with a static promoter or inducible promoter or by using the CRISPRi system [3,4,5,6,7,8]. To modulate protein translation efficiency, ribosome binding sites with appropriate strength can be designed or selected from a library [9,10,11]. Alternatively, translation can be down regulated using antisense RNA , small RNA [13, 14], or Tl-CRISPRi . The amount of protein can also be modulated at the post-translational level. Degradation tags such as ssrA and Pup, or degrons have been used in conjunction with their partner proteases to enhance the degradation of target proteins [16,17,18]. Recent studies have shown that by altering the components of the plasmid origin of replication, gene dosage can be regulated through dynamic control of plasmid copy number (PCN) [19,20,21]. This approach reduces the requirement for multiple cloning steps to determine the optimal expression levels of the genes residing in the same plasmid. By substituting the promoter of the plasmid replication initiation factor with one regulated by cuminic acid to regulate the inhibitory transcription factor of the promoter, cuminic acid-dependent PCN change was enabled . In the case of pUC19 origin, the PCN was manipulated by tuning the priming and inhibitory RNA working on the replication origin, and the system was applied to the violacein production . In addition, the PCN within ColE1 and p15A origin could be regulated in two plasmids system in which one plasmid regulated the transcription of either priming or inhibitory RNA of the origin on the other plasmid . However, all the aforementioned static or dynamic gene expression systems still necessitate the use of antibiotics for plasmid maintenance. The use of antibiotics in metabolite production has several disadvantages. Plasmid maintenance with antibiotics can lead to the emergence of antibiotic-resistant organisms and cause PCN heterogeneity . Especially when producing compounds intended for human contact or ingestion, reducing antibiotics during the production process is preferable to ensure safety. Previous studies demonstrated the stable maintenance of plasmids without antibiotics by applying toxin-antitoxin mechanism [24, 25], auxotrophic or essential gene complementation [26, 27], or repressor titration . Especially, STAPL system that maintains plasmids within cells without using antibiotics by relocating essential gene, translation initiation factor IF-1 (infA), to the plasmid and removing the gene from the chromosomes was developed . The study also discovered that the PCN shifts in response to static changes in infA transcription levels . The PCN and expression levels of the essential gene exhibited an inverse correlation. In this study, we developed a dynamic PCN controlling system maintained without antibiotics by dynamically modulating the transcription level of a plasmid-relocated essential gene, infA. The transcription level of infA is controlled by an inverting genetic circuit in which anhydrotetracycline (aTc) addition leads to the reduction in infA expression level. The system was designed to maintain maximum infA expression during the cloning step, keeping PCN low. For the production of target metabolite, PCN can be upregulated by adding an inducer to reduce infA expression. The developed system was applied to the optimization of itaconic acid production. Methods Strains, plasmids, primers, genetic manipulation Mach-T1R was used for the construction of plasmids used in this study. Otherwise, MG1655 was used. The strains and plasmids used in the study are listed in Table S1. The primers used for the construction of plasmids are listed in Table S2. Gene parts for tetR (TetR-HpaI-R, phlF-TetR-R), phlF (phlF-TetR-F, phlF-R1) (phlF-TetR-F, phlF-R2), gfp (opt-gfp-R, opt-gfp-F) (opt-gfp-R, PphlF-phlF-F), infA(InfA- PphlF -F, InfA-PstI-R) (InfA-PstI-R, PphlF -phlF-F), were amplified once or twice using PCR and assembled using NEBuilder HiFi DNA Assembly kit (New England Biolabs, Ipswich, MA, USA) with Bsu36I, HpaI cut pCDF-Duet vector. This vector was used as a template and modified accordingly for other plasmids used in this study. For the tight regulation of the dynamic control genetic circuit, BBa_B1006 (90% efficiency) was chosen as the terminator for phlF. In addition, the RBS of phlF was set to the expression level around 40,000 using the UTR designer . In order to knock out the infA gene on the chromosome, a plasmid containing infA and the induction module (tetR and phlF with tet promoter, infA with phlF promoter) was electroporated to MG1655_pSIM5. The editing template containing the kanamycin resistance gene and FRT sequences was prepared by PCR with primers (D-infA-F2, D-infA-B2) and electroporated to the lambda red competent cell. The cells undergone successful recombination were selected on LB plates containing spectinomycin and kanamycin. Specific fluorescence measurement Modified M9 medium (47.8 mM Na2HPO4, 22.0 mM KH2PO4, 18.7 mM NH4Cl, 8.6 mM NaCl, 2 mM MgSO4, and 0.1 mM CaCl2) containing 4 g/L glucose and 7.5 g/L casamino acids was used. The regulation range of phlF promoter in E. coli DCP strain (Supplementary Table. S1) was measured with Sense (Hidex, Turku, Finland). Overnight incubation was done in 5 mL medium in 100 mL test tube at 37 °C, 250 rpm. The incubated cells were diluted to OD600 0.1 and grown until OD600 exceeds 1.0. The cells were inoculated to OD600 0.05. When OD600 reached around 0.3, aTc was added to the final concentration of 0.01, 0.1, 1, 2. 3, 5, 10, 50, 100, 500, and 1000 ng/mL. The specific fluorescence was measured by the dividing fluorescence value by the measured OD600 at 10 h after aTc addition. The GFP measurement in our dynamic control of the PCN system was conducted in E. coli DCG strain (Supplementary Table. S1) using S3e Cell Sorter (Bio-Rad, California, USA). Overnight inoculation was prepared in 5 mL medium in a 100 mL test tube. The incubated cells were refreshed to OD600 0.1, grown to exponential phase, and inoculated to OD600 0.05. When OD600 reached around 0.3, aTc was added to the final concentrations of 0.05, 0.5, 5, 50 ng/mL in 5 mL. After 10 h of incubation from the aTc addition, the specific fluorescence was directly measured using flow cytometry. Quantification of the PCN The copy number was quantified using quantitative PCR (qPCR) following a previously used method and modified accordingly . Accupower 2X greenstar qPCR Master Mix (Bioneer, Daejon, Republic of Korea) was used for the reaction mixture. StepOnePlus Real-time PCR system (Thermo Fisher Scientific, Waltham, MA, USA) was used. For the standard curve development, MG1655 gDNA and pCDF-Duet vector were prepared. The rpoA flanking region was amplified with Flank_rpoA_eco_F and Flank_rpoA_eco_R, and was used as the gDNA template. The concentrations of pCDF-Duet vector and rpoA flanking region of gDNA were measured using Qubit assay. The primers qPCR_cloDF13_F, qPCR_cloDF13_R and qPCR_rpoA_eco_F, qPCR_rpoA_eco_R were used respectively for serially diluted pCDF-Duet and rpoA flanking region (Supplementary Fig. S4). Cell cultures diluted in DDW were boiled at 95 °C for 10 min and diluted to OD600 0.01. Each sample was amplified using pCDF-Duet vector primers and gDNA primers. PCN was calculated by dividing the plasmid molecules by gDNA molecules. Itaconic acid production The strain used for the production of itaconic acid (DCI (Supplementary Table. S1)) was grown in an M9 medium supplemented with 2 g/L yeast extract, 10 mL/L ATCC trace mineral solution, and 20 g/L glucose at 30 °C, 200 rpm. Kanamycin (50 g/mL) was used to prevent contamination. Each seed was inoculated in a 5 mL medium contained in a 100 mL test tube. Overnight cultured cells were diluted to OD600 0.1 in a 10 mL medium contained in a 300 mL flask. The cells were inoculated to OD600 0.05 in 25 mL of medium in a 300 mL flask, and aTc (final concentration: 0.05, 0.5, 5, 50, 500 ng/mL) was added when OD600 reached approximately 0.3. The pH titration was conducted every 6 h after the aTc addition, to approximately pH 7.0. All samples were prepared in biological triplicates. HPLC analysis For the quantitative analysis of itaconic acid, the supernatant of the cell culture was filtered and analyzed using HPLC (Shimadzu, Kyoto, Japan) with cation-exchange Aminex HPX-87 H column (Biorad, California, USA) . 5 mM sulfuric acid was used as the mobile phase. The flow rate was 0.6 mL/min and the oven temperature was maintained at 14 °C. Glucose was detected with a refractive index detector, and acetic acid and itaconic acid were detected with a UV–vis diode array detector (at 210 nm). For HPLC standards, itaconic acid was purchased from TCI America, glucose was purchased from Acros Organics, and acetic acid was purchased from Supelco. Results Design of dynamic PCN control system A genetic circuit which enables the dynamic PCN control within an origin of replication was devised (Fig. 1). The previous study discovered that the static change in the transcription level of infA led to the alteration in the PCN . According to the study, the PCN was inversely proportional to the transcription level of infA. For metabolite production, the expression level of the corresponding gene is usually maintained low during the cloning process and is induced during the production stage to minimize the burden posed on the cell. Therefore, we designed a system that keeps PCN low without an inducer and increases PCN at specific time points with the addition of an inducer. We employed an inverted induction module by which the inducer addition lowers the infA expression level, thereby upregulating PCN. Therefore, we combined two activating inducible systems to reverse the signal of the induction and repress infA expression. For future applications, it is sufficient to change only the primary sensing module while keeping the infA promoter unchanged. This approach will relieve us of the need for repeated optimization and validation of the infA promoter each time. We selected a regulatory system consisting of a PhlF promoter (PphlF) and tetA promoter (PtetA), in which the PhlF repressor is controlled by the PtetA and repressed by the TetR (Fig. 1). The addition of aTc activates the expression of phlF. Subsequently, PphlF driven infA expression is repressed . There is a strong correlation between infA and cell growth; not only is the transcription of infA associated with cell growth [33, 34], but infA is also essential for cell viability . As a result, a decrease in infA expression leads to an increase in PCN to maintain the necessary levels of infA. Since infA is a critical gene, insufficient expression hampers cell growth, causing an accumulation of plasmids within the cell. Conversely, since infA expression is not repressed in the absence of aTc, cells are expected to retain enough amount of translation initiation factor to survive without increasing the PCN (Fig. 1). Validation of the designed dynamic PCN control system The inverted induction module was first tested in DCP. The regulation range of PphlF by the addition of aTc was tested by placing gfp under the control of the promoter (Fig. 2A). The specific fluorescence was measured to deduce the regulation range of the inverted induction module. As the concentration of aTc increased, the specific fluorescence exhibited a marked decline beginning at 50 ng/mL, showing a 24-fold reduction. At higher concentrations of aTc, the specific fluorescence remained consistent, with the maximum reduction reaching 26-fold (Fig. 2B). However, cell growth was inhibited above 500 ng/mL aTc, which is known to have cytotoxicity (Fig. 2C). To avoid reducing cell growth by more than 25%, aTc concentrations below 500 ng/mL were used in the following experiments. Next, we examined how the PCN changed in response to the infA expression level (Fig. 2D). The PCN of DCG was indirectly measured by the constitutively expressed gfp. The specific fluorescence was measured by flow cytometry as it segregates each cell into individual droplets and yields a precise fluorescence each cell possesses . The control strain (SCG) was cultured in spectinomycin containing condition, as we aimed to compare our system to the conventional plasmid maintaining system. The specific fluorescence elevated as the concentration of aTc increased, which indirectly indicated the increase in the PCN. The DCG strain showed a 6.5-fold regulation range of specific fluorescence compared to the control strain, when aTc 500 ng/mL was added (Fig. 2E, Supplementary Fig. S1). The qPCR result demonstrated the actual PCN had a 22-fold regulation range compared to the control strain when aTc 50 ng/mL was added (Fig. 2F). The regulation range of PCN (9-196 copies, 22-fold) is much broader compared to the reported copy number range of CloDF13 (20–40 copies, 2-fold) . Therefore, it was confirmed that adjusting the expression level of infA can dynamically regulate PCN, thereby enabling control over the expression level of the target protein. The dynamic range of the specific fluorescence was less than that of the PCN. A plausible reason for this phenomenon would be the role of infA, which is a translation initiation factor. Too low infA expression due to the addition of a high concentration of aTc might impair the efficiency of translation initiation in the cell. This could be improved in further studies to encompass a wider range of target protein expression levels. Although the specific fluorescence was highest at 500 ng/mL of aTc, the PCN was lower than that at 50 ng/mL of aTc. PphlF was effectively turned off at aTc 50 ng/mL (Fig. 2B). The deviation from the expected increase in PCN at aTc 500 ng/mL may have resulted because this concentration exceeded the regulatory range. One hypothesis is that the repressor of primer (Rop) protein, which is encoded on a dynamically regulated plasmid, accumulates beyond a certain level like GFP. This may enhance the formation of kissing complexes or repress the transcription of RNAII thereby reducing the PCN. Yet, we are not certain of the exact mechanism. The major reason for the cell growth inhibition (Fig. 2G) was considered to be infA expression level reduction as the reduction in cell growth was two times greater than the aTc cytotoxicity effect (Fig. 2C). The real-time quantitative reverse transcription PCR (qRT-PCR) results indicated that cells maintained infA mRNA levels up to 5 ng/mL of aTc, but these levels decreased to 60% at 50 ng/mL (Fig. S2A), likely contributing to growth inhibition. Slight growth inhibition at 5 ng/mL aTc was due to increased PCN to restore infA levels. The expression level of the infA transcript from each plasmid (Fig. S2B) corresponded with the relative fluorescence intensity of DCP (Fig. 2B). This demonstrates that variations in infA expression levels affected both PCN and cell growth. Overall, the genetic circuit consisting of two inducible promoters allowed the adjustment of PCN by modulating the infA expression level through simple changes in inducer concentration. Application of the dynamic PCN control system in metabolite production Dynamic PCN control system can be applied to determine the optimal PCN for metabolite production. The optimal expression level of enzymes is often difficult to predict, depending on the metabolic pathway. A specific expression level, which may include low expression of the target gene, often results in higher titer [2, 39, 40]. The PCN control system minimizes the need for redundant cloning typically required to adjust PCNs, thereby simplifying the process of optimizing the expression level of the target gene. This optimization is crucial when the target gene competes with the metabolic pathways related to cell growth in the host bacteria. The system can be applied not only to cell growth-related pathways but also to other metabolic pathways that are difficult to delete or engineer to reduce flux. This approach ensures the efficient balance of gene expression and metabolic activities necessary for optimal bacterial function and chemical production. Additionally, our developed system can incorporate the optimization of PCN when there is a discrepancy between PCN and protein expression levels. We applied our system for itaconic acid production, a platform chemical for high-value products such as plastics and latex [41, 42]. The synthesis pathway of itaconic acid competes with the tricarboxylic acid (TCA) cycle through cis-aconitate. Cis-aconitate is a substrate for cis-aconitate decarboxylase (CAD) and is also an intermediate in the TCA cycle. Therefore, the optimal flux distribution is crucial to achieve a high titer of itaconic acid. The CAD, which enables the conversion from cis-aconitate to itaconic acid, was expressed under the control of medium strength promoter J23106 to avoid cellular burden. The gene was integrated into the plasmid, which included a dynamic PCN control cassette. Subsequently, the infA gene on the chromosome was deleted to yield the strain, DCI (Fig. 3A). Increase in the titers of acetate and itaconic acid was observed throughout the cultivation (Supplementary Fig. 3). The itaconic acid titer was not proportional to the pattern of PCN. However, we were able to determine the optimal copy number of DCI for maximizing itaconic acid production. The itaconic acid titer from the DCI strain was 5.3 times greater than that of the control strain (SCI). When compared to the STAPL system in the previous study, our DCI achieved 3.4 fold itaconic acid titer (Supplementary Figure. S3) . This significant increase was observed when the PCN ranged from 45 to 80, with the concentration of aTc maintained between 0.05 ng/mL and 5 ng/mL (Fig. 3B and C). To analyze the expression of CAD, we performed SDS-PAGE (Supplementary Figure S5). The expression of CAD mirrored the itaconic acid titer rather than the PCN, even though other proteins on the same plasmid (SmR, PhlF, and TetR) appeared to exhibit increased expression levels in accordance with the PCN (Supplementary Figure S5 and Fig. 3C). This result is consistent with previous literature demonstrating that an increase in CAD expression leads to an increased itaconic acid titer . We hypothesize that CAD expression is hindered by an increase in PCN, as it is linked to cell growth and competes in the TCA cycle, resulting in lower expression levels due to cellular burden. Although we do not fully understand the underlying mechanism, enzyme expression decrease at high plasmid copy number has been reported . The culture conditions such as media composition, volume, and temperature differed between the production of GFP and itaconic acid. The trends in cell growth in response to aTc addition were analyzed . When comparing the PCN in Figs. 2F and 3C, the control strain and the DCI strains at aTc concentrations of 0.05 and 0.5 ng/mL showed significantly greater PCN in DCI than in DCG. In contrast, at aTc concentrations of 5, 50, and 500 ng/mL, the PCN of DCI was similar to or slightly lower than that of DCG. This indicates that infA repression was comparatively less stringent at aTc concentrations of 5, 50, and 500 ng/mL. Consequently, the growth curves for DCI at 5, 50, and 500 ng/mL appeared to be elevated compared to those of DCG. To sum up, the dynamic PCN control system enabled the determination of the optimal expression of cad, which competes with the TCA cycle. When stronger CAD promoters are utilized or when CAD is colocalized with aconitase (Acn) , the titer of itaconic acid is expected to increase. This occurs as CAD fully redirects the metabolic flux from the TCA cycle toward the production of itaconic acid. The system will ease the prediction of optimal gene expression in metabolite production. Discussion Recently, several studies regarding the dynamic PCN control system have been developed [19,20,21]. These studies modulated the genetic parts involved in the mechanism of replication in the origin. In this study, we developed a dynamic PCN control system that is not restricted to a specific origin of replication. Instead, it may be universally applied by regulating PCN through the dynamic control of the transcriptional expression level of the essential gene, infA. The ColE1 origin of replication including CloDF13 is primarily controlled by the RNAII to RNAI ratio. RNAII initiates replication by binding to the replication start site, while RNAI inhibits this process by forming a complex with RNAII, assisted by the Rop protein. ColE1-like origins can experience runaway replication if the RNAII/RNAI ratio is disrupted . We hypothesized that reduced infA expression may decrease Rop levels, leading to increased PCN. Conversely, an excess of Rop could enhance complex formation, repress RNAII, and reduce PCN. The study observed PCN changes with dynamic alterations in infA expression, indicating that while infA does not directly interfere with the replication mechanism of CloDF13 origin, it influences PCN. Another hypothesis is that a decrease in infA expression leads to an increase in PCN to maintain essential protein levels. Insufficient expression of an essential gene hinders cell growth, resulting in an accumulation of plasmids within the cell. The qRT-PCR results support our concept: reducing infA expression from individual promoters led cells to increase the PCN to restore intracellular infA mRNA levels (Supplementary Figures S2A and S2B). The previous study has succeeded in stable plasmid maintenance without antibiotics in CloDF13 origin and pMB1 origin. This strategy may extend to other ColE1-like origins with similar replication mechanisms. Given that programmed infA reduction upregulates PCN, this system may be particularly more applicable for origins with medium or low copy numbers. As a member of the translation initiation factors, excessively reducing infA expression to increase PCN results in growth inhibition and reduction in translation initiation efficiency. For further studies, other essential genes, such as those encoding elongation factor P or elongation factor G, could be used to replace infA to improve the dynamic range of the target gene and to enable multi-plasmid usage [47, 48]. The dynamic PCN control system can be applied to efficiently presume the optimal gene expression level within a replication of origin, reducing the number of redundant cloning to tune the expression level of the genes by changing the promoter, ribosome binding site strength for each gene on the plasmid. When our system was applied, itaconic acid titer increased 3.4-fold compared to that of the previously reported STAPL strain . This demonstrated efficient dynamic flux distribution at the desirable time point by dynamically adjusting the PCN. The PCN control system, when combined with inducible promoters, can maximize the variation in target protein levels within the cell. In addition, optimizing the expression of multiple genes in synthetic pathways can be challenging due to the need to select specific inducible promoters and their corresponding inducers. The system can be applied to multiple genes on the same plasmid, achieving different expression levels. A single input signal can upregulate the entire pathway while maintaining the optimized expression ratio for each gene by modulating the PCN. This system can be further applied to various fields of metabolic engineering when the inverted induction module of the genetic circuit is replaced with other sensing systems such as quorum sensing system or metabolite responsive biosensors [49, 50]. In addition, the system has the potential to be widely used in various bacteria, the essential genes of which are revealed. These demonstrate the borad applicability of the PCN dynamic control system. Conclusions In this study, we have developed the dynamic PCN control system which does not require the use of antibiotics for the plasmid maintenance. Our system maintains the plasmid and modulates its copy number by dynamically controlling the transcription of the essential gene infA on the plasmid while removing it from the chromosome. The optimal gene expression level can be determined by adjusting the PCN within an origin of replication through varying inducer concentrations. 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Author information Authors and Affiliations Interdisciplinary Program in Bioengineering, Seoul National University, 1 Gwanak-ro, Gwanak-gu, Seoul, 08826, Republic of Korea Geunyung Park & Sang Woo Seo 2. Department of Chemical Engineering, Jeju National University, 102, Jejudaehak-ro, Jeju-si, Jeju-do, 63243, Korea Jina Yang 3. School of Chemical and Biological Engineering, Seoul National University, 1 Gwanak-ro, Gwanak-gu, Seoul, 08826, Republic of Korea Sang Woo Seo 4. Institute of Chemical Processes, Seoul National University, 1 Gwanak-ro, Gwanak-gu, Seoul, 08826, Republic of Korea Sang Woo Seo 5. Bio-MAX Institute, Seoul National University, 1 Gwanak-ro, Gwanak-gu, Seoul, 08826, Republic of Korea Sang Woo Seo 6. Institute of Bio Engineering, Seoul National University, 1 Gwanak-ro, Gwanak-gu, Seoul, 08826, Republic of Korea Sang Woo Seo Authors Geunyung Park View author publications Search author on:PubMed Google Scholar 2. Jina Yang View author publications Search author on:PubMed Google Scholar 3. Sang Woo Seo View author publications Search author on:PubMed Google Scholar Contributions GP, JY, and SWS conceived the project. GP designed and performed the experiments. GP, JY, and SWS conducted data analysis and interpretation and wrote the manuscript. JY and SWS supervised the project. All authors read and approved the final manuscript. Corresponding author Correspondence to Sang Woo Seo. Ethics declarations Ethics approval and consent to participate Not applicable. Consent for publication Not applicable. Competing interests The authors declare no competing interests. Additional information Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Electronic supplementary material Below is the link to the electronic supplementary material. Supplementary Material 1 Rights and permissions Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit Reprints and permissions About this article Cite this article Park, G., Yang, J. & Seo, S.W. Dynamic control of the plasmid copy number maintained without antibiotics in Escherichia coli. J Biol Eng 18, 71 (2024). Received: Accepted: Published: DOI: Share this article Anyone you share the following link with will be able to read this content: Provided by the Springer Nature SharedIt content-sharing initiative Keywords Plasmid copy number Dynamic control Essential gene Antibiotics-free Metabolic engineering Gene regulation
188085	https://pmc.ncbi.nlm.nih.gov/articles/PMC4512600/	Pericarditis Epistenocardica or Dressler Syndrome? An Autopsy Case - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Case Rep Med . 2015 Jul 9;2015:215340. doi: 10.1155/2015/215340 Search in PMC Search in PubMed View in NLM Catalog Add to search Pericarditis Epistenocardica or Dressler Syndrome? An Autopsy Case Alessandro Feola Alessandro Feola 1 Department of Experimental Medicine, Second University of Naples, Via Luciano Armanni 5, 80138 Naples, Italy Find articles by Alessandro Feola 1,, Noè De Stefano Noè De Stefano 2 Unit of Histology and Anatomical Pathology, A.O.R.N. “San Giuseppe Moscati”, Contrada Amoretta, 83100 Avellino, Italy Find articles by Noè De Stefano 2, Bruno Della Pietra Bruno Della Pietra 1 Department of Experimental Medicine, Second University of Naples, Via Luciano Armanni 5, 80138 Naples, Italy Find articles by Bruno Della Pietra 1 Author information Article notes Copyright and License information 1 Department of Experimental Medicine, Second University of Naples, Via Luciano Armanni 5, 80138 Naples, Italy 2 Unit of Histology and Anatomical Pathology, A.O.R.N. “San Giuseppe Moscati”, Contrada Amoretta, 83100 Avellino, Italy ✉ Alessandro Feola: alessandro.feola@unina2.it Academic Editor: Michael S. Firstenberg Received 2015 May 11; Accepted 2015 Jul 5; Issue date 2015. Copyright © 2015 Alessandro Feola 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. PMC Copyright notice PMCID: PMC4512600 PMID: 26240567 Abstract Postinfarction pericarditis can be classified as “early,” referred to as pericarditis epistenocardica, or “delayed,” referred to as Dressler syndrome. The incidence of postinfarction pericarditis has decreased to <5% since the introduction of reperfusion therapies and limitation of infarct size. We report on a 57-year-old man who suffered sudden cardiac death as a result of acute myocardial infarction. Autopsy revealed an area of previous infarction and fibrinous pericarditis related to the previous infarction, leading to a diagnosis of Dressler syndrome. 1. Introduction Acute pericarditis is defined as inflammation of the pericardium with production of fibrinous inflammatory exudate. It is diagnosed in approximately 0.1% of hospitalized patients and in 5% of patients admitted to the emergency department with noncardiac chest pain . Acute pericarditis has several potential causes. Idiopathic pericarditis is the most common cause; others include infection (viral, e.g., Coxsackie A/B; bacterial, e.g., Staphylococcus; mycobacterial, fungal, and parasitic), systemic autoimmune inflammatory diseases (e.g., connectivitis, vasculitis, rheumatoid fever, and granulomatosis), pathologies of the surrounding organs, cardiac damage (e.g., Dressler syndrome, postpericardiotomy syndrome, postheart transplant, and pulmonary embolism), and pericardial trauma. Less frequent causes include neoplastic pericarditis, radiation or drugs, congenital renal pathologies, and endocrine metabolic pathologies (e.g., myxedema and goiter) . At the clinical level, it is possible to distinguish among acute (onset <6 weeks), subacute (6 weeks–6 months), and chronic forms (>6 months) . We present a patient with pericarditis resulting in sudden cardiac death. 2. Case Presentation A 57-year-old man was found dead in the hallway of his house. His family members reported seeing him 24 hours earlier, when he appeared to be in good health, except for a slight pain in his right ankle. However, they also reported that he had complained of intense chest pain approximately 1 month before. His medical history consisted of untreated hypertension. An autopsy performed 72 hours after discovery was noted. All the organs appeared congested and macroscopic examination showed evidence of interlobular pleurisy. The heart after extraction and fixation in 10% buffered formalin weighed 724 g, measured 15 × 13 × 5 cm, and had a truncated conical shape with a brick-red color. The epicardium was characterized by a fine granular layer on its surface with a “bread and butter” appearance, particularly at the level of the atrial regions and the cardiac apex (Figures 1 and 2). The cardiac cavities were slightly dilated and lined by a smooth and shiny endocardium. The thicknesses of the left ventricle, right ventricle, and interventricular septum were 23 mm, 6 mm, and 22 mm, respectively. The atrioventricular valve apparatus appeared normal. The mitral valve was two fingers wide and the tricuspid valve was three fingers wide, both with elastic valve flaps and no signs of nodular formations or calcium deposits. The aortic valve flaps were slightly thickened and focally affected by a few yellowish plaques. In cross section, the myocardial tissue showed a soft-elastic consistency and was dark red in color, with darker variegations appearing compacted. The coronary sinus was free from obstruction. Fat striations were observed in the aortic sinus (sinus of Valsalva). The coronary vessels had focal atherosclerotic plaques resulting in reduced lumen diameter, particularly in the anterior descending branch of the left coronary artery. Cardiac tissue was prepared and embedded in paraffin wax, and 4 μ m thick cross sections were cut and stained with hematoxylin/eosin, by standard methods. Plurifocal signs of myocardial sclerosis were detected in the wall of the left ventricle and the interventricular septum (Figure 3). Few lymphocytic perimyofibrillar microinfiltrations were observed, as well as several nonspecific postmortem alterations, including fragmentatio cordis, and structural disarrangement of the myocytes. Localized lipofuscin pigment deposits were occasionally observed in the myocytes, mainly in the perinuclear area, and moderate adipose replacement with no fibrous aspects involving the right ventricular myocardium. No alterations were observed in the endocardium. The serous cavity demonstrated histological features that could be attributed to fibrinous pericarditis (Figures 4 and 5). Samples collected from the proximal coronary branches revealed coronarosclerosis, subendothelial intimal sclerosis, and sclerotic calcified plaques. There was no evidence of significant alterations of the subepicardial distal vessels in the sections examined. Biological samples collected at autopsy were subjected to systematic toxicological analysis using headspace gas chromatography (flame-ionization detection), gas chromatography-mass spectrometry, and liquid chromatography-tandem mass spectrometry but were negative for alcohol and for most common illicit drugs and pharmaceuticals. Figure 1. Open in a new tab Macroscopic view of the heart prior to formalin fixation. Figure 2. Open in a new tab Macroscopic view of areas with fibrinous pericarditis (following formalin fixation). Figure 3. Open in a new tab Hematoxylin/eosin-stained section of myocardium illustrating areas of myocardial sclerosis (×100). Figure 4. Open in a new tab Hematoxylin/eosin-stained view of fibrinous pericarditis (×40). Figure 5. Open in a new tab Hematoxylin/eosin-stained view of fibrinous pericarditis (×40). 3. Discussion The pericardium can respond to acute myocardial infarction (MI) in different ways, particularly by pericardial effusion or pericarditis . Postinfarction pericarditis can be classified as either “early,” referred to as pericarditis epistenocardica, or “delayed,” referred to as Dressler syndrome . Before the advent of thrombolysis, the reported incidence of clinical infarction-associated pericardial involvement was 7–23%, with much higher rates detected during autopsy . However, the incidence of postinfarction pericarditis decreased to <5% since the introduction of reperfusion therapies and the limitation of infarct size . Pericarditis epistenocardica, which initially manifests with pain and a pericardial rub, usually occurs within the first three days after a transmural infarction [5, 7]. Transmural infarctions result from transmural necrosis with inflammation affecting the adjacent visceral and parietal pericardium . In such cases, the classic electrocardiography (ECG) changes of pericarditis are usually not apparent, suggesting a relapse of the subepicardial lesion and/or an increase in ischemia . Although chest X-ray and echocardiography are not diagnostic in the event of a localized pericardial reaction, they become useful, especially the latter, in the event of a pericardial effusion. The diagnosis is based on clinical suspicion, fever, pleuritic chest pain, and the presence of an effusion by echocardiography . Dressler syndrome occurs after MI. Prior to the reperfusion era, its reported incidence was 1–5% of patients with acute MI . However, its advent of thrombolysis and widespread use of heparin have reduced the incidence of this syndrome . When present, it arises 2 weeks after an MI presumably due to an autoreactive immune mechanism similar to postcardiac injury syndrome [10, 13]. Typical symptoms include pleuritic chest pain with low-grade fever. A pericardial rub is present in 50% of patients . Chest X-ray may show a pleural effusion and/or enlargement of the cardiac silhouette. An ECG often demonstrates ST-elevation and T-wave changes typical of acute pericarditis . The two syndromes differ dramatically in several areas. Dressler syndrome is more paucisymptomatic with few manifestations, such as fever, malaise, and chest pain [15, 16]. Diagnostic ECG changes of pericarditis epistenocardica require a transmural MI in order to injure the visceral pericardium but Dressler syndrome does not . In the current case, death was caused by a hyperacute infarct. As there was no histological evidence of granulocytes at the myocardial level and only bands with myosclerosis, there was no evidence of pericarditis caused by the most recent infarction (Figure 3). However, the granulocytes were present at the level of the pericardium, evidence of pericarditis due to the previous infarction (Figures 4 and 5). In conclusion, Dressler syndrome appeared to be the most likely diagnosis, given the presence of fibrinous pericarditis with evidence of a previous MI and interlobular pleurisy. A pulmonary genesis of the pleurisy was excluded based on negative macroscopic and microscopic examination of the lungs. Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper. References 1.Khandaker M. H., Espinosa R. E., Nishimura R. A., et al. Pericardial disease: diagnosis and management. Mayo Clinic Proceedings. 2010;85(6):572–593. doi: 10.4065/mcp.2010.0046. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Fork Join optimization Ask Question Asked 11 years, 7 months ago Modified11 years, 6 months ago Viewed 2k times This question shows research effort; it is useful and clear 7 Save this question. Show activity on this post. What I want I want to work on optimization of fork/join algorithm. By optimization I mean just calculation of optimal number of threads, or if you want - calculation of SEQUENTIAL_THRESHOLD (see code below). java // PSEUDOCODE Result solve(Problem problem) { if (problem.size < SEQUENTIAL_THRESHOLD) return solveSequentially(problem); else { Result left, right; INVOKE-IN-PARALLEL { left = solve(extractLeftHalf(problem)); right = solve(extractRightHalf(problem)); } return combine(left, right); } } How do I imagine that For example, I want to calculate the product of big array. Then I just evaluate all components and get the optimal threads amount: SEQUENTIAL_THRESHOLD = PC IS / MC (just example) PC - number of processor cores; IS - constant, that indicates the optimal array size with one processor core and the simplest operation on data (for example reading); MC - multiply operation cost; Suppose MC = 15; PC = 4 and IS = 10000; SEQUENTIAL_THRESHOLD = 2667. Than if subtask-array is bigger than 2667 I'll fork it. Broad questions Is it possible to make SEQUENTIAL_THRESHOLD formula in such way? Is it possible to accomplish the same for more complex computation: not only for operations on arrays/collections and sorting? Narrow question: Do already exist some investigations about calculation of SEQUENTIAL\_THRESHOLD for arrays/collections/sorting? How do they accomplish that? Updated 07 March 2014: If there is no way to write a single formula for threshold calculation, can I write an util which will perform predefined tests on PC, and than gets the optimal threshold? Is that also impossible or not? What can Java 8 Streams API do? Can it help me? Does Java 8 Streams API eliminate a need in Fork/Join? java multithreading concurrency java-8 fork-join Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Mar 11, 2014 at 11:28 VB_VB_ asked Mar 5, 2014 at 7:32 VB_VB_ 45.9k 45 45 gold badges 162 162 silver badges 313 313 bronze badges 4 1 "The ideal threshold for choosing between sequential and parallel execution is a function of the cost of coordinating the parallel tasks. If coordination costs are zero, a larger number of finer-grained tasks tend to offer better parallelism; the lower the coordination costs, the finer-grained we can go before we need to switch to a sequential approach." - Quote from the website where the pseudocode is from: ibm.com/developerworks/library/j-jtp11137 ;)Marco13 –Marco13 2014-03-05 09:18:25 +00:00 Commented Mar 5, 2014 at 9:18 In other words, don’t use the number of CPU cores as the number of available cores can change at any time. The threshold should just be large enough so that the overhead of splitting does not matter compared to the problem size. The more subtasks the more freedom for the Executor/scheduler to adapt to the real system’s load situation.Holger –Holger 2014-03-05 09:48:55 +00:00 Commented Mar 5, 2014 at 9:48 1 This is a typical question for map-reduce tasks. You should check out the Streams API of Java 8. It takes care of the near optimal execution. You should not be concerned too much about reaching the optimum as long as you're not doing HPC. You want your desktop remain responsible if the computation happens there, even if it takes a little longer.allprog –allprog 2014-03-05 13:37:02 +00:00 Commented Mar 5, 2014 at 13:37 @allprog I'm asking about HPC :) More details about Streams API please. Docs doesn't provide enough information about how does Java 8 parallelism optimize itself.VB_ –VB_ 2014-03-07 17:00:05 +00:00 Commented Mar 7, 2014 at 17:00 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 5 Save this answer. Show activity on this post. There is absolutely, positively no way to calculate a proper threshold unless you are intimate with the execution environment. I maintain a fork/join project on sourceforge.net and this is the code I use in most built-in-functions: ```java private int calcThreshold(int nbr_elements, int passed_threshold) { // total threads in session // total elements in array int threads = getNbrThreads(); int count = nbr_elements + 1; // When only one thread, it doesn't pay to decompose the work, // force the threshold over array length if (threads == 1) return count; / Whatever it takes / int threshold = passed_threshold; // When caller suggests a value if (threshold > 0) { // just go with the caller's suggestion or do something with the suggestion } else { // do something usful such as using about 8 times as many tasks as threads or // the default of 32k int temp = count / (threads << 3); threshold = (temp < 32768) ? 32768 : temp; } // endif // whatever return threshold; } ``` Edit on 9 March: How can you possibly have a general utility that can know not only the processor speed, memory available, number of processors, etc. (the physical environment) but the intention of the software? The answer is you cannot. Which is why you need to develop a routine for each environment. The above method is what I use for basic arrays (vectors.) I use another for most matrix processing: ```java // When very small, just spread every row if (count < 6) return 1; // When small, spread a little if (count < 30) return ((count / (threads << 2) == 0)? threads : (count / (threads << 2))); // this works well for now return ((count / (threads << 3) == 0)? threads : (count / (threads << 3))); ``` As far as Java8 streams: They use the F/J framework under the hood and you cannot specify a threshold. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Mar 10, 2014 at 15:03 assylias 330k 83 83 gold badges 680 680 silver badges 806 806 bronze badges answered Mar 5, 2014 at 15:22 edharnededharned 1,904 1 1 gold badge 19 19 silver badges 20 20 bronze badges Comments Add a comment This answer is useful 3 Save this answer. +50 This answer has been awarded bounties worth 50 reputation by VB_ Show activity on this post. You cannot boil this down to a simple formula for several reasons: Each PC will have vastly different parameters depending not only on the core, but also on other factors like RAM timing or background tasks. Java itself is optimizing loops on the fly during execution. So a momentary perfect setting could be sub-optimal a few seconds later. Or worse: the adjustment could prevent perfect optimization all together. The only way to go that I can see is to dynamically adjust the values in some form of AI or genetic algorithm. However that includes that the program frequently checks non-optimal settings just to determine whether or not the current setting is still the best. So it is questionable if the speed gained is actually higher than the speed lost for trying other settings. In the end probably only a solution during an initial learning phase, while further executions then use those trained values as fixed numbers. As this not only costs time but also greatly increases the code complexity, I don't think this is an option for most programs. Often it is more beneficial to not even use Fork-Join in the first place, as there are many other parallelization options that might better suit the problem. An idea for a "genetic" algorithm would be to measure the loop efficiency each run, then have a background hash-map loop-parameters -> execution time that is constantly updated, and the fastest setting is selected for most of the runs. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Mar 5, 2014 at 15:16 TwoTheTwoThe 14.4k 6 6 gold badges 35 35 silver badges 56 56 bronze badges 4 Comments Add a comment VB_ VB_Over a year ago yes, I can make something like loop-parameters -> execution time. But than I should calculate that table for each PC configuration, it's impossible. How can I link all that with PC configs or maybe wich parameters should I consider? 2014-03-07T16:26:33.157Z+00:00 0 Reply Copy link VB_ VB_Over a year ago And is there any services with 16+ cores where I can test performance of my fork/join algorithms? 2014-03-07T16:38:23.353Z+00:00 0 Reply Copy link TwoThe TwoTheOver a year ago You cannot pre-calculate those parameters, you need to do it on the actual machine. So either have something like a calibration phase added to your code to determine those parameters, or - if you only have a small set of machines - do some measurements by hand before the actual execution. 2014-03-07T16:46:14.217Z+00:00 0 Reply Copy link VB_ VB_Over a year ago look at updated version of my question please. Also I'm interresting to make something universal, not for determined range of machines 2014-03-07T16:51:03.497Z+00:00 0 Reply Copy link Add a comment This answer is useful 1 Save this answer. Show activity on this post. This is a very interesting problem to investigate. I have written this simple code to test the optimal value of sequential threshold. I was unable to reach any concrete conclusions though, most probably because I am running it on a old laptop with only 2 processors. The only consistent observation after many runs was that the time taken drops rapidly till sequential threshold of 100. Try running this code and let me know what you find. Also at the bottom I have attached a python script for plotting the results so that we can visually see the trend. ```java import java.io.FileWriter; import java.util.concurrent.ForkJoinPool; import java.util.concurrent.RecursiveAction; public class Testing { static int SEQ_THRESHOLD; public static void main(String[] args) throws Exception { int size = 100000; int[] v1 = new int[size]; int[] v2 = new int[size]; int[] v3 = new int[size]; for (int i = 0; i < size; i++) { v1[i] = i; // Arbitrary initialization v2[i] = 2 i; // Arbitrary initialization } FileWriter fileWriter = new FileWriter("OutTime.dat"); // Increment SEQ_THRESHOLD and save time taken by the code to run in a file for (SEQ_THRESHOLD = 10; SEQ_THRESHOLD < size; SEQ_THRESHOLD += 50) { double avgTime = 0.0; int samples = 5; for (int i = 0; i < samples; i++) { long startTime = System.nanoTime(); ForkJoinPool fjp = new ForkJoinPool(); fjp.invoke(new VectorAddition(0, size, v1, v2, v3)); long endTime = System.nanoTime(); double secsTaken = (endTime - startTime) / 1.0e9; avgTime += secsTaken; } fileWriter.write(SEQ_THRESHOLD + " " + (avgTime / samples) + "\n"); } fileWriter.close(); } } class VectorAddition extends RecursiveAction { int[] v1, v2, v3; int start, end; VectorAddition(int start, int end, int[] v1, int[] v2, int[] v3) { this.start = start; this.end = end; this.v1 = v1; this.v2 = v2; this.v3 = v3; } int SEQ_THRESHOLD = Testing.SEQ_THRESHOLD; @Override protected void compute() { if (end - start < SEQ_THRESHOLD) { // Simple vector addition for (int i = start; i < end; i++) { v3[i] = v1[i] + v2[i]; } } else { int mid = (start + end) / 2; invokeAll(new VectorAddition(start, mid, v1, v2, v3), new VectorAddition(mid, end, v1, v2, v3)); } } } ``` and here is the Python script for plotting results: ```java from pylab import threshold = loadtxt("./OutTime.dat", delimiter=" ", usecols=(0,)) timeTaken = loadtxt("./OutTime.dat", delimiter=" ", usecols=(1,)) plot(threshold, timeTaken) show() ``` Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Mar 5, 2014 at 10:10 Sourabh BhatSourabh Bhat 1,903 17 17 silver badges 22 22 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! 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188087	https://blog.hubspot.com/sales/what-is-a-founder	Home Sales The Business Definition of a Founder The Business Definition of a Founder Written by: Lestraundra Alfred FREE BUSINESS PLAN TEMPLATE 2 Essential Templates For Starting Your Business Download the Free Template Updated: In many young organizations, the company founder or a co-founder serves as the CEO. However, just because the roles are commonly held by the same individual does not mean they are interchangeable. It also doesn’t mean this setup is ideal for all businesses. During the early stages, many companies operate in an “all hands on deck” manner, meaning the team members on board handle all aspects of the business, including leadership. Before we examine how the founder and CEO roles differ from one another, let’s discuss what it means to be a founder of a company. What is a founder? In business, a founder is an individual who forms and establishes a business or organization. The founder is typically responsible for setting the mission and vision of a company. Essentially, a founder takes a business from an idea to an entity. As mentioned above, in the early stages of a company, team members wear many hats ensuring the work gets done to keep the company going as it grows and scales. In many situations, this means the individual who created the business is often tasked with managing the overall operations of the company as CEO. Founder vs. CEO The biggest difference between a founder and a CEO is whether the role is active or passive. Even when a founding individual leaves the company and is no longer serving an active role, they can still be considered a founder due to their work establishing the company. Additionally, much of the work for founders occurs at the beginning stages of the business. On the other hand, the CEO is an active role. If someone has the title of CEO, that means they are currently serving as the head of the company even if they weren’t responsible for establishing the entity. While it is common for an individual to serve as both founder and CEO, that is not always the case. Even if one person does have both roles, each title has its own set of tasks. Now that we understand how the role of founder differs from the CEO at a basic level, let’s discuss the different responsibilities of each role. Founder Role and Responsibilities 1. Develop a Business Plan Most successful businesses begin with a well-thought-out business plan, and in many cases, it’s up to the company’s founder to ensure a business plan has been created. The key elements of a an effective business plan include the executive summary, a description of the company’s business model, market analysis, the products/services your company offers, operational plan, the marketing and sales strategy, and a detailed financial plan. 2. Establish Mission and Vision Startup companies are often established from a specific product idea or service the founder wants to offer customers. Along with determining what products to offer, founders are often tasked with determining the company’s mission and vision to keep their employees and team members aligned and on the same page regarding how they best serve their customers. 3. Form the Board of Directors In the beginning stages of a company, the founder is often tasked with determining what kind of governing body or board the business should have, along with who should be on it. Once the board is put in place, the company’s founder may oversee the relationship between the board and the company. 4. Recruit Employees and Executive Team When establishing a new business, the founder is also responsible for assembling a team that can bring their vision to life. This typically begins with an executive team that can oversee critical aspects of the business, as well as support staff and employees who are responsible for product execution and delivering the final offering to the customer. 5. Initial Funding The company’s founder is also responsible for any initial funding to get the business off the ground. This could mean applying for loans or financing, applying for grants, seeking venture capital, or using their own personal assets to fund the initial business costs until the company begins earning enough revenue to cover costs. CEO Role and Responsibilities Now let’s examine some of the roles and responsibilities of the CEO. When we reviewed the job of a founder, many of the tasks focused on the beginning stages of the business. The role of the CEO, on the other hand, has roles and responsibilities that extend throughout the lifespan of the company. While the specific role of the CEO can vary depending on the size and nature of the company, here are some common responsibilities that are owned by the company’s CEO. 1. Public Spokesperson for the Company The CEO is often tasked with being the public face of the company. That means the CEO is responsible for taking media requests and attending industry and community events on behalf of the company they work for. Their personal brand becomes closely aligned with the company’s reputation. 2. Oversee Company Operations While many companies have a Chief Operating Officer in place who manages the team that executes operations, the CEO is still ultimately responsible for making decisions on how the company operates and best serves its customers. That could entail making key hiring decisions related to company operations, and updating strategic plans to ensure the company continues performing well and meeting its goals. 3. Maintain Relationship with the Board of Directors Though the founder is responsible for determining what kind of governing board oversees the company, the CEO is tasked with maintaining these relationships and reports directly to the board. In many companies, the CEO goes to the board for guidance around major business decisions. The CEO also communicates business updates and milestones to the board to ensure company performance is on track. 4. Track Company Performance At a high level, the CEO is also responsible for their organization’s performance and ability to meet its major goals. If for some reason teams within the company are not meeting their goals, the CEO should provide strategic guidance on how to do so, or as mentioned above, should solicit advice on how to do so from the board. Should a company founder and CEO be the same person? Now you understand what founders and CEOs do, respectively. That leaves us with one last question: should the founder and CEO of a company be the same person? There’s no clear answer — it largely depends on the company in question and of the skillset of the founder and CEO. In some instances, an individual can serve as founder and CEO of a business during the startup stages and decides to bring on a new CEO to manage the business after reaching the sustainment stage. In other instances, an individual can be the founder of a company, and quickly determine their skill set is better served in an area outside of company leadership, prompting them to hire a CEO who may be a better fit. Whether a founder chooses to take the role of CEO of their company comes down to resources, their individual strengths, and what is best for their business, especially in the startup stages. To learn more about the different stages of a startup, check out this post. Don't forget to share this post! 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188088	https://ptacts.uspto.gov/ptacts/public-informations/petitions/1548489/download-documents?artifactId=wHBNSxCeMtlQ4RJTfnA57KZz2314ySM1jqmeWbSNzxCOxPZzNq6hTuY	POWER ELECTRONICS iSolarEdge, Ex. 1212 ABOUT THE AUTHORS Ned Mohan is a professor in the Department of Electrical Engineering at the University of Minnesota, where he holds the Oscar A. Schott Chair in Power Electronics. He has worked on several power electronics projects sponsored by the industry and the electric power utilities, including the Electric Power Research Institute. He has numerous pub-lications and patents in this field. Tore M. Undeland is a Professor in Power Electronics in the Faculty of Electrical Engineering and Computer Science at the Norwegian Institute of Technology. He is also Scientific Advisor to the Norwegian Electric Power Research Institute of Electricity Supply. He has been a visiting scientific worker in the Power Electronics Converter Department of ASEA in Vaasteras, Sweden, and at Siemens in Trondheim, Norway, and a visiting professor in the Department of Electrical Engineering at the University of Minnesota. He has worked on many industrial research and development projects in the power electronics field and has numerous publications. William P. Robbins is a professor in the Department of Electrical Engineering at the University of Minnesota. Prior to joining the University of Minnesota, he was a research engineer at the Boeing Company. He has taught numerous courses in electronics and semiconductor device fabrication. His research interests are in ultrasonics, pest insect detection via ultrasonics, and micromechanical devices, and he has numerous publications in this field. ii POWER ELECTRONICS Converters, Applications, and Design SECOND EDITION NED MOHAN Department of Electrical Engineering University of Minnesota Minneapolis, Minnesota TORE M. UNDELAND Faculty of Electrical Engineering and Computer Science Norwegian Institute of Technology Trondheim, Norway WILLIAM P. ROBBINS Department of Electrical Engineering University of Minnesota Minneapolis, Minnesota JOHN WILEY & SONS, INC. New York Chichester Brisbane Toronto Singapore iii Acquisitions Editor Developmental Editor Marketing Manager Senior Production Editor Text Designer Cover Designer Manufacturing Manager Illustration Coordinator Steven M. Elliot Sean M. Culhane Susan Elbe Savoula Amanatidis Lynn Rogan David Levy Lori Bulwin Jaime Perea This book was typeset in Times Roman by The Clarinda Company, and printed and bound by Hamilton Printing Company. The cover was printed by NEBC. Recognizing the importance of preserving what has been written, it is a policy of John Wiley & Sons, Inc. to have books of enduring value published in the United States printed on acid-free paper, and we exert our best efforts to that end. PSpice is a registered trademark of MicroSim Corporation. MATLAB is a registered trademark of The MathWorks, Inc. Copyright © 1989, 1995 by John Wiley & Sons, Inc. All rights reserved. Published simultaneously in Canada. Reproduction or translation of any part of this work beyond that permitted by Sections 107 and 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Requests for permission or further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. Library of Congress Cataloging in Publiclllion Data: Mohan, Ned. Power electronics: converters, applications, and design / Ned Mohan, Tore M. Undeland, William P. Robbins.-2nd ed. p. cm. Includes bibliographical references and indexes. ISBN 0-471-58408-8 (cloth) I. Power electronics. 2. Electric current converters. 3. Power semiconductors. I. Undeland, TOre M. 11. Robbins, William P. III. Title. TK7881.15.M64 1995 621.317-dc20 Printed in the United States of America. 10 9 8 7 6 5 4 3 2 I94-21158 CIP iv To OUf Families . . . Mary, Michael, and Tara Mona, Hilde, and Arne Joanne and Jon vvi PREFACE SECOND EDITION The first edition of this book was published in 1989. The basic intent of this edition remains the same; that is, as a cohesive presentation of power electronics fundamentals for applications and design in the power range of 500 kW or less, where a huge market exists and where the demand for power electronics engineers is likely to be. Based on the comments collected over a five-year period, we have made a number of substantial changes to the text. The key features are as follows: • An introductory chapter has been added to provide a review of basic electrical and magnetic circuit concepts, making it easier to use this book in introductory power electronics courses. • A chapter on computer simulation has been added that describes the role of com-puter simulations in power electronics. Examples and problems based on PSpice@ and MATLAB@ are included. However, we have organized the material in such a way that any other simulation package can be used instead or the simulations can be skipped altogether. • Unlike the first edition, the diode rectifiers and the phase-controlled thyristor con-verters are covered in a complete and easy-to-follow manner. These two chapters now contain 56 problems. • A new chapter on the design of inductors and transformers has been added that describes easy-to-understand concepts for step-by-step design procedures. This material will be extremely useful in introducing the design of magnetics into the curriculum. • A new chapter on heat sinks has been added. ORGANIZATION OF THE BOOK This book is divided into seven parts. Part 1 presents an introduction to the field of power electronics, an overview of power semiconductor switches, a review of pertinent electric and magnetic circuit concepts, and a generic discussion of the role of computer simula-tions in power electronics. Part 2 discusses the generic converter topologies that are used in most applications. The actual semiconductor devices (transistors, diodes, and so on) are assumed to be ideal, thus allowing us to focus on the converter topologies and their applications. Part 3 discusses switch-mode dc and uninterruptible power supplies. Power supplies represent one of the major applications of power electronics. vii viii PREFACE Part 4 considers motor drives, which constitute another major applications area. Part 5 includes several industrial and commercial applications in one chapter. An-other chapter describes various high-power electric utility applications. The last chapter in this part of the book examines the harmonics and electromagnetic interference concerns and remedies for interfacing power electronic systems with the electric utilities. Part 6 discusses the power semiconductor devices used in power electronic converters including diodes, bipolar junction thyristors, metaI-oxide-semiconductor (MaS) field effect transistors, thyristors, gate tum-off thyristors, insulated gate bipolar transistors, and MaS-controlled thyristors. Part 7 discusses the practical aspects of power electronic converter design including snubber circuits, drive circuits, circuit layout, and heat sinks. An extensive new chapter on the design of high-frequency inductors and transfonners has been added. PSPICE SIMULATIONS FOR TEACHING AND DESIGN As a companion to this book, a large number of computer simulations are available directly from Minnesota Power Electronics Research and Education, P.O. Box 14503, Minneapolis, MN 55414 (Phone/Fax: 612-646-1447) to aid in teaching and in the design of power electronic systems. The simulation package comes complete with a diskette with 76 simulations of power electronic converters and systems using the classroom (evalua-tion) version of PSpice for IBM-PC-compatible computers, a 261-page detailed manual that describes each simulation and a number of associated exercises for home assignments and self-learning, a 5-page instruction set to illustrate PSpice usage using these simula-tions as examples, and two high-density diskettes containing a copy of the classroom (evaluation) version of PSpice. This package (for a cost of $395 plus a postage of $4 within North America and $25 outside) comes with a site license, which allows it to be copied for use at a single site within a company or at an educational institution in regular courses given to students for academic credits. SOLUTIONS MANUAL As with the first edition of this book, a solutions manual with completely worked-out solutions to all the problems is available from the publisher. ACKNOWLEDGMENTS We wish to thank all the instructors who have allowed us this opportunity to write the second edition of our book by adopting its first edition. Their comments have been most useful. We are grateful to Professors Peter Lauritzen of the University of Washington, Thomas Habetler of the Georgia Institute of Technology, Daniel Chen of the Virginia Institute of Technology, Alexander Emanuel of the Worcester Polytechnic Institute, F. P. Dawson of the University of Toronto, and Marian Kazimierczuk of the Wright State University for their helpful suggestions in the second edition manuscript. We express our sincere appreciation to the Wiley editorial staff, including Steven Elliot, Sean Culhane, Lucille Buonocore, and Savoula Amanatidis, for keeping us on schedule and for many spirited discussions. Ned Mohan Tore M. Undeland William P. Robbins CONTENTS PART 1 INTRODUCTION 1 Chapter 1 Power Electronic Systems 31-1 Introduction 31-2 Power Electronics versus Linear Electronics 41-3 Scope and Applications 71-4 Classification of Power Processors and Converters 91-5 About the Text 12 1-6 Interdisciplinary Nature of Power Electronics 13 1-7 Convention of Symbols Used 14 Problems 14 References 15 Chapter 2 Overview of Power Semiconductor Switches 16 2-1 Introduction 16 2- 2 Diodes 16 2-3 Thyristors 18 2-4 Desired Characteristics in Controllable Switches 20 2-5 Bipolar Junction Transistors and Monolithic Darlingtons 24 2-6 Metal- Oxide- Semiconductor Field Effect Transistors 25 2-7 Gate-Tum-Off Thyristors 26 2-8 Insulated Gate Bipolar Transistors 27 2-9 MOS-Controlled Thyristors 29 2-10 Comparison of Controllable Switches 29 2-11 Drive and Snubber Circuits 30 2-12 Justification for Using Idealized Device Characteristics 31 Summary 32 Problems 32 References 32 Chapter 3 Review of Basic Electrical and Magnetic Circuit Concepts 33 3-1 Introduction 33 3-2 Electric Circuits 33 3-3 Magnetic Circuits 46 Summary 57 Problems 58 References 60 ix x CONTENTS Chapter 4 Computer Simulation of Power Electronic Converters and Systems 61 4-1 Introduction 61 4-2 Challenges in Computer Simulation 62 4-3 Simulation Process 62 4-4 Mechanics of Simulation 64 4-5 Solution Techniques for Time-Domain Analysis 65 4-6 Widely Used, Circuit-Oriented Simulators 69 4-7 Equation Solvers 72 Summary 74 Problems 74 References 75 PART 2 GENERIC POWER ELECTRONIC CIRCUITS 77 Chapter 5 Line-Frequency Diode Rectifiers: Line-Frequency ac -+ Uncontrolled dc 79 5-1 Introduction 79 5-2 Basic Rectifier Concepts 80 5-3 Single-Phase Diode Bridge Rectifiers 82 5-4 Voltage-Doubler (Single-Phase) Rectifiers 100 5-5 Effect of Single-Phase Rectifiers on Neutral Currents in Three-Phase, Four-Wire Systems 101 5-6 Three-Phase, Full-Bridge Rectifiers 103 5-7 Comparison of Single-Phase and Three-Phase Rectifiers 112 5-8 Inrush Current and Overvoltages at Tum-On 112 5-9 Concerns and Remedies for Line-Current Harmonics and Low Power Factor 113 Summary 113 Problems 114 References 116 Appendix 117 Chapter 6 Line-Frequency Phase-Controlled Rectifiers and Inverters: Line-Frequency ac - Controlled dc 121 6-1 Introduction 121 6-2 Thyristor Circuits and Their Control 122 6-3 Single-Phase Converters 126 6-4 Three-Phase Converters 138 6-5 Other Three-Phase Converters 153 Summary 153 Problems 154 References 157 Appendix 158 Chapter 7 dc-dc Switch-Mode Converters 161 7-1 Introduction 161 7-2 Control of dc-de Converters 162 CONTENTS xi 7-3 Step-Down (Buck) Converter 164 7-4 Step-Up (Boost) Converter 172 7-5 Buck-Boost Converter 178 7-6 CUk dc-dc Converter 184 7-7 Full Bridge dc-dc Converter 188 7-8 dc-dc Converter Comparison 195 Summary 196 Problems 197 References 199 Chapter 8 Switch-Mode dc-ac Inverters: dc ~ Sinusoidal ac 200 8-1 Introduction 200 8-2 Basic Concepts of Switch-Mode Inverters 202 8-3 Single-Phase Inverters 211 8-4 Three-Phase Inverters 225 8-5 Effect of Blanking Time on Output Voltage in PWM Inverters 236 8-6 Other Inverter Switching Schemes 239 8-7 Rectifier Mode of Operation 243 Summary 244 Problems 246 References 248 Chapter 9 Resonant Converters: Zero-Voltage and/or Zero-Current Switchings 249 9-1 Introduction 249 9-2 Classification of Resonant Converters 252 9-3 Basic Resonant Circuit Concepts 253 9-4 Load-Resonant Converters 258 9-5 Resonant-Switch Converters 273 9-6 Zero-Voltage-Switching, Clamped-Voltage Topologies 280 9-7 Resonant-dc-Link Inverters with Zero-Voltage Switchings 287 9-8 High-Frequency-Link Integral-Half-Cycle Converters 289 Summary 291 Problems 291 References 295 PART 3 POWER SUPPLY APPLICATIONS 299 Chapter 10 Switching dc Power Supplies 301 10-1 Introduction 30 I10-2 Linear Power Supplies 301 10-3 Overview of Switching Power Supplies 302 10-4 dc-dc Converters with Electrical Isolation 304 10-5 Control of Switch-Mode dc Power Supplies 322 10-6 Power Supply Protection 341 10-7 Electrical Isolation in the Feedback Loop 344 10-8 Designing to Meet the Power Supply Specifications 346 Summary 349 xii CONTENTS Problems 349 References 351 Chapter 11 Power Conditioners and Uninterruptible Power Supplies 11-1 Introduction 354 11-2 Power Line Disturbances 354 11-3 Power Conditioners 357 11-4 Uninterruptib1e Power Supplies (UPSs) 358 Summary 363 Problems 363 References 364 PART 4 MOTOR DRIVE APPLICATIONS Chapter 12 Introduction to Motor Drives 12-1 Introduction 367 12-2 Criteria for Selecting Drive Components 368 Summary 375 Problems 376 References 376 Chapter 13 de Motor Drives 13-1 Introduction 377 13-2 Equivalent Circuit of dc Motors 377 13-3 Permanent-Magnet dc Motors 380 13-4 dc Motors with a Separately Excited Field Winding 381 13-5 Effect of Armature Current Waveform 382 13-6 dc Servo Drives 383 13-7 Adjustable-Speed dc Drives 391 Summary 396 Problems 396 References 398 Chapter 14 Induction Motor Drives 14-1 Introduction 399 14-2 Basic Principles of Induction Motor Operation 400 14-3 Induction Motor Characteristics at Rated (Line) Frequency and Rated Voltage 405 14-4 Speed Control by Varying Stator Frequency and Voltage 406 14-5 Impact of Nonsinusoidal Excitation on Induction Motors 415 14-6 Variable-Frequency Converter Classifications 418 14-7 Variable-Frequency PWM-VSI Drives 419 14-8 Variable-Frequency Square-Wave VSI Drives 425 14-9 Variable-Frequency CSI Drives 426 14-10 Comparison of Variable-Frequency Drives 427 354 365 367 377 399 14-11 Line-Frequency Variable-Voltage Drives 428 14-12 Reduced Voltage Starting ("Soft Start") of Induction Motors 14-13 Speed Control by Static Slip Power Recovery 431 Summary 432 Problems 433 References 434 Chapter 15 Synchronous Motor Drives 15-1 Introduction 435 15-2 Basic Principles of Synchronous Motor Operation 435 15-3 Synchronous Servomotor Drives with Sinusoidal Waveforms 15-4 Synchronous Servomotor Drives with Trapezoidal Waveforms 15-5 Load-Commutated Inverter Drives 442 15-6 Cycloconverters 445 Summary 445 Problems 446 References 447 PART 5 OTHER APPLICATIONS Chapter 16 Residential and Industrial Applications 16-1 Introduction 451 16-2 Residential Applications 451 16-3 Industrial Applications 455 Summary 459 Problems 459 References 459 Chapter 17 Electric Utility Applications 17-1 Introduction 460 17-2 High-voltage dc Transmission 460 17-3 Static var Compensators 471 CONTENTS xiii 430 439 440 435 449 451 460 17-4 Interconnection of Renewable Energy Sources and Energy Storage Systems to the Utility Grid 475 17-5 Active Filters 480 Summary 480 Problems 481 References 482 Chapter 18 Optimizing the Utility Interface with Power Electronic Systems 483 18-1 Introduction 483 18-2 Generation of Current Harmonics 484 18-3 Current Harmonics and Power Factor 485 18-4 Harmonic Standards and Recommended Practices 485 18-5 Need for Improved Utility Interface 487 xiv CONTENTS 18-6 Improved Single-Phase Utility Interface 488 18-7 Improved Three-Phase Utility Interface 498 18-8 Electromagnetic Interference 500 Summary 502 Problems 503 References 503 PART 6 SEMICONDUCTOR DEVICES 505 Chapter 19 Basic Semiconductor Physics 507 19-1 Introduction 507 19-2 Conduction Processes in Semiconductors 507 19-3 pn Junctions 513 19-4 Charge Control Description of pn-Junction Operation 518 19-5 Avalanche Breakdown 520 Summary 522 Problems 522 References 523 Chapter 20 Power Diodes 524 20-1 Introduction 524 20-2 Basic Structure and /- V Characteristics 524 20-3 Breakdown Voltage Considerations 526 20-4 On-State Losses 531 20-5 Switching Characteristics 535 20-6 Schottky Diodes 539 Summary 543 Problems 543 References 545 Chapter 21 Bipolar Junction Transistors 546 21-1 Introduction 546 21-2 Vertical Power Transistor Structures 546 21-3 /-V Characteristics 548 21-4 Physics of BJT Operation 550 21-5 Switching Characteristics 556 21-6 Breakdown Voltages 562 21-7 Second Breakdown 563 21-8 On-State Losses 565 21-9 Safe Operating Areas 567 Summary 568 Problems 569 References 570 Chapter 22 Power MOSFETs 571 22-1 Introduction 571 22-2 Basic Structure 571 CONTENTS xv 22-3 1-V Characteristics 574 22-4 Physics of Device Operation 576 22-5 Switching Characteristics 581 22-6 Operating Limitations and Safe Operating Areas 587 Summary 593 Problems 594 References 595 Chapter 23 Thyristors 596 23-1 Introduction 596 23-2 Basic Structure 596 23-3 1-V Characteristics 597 23-4 Physics of Device Operation 599 23-5 Switching Characteristics 603 23-6 Methods of Improving dildt and dv/dt Ratings 608 Summary 6/0 Problems 611 References 612 Chapter 24 Gate Tum-Off Thyristors 613 24-1 Introduction 613 24-2 Basic Structure and 1-V Characteristics 613 24-3 Physics of Turn-Off Operation 614 24-4 GTO Switching Characteristics 616 24-5 Overcurrent Protection of GTOs 623 Summary 624 Problems 624 References 625 Chapter 25 Insulated Gate Bipolar Transistors 626 25-1 Introduction 626 25-2 Basic Structure 626 25-3 1-V Characteristics 628 25-4 Physics of Device Operation 629 25-5 Latchup in IGBTs 631 25-6 Switching Characteristics 634 25-7 Device Limits and SOAs 637 Summary 639 Problems 639 References 640 Chapter 26 Emerging Devices and Circuits 641 26-1 Introduction 641 26-2 Power Iunction Field Effect Transistors 641 26-3 Field-Controlled Thyristor 646 26-4 IFET-Based Devices versus Other Power Devices 648 26-5 MOS-Controlled Thyristors 649 xvi CONTENTS 26-6 Power Integrated Circuits 656 26-7 New Semiconductor Materials for Power Devices 661 Summary 664 Problems 665 References 666 PART 7 PRACTICAL CONVERTER DESIGN CONSIDERATIONS 667 Chapter 27 Snubber Circuits 669 27-1 Function and Types of Snubber Circuits 669 27-2 Diode Snubbers 670 27-3 Snuber Circuits for Thyristors 678 27-4 Need for Snubbers with Transistors 680 27-5 Turn-Off Snubber 682 27-6 Overvoltage Snubber 686 27-7 Turn-On Snubber 688 27 -8 Snubbers for Bridge Circuit Configurations 691 27-9 GTO Snubber Considerations 692 Summary 693 Problems 694 References 695 Chapter 28 Gate and Base Drive Circuits 696 28-1 Preliminary Design Considerations 696 28-2 dc-Coupled Drive Circuits 697 28-3 Electrically Isolated Drive Circuits 703 28-4 Cascode-Connected Drive Circuits 710 28-5 Thyristor Drive Circuits 712 28-6 Power Device Protection in Drive Circuits 717 28-7 Circuit Layout Considerations 722 Summary 728 Problems 729 References 729 Chapter 29 Component Temperature Control and Heat Sinks 730 29-1 Control of Semiconductor Device Temperatures 730 29-2 Heat Transfer by Conduction 731 29-3 Heat Sinks 737 29-4 Heat Transfer by Radiation and Convection 739 Summary 742 Problems 743 References 743 Chapter 30 Design of Magnetic Components 744 30-1 Magnetic Materials and Cores 744 30-2 Copper Windings 752 CONTENTS xvii 30-3 Thennal Considerations 754 30-4 Analysis of a Specific Inductor Design 756 30-5 Inductor Design Procedures 760 30-6 Analysis of a Specific Transfonner Design 767 30-7 Eddy Currents 771 30-8 Transfonner Leakage Inductance 779 30-9 Transfonner Design Procedure 780 30-10 Comparison of Transfonner and Inductor Sizes 789 Summary 789 Problems 790 References 792 Index 793 8-3 SINGLE-PHASE INVERTERS 211 8-3 SINGLE-PHASE INVERTERS 8-3-1 HALF-BRIDGE INVERTERS (SINGLE PHASE) Figure 8-10 shows the half-bridge inverter. Here, two equal capacitors are connected in series across the dc input and their junction is at a midpotential, with a voltage ~Vd across each capacitor. Sufficiently large capacitances should be used such that it is reasonable to assume that the potential at point 0 remains essentially constant with respect to the negative dc bus N. Therefore, this circuit configuration is identical to the basic one-leg inverter discussed in detail earlier, and v0 = V Ao. Assuming PWM switching, we find that the output voltage waveform will be exactly as in Fig. 8-5h. It should be noted that regardless of the switch states, the current between the two capacitors C + and C _ (which have equal and very large values) divides equally. When T + is on, either T + or D + conducts depending on the direction of the output current, and io splits equally between the two capacitors. Similarly, when the switch T _ i~ in its on state, either T _ or D_ conducts depending on the direction of io' and io splits equally between the two capacitors. Therefore, the capacitors C+ and C_ are "effectively" connected in parallel in the path of io . This also explains why the junction 0 in Fig. 8-10 stays at midpotential. Since io must flow through the parallel combination of C + and C _, io in steady state cannot have a dc component. Therefore, these capacitors act as dc ,blocking capacitors, thus eliminating the problem of transformer saturation from the primary side, if a trans- former is used at the output to provide electrical isolation. Since the current in the primary winding of such a transformer would not be forced to zero with each switching, the transformer leakage inductance energy does not present a problem to the switches. In a half-bridge inverter, the peak voltage and current ratings of the switches are as follows: (8-15) and (8-16) 8-3-2 FULL-BRIDGE INVERTERS (SINGLE PHASE) A full-bridge inverter is shown in Fig. 8-11. This inverter consists of two one-leg inverters of the type discussed in Section 8-2 and is preferred over other arrangements in higher power ratings. With the same dc input voltage, the maximum output voltage of the A + Figure 8-10 Half-bridge inverter. 212 CHAPTER 8 SWITCH-MODE de-ae INVERTERS: de ++ SINUSOIDAL ae id -+ + Voj DA+ "2 A Vd 0 + ~ DA- 2Figure 8-11 Single-phase full-bridge inverter. full-bridge inverter is twice that of the half-bridge inverter. This implies that for the same power, the output current and the switch currents are one-half of those for a half-bridge inverter. At high power levels, this is a distinct advantage, since it requires less paralleling of devices. 8-3-2-1 PWM with Bipolar Voltage Switching This PWM scheme was first discussed in connection with the full-bridge dc- dc convert- ers in Chapter 7. Here, the diagonally opposite switches (T A +, TB -) and (T A -, TB +) frona the two legs in Fig. 8-11 are switched as switch pairs 1 and 2, respectively. With this type of PWM switching, the output voltage waveform of leg A is identical to the output cI the basic one-leg inverter in Section 8-2, which is determined in the same manner by comparison .<:>f Vcontrol and Vtri in Fig. 8-12a. The output of inverter leg B is negative cI I "".- ~ o t I I I y V I I I ~ ,., r- ~. o ~ Vcontro! ~~~ I t'. I I YI I :~ Hi) I I I I 1- .. ~ t" .. ~. Vtri A A A A A ~ A .... ,/ ..... ,...... , ".,."'- V \ fa) Vol V -T / Vd / , I r. < ~ '-, r". .... - Vd - .- -.1 - ~ ""- (6) Figure 8-12 PWM with bipolar voltage switching. 8-3 SINGLE-PHASE INVERTERS 213 the leg A output; fOT example, when TA+ is on and VAo is equal to +~Vd' TB - is also on and VBo = -~Vd' Therefore (8-17) and (8-18) The Vo wavefonn is shown in Fig. 8-12b. The analysis carried out in Section 8-2 for the basic one-leg inverter completely applies to this type of PWM switching. Therefore, the peak of the fundamental-frequency component in the output voltage (Vol) can be obtained from Eqs. 8-7, 8-12, and S-ils as (ma s 1.0) (8-19) and (m a 1.0) (8-20) In Fig. 8-12b, we observe that the output voltage Vo switches between -Vd and + Vd voltage levels. That is the reason why this type of switching is called a PWM with bipolar voltage switching. The amplitudes of harmonics in the output voltage can be obtained by using Tab,e 8-1, as illustrated by the following example . • Example 8-2 In the full-bridge converter circuit of Fig. 8-11, Vd = 300 V, ma = 0.8, mf = 39, and the fundamental frequency is 47 Hz. Calculate the rms values of the fundamental-frequency voltage and some of the dominant harmonics in the output voltage v0 if a PWM bipolar voltage-switching scheme is used. Solution From Eq. 8-18, the hannonics in Vo can be obtained by multiplying the hannonics in Table 8-1 and Example 8-1 by a factor of 2. Therefore from Eq. 8-11, the nns voltage at any hannonic h is given as (V) = 1 . 2 • Vd eVAO)h = ~ (VAO)h oh V2 2 Vi2 V2 VJ2 = 212 13 (VAO)h . VJ2 Therefore, the rms voltages are as follows: Fundamental: Vol = 212.13 x 0.8 = 169.7 V at 47 Hz (Voh7 = 212.13 x 0.22 = 46.67 V at 1739 Hz (Voh9 = 212.13 x 0.818 ;: 173.52 V at 1833 Hz (Vo)41 ;: 212.13 x 0.22 ;: 46.67 V at 1927 Hz (Voh, = 212.13 x 0.314 = 66.60 V at 3619 Hz (Voh9 = 212.13 x 0.314 = 66.60 V at 3713 Hz etc. (8-21) • tIc-Side C~nt ;4' It is infonnative to look at the dc-side current id in the PWM biopolar voltage-switching scheme. For simplicity, fictitious L-C high-frequency filters will be used at the dc side as well as at the ac side, as shown in Fig. 8-13. The switching frequency is assumed to be very high, approaching infinity. Therefore, to filter out the high-switching-frequency compo-214 CHAPTER 8 SWITCH-MODE de-ae INVERTERS: de - SINUSOIDAL ae +Switch-mode id inverter - f. Figure 8- 13 Inverter with "fictitious" filters. nents in v0 and id , the filter components L and C required in both ac- and dc-side filters approach zero. This implies that the energy stored in the filters is negligible. Since the converter itself has no energy storage elements, the instantaneous power input must equal the instantaneous power output. Having made these assumptions, Vo in Fig. 8-13 is a pure sine wave at the funda-mental output frequency 00" (8-22) If the load is as shown in Fig. 8-13, where eo is a sine wave at frequency 00" then the output current would also be sinusoidal and would lag v0 for an inductive load such as an ac motor: (8-23) where is the angle by which io lags vo. On the dc side, the L-C filter will filter the high-switching-frequency components in id , and i; would only consist of the low-frequency and dc components. Assuming that no energy is stored in the filters, Vdi:<t) = vo(t)io(t) = V2Vosin w,tV2losin(w)t - <f» (8-24) Therefore where and .• Volo Volo . I~t) = Vd cos - Vd cos(2w,t - <f» = Id + Id2 = Id - V2ld 2cos(2w,t - <f» 1 Volo Id2 = V2 Vd (8-25) (8-26) (8-27) (8-28) Equation 8-26 for i; shows that it consists of a dc component Id' which is responsible for the power transfer from Vd on the dc side of the inverter to the ac side. Also, i; contains a sinusoidal component at twice the fundamental frequency. The inverter input current id consists of i; and the high-frequency components due to inverter switchings, as shown in Fig. 8-14. B-3 SINGLE-PHASE INVERTERS 215 o Vo (filtered) r- r- r- - r-- r-- r-- ...- I 'T r" ... -...... '" ... ~ r~ Vd /: rr' I{ ~' !. i"" -Vd ..... 1""" ., --L r-.... '-- '-- .... Figure 8-14 The dc-side current in a single-phase inverter with PWM bipolar voltage switching. In practical systems, the previous assumption of a constant dc voltage as the input to the inverter is not entirely valid. Normally, this dc voltage is obtained by rectifying the ac utility line voltage. A large capacitor is used across the rectifier output terminals to filter the dc voltage. The ripple in the capacitor voltage, which is also the dc input voltage to the inverter, is due to two reasons: (1) The rectification of the line voltage to produce dc does not result in a pure dc as discussed in Chapters S and 6, dealing with the line-frequency rectifiers. (2) As shown earlier by Eq. 8-26, the current drawn by asingle-phase inverter from the dc side is not a constant dc but has a second harmonic component (of the fundamental frequency at the inverter output) in addition to the high-switching-frequency components. The second harmonic current component results in a ripple in the capacitor voltage, although the voltage ripple due to the high switching frequencies is essentially negligible. 8-3-2-2 PWM with Unipolar Voltage Switching In PWM with unipolar voltage switching, the switches in the two legs of the full-bridge inverter of Fig. 8-11 are not switched simultaneously, as in the previous PWM scheme. Here, the legs A and B of the full-bridge inverter are controlled separately by comparing Vtri with Vcontrol and -Vcontrol> respectively. As shown in Fig. 8-1Sa, the comparison of Vcontrol with the triangular waveform results in the following logic signals to control the switches in leg A: Vcontrol Vtri: Vcontrol < Vtri: TA + on and VAN = Vd TA - on and VAN = 0 (8-29) The output voltage of inverter leg A with respect to the negative dc bus N is shown in Fig. 8-ISh. For controlling the leg B switches, -vcontrol is compared with the same triangular waveform, which yields the following: (-Vcontrol) > Vlri: (-Vcontrol) < Vlri: TB + on and VBN = Vd TB - on and VBN = 0 (8-30) 216 CHAPTER 8 SWITCH-MODE de-ac INVERTERS: de - SINUSOIDAL ac TB+~ on TA+ / _on (- Voonlrol) > VIr; Voanlrol > Vlri ( - voanlroi> (a) ,~~ nDODO ~ ~ ~ ~ ~ ~ [1. (b) 'B:~ ~ ~ ~ ~ ~ nnnnno [1. Va )(= VAN - vBN 1.0 0.8 0.6 0.4 o I- .... "" ~ - (e) ... , k'V"l " ~' '" ~ "" (d) 0.2 r kO~----~~ __ -LI _ ~I'~~ __ ~I~f~~I~~ ___ II~I _ ~1 __ ~II~ __ ~'-Lf~f.lf~f~I~.A m, 12m, 3m, 4m, (2m, - 1) (2m, + 1) (h.rmonics of 'I) {e} Figure 8-15 PWM with unipolar voltage switching (single phase). Because of the feedback diodes in andparallel with the switches, the forego-ing voltages given by Eqs. 8-29 and 8-30 are independent of the direction of the output current io • The wavefonns of Fig. 8-15 show that there are four combinations of switch on-states and the corresponding voltage levels: 8-3 SINGLE-PHASE INVERTERS 217 1. TA +, TB - on: VAN :::::: Vd, V BN :::::: 0; Vo:::::: Vd T A -, TB + on: VAN :::::: 0, V BN :::::: Vd; vo:::::: -Vd T A +, TB + on: VAN :::::: V d , V BN :::::: Vd ; Vo:::::: 0 (8-31) TA -, TB - on: VAN :::::: 0, V BN :::::: 0; vo:::::: 0 We notice that when both the upper switches are on, the output voltage is zero. The output current circulates in a loop through TA + and DB + or D A + and TB + depending on the direction of io • During this interval, the input current id is zero. A similar condition occurs when both bottom switches TA - and TB - are on. In this type of PWM scheme, when a switching occurs, the output voltage changes between zero and + Vd or between zero and - Vd voltage levels. For this reason, this type of PWM scheme is called PWM with a unipolar voltage switching, as opposed to the PWM with bipolar (between + Vd and - Vd) voltage-switching scheme described earlier. This scheme has the advantage of "effectively" doubling the switching frequency as far as the output harmonics are concerned, compared to the bipolar voltage-switching scheme. Also, the voltage jumps in the output voltage at each switching are reduced to Vd , as compared to 2V d in the previous scheme. The advantage of "effectively" doubling the switching frequency appears in the harmonic spectrum of the output voltage waveform, where the lowest harmonics (in the idealized circuit) appear as sidebands of twice the switching frequency. It is easy to understand this if we choose the frequency modulation ratio mf to be even (mf should be odd for PWM with bipolar voltage switching) in a single-phase inverter. The voltage waveforms VAN and VBN are displaced by 180 0 of the fundamental frequency II with respect to each other. Therefore, the harmonic components at the switching frequency in VAN and VBN have the same phase (c1>AN - c1>BN:::::: 180 0 • mf:::::: 0°, since the waveforms are 180° displaced and mf is assumed to be even). This results in the cancellation of the harmonic component at the switching frequency in the output voltage V 0 :::::: VAN - V BN • In addition, the sidebands of the switching-frequency harmonics disappear. In a similar manner, the other dominant harmonic at twice the switching frequency cancels out, while its sidebands do not. Here also (ma :s; 1.0) (8-32) and (m a > 1.0) (8-33) • Example 8·3 In Example 8-2, suppose that a PWM with unipolar voltage-switching scheme is used, with mf :::::: 38. Calculate the rms values of the fundamental-frequency voltage and some of the dominant harmonics in the output voltage. Solution Based on the .. discussion of unipolar voltage switching, the harmonic order h can be written as (8-34) where the harmonics exist as sidebands around 2mf and the multiples of 2m f . Since h is odd, kin Eq. 8-34 attains only odd values. From Example 8-2 (VAoh (Vo)h:::::: 212.13 Vd /2 (8-35) 218 CHAPTER 8 SWITCH-MODE de-ae INVERTERS: de ....,. SINUSOIDAL ae Using Eq. 8-35 and Table 8-1, we find that the rms voltages are as follows: At fundamental or 47 Hz: Vol = 0,8 x 212.13 = 169.7 V At h =. 2mf - 1 == 75 or 3525 Hz: {Vohs = 0.314 x 212.13 = 66.60 V At h = 2mf + 1 :; 77 or 3619 Hz: {Voh7 = 0.314 X 212.13 = 66.60 V etc. Comparison of the unipolar voltage switching with the bipolar voltage switching of Example 8-2 shows that, in both cases, the fundamental-frequency voltages are the same for equal mao However, with unipolar voltage switching, the dominant harmonic voltages centered around mf disappear, thus resulting in a significantly lower harmonic content. • dc-Side Current id • Under conditions similar to those in the circuit of Fig. 8-13 for the PWM with bipolar voltage switching, Fig. 8-16 shows the dc-side current id for the PWM unipolar voltage-switching scheme, where mf = 14 (instead of mf = 15 for the bipolar voltage switching). By comparing Figs. 8-14 and 8-16, it is clear that using PWM with unipolar voltage switching results in a smaller ripple in the current on the dc side of the inverter. 8-3-2-3 Square-Wave Operation The full-bridge inverter can also be operated in a square-wave mode. Both types of PWM discussed earlier degenerate into the same square-wave mode of operation, where the switches (TA+' TB -) and (T B +, TA -) are operated as two pairs with a duty ratio of 0.5. As is the case in the square-wave mode of operation, the output voltage magnitude given below is regulated by controlling the input dc voltage: ~ 4 VI = -Vd o 11 8-3-2-4 Output Control by Voltage CaneeUation (8-36) This type of control is feasible only in a single-phase, full-bridge inverter circuit. It is based on the combination of square-wave switching and PWM with a unipolar voltage switching. In the circuit of Fig. 8-17a, the switches in the two inverter legs are controlled separately (similar to PWM unipolar voltage switching). But all switches have a duty raio of 0.5, similar to a square-wave control. This results in waveforms forv AN and vBN shown in Fig. 8-17b, where the waveform overlap angle a can be controlled. During this overlap interval, the output voltage is zero as a consequence of either both top switches or both Figure 8-16 The dc-side current in a single-phase inverter with PWM unipolar voltage switching. 8-3 SINGLE-PHASE INVERTERS 219 VAN +J i" -+ 11 .. va -1 a/-- r f--lso°----j VBN I J I I N I(a) f--1SO°--l Va 1.0,--_~ I" - a"l (ISO )0 0.8 o I TI Vel -crf.o- : r 1-8-1 1. \ -Vel --~ IPilO - a)~ (e) fl = (1(10 2 .. )0 = (90 _ ;)" (b) F~e 8-17 Full-bridge, single-phase inverter control by voltage cancellation: (a) power c:ircuit; (b) wavefonn5; (c) normalized fundamental and harmonic voltage output and total hannonic distortion as a function of a. bottom switches being on. With a = 0, the output waveform is similar to a square-wave inverter with the maximum possible fundamental output magnitude. It is easier to derive the fundamental and the harmonic frequency components of the output voltage in terms of ~ =: 90° - ~, as is shown in Fig. 8-17b: • 2 fTrl2 (Vo)1I = - vocos(hO) dO 11' -'1'(12 = ~ f~ VdCOS(h9) de 11' _p A 4 :,(V o)1I = lI'h Vdsin(h~) (8-37) where ~ = 90° - ~a and h is an odd integer. Figure 8-17e shows the variation in the fundamental-frequency component as well as the harmonic voltages as a function of a. These are normalized with respect to the fundamental-frequency component for the square-wave (a == 0) operation. The total harmonic distortion, which is the ratio of the rms value of the harmonic distortion to the rms value of the fundamental-frequency component, is also plotted as a function of a. Because of a large distortion, the curves are shown as dashed for large values of a. 8 =CHAPTER 8 SWITCH-MODE de-ac INVERTERS: de - SI!'.1]SOIDAL ae 8-3-2-5 Switch Utilization in Full-Bridge Inverters Similar to a half-bridge inverter, if a transformer is utilized at the output of a full-bridge inverter, the transformer leakage inductance does not present a problem to the switches. Independent of the type of control and the switching scheme used, the peak switch voltage and current ratings required in a full-bridge inverter are as follows: VT = Vd (8-38) and (8-39) 8-3-2-6 Ripple in the Single-Phase Inverter Output The ripple in a repetitive waveform refers to the difference between the instantaneous values of the waveform and its fundamental-frequency component. Figure 8-18a shows a single-phase switch-mode inverter. It is assumed to be sup-plying an induction motor load, which is shown by means of a simplified equivalent circuit with a counter electromotive force (emf) eo' Since eo(t) is sinusoidal, only the sinusoidal (fundamental-frequency) components of the inverter output voltage and current are responsible for the real power transfer to the load. We can separate the fundamental-frequency and the ripple components in Vo and io by applying the principle of superposition to the linear circuit of Fig. 8-18a. Let v0 = Vol + vripple and io = io1 + iripple' Figures 8-18b, c show the circuits at the fundamental frequency and at the ripple frequency, respectively, where the ripple frequency compo-nent consists of sub-components at various harmonic frequencies. Therefore, in a phasor form (with the fundamental-frequency components designated by subscript 1) as shown in Fig. 8-18d, Single-phase inverter (b) (a) (8-40) (c) (el) Figure 8-18 Single-phase inverter: (a) circuit; (b) fundamental-frequency components; (c) ripple frequency components; (rf) fundamental-frequency phasor diagram. 8-3 SINGLE-PHASE INVERTERS 221 Since the superposition principle is valid here, all the ripple in Vo appears across L, where Vripple(t) :::: Vo - Vol The output current ripple can be calculated as I f' iripp1e(t) = L 0 Vripple(') d, + k where k is a constant and , is a variable of integration. (8-41) (8-42) With a properly selected time origin t :::: 0, the constant k in Eq. 8-42 wiII be zero. Therefore, Eqs. 8-41 and 8-42 show that the current ripple is independent of the power being transferred to the load. As an example, Fig. 8-19a shows the ripple current for a square-wave inverter output. Figure 8-l9b shows the ripple current in a PWM bipolar voltage switching. In drawing Figs. 8-19a and 8-l9b, the fundamental-frequency components in the inverter output voltages are kept equal in magnitude (this requires a higher value of Vd in the PWM inverter). The PWM inverter results in a substantially smaller peak ripple current com- pared to the square-wave inverter. This shows the advantage of pushing the harmonics in the inverter output voltage to as high frequencies as feasible, thereby reducing the losses in the load by reducing the output current harmonics. This is achieved by using higher inverter switching frequencies, which would result in more frequent switchngs and hence higher switching losses in the inverter. Therefore, from the viewpoint of the overall system energy efficiency, a compromise must be made in selecting the inverter switching frequency. 8-3-3 PUSH-PULL INVERTERS Figure 8-20 shows a push-pull inverter circuit. It requires a transformer with a center-tapped primary. We will initially assume that the output current io flows continuously. With this assumption, when the switch T) is on (and T2 is off), T) would conduct for a positive value of io ' and Dl would conduct for a negative value of io ' Therefore, regardless of the direction of io ' Vo :::: Vd1n, where n is the transformer turns ratio between the primary half and the secondary windings, as shown in Fig. 8-20. Similarly, when T2 is on (and T) is off), Vo :::: - Vd1n. A push-pull inverter can be operated in a PWM or asquare-wave mode and the waveforms are identical to those in Figs. 8-5 and 8-12 for half-bridge and full-bridge inverters. The output voltage in Fig. 8-20 equals and • Vd Vol :::: ma-; Vd • 4 Vd -<V)<--n 0 1T n(ma ~ 1.0) (m a 1.0) In a push-pull inverter, the peak switch voltage and current ratings are (8-43) (8-44) (8-45) The main advantage of the push - pull circuit is that no more than one switch in series conducts at any instant of time. This can be important if the dc input to the converter is from a low-voltage source, such as a battery, where the voltage drops across more than one switch in series would result in a significant reduction in energy efficiency. Also, the Vo vol .," Le .. ........ 71 Or- 4", , .," / ---Vripple = Vo - vol o~ /1 -- r.. ~~"'J (a) Vo v - ~ .... -, k" .~ o ~ " ~ .... l,..-- ~. "'"--- v ripple = Vo - vol i'. / o r--.r---., ,...,~ "' ..... ~, I--" ..... ""'" 1\ II •• ~ A " A II. ' -'l\fl A~ -YV{'J 0"" (b) Figure 8-19 Ripple in the inverter output: (a) square-wave switching; (b) PWM bipolar voltage switching. §rrl =0:> ~::r:: ~ g rrl Q.. (") I i!i ~rrl :j rrl =~ ~ t r!l ~en 2 I:' ~ 0: (") + vd -::r: Dl Tl T2 ~2 1 18-3 SINGLE-PHASE INVERTERS 223 II I f 1 J # \ \~ [ I,,;1 Figure 8-20 Push-pull inverter (single phase). control drives for the two switches have a common ground. It is, however, difficult to avoid the dc saturation of the transfonner in a push-pull inverter. The output current, which is the secondary current of the transfonner, is a slowly varying current at the fundamental output frequency. It can be assumed to be a constant during a switching interval. When a switching occurs, the current shifts from one half to the other half of the primary winding. This requires very good magnetic coupling between these two half-windings in order to reduce the energy associated with the leakage induc-tance of the two primary windings. This energy will be dissipated in the switches or in snubber circuits used to protect the switches. This is a general phenomenon associated with all converters (or inverters) with isolation where the current in one of the windings is forced to go to zero with every switching. This phenomenon is very important in the design of such converters. In a pulse-width-modulated push-pull inverter for producing sinusoidal output (un-like those used in switch-mode dc power supplies), the transfonner must be designed for the fundamental output frequency. The number of turns will therefore be high compared to a transfonner designed to operate at the switching frequency in a switch-mode dc power supply. This will result in a high transfonner leakage inductance, which is proportional to the square of the number of turns, provided all other dimensions are kept constant. This makes it difficult to operate a sine-wave-modulated PWM push-pull inverter at switching frequencies higher than approximately 1 kHz. 8-3-4 SWITCH UTILIZATION IN SINGLE-PHASE INVERTERS Since the intent in this section is to compare the utilization of switches in various single-phase inverters, the circuit conditions are idealized. We will assume that Vd,max is the highest value of the input voltage, which establishes the switch voltage ratings. In the PWM mode, the input remains constant at Vd,max' In the square-wave mode, the input voltage is decreased below Vd,max to decrease the output voltage from its maximum value. Regardless of the PWM or the square-wave mode of operation, we assume that there is enough inductance associated with the output load to yield a purely sinusoidal current (an idealized condition indeed for a square-wave output) with an nns value of lo.max at the maximum load. If the output current is assumed to be purely sinusoidal, the inverter nns volt-ampere output at the fundamental frequency equals Vol o• max at the maximum rated output, where the subscript 1 designates the fundamental-frequency component of the inverter output. 224 CHAPTER 8 SWITCH-MODE dC-8c INVERTERS: dc ++ SINUSOIDAL 8C With VT and IT as the peak voltage and current ratings of a switch, the combined utili-zation of all the switches in an inverter can be defined as V I Switch utilization ratio = OIVO~ q T T where q is the number of switches in an inverter. (8-46) To compare the utilization of switches in various single-phase inverters, we will initially compare them for a square-wave mode of operation at the maximum rated output. (The maximum switch utilization occurs at Vd = Vd•max ') Push-Pull Inverter VT = 2V d ,max Ir = V2/ o ,max n4 Vdmax V -----'-ol,max - 11'V2 n q=2 (n = turns ratio, Fig. 8-20) I (8-47) :.Maximum switch utilization ratio = 211' = 0.16 (8-48) Half-Bridge Inverter 4 Vd max V -----'-ol.max - 11'V2 2 q = 2 (8-49) I :.Maximum switch utilization ratio = 211' = 0.16 Full-Bridge Inverter 4 VT = Vd,max Ir = V2l o ,max = 11'V2 Vd,max q = 4 1 :.Maximum switch utilization ratio = 211' = 0.16 This shows that in each inverter, the switch utilization is the same with 1 Maximum switch utilization ratio = 211' = 0.16 (8-50) (8-51) (8-52) (8-53) In practice, the switch utilization ratio would be much smaller than 0.16 for the following reasons: (1) switch ratings are chosen conservatively to provide safety margins; (2) in determining the switch current rating in a PWM inverter, one would have to take into account the variations in the input dc voltage available; and (3) the ripple in the output current would influence the switch current rating. Moreover, the inverter may be required to supply a short-term overload. Thus, the switch utilization ratio, in practice, would be substantially less than the 0.16 calculated. At the lower output volt-amperes compared to the maximum rated output, the switch utilization decreases linearly. It should be noted that using a PWM switching with ma !S; 1.0, this ratio would be smaller by a factor of (1I'/4)m a as compared to the square-wave switching: M . . h '1" . 1 11' 1axlmum SWltc utllzatlon ratio = 211' "4 ma = gma (8-54) (PWM, ma !S; 1.0) Therefore, the theoretical maximum switch utilization ratio in a PWM switching is only 0.125 at ma = 1, as compared with 0.16 in a square-wave inverter. 8-4 THREE-PHASE INVERTERS 225 • Example 84 In a single-phase full-bridge PWM inverter, Vd varies in a range of 295-325 V. The output voltage is required to be constant at 200 V (nns) , and the maximum load current (assumed to be sinusoidal) is IO A (nns). Calculate the combined switch utilization ratio (under these idealized conditions, not accounting for any overcur-rent capabilities). Solution In this inverter VT = Vd,max = 325 V IT = Vilo = Vi x 10 == 14.14 q = no. of switches = 4 The maximum output volt-ampere (fundamental frequency) is Vollo,max = 200 X 10 == 2000 VA Therefore, from Eq. 8-46 V I 2000 Switch utilization ratio == ;~~;; == 4 X 325 X 14.14 = 0.11 1-4 THREE-PHASE INVERTERS (8-55) • In applications such as uninterruptible ac power supplies and ac motor drives, three-phase inverters are commonly used to supply three-phase loads. It is possible to supply a three-phase load by means of three separate single-phase inverters, where each inverter produces an output displaced by 120° (of the fundamental frequency) with respect to each other. Though this arrangement may be preferable under certain conditions, it requires either a three-phase output transfonner or separate access to each of the three phases of the load. In practice, such access is generally not available. Moreover, it requires 12 switches. The most frequently used three-phase inverter circuit consists of three legs, one for each phase, as shown in Fig. 8-21. Each inverter leg is similar to the one used for describing the basic one-leg inverter in Section 8-2. Therefore, the output of each leg, for example VAN (with respect to the negative dc bus), depends only on Vd and the switch i4 + ~ 2DTB+ A+ Vd Vd 2 DTe- B-Dc-DTB- A- AB c Figure 8-21 Three-phase inverter. CHAPTER 8 SWITCH-MODE de-ae INVERTERS: de - SINUSOIDAL ae status; the output voltage is independent of the output load current since one of the two switches in a leg is always on at any instant. Here, we again ignore the blanking time required in practical circuits by assuming the switches to be ideal. Therefore, the inverter output voltage is independent of the direction of the load current. 8-4-1 PWM IN THREE-PHASE VOLTAGE SOURCE INVERTERS Similar to the single-phase inverters, the objective in pulse-width-modulated three-phase inverters is to shape and control the three-phase output voltages in magnitude and fre-quency with an essentially constant input voltage Yd' To obtain balanced three-phase output voltages in a three-phase PWM inverter, the same triangular voltage waveform is compared with three sinusoidal control voltages that are 120° out of phase, as shown in Fig. 8-22a (which is drawn for mf = IS). It should also be noted from Fig. 8-22b that an identical amount of average dc component is present in the output voltages vAN and VBN' which are measured with respect to the negative dc bus. These dc components are canceled out in the line-to-line voltages, for example in vAB shown in Fig. 8-22b. This is similar to what happens in a single-phase full-bridge inverter utilizing a PWM switching. In the three-phase inverters, only the harmonics in the line-to-line voltages are of concern. The harmonics in the output of anyone of the legs, for example vAN in Fig. 8-22b, are identical to the harmonics in VAo in Fig. 8-S, where only the odd harmonics exist as sidebands, centered around mf and its multiples, provided mf is odd. Only considering the harmonic at mf(the same applies to its odd multiples), the phase difference between the mf harmonic in VAN and VBN is (120 mft. This phase difference will be equivalent to zero (a mUltiple of 360°) if mf is odd and a multiple of 3. As a consequence, the harmonic at mf is suppressed in the line-to-line voltage vAB' The same argument applies in the suppression of harmonics at the odd multiples of mf if mf is chosen to be an odd multiple of 3 (where the reason for choosing mf to be an odd multiple of 3 is to keep mf odd and, hence, eliminate even harmonics). Thus, some of the dominant harmonics in the one-leg inverter can be eliminated from the line-to-line voltage of a three-phase inverter. PWM considerations are summarized as follows: I. For low values of mf' to eliminate the even harmonics, a synchronized PWM should be used and!7'f should be an odd integer. Moreover, mf should be amultiple of 3 to cancel out the most dominant harmonics in the line-to-line voltage. For large values ofmf. the comments in Section 8-2-1-2 for a single-phase PWM apply. During overmodulation (ma > 1.0), regardless of the value of mf' the conditions pertinent to a small mf should be observed. 8-4-1-1 Linear Modulation (ma ::s 1.0) In the linear region (ma ::s 1.0), the fundamental-frequency component in the output voltage varies linearly with the amplitude modulation ratio ma' From Figs. 8-Sb and 8-22b, the peak value of the fundamental-frequency component in one of the inverter legs is (8-S6) 1,0 0,8 0,6 0,4 0,2 ° 8-4 THREE-PHASE INVERTERS 227 (4) VAN .... ~- r-- TVal o I t VSN ] ~ ~ ~ ~ ~ nnmnn ~ ~ ~ [r, ° I.. ¥ f I L" II m,~ ~ (m, + 2) vFundamental VLLI 1 ~ .... .~ j., i-' (0) m" = 0,8. m, = 15 " I I ILl" Il 2m,\;( (2m, + 1) Harmonics of ft (e) I V ~ I" Figure 8-22 Three-phase PWM waveforms and harmonic spectrum. 228 CHAPTER 8 swrrcI PdODE dC-8c INVERTERS: dc ..... SINCSOIDAL 8C Therefore, the line-to-line rms voltage at the fundamental frequency, due to 120 0 phase displacement between phase voltages, can be written as VLL v'3 A (Iine-Iin~, rms) = "Vi (V ANh V3 = 2y'2 m a Vd = 0.612maVd (ma $ 1.0) (8-57) The harmonic components of the line-to-line output voltages can be calculated in a similar manner from Table 8-1, recognizing that some of the harmonics are canceled out in the line-to-line voltages. These rms harmonic voltages are listed in Table 8-2. 8-4-1-2 Ovennodulation (ma > 1.0) In PWM overmodulation, the peak of the control voltages are allowed to exceed the peak of the triangular waveform. Unlike the linear region, in this mode of operation the fundamental-frequency voltage magnitude does not increase proportionally with mao This is shown in Fig. 8-23, where the rms value of the fundamental-frequency line-to-Iine voltage VLLI is plotted as a function of mao Similar to a single-phase PWM, for sufficiently large values of m a , the PWM degenerates into a square-wave inverter waveform. This results in the maximum value of VLL J equal to 0.78Vd as explained in the next section. In the overmodulation region compared to the region with ma $ 1.0, more sideband harmonics appear centered around the frequencies of harmonics mf and its multiples. However, the dominant harmonics may not have as large an amplitude as with ma $ 1.0. Therefore, the power loss in the load due to the harmonic frequencies may not be as high in the ovennodulation region as the presence of additional sideband harmonics would suggest. Depending on the nature of the load and on the switching frequency, the losses due to these harmonics in overmodulation may be even less than those in the linear region of the PWM. Table 8-2 Generalized Hannonics of Vn for a Large and Odd mf That Is a Multiple of 3. ~ 0.2 0.4 0.6 0.8 1.0 10.122 0.245 0.367 0.490 0.612 mf ± 2 0.010 0.037 0.080 0.135 0.195 mf ± 4 0.005 0.011 2m f ± 1 0.116 0.200 0.227 0.192 0.111 2m f ± 5 0.008 0.020 3m f ± 2 0.027 0.085 0.124 0.108 0.038 3m f ± 4 0.007 0.029 0.064 0.0% 4m f ± 1 0.100 0.096 0.005 0.064 0.042 4m f ± 5 0.021 0.051 0.073 4m f ± 7 0.010 0.030 Note: (V LLhlV dare tabulated as a function of rna where (V LL)h are the nns values of the harmonic voltages. 8-4 THREE-PHASE INVERTERS 229 Square-wave :ll '" 0.78 - --- - -:.:;;--..,..--r .J1. '" 0.612 - 2../2 I--f-L....,I--------+-O-Square-wave IOvermodulationl II II II II 0·L......-~---~,.---· "'. Figure 8-23 Three-phase inverter; VLL,(nns)/V d as a function of rna' 8-4-2 SQUARE-WAVE OPERATION IN THREE-PHASE INVERTERS If the input dc voltage Vd is controllable, the inverter in Fig. 8-24a can be operated in a square-wave mode. Also, for sufficiently large values of rna' PWM degenerates into square-wave operation and the voltage waveforms are shown in Fig. 8-24h. Here, each switch is on for 180 0 (i.e., its duty ratio is 50%). Therefore, at any instant of time, three switches are on. In the square-wave mode of operation, the inverter itself cannot control the magnitude of the output ac voltages. Therefore, the dc input voltage must be controlled in order to control the output in magnitude. Here, the fundamental-frequency line-to-line rms voltage component in the output can be obtained from Eq. 8-13 for the basic one-leg inverter operating in a square-wave mode: Vu. V3 4 Vd ,== __ _(rms) V211" 2 V6 ==-Vd 11" = 0. 78V d (8-58) The line-to-line output voltage waveform does not depend on the load and contains harmonics (6n ± 1; n == 1, 2, ... ), whose amplitudes decrease inversely proportional to their harmonic order, as shown in Fig. 8-24c: where 0.78 Vu. == --Vd • hh = 6n ± 1 (n == 1,2,3, ... )(8-59) 230 CHAPTER 8 SWITCH-MODE dc-ftc INVERTERS: de - SINUSOIDAL ftC + 1.21 1.0 0.8 0.6 0.4 0.2 .A. (a) DC- B c harmonics 0L-1~~3~~~~~~~~~~~~ mh (c) VBN I 1---180°---i I TB+ IOL_T-=B:-- L-_ --L ____ " "'I( vCN oBLT_C_+-L. __ T~C:..-__ '__T_C_+ ___ " "'It (6) Figure 8-24 Square-wave inverter (three phase). It should be noted that it is not possible to control the outRlltl!lagnitude l!1 a three- phase, square-wave inverter by means of voltage cancellation as described in Section 8-3-2-4. 8-4-3 SWITCH UTILIZATION IN THREE-PHASE INVERTERS We will assume that Vd •max is the maximum input voltage that remains constant during PWM and is decreased below this level to control the output voltage magnitude in a square-wave mode. We will also assume that there is sufficient inductance associated with the load to yield a pure sinusoidal output current with an rms value of lo,max (both in the PWM and the square-wave mode) at maximum loading. Therefore, each switch would have the following peak. ratings: (8-60) and (8-61) If VIL, is the rms value of the fundamental-frequency line-to-line voltage component, the three-phase output volt-amperes (rms) at the fundamental frequency at the rated out-put is (VA)3_phase = 0v IL ,/o,max (8-62) + N 8-4 THREE-PHASE INVERTERS 231 Therefore, the total switch utilization ratio of all six switches combined is S 'h 'I" , (V Ah-phue WltC uti IzatIon ratio = 6V Th _ V3V u ,Io.maA 6V d •max 0l o •max I Vu , = 2V6 Vd •max (8-63) In the PWM linear region (ma :S 1,0) using Eq. 8-57 and noting that the maximum switch utilization occurs at Vd = Vd. max , Maximum switch utilization ratio I V3 (PWM) = -2V6-6 2-0-2 ma = !m 8 a (8-64) In the square-wave mode, this ratio is 1/211' = 0.16 compared to a maximum of 0.125 for a PWM linear region with ma = 1.0. In practice, the same derating in the switch utilization ratio applies as discussed in Section 8-3-4 for single-phase inverters. Comparings Eqs. 8-54 and 8-64, we observe that the maximum switch utilization ratio is the same in a three-phase, three-leg inverter as in a single-phase inverter. In other words, using the switches with identical ratings, a three-phase inverter with 50% increase in the number of switches results in a 50% increase in the output volt-ampere, compared to a single-phase inverter. 8-4-4 RIPPLE IN THE INVERTER OUTPUT Figure 8-25a shows a three-phase, three-leg, voltage source, switch-mode inverter in a block diagram form. It is assumed to be supplying a three-phase ac motor load. Each phase of the load is shown by means of its simplified equivalent circuit with respect to the load neutral n. The induced back-emfs eA(t), eB(t), and ec(t) are assumed to be sinusoidal . A Three-phase three-leg !---oOB inverter ...... -ooc -ic (0) ..,.--- ...... ,. R = 0 ...... , /L"/ # ,/ ,I \ ## ( \ I II I \I(LOId neutral) J \II\I/ ,I / " I // ..... '" ..... ---""laid Fl8W'e 8-25 Three-phase inverter: (a) circuit diagram; (b) phasor diagram (fundamental &equency), CHAPTER 8 SWITCH-MODE dc-ac INVERTERS: dc - SINUSOIDAL ac Under balanced operating conditions, it is possible to express the inverter phase output voltages vAN. and so on (with respect to the load neutral n), in tenns of the inverter output voltages with respect to the negative dc bus N: Each phase voltage can be written as di k Vkn = L dt + ekn In a three-phase, three-wire load and (k = A, B, C) (k = A, B, C) d d/iA + iB + id = 0 (8-65) (8-66) (8-()7a) (8-67b) Similarly, under balanced operating conditions, the three back-emfs are a balanced three-phase set of voltages, and therefore (8-68) From the foregoing equations, the following condition for the inverter voltages can be written: (8-69) Using Eqs. 8-65 through 8-69, VnN = ~(v AN + VBN + VCN) (8-70) Substituting V nN from Eq. 8-70 into Eq. 8-65, we can write the phase-to-neutral voltage for phase A as (8-71) Similar equations can be written for phase B and C voltages. Similar to the discussion in Section 8-3-2-6 for the ripple in the single-phase inverter output, only the fundamental-frequency components of the phase voltage vAll, and the output current i A1 are responsible for the real power transfonner since the back-emf eA(t) is assumed to be sinusoidal and the load resistance is neglected. Therefore, in a phasor fonn as shown in Fig. 8-25b VAil, = EA + jWlLIA I (8-72) By using the principle of superposition, all the ripple in vAn appears across the load inductance L. Using Eq. 8-71, the wavefonn for the phase-to-Ioad-neutral voltage VAil is shown in Figs. 8-260 and 8-26b for square-wave and PWM operations, respectively. Both inverters have identical magnitudes of the fundamental-frequency voltage component VAil" which requires a higher Vd in the PWM operation. The voltage ripple Vrippl e (=VAII -VAll,) is the ripple in the phase-to-neutral voltage. Assuming identical loads in these two cases, the output current ripple is obtained by using Eq. 8-42 and plotted in Fig. 8-26. This current ripple is independent of the eower beinK transferred, that is, the current ripple would be the same so long as for a given load inductance L, the ripple in the inverter output voltage remains constant in magnitude and frequency. This comparison indicates that for large values of ml' the currel}tripplC? il! the Y}YM inverter_ will~ significantly lower compared to a. square-wave inverter. lIA" VAn i Vd ir;pple. A iripple. A Q:> ..i:. ~ fa) fb) ~ Figure 8·26 Phase-to-Ioad-neutral variables of a three-phase inverter: (a) square wave; (6) PWM. ~ ~ ~ e'Xl ~234 CHAPTER 8 SWITCH-MODE de-ac INVERTERS: de - SINUSOIDAL ae 8-4-5 dc-SIDE CURRENT id Similar to the treatment of a single-phase inverter, we now look at the voltage and current wavefonns associated with the dc side of a pulse-width-modulated, three-phase inverter. The input voltage Vd is assumed to be dc without any ripple. If the switching frequency in Fig. 8-25a is assumed to approach infinity, then similar to Fig. 8-13, a fictitious filter with negligible energy storage can be inserted on the ac side and the current at the inverter output will be sinusoidal with no ripple. Because of the assumption of no energy storage in the fictitious ac-side filter, the instantaneous ac power output can be expressed in tenns fundamental-frequency output voltages and currents. Similarly, on the dc side, a fictitious filter with no energy storage, as shown in Fig. 8-13, can be assumed. Then, the high- switching-frequency components in id are filtered. Now equating the instantaneous power input to the instantaneous power output, we get Vdi; = vA",(t)iA(t) + vs"Jt)is(t) + vc",(t)idt) (8-73) In a balanced steady-state operation, the three phase quantities are displaced by 120° with respect to each other. Assuming that is the phase angle by which a phase current lags the inverter phase voltage and V2Vo and V2Io are the amplitudes of the phase voltages and currents, respectively, yields • 2Volo [ id = -v;- cos Wit COS(Wlt - <1» + COS(Wlt - 1200)cos(wlt - 120° - <1» +COS(Wlt + 120°) COS(Wlt + 120° - <1»] 3Volo = -v;-cos = Id (a dc quantity) (8-74) The foregoing analysis shows that i; is a dc quantity, unlike in the single-phase inverter, where i; contained a component at twice the output frequency. However, ill consists of high-frequency switching components as shown in Fig. 8-27, in addition to I;. These high-frequency components, due to their high frequencies, would have a neg-ligible effect on the capacitor voltage Vd • 8-4-6 CONDUCTION OF SWITCHES IN THREE-PHASE INVERTERS We discussed earlier that the output voltage does not depend on the load. However, the duration of each switch conduction is dependent on the power factor of the load. 8-4-6-1 Square-Wave Operation Here, each switch is in its on state for 180°. To determine the switch conduction interval, a load with a fundamental-frequency displacement angle of 30° (lagging) is assumed (as ~ '" ) ~ '" J~ I ~ ~ jI ) r II t Ir II J~ = i; 1 oFigure 8-27 Input dc current in a three-phase inverter. Devices conducting: D A + VAN 8-4 THREE-PHASE INVERTERS 235 " (a) Figure 8-28 Square-wave inverter: phase A wavefonns. an example). The waveforms are shown in Fig. 8-28 for one of the three phases. The phase-to-neutral voltages VAn and VAn, are shown in Fig. 8-28a. In Fig. 8-28b, VAN (with respect to the negative dc bus), iA , and its fundamental component iA , are plotted. Even though the switches TA + and TA - are in their on state for 180°, due to the lagging power factor of the load, their actual conduction intervals are smaller than 180°. It is easy to interpret that as the power factor (lagging) of the load decreases, the diode conduction intervals will increase and the switch conduction intervals will decrease. On the other hand, with a purely resistive load, theoretically the feedback diodes would not conduct at all. 8-4-6-2 PWM Operation The voltage and current waveforms associated with a PWM inverter are shown in Fig. 8-29. Here, as an example, the load displacement power factor angle is assumed to be 30° (lagging). Also, the output current is assumed to be a perfect sinusoid. In Figs. 8-29a through 8-29c, the phase to the negative dc bus voltages and the phase current (VAn> iA • etc.) are plotted for approximately one-fourth of the fundamental-frequency cycle. By looking at the devices conducting in Figs. 8-29a through 8-29c, we notice that there are intervals during which the phase currents iA , iB , and ic flow through only the devices connected to the positive dc bus (i.e., three out ofT A +, D A +, TB +, D B +, Tc +, and Dc+)' This implies that during these intervals, all three phases of the load are short-circuited and there is no power input from the dc bus (i.e .• id = 0), as shown in Fig. 8-30a. Similarly, there are intervals during which all conducting devices are connected to the negative dc bus resulting in the circuit of Fig. 8-30b. The output voltage magnitude is controlled by controlling the duration of these sh~t1::Qrc_pitiotervals. Such intervals of three-phase short circuit do not exist in a square-wave mode of operation. Therefore, the output voltage magnitude in an inverter operating in a square-wave mode must be controlled by controlling the input voltage Vd' 236 CHAPTER 8 SWITCH-MODE de-ae IWERTERS: de - SINUSOIDAL ae + Vd (a) (b) (e) VAN VBN VCN Or-~I ~I--~~~+-~~~~---u----u----p'--------, I I I I I~ (T A -. TB _. Dc-> conducting I<'igure 8-29 PWM inverter wavefonns: load power factor angle = 30° (lag). id = 0 ill = 0 - -+IiB - I ic Vd - (a) fb) Figure 8-30 Short-circuit states in a three-phase PWM inverter. 8-5 EFFECT OF BLANKING TIME ON VOLTAGE IN PWM INVERTERS The effect of blanking time on the output voltage is described by means of one leg of asingle-phase or a three-phase full-bridge inverter, as shown in Fig. 8-31a. In the previous discussion, the switches were assumed to be ideal, which allowed the status of the two switches in an inverter leg to change simultaneously from on to off and vice versa. Concentrating on one switching time period, Vcontrol is a constant dc voltage, as explained in Fig. 8-6; its comparison with a triangular waveform Vni determines the switching CHAPTER 28 GATE AND BASE DRIVE CIRCUITS 28-1 PRELIMINARY DESIGN CONSIDERATIONS 696 The primary function of a drive circuit is to switch a power semiconductor device from the off state to the on state and vice versa. In most situations the designer seeks a low cost drive circuit that minimizes the turn-on and turn-off times so that the power device spends little time in traversing the active region where the instantaneous power dissipation is large. In the on state the drive circuit must provide adequate drive power (e.g., base current to a BIT or gate-source voltage to a MOSFET) to keep the power switch in the on state where the conduction losses are low. Very often the drive circuit must provide reverse bias to the power switch control terminals to minimize turn-off times and to ensure that the device remains in the off-state and is not triggered on by stray transient signals generated by the switchings of other power devices. The signal processing and control circuits that generate the logic-level control signals used to turn the power switch on and off are not considered part of the drive circuit. The drive circuit is the interface between the control circuit and the power switch. The drive circuit amplifies the control signals to levels required to drive the power switch and provides electrical isolation when required between the power switch and the logic-level signal processing/control circuits. Often the drive circuit has significant power capabilities compared to the logic-level control/signal processing circuits. For example, power BITs have low values of beta, typically 5-10, so that the base current supplied by the drive circuit is often a significant fraction of the total load current. The basic topology of the drive circuit is dictated by three functional considerations. First, is the output signal provided by the drive circuit unipolar or bipolar? Unipolar signals lead to simpler drive circuits, but bipolar signals are needed for rapid turn-on and tum-off of the power switch. Second, can the drive signals be directly coupled to the power switch, or is electrical isolation required between the logic-level control circuits and the power device? Most electrically isolated drive circuits will require isolated dc power supplies. Third, is the output of the drive circuit connected in parallel with the power switch (the usual situation) or in series with the switch (cascode connection)? Additional functionality may be required of the drive circuit, which will further influence the topological details of the circuit. Provisions may be included in the drive circuit design for protection of the power switch from overcurrents. Then communication between the drive circuit and the control circuit is needed. In bridge circuits, the drive circuit must often provide blanking times for the power switch. Incorporation of these types of functionality requires design inputs to both the drive circuit and the logic-level 28-2 dc-COlWLED DRIVE CIRCUITS 697 control circuit. Waveshaping of the drive circuit output may also be included to improve the power switch performance. The specific details of component values to be used in a drive circuit will vary depending on the characteristics of the power switch being driven. For example, BJT drive circuits must provide a relatively large output current (the base current of the power BIT) for the duration of the BIT on-state time interval, whereas MOSFET drive circuits need only provide an initial large current as the device turns on and for the rest of the on-state time interval merely provide a large gate-source voltage at low current levels. It is a good idea to consider how the drive circuit will be configured on a circuit board even at the earliest stages in the design process. The placement of components to mini-mize stray inductance and to minimize susceptibility to switching noise may affect the choice of topology for the drive circuit. 28-2 dc-COUPLED DRIVE CIRCUITS 28-2-1 dc-COUPLED DRIVE CIRCUITS WITH UNIPOLAR OUTPUT A very simple base drive circuit suitable for converters with a single-switch topology is shown in Fig. 28-1. At turn-on, the pnp driver transistor is turned on by saturating one of the internal transistors in the comparator (type 311, for example). This provides a base current for the main power BJT that can be calculated by noting in the circuit of Fig. 28-1 a (28-1) and VB£(oo) IB(oo) = II - ~ (28-2) For the specified maximum collector current Ic that the application demands of the transistor, the necessary base current IB(oo) and corresponding VBE(oo) can be found from the power transistor data sheets. Similarly, the VCE(sat) for the pnp transistor in the base drive circuit can be obtained from its data sheets. In selecting R I , R2 , and VBB , it should be recognized that a small R2 will allow a faster turn-off but will also cause the power dissipation in the drive circuit to be large. The approximate turn-off waveforms are shown in Fig. 28-lb where vBE is shown as larger during the on state compared to the storage interval. A step-by-step design procedure is shown below. I. Based on the turn-off speed required, the negative base current, IB,storage during the storage time is estimated. From this, R2 in Fig. 28-la can be calculated as R2 = VBE,storage (28-3) IB,storage 2. Knowing the required on-state base current IB(oo) and the corresponding VBE(oo) and R2 from the previous step, 1\ becomes I_I VBE(on) 1 -B(on) R2 (28-4) 3. Two unknowns remain in Eq. 28-1, VBB and R I . The on-state losses in the drive circuit are approximately equal to VBB/ I , which suggests that V BB should be small. On the other hand, to reduce the influence of variations in VBE(on)' V BB should be CHAPTER 28 GATE A.."ID BASE DRIVE CIRCUITS IB Comparator (a) I i'-..... TBon--t. iB 0 V- (b) Figure 28-t (a) Simple base current drive circuit for a power BIT and (b) the associated current and voltage waverfonns at turn-off. large. In practice, a VBB of about 8 V is optimum. With VBB = 8 V, RI can then be estimated using Eq. 28-1. This base drive circuit should not be used in pulse-width-modulated bridge converter circuits for reasons that will be discussed shortly. A simple MOSFET gate drive circuit with only one switch to control the gate current is shown in Fig. 28-2, where the output transistor of a comparator (e.g. LM311) controls the MOSFET. When the output transistor is off, the MOSFET is on and vice versa. When the comparator is on, it must sink a current VGGIR I, and to avoid large losses in the drive circuit, RI should be large. This will slow down the MOSFET tum-on time. This means that the drive circuit is only suitable for low switching speed applications. The inadequacy of this circuit can be overcome by the MOSFET gate drive circuit shown in Fig. 28-3 where two switches are used in a totem-pole arrangement with the comparator (type 311) controlling the npn-pnp totem-pole stack. Here, to tum the MOS- FET on, the output transistor of the comparator turns off, thus turning the npn BIT on, which provides a positive gate voltage to the MOSFET. At the tum-off of the MOSFET, From + control Comparator (LM311) 28-2 dc-COUPLED DRIVE CIRCUITS 699 circuit 0----1 From control circuit +From control circuit IC D50026 or UC1706/07 (a) (b) Figure 28-2 Simple MOSFET gate drive circuit suitable for low-speed and low-switching-frequency applications. Figure 28-3 A MOSFET gate drive circuit with a totem-pole configuration for faster tum-off times: (a) discrete totem-pole gate drive circuit; (b) integrated circuit totem pole gate drive circuit. the gate is shorted to the source through RG and the pnp transistor. Since no steady-state current flows through RG in contrast to R I described in the previous paragraph, RG can be chosen to be much smaller in value, which results in much faster tum-on and turn-off times. Very often, instead of using discrete components, similar performance can be obtained, as is shown in Fig. 28-3b by using buffer ICs such as CMOS 4049 or 4050 if a low gate current is needed or a DSOO26 or UC 1707, which can source or sink currents in excess of 1 A. 700 CHAPTER 28 GATE AND BASE DRIVE CIRCUITS 28-2-2 dc-COUPLED DRIVE CIRCUITS WITH BIPOLAR OUTPUT In order to operate power semiconductor devices at high switching frequencies, drive circuits must be designed to turn-off the devices as rapidly as they turn on. The descrip-tions of the switching characteristics of BITs, MOSFETs, IGBTs, and other devices clearly illustrated the need for a reverse bias to be applied to the control terminals of the power switch in order to affect a rapid tum-off. Drive circuits with unipolar outputs are unable to provide the needed reverse bias and thus are incapable of providing fast turn-off of power devices. In order to provide a reverse bias to the control terminals of the power device, the drive circuit must have a bipolar output (an output that can be either positive or negative). This in tum requires that the drive circuit be biased by a negative power supply as wen as a positive power supply. The BIT base drive circuit shown in Fig. 28-4 where both a positive and negative voltage supply with respect to the emitter are used provides a fast tum-off. For the tum-on From control circuit +p 1---- 1 1 :::;::: Con 1 1L___ _ VBB_o-~~~--------------~~--~ (a) Reference t----I Control (b) 1 ~Rl 1To the output ---base current amplifier Figure 28-4 (a) A BIT base current drive circuit with both positive and negative voltages with respect to the BIT emitter for faster tum-off of the power device. (b) A pre-converter circuit should be used if the input from the control circuit has signal only between Jl BB + and ground. 28-2 dc-COL1PLED DRIVE CIRCOITS 701 interval, the output transistor of the comparator turns off, thus turning the transistor TB + on. The on-state base current is (28-5) Arguments similar to step 3 of the previous BJT drive circuit design apply in the selection of V BB+ and R B • The optional capacitor Con shown as dashed, acts as a speed-up capacitor by providing a large transient base current to the power transistor at the instant of turn-on to speed up the turn-on sequence. For turning the BJT off, the internal output transistor of the comparator turns on, thus turning the pnp transistor TB - on (and automatically turns the npn transistor TB + off). For a fast tum-off, no external resistance is used in series with TB _. The magnitude of the negative voltage must be less than the BE breakdown voltage of the BJT, which is given on the data sheets and is normally in the 5- 7 V range. The switching waveforms will be similar to those described in Section 21-5 if the BJT is used in a similar circuit. If the BJT has a tendency for collector current tailing due to a too rapid tum-off of the BE junction compared to the CB junction, as described in Section 21-5, then a resistor or, if necessary, an inductor can be added in the tum-off base drive between points A and the emitter of TB - in Fig. 28-4a. If the control signal is supplied by a logic circuit that is connected between VBB + and the emitter of the BJT, then the reference input to the comparator should be at the mid-potential between V BB + and the BIT emitter terminal, as is shown in Fig. 28-4b where R4 = Rs. The modifications shown in Fig. 28-5a further enhance the BJT turn-off performance of the drive circuit of Fig. 28-4. An antisaturation diode Das is added to keep the BJT voltage VCE slightly above its saturation value VCE(sat). This can be seen in Fig. 28-5a where VAE = VBE(on) + V DI = VCE(on) + V Das (28-6) and therefore, VBE(on) = VCE(on) (28-7) since V DI = V Das • Since VBE(On) is in general larger than VCE(sat)' the presence of the antisaturation diode keeps the transistor slightly out of saturation, thus reducing the storage time at the expense of increased on-state losses in the BIT. Therefore the anti-saturation diode should only be used if the capability to use the BJT in a high switching frequency application is required. If still faster turn-off switching is needed, the on-state voltage VCE(sat) can be adjusted by putting one or more diodes in series with D 1• In the circuit of Fig. 28-5a, the diode D2 is needed to provide a path for the negative base current. D as should be a fast recovery diode with a reverse recovery time smaller than the storage time of the BIT. Moreover, its reverse voltage rating must be similar to the off-state voltage rating of the power transistor. An improved version of the circuit of Fig. 28-5a is shown in 28-5b where the power loss in the positive portion of the base drive circuit is reduced compared to the original circuit. Here the anti saturation diode adjusts the base current of the drive transistor TB + such that TB + operates in the active mode and the current drawn from V BB + now is only equal to the actuall B needed to barely saturate the BIT. Moreover, the current rating required of D as is reduced. A small resistance in series with the antisaturation diode can CHAPTER 28 GATE AND BASE DRIVE CIRCUITS D,.. BJT -t--1E Vss- (a) D,.. BJT _....-..... E Vss- (b) Figure 28-5 (a) Base drive circuit with antisatura-tion to minimize the storage time of the BIT and thus the tum-off time. The modifications in (b) permit the antisaturation diode to have a lower current rating compared to the situation in (a). significantly help reduce oscillations at tum-on. Since Ts+ operates in the active region, it must be mounted on a small heat sink. A drive circuit for MOSFETs that provides positive gate voltages at tum on and negative gate voltages at tum off by means of a split power supply with respect to the MOSFET source is shown in Figs. 28-00 through 28-6c. If the control signal is supplied by logic circuit that is connected between VGG+ and the source of the MOSFET, then the reference input to the comparator should be shifted to be at the mid-potential between VGG+ and the source of the MOSFET using a pre-converter circuit similar to that in Fig. 28-4b. 28-3 ELECTRICALLY ISOLATED DRIVE CIRCUITS 703 VGG +-_--..., VGG---~------------~ (a) (b) VGG +----...-------, Ie VGG-------~--------~ Figure 28-6 Various gate drive circuits using split dc power supplies for providing an n-channel MOSFET with positive gate-source voltages at turn-on and negative gate drive at turn-off. 28-3 ELECTRICALLY ISOLATED DRIVE CIRCUITS 28-3-1 NEED FOR AND TYPES OF ELECTRICAL ISOLATION Very often, there is a need for electrical isolation between the logic-level control signals and the drive circuits. This is illustrated in Fig. 28-7 for the case of a power BJT half-bridge converter having a single-phase ac supply as its input where one of the power tenninals is a grounded neutral wire. Now the positive dc bus is close to the ground potential during the negative half cycle of vs' and the negative dc bus is near ground potential during the positive half cycle of VS. Under these conditions the emitter tenninals of both BJTs must be treated as "hot" with respect to power neutral. The logic-level control signals are nonnally referenced with respect to logic ground, which is at the same potential as the power neutral since the logic circuits are connected to the neutral by means of a safety ground wire. The basic ways to provide electrical isolation are either by optocouplers, fiber optics, or by transfonners. The optocoupler shown in Fig. 28-8 consists of a light-emitting diode (LED), the output transistor, and a built-in Schmitt trigger. A positive signal from the control logic causes the LED to emit light that is focused on the optically sensitive base region of a photo transistor. The light falling on the base region generates a significant number of electron-hole pairs in the base region that causes the photo transistor to turn on. The resulting drop in voltage at the photo transistor collector causes the Schmitt trigger to change state. The output of the Schmitt trigger is the optocoupler output and can be used as the control input to the isolated drive circuit. The capacitance between the LED and the 704 CHAPTER 28 GATE AND BASE DRIVE CIRCUITS ;--- [ Isolated auxiliary power to base [ drive circuits Antisaturation diodes and/or r- overcurrent protection c0j"ections ~< Signal Base Dr isolation Drive T+ Circuit LOIIic and ...L. 1 ""' ,........ Control -r Vd -'- R-Electronics Signal Base Drive Dr isolation Circuit + II, Control ..L 1inputs _ ... Safety ":" ground Figure 28-7 Power BIT base drive system showing the need for electrical isolation between the base drive circuitry and the logic level control circuitry. S;~'If~!i;:: controllOll~ ~ ..... Photo-transistor Control 1000ic ground VBB+ Isolated dc supply Input to remainder of isolated drive circuit Power switch reference node (emitter or source) Figure 28-8 Schematic of an optocoupler used to couple signals to a floating (electrically isolated) drive circuit from a control circuit referenced with respect to the control logic ground (and power system neutral). base of the receiving transistor within the optocoupler should be as small as possible to avoid retriggering at both tum-on and tum-off of the power transistor due to the jump in the potential between the power transistor emitter reference point and the ground of the control electronics. To reduce this problem, optocouplers with electrical shields between the LED and the receiver transistor should be used. As an alternative, fiber optic cables can be used to completely eliminate this re-triggering problem and to provide very high electrical isolation and creepage distance. When using fiber optic cables, the LED is kept on the printed circuit board of the control 28-3 ELECTRIC ALL Y ISOLATED DRIVE CIRCUITS 705 electronics, and the optical fiber transmits the signal to the receiver transistor, which is put on the drive circuit printed circuit board. Instead of using optocouplers or fiber optic cables, the control signal can be coupled to the electrically isolated drive circuit by means of a transformer as is shown in Fig. 28-9a. If the switching frequency is high (several tens of kilohertz or more) and the duty ratio D varies only slightly around 0.5, a baseband control signal of appropriate magni-tude can be applied directly to the primary of a relatively small and lightweight pulse transformer as implied in Fig. 28-9a, and the secondary output can be used to either directly drive the power switch or used as the input to an isolated drive circuit. As the switching frequency is decreased below the tens of kilohertz range, a baseband control Veon!,ol Input to remainder II • Power sWitch reference C of isolated drrve node (BJT emitter, MOSFET source) Logic ground (aJ Signal isolation (b) Fast signal diodes I I I I hIl 11: I LJLJ H .. t .. t .. t Input to isolated drive circuit C Rdischa,ge Oscillator output Veon!,ol Transformer voltage I I I I I ~Comparator I I I I I I I reference level I I I I I I I ':3 k Input to .. t comparator of drive circuit Figure 28-9 Transfonner coupling of control signals from control circuits to electrically isolated drive circuits. In (a) the baseband control signal is directly connected to the transfonner primary. In (b) the control signal modulates a high-frequency carrier that is then applied to the primary of a small high-frequency signal transfonner. The waveforms associated with (b) are shown in (e). CHAPTER 28 GATE AND BASE DRIVE CIRCUITS signal directly applied to the transformer primary becomes impractical because the size and weight of the transformer becomes increasing larger. Modulation of a high-frequency carrier by a low-frequency control signal enables a small high-frequency pulse transformer to be used for even low-frequency control signals. In Fig. 28-9b the control signal modulates a high-frequency (e.g., 1 MHz) oscillator output before being applied to the primary of a high-frequency signal transformer. Since a high-frequency transformer can be made quite small, it is easy to avoid stray capaci-tances between the input and the output windings, and the transformer will be inexpen-sive. The transformer secondary output is rectified and filtered and then applied to the comparator and the rest of the isolated drive circuit. The waveforms for this modulation scheme are shown in Fig. 28-9c. 28-3-2 OPTOCOUPLER ISOLATED DRIVE CIRCUITS In optocoupler-isolated drive circuits, the optocoupler itself is the interface between the output of the control circuit and the input of the isolated drive circuit. The input side of the optocoupler is directly coupled to the control circuit and the output side of the optocoupler is directly connected to the isolated drive circuit. The topology of the isolated drive circuit between the output of the optocoupler and the control terminal of the power switch can take many different forms. An optocoupler-isolated drive circuit for a power BJT is shown in Fig. 28-10. The drive circuit has a bipolar output so that rapid tum-on and tum-off of the BIT can be achieved. An npn-pnp totem-pole circuit couples the appropriate dc voltage to the base of the power BIT to tum it on or off as required. The isolated split dc power supplies are implemented by the circuit segment in the lower left side of Fig. 28-10. $ilnal from control electronics Opt~w~ler 1--- -t-----. I I L---~--~~--~------------------~~-oVBB_ Figure 28-10 Optocoupler isolation of base drive circuits. 28-3 ELECTRICALLY ISOLATED DRIVE CIRCUITS 707 Optocoupler-isolated drive circuits can also be used with power MOSFETs and IGBTs. The circuit shown in Fig. 28-11 uses a high common-mode noise immunity optocoupler (HPCL-4503) and a high-speed driver (IXLD4425) with a 3-A output capa-bility. The drive circuit uses a single-ended floating 15-V supply and provides a ±15-V output voltage for high noise immunity and fast switching to drive the gate of a power MOSFET or IGBT. The integrated high-speed driver circuit connects the gate of the power device to the IS-V bus bar while it simultaneously connects the source to the negative side of the bias supply in order to turn the power device on. To turn the power device off, the drive circuit connects the gate to the negative side of the single-ended supply while it connects the source to the + IS-V bus bar. 28-3-3 TRANSFORMER-ISOLATED DRIVE CIRCUITS PROVIDING BOTH SIGNAL AND POWER The use of transfonners for electrically isolating the drive circuit from the control circuit introduces a great deal of flexibility into the design of the drive circuit. If floating dc power supplies are available, the transfonner coupling scheme with a modulated carrier shown in Fig. 28-9b can be used as a replacement for the optocoupler in the circuits shown in Figs. 28-10 and 28-11. However, the same transfonner used to transfer the control signal from the control circuits to the isolated drive circuit can also be used to provide the isolated dc bias power as well and do away with a separate transfonner for the isolated dc supplies. Consider the BIT base drive circuit shown in Fig. 28-12. This is an isolated base drive in which the base current is made to be proportional to the collector current. Here the need for an auxiliary dc power supply with respect to the emitter terminal is avoided. The transfonner is a combination of a flyback converter transfonner and a current transfonner. When the drive transistor Tl is on, the BJT is off and vice versa. When Tl is conducting and the BIT is off, the transfonner core is magnetized to the limit of saturation with ip = VBB+I Rp. Due to the stored energy in the slightly gapped transfonner core, turning Tl off forces a current to flow in the second winding as in a flyback converter, thus resulting in a positive base current to the BJT. This causes the BIT to start conducting, and its base current is mainly provided by the transfonner action between windings 2 and 3, causing iB = N3ic1N2' During the off interval of Tl' the voltage across the capacitor C p discharges to zero due to the resistance Rp- Therefore, when Tl is turned on in order to turn the BJT off, a voltage essentially equal to V BB + is applied across winding 1 causing ip to be large. Signal from control r----------electronics I \: I~ I +15V JI~ I II ~ - - - - - - - - - --L----------... ----J!-+----.... -I~XL-JD4425 -=- Optocoupler (HCPL-4503) Figure 28-11 An optocoupler isolated drive circuit suitable (or driving MOSFETs and IGBTs. The circuit providing the isolated single-ended 15 V bias supply is not shown for simplicity. 708 CHAPTER 28 GATE AND BASE DRIVE CIRCUITS • 4011 Vcontroi Rz DB Vcontrol _____ ..I! vQ _____ --I vls _____ .... vZs----..1 Cz + vIs lie Power transistor 7555 II Figure 28·12 Proportional "tlyback" BIT base drive. C1 + v2s S Figure 28·13 Transformer-isolated MOSFET gate drive circuit using a high-frequency modulated carrier so that the MOSFET can be held on for long periods. No auxiliary dc power supplies are needed since both the control signal and bias power come through the transformer. 28-4 CASCODE-CONNECTED DRIVE CIRCUITS 709 During the turn-off of the BIT, its base current is given as iB = N3iClN2 - NlipN2 (28-8) The drive circuit must be designed so that the base current during turn-off is negative and of adequate magnitude and duration. This drive configuration is best suited for high-frequency applications where the variations in the duty cycle are limited. If in a given application, the MOSFET to be controlled is to be on for a long time, the circuit shown in Fig. 28-13 can be used. In this circuit the control voltage is modu-lated by a high-frequency oscillator output before being applied to the buffer circuits. Now a high-frequency ac signal appears across the transformer primary when the con-trol voltage is high, thus charging the energy storage capacitance C 1 and the capacitance C 2 at the input to the 7555 IC, which is used here as a buffer and a Schmitt trigger be-cause of its low power consumption. With the input to the 7555 low, it provides a posi-tive voltage to the MOSFET gate, thus turning it on as is shown in Fig. 28-13. At turn off, the control voltage goes low and the voltage across the transformer primary goes to zero. Now C 2 discharges through R2 and the input voltage to the 7555 goes high, which causes its output voltage to go low, thus turning the MOSFET off. The diode DB is used to prevent the energy stored in the capacitance C 1 from discharging into the resistance R2 • In pulse-width-modulated inverters, such as motor drives and UPS, there is a need for a smooth transition in the duty ratio D from a finite value to either zero or one. The use of a phase shift resonant controller, UC3875, combined with two transformers, a demod-ulator, and some buffers will make a good gate drive circuit for a MOSFET or an IGBT as is shown in Fig. 28-14. As in the drive circuit of Fig. 28-13, the transformers may have a relatively large leakage inductance. This makes it easy to produce noise immune trans-formers with a high isolation test voltage. .!! ~c: 0 u -c: '" c: 5l ~ ;: .r: on .. on '" .r: 0. Ii) .... <XI rt') u :::J II II Exclusive OR demodulator Gate Source Figure 28-14 A transformer-coupled MOSFET or lGBT gate drive circuit, which has a smooth transition in the duty ratio D from zero to one.
188089	https://online.stat.psu.edu/stat414/lesson/13/13.1	STAT 414 Introduction to Probability Theory User Preferences Content Preview Arcu felis bibendum ut tristique et egestas quis: Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris Duis aute irure dolor in reprehenderit in voluptate Excepteur sint occaecat cupidatat non proident Lorem ipsum dolor sit amet, consectetur adipisicing elit. Odit molestiae mollitia laudantium assumenda nam eaque, excepturi, soluta, perspiciatis cupiditate sapiente, adipisci quaerat odio voluptates consectetur nulla eveniet iure vitae quibusdam? Excepturi aliquam in iure, repellat, fugiat illum voluptate repellendus blanditiis veritatis ducimus ad ipsa quisquam, commodi vel necessitatibus, harum quos a dignissimos. Keyboard Shortcuts Help : F1 or ? Previous Page : ← + CTRL (Windows) : ← + ⌘ (Mac) Next Page : → + CTRL (Windows) : → + ⌘ (Mac) Search Site : CTRL + SHIFT + F (Windows) : ⌘ + ⇧ + F (Mac) Close Message : ESC 13.1 - Histograms Example 13-1 Section The material on this page should look awfully familiar as we briefly investigated histograms in the first lesson of the course. We review them again briefly here. The following numbers are the measured nose lengths (in millimeters) of 60 students: \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- \| 38 \| 50 \| 38 \| 40 \| 35 \| 32 \| 45 \| 50 \| 40 \| 32 \| 40 \| 47 \| 70 \| 55 \| 51 \| \| 43 \| 40 \| 45 \| 45 \| 55 \| 37 \| 50 \| 45 \| 45 \| 55 \| 50 \| 45 \| 35 \| 52 \| 32 \| \| 45 \| 50 \| 40 \| 40 \| 50 \| 41 \| 41 \| 40 \| 40 \| 46 \| 45 \| 40 \| 43 \| 45 \| 42 \| \| 45 \| 45 \| 48 \| 45 \| 45 \| 35 \| 45 \| 45 \| 40 \| 45 \| 40 \| 40 \| 45 \| 35 \| 52 \| Recall that although the numbers look discrete, they are technically continuous. The measuring tools, which consisted of a piece of string and a ruler, were the limiting factors in getting more refined measurements. In most cases, it appears as if nose lengths come in five-millimeter increments... 35, 40, 45, 55... but that's, again, just measurement error. In order to create a histogram of these continuous measurements, we will use the following guidelines. To create a histogram of continuous data Section First, you have to group the data into a set of classes, typically of equal length. There are many, many sets of rules for defining the classes. For our purposes, we'll just rely on our common sense — having too few classes is as bad as having too many. Determine the number, (n), in the sample. Define (k) class intervals ((c_0, c_1], (c_1, c_2], \ldots, (c_{k-1}, c_k]) . Determine the frequency, (f_i), of each class (i). Calculate the relative frequency (proportion) of each class by dividing the class frequency by the total number in the sample — that is, (\frac{f_i}{n}). For a frequency histogram: draw a rectangle for each class with the class interval as the base and the height equal to the frequency of the class. For a relative frequency histogram: draw a rectangle for each class with the class interval as the base and the height equal to the relative frequency of the class. For a density histogram: draw a rectangle for each class with the class interval as the base and the height equal to (h(x)=\dfrac{f_i}{n(c_i-c_{i-1})}) Example 13-1 Continued Section Here's what the work would like for our nose length example if we used 5 mm classes centered at 30, 35, ... 70: \| Class Interval \| Tally \| Frequency \| Relative Frequency \| Density Height \| \| 27.5-32.5 \| \|\| \| 2 \| 0.033 \| 0.0066 \| \| 32.5-37.5 \| \|\|\|\|\| \| 5 \| 0.083 \| 0.0166 \| \| 37.5-42.5 \| \|\|\|\|\| \|\|\|\|\| \|\|\|\|\| \|\| \| 17 \| 0.283 \| 0.0566 \| \| 42.5-47.5 \| \|\|\|\|\| \|\|\|\|\| \|\|\|\|\| \|\|\|\|\| \| \| 21 \| 0.350 \| 0.0700 \| \| 47.5-52.5 \| \|\|\|\|\| \|\|\|\|\| \| \| 11 \| 0.183 \| 0.0366 \| \| 52.5-57.5 \| \|\|\| \| 3 \| 0.183 \| 0.0366 \| \| 57.5-62.5 \| \| 0 \| 0 \| 0 \| \| 62.5-67.5 \| \| 0 \| 0 \| 0 \| \| 67.5-72.5 \| \| \| 1 \| 0.017 \| 0.0034 \| \| \| \| 60 \| 0.999 (rounding) \| And, here is what the density histogram would like: Note that a density histogram is just a modified relative frequency histogram. A density histogram is defined so that: the area of each rectangle equals the relative frequency of the corresponding class, and the area of the entire histogram equals 1. Empirical Rule Section We've previously learned that the sample mean can be thought of as the "center" of a set of data, while the sample standard deviation indicates "how spread out" the data are from the sample mean. Now, if a histogram is "mound-shaped" or "bell-shaped," then we can use the sample mean, sample standard deviation, and what is called the Empirical Rule to determine three intervals for which we would expect approximately 68%, 95%, and 99.7% of the data to fall. The Empirical Rule tells us that if a histogram is at least approximately bell-shaped, then: Approximately 68% of the data are in the interval: ((\bar{x}-s,\bar{x}+s)) 2. Approximately 95% of the data are in the interval: ((\bar{x}-2s,\bar{x}+2s)) 3. Approximately 99.7% of the data are in the interval: ((\bar{x}-3s,\bar{x}+3s)) Example 13-2 Section The federal government's average income from federal income taxes (on a per capita basis) for each of the 50 states in fiscal year 1991 is \$1252.44 with a standard deviation of \$393.75. Assuming the data are approximately bell-shaped, use the Empirical Rule to determine three intervals for which we would expect approximately 68%, 95%, and 99.7% of the data to fall. Solution The Empirical Rule tells us that we can expect 68% of the per capita taxes to fall between: (\bar{x}-s=\$ 1252.44-\$ 393.75=\$ 858.69) and (\bar{x}+s=\$ 1252.44+\$ 393.75=\$ 1646.19) The Empirical Rule also tells us that we can expect 95% of the per capita taxes to fall between: (\bar{x}-2s=\$ 1252.44-2(\$ 393.75)=\$ 464.94) and (\bar{x}+2s=\$ 1252.44+2(\$ 393.75)=\$ 2039.94) The Empirical Rule also tells us that we can expect 99.7% (virtually all!) of the per capita taxes to fall between: (\bar{x}-3s=\$ 1252.44-3(\$ 393.75)=\$ 71.19) and (\bar{x}+3s=\$ 1252.44+3(\$ 393.75)=\$ 2433.69)
188090	https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/11%3A_Solutions/11.5%3A_Phase_Equilibrium_in_Solutions_-_Nonvolatile_Solutes	11.5.1 11.5.1 11.5.3 11.5.3 11.5.4 11.5.4 11.5.5 11.5.5 11.5.6 11.5.6 11.5.7 11.5.7 11.5.8 11.5.8 NaCl Skip to main content 11.5: Phase Equilibrium in Solutions - Nonvolatile Solutes Last updated : May 13, 2023 Save as PDF 11.4: Reaction Stoichiometry in Solutions: Oxidation-Reduction Titrations 11.6: Phase Equilibrium in Solutions - Volatile Solutes Page ID : 41596 ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives To describe the relationship between solute concentration and the physical properties of a solution. To understand that the total number of nonvolatile solute particles determines the decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus the pure solvent. Many of the physical properties of solutions differ significantly from those of the pure substances discussed in earlier chapters, and these differences have important consequences. For example, the limited temperature range of liquid water (0°C–100°C) severely limits its use. Aqueous solutions have both a lower freezing point and a higher boiling point than pure water. Probably one of the most familiar applications of this phenomenon is the addition of ethylene glycol (“antifreeze”) to the water in an automobile radiator. This solute lowers the freezing point of the water, preventing the engine from cracking in very cold weather from the expansion of pure water on freezing. Antifreeze also enables the cooling system to operate at temperatures greater than 100°C without generating enough pressure to explode. Changes in the freezing point and boiling point of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are called colligative properties (from the Latin colligatus, meaning “bound together” as in a quantity). As we will see, the vapor pressure and osmotic pressure of solutions are also colligative properties. Counting Concentrations and Effective Concentrations When we determine the number of particles in a solution, it is important to remember that not all solutions with the same molarity contain the same concentration of solute particles. Consider, for example, 0.01 M aqueous solutions of sucrose, NaClNaCl, and CaCl2CaCl2. Because sucrose dissolves to give a solution of neutral molecules, the concentration of solute particles in a 0.01 M sucrose solution is 0.01 M. In contrast, both NaClNaCl and CaCl2CaCl2 are ionic compounds that dissociate in water to yield solvated ions. As a result, a 0.01 M aqueous solution of NaClNaCl contains 0.01 M Na+Na+ ions and 0.01 M Cl−Cl− ions, for a total particle concentration of 0.02 M. Similarly, the CaCl2CaCl2 solution contains 0.01 M Ca2+Ca2+ ions and 0.02 M Cl−Cl− ions, for a total particle concentration of 0.03 M. These values are correct for dilute solutions, where the dissociation of the compounds to form separately solvated ions is complete. At higher concentrations (typically >1 M), especially with salts of small, highly charged ions (such as Mg2+Mg2+ or Al3+Al3+), or in solutions with less polar solvents, dissociation to give separate ions is often incomplete. The sum of the concentrations of the dissolved solute particles dictates the physical properties of a solution. In the following discussion, we must therefore keep the chemical nature of the solute firmly in mind. A greater discussion of this is below. Vapor Pressure of Solutions and Raoult’s Law For Nonvolatile Solutes Adding a nonvolatile solute, one whose vapor pressure is too low to measure readily, to a volatile solvent decreases the vapor pressure of the solvent. We can understand this phenomenon qualitatively by examining Figure 11.5.111.5.1, which is a schematic diagram of the surface of a solution of glucose in water. In an aqueous solution of glucose, a portion of the surface area is occupied by nonvolatile glucose molecules rather than by volatile water molecules. As a result, fewer water molecules can enter the vapor phase per unit time, even though the surface water molecules have the same kinetic energy distribution as they would in pure water. At the same time, the rate at which water molecules in the vapor phase collide with the surface and reenter the solution is unaffected. The net effect is to shift the dynamic equilibrium between water in the vapor and the liquid phases, decreasing the vapor pressure of the solution compared with the vapor pressure of the pure solvent. Figure 11.5.211.5.2 shows two beakers, one containing pure water and one containing an aqueous glucose solution, in a sealed chamber. We can view the system as having two competing equilibria: water vapor will condense in both beakers at the same rate, but water molecules will evaporate more slowly from the glucose solution because fewer water molecules are at the surface. Eventually all of the water will evaporate from the beaker containing the liquid with the higher vapor pressure (pure water) and condense in the beaker containing the liquid with the lower vapor pressure (the glucose solution). If the system consisted of only a beaker of water inside a sealed container, equilibrium between the liquid and vapor would be achieved rather rapidly, and the amount of liquid water in the beaker would remain constant. If the particles of a solute are essentially the same size as those of the solvent and both solute and solvent have roughly equal probabilities of being at the surface of the solution, then the effect of a solute on the vapor pressure of the solvent is proportional to the number of sites occupied by solute particles at the surface of the solution. Doubling the concentration of a given solute causes twice as many surface sites to be occupied by solute molecules, resulting in twice the decrease in vapor pressure. The relationship between solution composition and vapor pressure is therefore PA=χAP0A PA=χAP0A(11.5.1) where PAPA is the vapor pressure of component A of the solution (in this case the solvent), χAχA is the mole fraction of AA in solution, and P0AP0A is the vapor pressure of pure AA. Equation 11.5.111.5.1 is known as Raoult’s law, after the French chemist who developed it. If the solution contains only a single nonvolatile solute (BB), then we can relate the mole fraction of the solvent to the solute χA+χB=1 χA+χB=1 and we can substitute χA=1−χBχA=1−χB into Equation 11.5.111.5.1 to obtain PA=(1−χB)P0A=P0A−χBP0A PA=(1−χB)P0A=P0A−χBP0A Rearranging and defining ΔPA=P0A−PAΔPA=P0A−PA, we obtain a relationship between the decrease in vapor pressure and the mole fraction of nonvolatile solute: P0A−PA=ΔPA=χBP0A P0A−PA=ΔPA=χBP0A(11.5.2)(11.5.3) We can solve vapor pressure problems in either of two ways: by using Equation 11.5.111.5.1 to calculate the actual vapor pressure above a solution of a nonvolatile solute, or by using Equation 11.5.311.5.3 to calculate the decrease in vapor pressure caused by a specified amount of a nonvolatile solute. Example 11.5.111.5.1: Anti-Freeze Ethylene glycol (HOCH2CH2OHHOCH2CH2OH), the major ingredient in commercial automotive antifreeze, increases the boiling point of radiator fluid by lowering its vapor pressure. At 100°C, the vapor pressure of pure water is 760 mmHg. Calculate the vapor pressure of an aqueous solution containing 30.2% ethylene glycol by mass, a concentration commonly used in climates that do not get extremely cold in winter. Given: identity of solute, percentage by mass, and vapor pressure of pure solvent Asked for: vapor pressure of solution Strategy: Calculate the number of moles of ethylene glycol in an arbitrary quantity of water, and then calculate the mole fraction of water. Use Raoult’s law (Equation 11.5.111.5.1) to calculate the vapor pressure of the solution. Solution: A A 30.2% solution of ethylene glycol contains 302 g of ethylene glycol per kilogram of solution; the remainder (698 g) is water. To use Raoult’s law to calculate the vapor pressure of the solution, we must know the mole fraction of water. Thus we must first calculate the number of moles of both ethylene glycol (EG) and water present: molesEG=(302g)(1mol62.07g)=4.87molEG molesEG=(302g)(1mol62.07g)=4.87molEG molesH2O=(698g)(1mol18.02g)=38.7molH2O molesH2O=(698g)(1mol18.02g)=38.7molH2O The mole fraction of water is thus χH2O=38.7molH2O38.7molH2O+4.87molEG=0.888 χH2O=38.7molH2O38.7molH2O+4.87molEG=0.888 B From Raoult’s law (Equation 11.5.111.5.1), the vapor pressure of the solution is PH2O=(χH2O)(P0H2O)=(0.888)(760mmHg)=675mmHg PH2O=(χH2O)(P0H2O)=(0.888)(760mmHg)=675mmHg Alternatively, we could solve this problem by calculating the mole fraction of ethylene glycol and then using Equation 11.5.311.5.3 to calculate the resulting decrease in vapor pressure: χEG=4.87molEG4.87molEG+38.7molH2O=0.112 χEG=4.87molEG4.87molEG+38.7molH2O=0.112 ΔPH2O=(χEG)(P0H2O)=(0.112)(760mmHg)=85.1mmHg ΔPH2O=(χEG)(P0H2O)=(0.112)(760mmHg)=85.1mmHg PH2O=P0H2O−ΔPH2O=760mmHg−85.1mmHg=675mmHg PH2O=P0H2O−ΔPH2O=760mmHg−85.1mmHg=675mmHg The same result is obtained using either method. Exercise 11.5.111.5.1 Seawater is an approximately 3.0% aqueous solution of NaClNaCl by mass with about 0.5% of other salts by mass. Calculate the decrease in the vapor pressure of water at 25°C caused by this concentration of NaClNaCl, remembering that 1 mol of NaClNaCl produces 2 mol of solute particles. The vapor pressure of pure water at 25°C is 23.8 mmHg. Answer : 0.45 mmHg. This may seem like a small amount, but it constitutes about a 2% decrease in the vapor pressure of water and accounts in part for the higher humidity in the north-central United States near the Great Lakes, which are freshwater lakes. The decrease therefore has important implications for climate modeling. Boiling Point Elevation Recall that the normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm. If a nonvolatile solute lowers the vapor pressure of a solvent, it must also affect the boiling point. Because the vapor pressure of the solution at a given temperature is less than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent. Thus the boiling point of a solution is always greater than that of the pure solvent. We can see why this must be true by comparing the phase diagram for an aqueous solution with the phase diagram for pure water (Figure 11.5.411.5.4). The vapor pressure of the solution is less than that of pure water at all temperatures. Consequently, the liquid–vapor curve for the solution crosses the horizontal line corresponding to P=1atmP=1atm at a higher temperature than does the curve for pure water. The boiling point of a solution with a nonvolatile solute is always greater than the boiling point of the pure solvent. The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure. As we have just discussed, the decrease in the vapor pressure is proportional to the concentration of the solute in the solution. Hence the magnitude of the increase in the boiling point must also be proportional to the concentration of the solute (Figure 11.5.511.5.5). We can define the boiling point elevation (ΔTbΔTb) as the difference between the boiling points of the solution and the pure solvent: ΔTb=Tb−T0b ΔTb=Tb−T0b(11.5.4) where TbTb is the boiling point of the solution and T0bT0b is the boiling point of the pure solvent. We can express the relationship between ΔTbΔTb and concentration as follows ΔTb=mKb ΔTb=mKb(11.5.5) where m is the concentration of the solute expressed in molality, and KbKb is the molal boiling point elevation constant of the solvent, which has units of °C/m. Table 11.5.111.5.1 lists characteristic KbKb values for several commonly used solvents. For relatively dilute solutions, the magnitude of both properties is proportional to the solute concentration. Table 11.5.111.5.1: Boiling Point Elevation Constants (Kb) and Freezing Point Depression Constants (Kf) for Some Solvents \| Solvent \| Boiling Point (°C) \| Kb (°C/m) \| Freezing Point (°C) \| Kf (°C/m) \| \| acetic acid \| 117.90 \| 3.22 \| 16.64 \| 3.63 \| \| benzene \| 80.09 \| 2.64 \| 5.49 \| 5.07 \| \| d-(+)-camphor \| 207.4 \| 4.91 \| 178.8 \| 37.8 \| \| carbon disulfide \| 46.2 \| 2.42 \| −112.1 \| 3.74 \| \| carbon tetrachloride \| 76.8 \| 5.26 \| −22.62 \| 31.4 \| \| chloroform \| 61.17 \| 3.80 \| −63.41 \| 4.60 \| \| nitrobenzene \| 210.8 \| 5.24 \| 5.70 \| 6.87 \| \| water \| 100.00 \| 0.51 \| 0.00 \| 1.86 \| The concentration of the solute is typically expressed as molality rather than mole fraction or molarity for two reasons. First, because the density of a solution changes with temperature, the value of molarity also varies with temperature. If the boiling point depends on the solute concentration, then by definition the system is not maintained at a constant temperature. Second, molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value (a mole fraction can be only between zero and one). Using molality allows us to eliminate nonsignificant zeros. According to Table 11.5.111.5.1, the molal boiling point elevation constant for water is 0.51°C/m. Thus a 1.00 m aqueous solution of a nonvolatile molecular solute such as glucose or sucrose will have an increase in boiling point of 0.51°C, to give a boiling point of 100.51°C at 1.00 atm. The increase in the boiling point of a 1.00 m aqueous NaClNaCl solution will be approximately twice as large as that of the glucose or sucrose solution because 1 mol of NaClNaCl produces 2 mol of dissolved ions. Hence a 1.00 m NaClNaCl solution will have a boiling point of about 101.02°C. Example 11.5.311.5.3 In Example 11.5.111.5.1, we calculated that the vapor pressure of a 30.2% aqueous solution of ethylene glycol at 100°C is 85.1 mmHg less than the vapor pressure of pure water. We stated (without offering proof) that this should result in a higher boiling point for the solution compared with pure water. Now that we have seen why this assertion is correct, calculate the boiling point of the aqueous ethylene glycol solution. Given: composition of solution Asked for: boiling point Strategy: Calculate the molality of ethylene glycol in the 30.2% solution. Then use Equation 11.5.511.5.5 to calculate the increase in boiling point. Solution: From Example 11.5.111.5.1, we know that a 30.2% solution of ethylene glycol in water contains 302 g of ethylene glycol (4.87 mol) per 698 g of water. The molality of the solution is thus molality of ethylene glycol=(4.87mol698gH2O)(1000g1kg)=6.98m molality of ethylene glycol=(4.87mol698gH2O)(1000g1kg)=6.98m From Equation 11.5.511.5.5, the increase in boiling point is therefore ΔTb=mKb=(6.98m)(0.51°C/m)=3.6°C ΔTb=mKb=(6.98m)(0.51°C/m)=3.6°C The boiling point of the solution is thus predicted to be 104°C. With a solute concentration of almost 7 m, however, the assumption of a dilute solution used to obtain Equation 11.5.511.5.5 may not be valid. Exercise 11.5.311.5.3 Assume that a tablespoon (5.00 g) of NaClNaCl is added to 2.00 L of water at 20.0°C, which is then brought to a boil to cook spaghetti. At what temperature will the water boil? Answer : 100.04°C, or 100°C to three significant figures. (Recall that 1 mol of NaCl produces 2 mol of dissolved particles. The small increase in temperature means that adding salt to the water used to cook pasta has essentially no effect on the cooking time.) Freezing Point Depression The phase diagram in Figure 11.5.411.5.4 shows that dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point. The solid–liquid curve for the solution crosses the line corresponding to P=1atmP=1atm at a lower temperature than the curve for pure water. We can understand this result by imagining that we have a sample of water at the normal freezing point temperature, where there is a dynamic equilibrium between solid and liquid. Water molecules are continuously colliding with the ice surface and entering the solid phase at the same rate that water molecules are leaving the surface of the ice and entering the liquid phase. If we dissolve a nonvolatile solute such as glucose in the liquid, the dissolved glucose molecules will reduce the number of collisions per unit time between water molecules and the ice surface because some of the molecules colliding with the ice will be glucose. Glucose, though, has a very different structure than water, and it cannot fit into the ice lattice. Consequently, the presence of glucose molecules in the solution can only decrease the rate at which water molecules in the liquid collide with the ice surface and solidify. Meanwhile, the rate at which the water molecules leave the surface of the ice and enter the liquid phase is unchanged. The net effect is to cause the ice to melt. The only way to reestablish a dynamic equilibrium between solid and liquid water is to lower the temperature of the system, which decreases the rate at which water molecules leave the surface of the ice crystals until it equals the rate at which water molecules in the solution collide with the ice. By analogy to our treatment of boiling point elevation,the freezing point depression (ΔTfΔTf) is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution: ΔTf=T0f−Tf ΔTf=T0f−Tf(11.5.6) where T0fT0f is the freezing point of the pure solvent and TfTf is the freezing point of the solution. The order of the terms is reversed compared with Equation 11.5.411.5.4 to express the freezing point depression as a positive number. The relationship between ΔTfΔTf and the solute concentration is given by an equation analogous to Equation 11.5.511.5.5: ΔTf=mKf ΔTf=mKf(11.5.7) where mm is the molality of the solution and KfKf is the molal freezing point depression constant for the solvent (in units of °C/m). Like KbKb, each solvent has a characteristic value of KfKf (see Table 11.5.111.5.1). Freezing point depression depends on the total number of dissolved nonvolatile solute particles, just as with boiling point elevation. Thus an aqueous NaClNaCl solution has twice as large a freezing point depression as a glucose solution of the same molality. People who live in cold climates use freezing point depression to their advantage in many ways. For example, salt is used to melt ice and snow on roads and sidewalks, ethylene glycol is added to engine coolant water to prevent an automobile engine from being destroyed, and methanol is added to windshield washer fluid to prevent the fluid from freezing. The decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus a pure liquid all depend on the total number of dissolved nonvolatile solute particles. Example 11.5.411.5.4: Salting the Roads In colder regions of the United States, NaClNaCl or CaCl2CaCl2 is often sprinkled on icy roads in winter to melt the ice and make driving safer. Use the data in the Figure below to estimate the concentrations of two saturated solutions at 0°C, one of NaClNaCl and one of CaCl2CaCl2, and calculate the freezing points of both solutions to see which salt is likely to be more effective at melting ice. Given: solubilities of two compounds Asked for: concentrations and freezing points Strategy: Estimate the solubility of each salt in 100 g of water from the figure. Determine the number of moles of each in 100 g and calculate the molalities. Determine the concentrations of the dissolved salts in the solutions. Substitute these values into Equation 11.5.711.5.7 to calculate the freezing point depressions of the solutions. Solution: A. From Figure above, we can estimate the solubilities of NaClNaCl and CaCl2CaCl2 to be about 36 g and 60 g, respectively, per 100 g of water at 0°C. The corresponding concentrations in molality are mNaCl=(36gNaCl100gH2O)(1molNaCl58.44gNaCl)(1000g1kg)=6.2m mNaCl=⎛⎝36gNaCl100gH2O⎞⎠⎛⎝1molNaCl58.44gNaCl⎞⎠(1000g1kg)=6.2m mCaCl2=(60gCaCl2100gH2O)(1molCaCl2110.98gCaCl2)(1000g1kg)=5.4m mCaCl2=⎛⎝60gCaCl2100gH2O⎞⎠⎛⎝1molCaCl2110.98gCaCl2⎞⎠(1000g1kg)=5.4m The lower formula mass of NaCl more than compensates for its lower solubility, resulting in a saturated solution that has a slightly higher concentration than CaCl2. B. Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of NaCl and CaCl2, respectively, the actual concentrations of the dissolved species in the two saturated solutions are 2 × 6.2 m = 12 m for NaCl and 3 × 5.4 m = 16 m for CaCl2. The resulting freezing point depressions can be calculated using Equation 11.5.7: NaCl:ΔTf=mKf=(12m)(1.86°C/m)=22°C CaCl2:ΔTf=mKf=(16m)(1.86°C/m)=30°C Because the freezing point of pure water is 0°C, the actual freezing points of the solutions are −22°C and −30°C, respectively. Note that CaCl2 is substantially more effective at lowering the freezing point of water because its solutions contain three ions per formula unit. In fact, CaCl2 is the salt usually sold for home use, and it is also often used on highways. Because the solubilities of both salts decrease with decreasing temperature, the freezing point can be depressed by only a certain amount, regardless of how much salt is spread on an icy road. If the temperature is significantly below the minimum temperature at which one of these salts will cause ice to melt (say −35°C), there is no point in using salt until it gets warmer Exercise 11.5.4 Calculate the freezing point of the 30.2% solution of ethylene glycol in water whose vapor pressure and boiling point we calculated in Examples 11.5.1 and 11.5.3. Answer : −13.0°C Example 11.5.5 Arrange these aqueous solutions in order of decreasing freezing points: 0.1 m KCl, 0.1 m glucose, 0.1 m SrCl2, 0.1 m ethylene glycol, 0.1 m benzoic acid, and 0.1 m HCl. Given: molalities of six solutions Asked for: relative freezing points Strategy: Identify each solute as a strong, weak, or nonelectrolyte, and use this information to determine the number of solute particles produced. Multiply this number by the concentration of the solution to obtain the effective concentration of solute particles. The solution with the highest effective concentration of solute particles has the largest freezing point depression. Solution: A Because the molal concentrations of all six solutions are the same, we must focus on which of the substances are strong electrolytes, which are weak electrolytes, and which are nonelectrolytes to determine the actual numbers of particles in solution. KCl, SrCl2, and HCl are strong electrolytes, producing two, three, and two ions per formula unit, respectively. Benzoic acid is a weak electrolyte (approximately one particle per molecule), and glucose and ethylene glycol are both nonelectrolytes (one particle per molecule). B The molalities of the solutions in terms of the total particles of solute are: KCl and HCl, 0.2 m; SrCl2, 0.3 m; glucose and ethylene glycol, 0.1 m; and benzoic acid, 0.1–0.2 m. Because the magnitude of the decrease in freezing point is proportional to the concentration of dissolved particles, the order of freezing points of the solutions is: glucose and ethylene glycol (highest freezing point, smallest freezing point depression) > benzoic acid > HCl = KCl > SrCl2. Exercise 11.5.5 Arrange these aqueous solutions in order of increasing freezing points: 0.2 m NaCl, 0.3 m acetic acid, 0.1 m CaCl2, and 0.2 m sucrose. Answer : 0.2 m NaCl (lowest freezing point) < 0.3 m acetic acid ≈ 0.1 m CaCl2 < 0.2 m sucrose (highest freezing point) Colligative properties can also be used to determine the molar mass of an unknown compound. One method that can be carried out in the laboratory with minimal equipment is to measure the freezing point of a solution with a known mass of solute. This method is accurate for dilute solutions (≤1% by mass) because changes in the freezing point are usually large enough to measure accurately and precisely. By comparing Kb and Kf values in Table 11.5.1, we see that changes in the boiling point are smaller than changes in the freezing point for a given solvent. Boiling point elevations are thus more difficult to measure precisely. For this reason, freezing point depression is more commonly used to determine molar mass than is boiling point elevation. Because of its very large value of Kf (37.8°C/m), d-(+)-camphor (Table 11.5.1) is often used to determine the molar mass of organic compounds by this method. Example 11.5.6: Sulfur A 7.08 g sample of elemental sulfur is dissolved in 75.0 g of CS2 to create a solution whose freezing point is −113.5°C. Use these data to calculate the molar mass of elemental sulfur and thus the formula of the dissolved Sn molecules (i.e., what is the value of n?). Given: masses of solute and solvent and freezing point Asked for: molar mass and number of S atoms per molecule Strategy: Use Equation 11.5.6, the measured freezing point of the solution, and the freezing point of CS2 from Table 11.5.1 to calculate the freezing point depression. Then use Equation 11.5.7 and the value of Kf from Table 11.5.1 to calculate the molality of the solution. From the calculated molality, determine the number of moles of solute present. Use the mass and number of moles of the solute to calculate the molar mass of sulfur in solution. Divide the result by the molar mass of atomic sulfur to obtain n, the number of sulfur atoms per mole of dissolved sulfur. Solution: A The first step is to calculate the freezing point depression using Equation 11.5.6: ΔTf=T0f−Tf=−112.1°C−(−113.5°C)=1.4°C Then Equation 11.5.7 gives m=ΔTfKf=1.4°C3.74°C/m=0.37m B The total number of moles of solute present in the solution is moles solute=(0.37molkg)(75.0g)(1kg1000g)=0.028mol C We now know that 0.708 g of elemental sulfur corresponds to 0.028 mol of solute. The molar mass of dissolved sulfur is thus molar mass=7.08g0.028mol=260g/mol The molar mass of atomic sulfur is 32 g/mol, so there must be 260/32 = 8.1 sulfur atoms per mole, corresponding to a formula of S8. Exercise 11.5.6 One of the byproducts formed during the synthesis of C60 is a deep red solid containing only carbon. A solution of 205 mg of this compound in 10.0 g of CCl4 has a freezing point of −23.38°C. What are the molar mass and most probable formula of the substance? Answer : 847 g/mol; C70 Osmotic Pressure Osmotic pressure is a colligative property of solutions that is observed using a semipermeable membrane, a barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions. The net flow of solvent through a semipermeable membrane is called osmosis (from the Greek osmós, meaning “push”). The direction of net solvent flow is always from the side with the lower concentration of solute to the side with the higher concentration. Osmosis can be demonstrated using a U-tube like the one shown in Figure 11.5.6, which contains pure water in the left arm and a dilute aqueous solution of glucose in the right arm. A net flow of water through the membrane occurs until the levels in the arms eventually stop changing, which indicates that equilibrium has been reached. The osmotic pressure (Π) of the glucose solution is the difference in the pressure between the two sides, in this case the heights of the two columns. Although the semipermeable membrane allows water molecules to flow through in either direction, the rate of flow is not the same in both directions because the concentration of water is not the same in the two arms. The net flow of water through the membrane can be prevented by applying a pressure to the right arm that is equal to the osmotic pressure of the glucose solution. Just as with any other colligative property, the osmotic pressure of a solution depends on the concentration of dissolved solute particles. Osmotic pressure obeys a law that resembles the ideal gas equation: Π=nRTV=MRT where M is the number of moles of solute per unit volume of solution (i.e., the molarity of the solution), R is the ideal gas constant, and T is the absolute temperature. As shown in Example 11.5.7, osmotic pressures tend to be quite high, even for rather dilute solutions. Example 11.5.7 When placed in a concentrated salt solution, certain yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding medium. Suppose that the yeast cells are placed in an aqueous solution containing 4.0% NaCl by mass; the solution density is 1.02 g/mL at 25°C. Calculate the osmotic pressure of a 4.0% aqueous NaCl solution at 25°C. If the normal osmotic pressure inside a yeast cell is 7.3 atm, corresponding to a total concentration of dissolved particles of 0.30 M, what concentration of glycerol must the cells synthesize to exactly balance the external osmotic pressure at 25°C? Given: concentration, density, and temperature of NaCl solution; internal osmotic pressure of cell Asked for: osmotic pressure of NaCl solution and concentration of glycerol needed Strategy: Calculate the molarity of the NaCl solution using the formula mass of the solute and the density of the solution. Then calculate the total concentration of dissolved particles. Use Equation 11.5.8 to calculate the osmotic pressure of the solution. Subtract the normal osmotic pressure of the cells from the osmotic pressure of the salt solution to obtain the additional pressure needed to balance the two. Use Equation 11.5.8 to calculate the molarity of glycerol needed to create this osmotic pressure. Solution: A The solution contains 4.0 g of NaCl per 100 g of solution. Using the formula mass of NaCl (58.44 g/mol) and the density of the solution (1.02 g/mL), we can calculate the molarity: MNaCl=molesNaClliter solution=(4.0gNaCl58.44g/molNaCl)(1100gsolution)(1.02gsolution1.00mLsolution)(1000mL1L)=0.70MNaCl Because 1 mol of NaCl produces 2 mol of particles in solution, the total concentration of dissolved particles in the solution is (2)(0.70 M) = 1.4 M. B Now we can use Equation 11.5.8 to calculate the osmotic pressure of the solution: Π=MRT=(1.4mol/L)0.0821(L⋅atm)/(K⋅mol)=34atm C If the yeast cells are to exactly balance the external osmotic pressure, they must produce enough glycerol to give an additional internal pressure of (34 atm − 7.3 atm) = 27 atm. Glycerol is a nonelectrolyte, so we can solve Equation 11.5.8 for the molarity corresponding to this osmotic pressure: M=ΠRT=27atm0.0821(L⋅atm)/(K⋅mol)=1.1Mglycerol In solving this problem, we could also have recognized that the only way the osmotic pressures can be the same inside the cells and in the solution is if the concentrations of dissolved particles are the same. We are given that the normal concentration of dissolved particles in the cells is 0.3 M, and we have calculated that the NaCl solution is effectively 1.4 M in dissolved particles. The yeast cells must therefore synthesize enough glycerol to increase the internal concentration of dissolved particles from 0.3 M to 1.4 M—that is, an additional 1.1 M concentration of glycerol. Exercise 11.5.7 Assume that the fluids inside a sausage are approximately 0.80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100°C to learn why experienced cooks pierce the semipermeable skin of sausages before boiling them. Answer : 24 atm Because of the large magnitude of osmotic pressures, osmosis is extraordinarily important in biochemistry, biology, and medicine. Virtually every barrier that separates an organism or cell from its environment acts like a semipermeable membrane, permitting the flow of water but not solutes. The same is true of the compartments inside an organism or cell. Some specialized barriers, such as those in your kidneys, are slightly more permeable and use a related process called dialysis, which permits both water and small molecules to pass through but not large molecules such as proteins. The same principle has long been used to preserve fruits and their essential vitamins over the long winter. High concentrations of sugar are used in jams and jellies not for sweetness alone but because they greatly increase the osmotic pressure. Thus any bacteria not killed in the cooking process are dehydrated, which keeps them from multiplying in an otherwise rich medium for bacterial growth. A similar process using salt prevents bacteria from growing in ham, bacon, salt pork, salt cod, and other preserved meats. The effect of osmotic pressure is dramatically illustrated in Figure 11.5.7, which shows what happens when red blood cells are placed in a solution whose osmotic pressure is much lower or much higher than the internal pressure of the cells. In addition to capillary action, trees use osmotic pressure to transport water and other nutrients from the roots to the upper branches. Evaporation of water from the leaves results in a local increase in the salt concentration, which generates an osmotic pressure that pulls water up the trunk of the tree to the leaves. Finally, a process called reverse osmosis can be used to produce pure water from seawater. As shown schematically in Figure 11.5.8, applying high pressure to seawater forces water molecules to flow through a semipermeable membrane that separates pure water from the solution, leaving the dissolved salt behind. Large-scale desalinization plants that can produce hundreds of thousands of gallons of freshwater per day are common in the desert lands of the Middle East, where they supply a large proportion of the freshwater needed by the population. Similar facilities are now being used to supply freshwater in southern California. Small, hand-operated reverse osmosis units can produce approximately 5 L of freshwater per hour, enough to keep 25 people alive, and are now standard equipment on US Navy lifeboats. Colligative Properties of Electrolyte Solutions Thus far we have assumed that we could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the actual concentration of dissolved particles in an electrolyte solution. We have used this simple model to predict such properties as freezing points, melting points, vapor pressure, and osmotic pressure. If this model were perfectly correct, we would expect the freezing point depression of a 0.10 m solution of sodium chloride, with 2 mol of ions per mole of NaCl in solution, to be exactly twice that of a 0.10 m solution of glucose, with only 1 mol of molecules per mole of glucose in solution. In reality, this is not always the case. Instead, the observed change in freezing points for 0.10 m aqueous solutions of NaCl and KCl are significantly less than expected (−0.348°C and −0.344°C, respectively, rather than −0.372°C), which suggests that fewer particles than we expected are present in solution. The relationship between the actual number of moles of solute added to form a solution and the apparent number as determined by colligative properties is called the van ’t Hoff factor (i) and is defined as follows: i=apparent number of particles in solution number of moles of solute dissolved Named for Jacobus Hendricus van ’t Hoff (1852–1911), a Dutch chemistry professor at the University of Amsterdam who won the first Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions. As the solute concentration increases, the van ’t Hoff factor decreases. The van ’t Hoff factor is therefore a measure of a deviation from ideal behavior. The lower the van ’t Hoff factor, the greater the deviation. As the data in Table 11.5.2 show, the van ’t Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions apparently contain fewer particles than predicted by the number of ions per formula unit. As the concentration of the solute increases, the van ’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution. Table 11.5.2: van ’t Hoff Factors for 0.0500 M Aqueous Solutions of Selected Compounds at 25°C \| Compound \| i (measured) \| i (ideal) \| \| glucose \| 1.0 \| 1.0 \| \| sucrose \| 1.0 \| 1.0 \| \| NaCl \| 1.9 \| 2.0 \| \| HCl \| 1.9 \| 2.0 \| \| MgCl2 \| 2.7 \| 3.0 \| \| FeCl3 \| 3.4 \| 4.0 \| \| Ca(NO3)2 \| 2.5 \| 3.0 \| \| AlCl3 \| 3.2 \| 4.0 \| \| MgSO4 \| 1.4 \| 2.0 \| Instead, some of the ions exist as ion pairs, a cation and an anion that for a brief time are associated with each other without an intervening shell of water molecules (Figure 11.5.9). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as Mg2+, Al3+, SO2−4, and PO3−4 have a greater tendency to form ion pairs because of their strong electrostatic interactions. The actual number of solvated ions present in a solution can be determined by measuring a colligative property at several solute concentrations. Example 11.5.8: Iron Chloride in Water A 0.0500 M aqueous solution of FeCl3 has an osmotic pressure of 4.15 atm at 25°C. Calculate the van ’t Hoff factor i for the solution. Given: solute concentration, osmotic pressure, and temperature Asked for: van ’t Hoff factor Strategy: Use Equation 11.5.8 to calculate the expected osmotic pressure of the solution based on the effective concentration of dissolved particles in the solvent. Calculate the ratio of the observed osmotic pressure to the expected value. Multiply this number by the number of ions of solute per formula unit, and then use Equation 11.5.9 to calculate the van ’t Hoff factor. Solution: A If FeCl3 dissociated completely in aqueous solution, it would produce four ions per formula unit (one Fe3+(aq) and three (\ce{Cl^{-}(aq)} ions) for an effective concentration of dissolved particles of 4 × 0.0500 M = 0.200 M. The osmotic pressure would be Π=MRT=(0.200mol/L)0.0821(L⋅atm)/(K⋅mol)=4.89atm B The observed osmotic pressure is only 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is 4.15 atm/4.89 atm = 0.849, which indicates that the solution contains (0.849)(4) = 3.40 moles of particles per mole of FeCl3 dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm: 4.15atm=M(0.0821(L⋅atm)(K⋅mol))(298K) or after rearranging M=0.170mol The ratio of this value to the expected value of 0.200 M is 0.170 M/0.200 M = 0.850, which again gives us (0.850)(4) = 3.40 mole of particles per mole of FeCl3 dissolved. From Equation 11.5.9, the van ’t Hoff factor for the solution is i=3.40 particles observed1 formula unitFeCl3=3.40 Exercise 11.5.8: Magnesium Chloride in Water Calculate the van ’t Hoff factor for a 0.050 m aqueous solution of MgCl2 that has a measured freezing point of −0.25°C. Answer : 2.7 (versus an ideal value of 3). Summary The colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as Raoult’s law. The boiling point elevation (ΔTb) and freezing point depression (ΔTf) of a solution are defined as the differences between the boiling and freezing points, respectively, of the solution and the pure solvent. Both are proportional to the molality of the solute. When a solution and a pure solvent are separated by a semipermeable membrane, a barrier that allows solvent molecules but not solute molecules to pass through, the flow of solvent in opposing directions is unequal and produces an osmotic pressure, which is the difference in pressure between the two sides of the membrane. Osmosis is the net flow of solvent through such a membrane due to different solute concentrations. Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through. In more concentrated solutions, or in solutions of salts with highly charged ions, the cations and anions can associate to form ion pairs, which decreases their effect on the colligative properties of the solution. The extent of ion pair formation is given by the van ’t Hoff factor (i), the ratio of the apparent number of particles in solution to the number predicted by the stoichiometry of the salt. Henry’s law: C=kP Raoult’s law: PA=χAP0A vapor pressure lowering: P0A−PA=ΔPA=χBP0A vapor pressure of a system containing two volatile components: Ptot=χAP0A+(1−χA)P0B boiling point elevation: ΔTb=mKb freezing point depression: ΔTf=mKf osmotic pressure: Π=nVRT=MRT van ’t Hoff factor: i=apparent number of particles in solution number of moles of solute dissolved 11.4: Reaction Stoichiometry in Solutions: Oxidation-Reduction Titrations 11.6: Phase Equilibrium in Solutions - Volatile Solutes
188091	https://www.youtube.com/watch?v=IKp76EepRRM	Algebra 19 - Linear Equations y = mx MyWhyU 191000 subscribers 611 likes Description 128740 views Posted: 22 May 2013 Equations of the form y = mx describe lines in the Cartesian plane which pass through the origin. The fact that many functions are linear when viewed on a small scale, is important in branches of mathematics such as calculus. 38 comments Transcript: Hello. I'm Professor Von Schmohawk and welcome to Why U. We have seen that many functions encountered in mathematics are real-valued functions of a real variable. These functions take their input values from a real variable, called the "independent variable" and for each valid input value, produce one specific real output value. When "x" is used as the input variable the output value is typically assigned to the variable "y" called the "dependent variable". These values of x and y can be visually represented on the Cartesian plane. This visual representation is called the "graph" of the function and the process of creating it is called "graphing" the function. Real-valued functions of a real variable are typically described by an equation. For example, the equation which describes the function represented by this graph is "y equals two x". This function is said to be "linear" since its graph is a line in the Cartesian plane. Linear functions are very simple but extremely important in mathematics. This is not only because many natural phenomena can be modeled by linear equations but also because more complicated functions often appear linear when viewed on a small enough scale. Notice that as we examine smaller and smaller regions of this curve the graph looks more and more linear. The fact that many functions appear linear on a small scale is the logical basis of branches of mathematics such as calculus. It is this same principle which allows us to treat the Earth as a flat plane locally even though on a large scale its surface is actually curved. Lines are simple geometric shapes and so the equations which describe them are also simple. As an example, let's create an equation which describes this linear graph. Notice that the x and y coordinates of each point on this line have the same value. For instance, when x has a value of five y also has a value of five. This equality between x and y applies to every point on this line. Therefore, the equation "y equals x" describes this function. This equation states the condition that the x and y coordinates of every point on the graph have the same value. We say that the coordinates of any point on this line "satisfy" this equation. We can verify this by replacing the variables x and y in the equation by the coordinates of any individual point. If this creates a true statement then the coordinates satisfy the equation and the point lies on the graph. For instance, the coordinates (3,3) satisfy this equation since replacing both x and y by three creates the true statement "three equals three". Therefore, the point whose coordinates are (3,3) lies on this line. On the other hand, we can confirm that the point (4,3) does not lie on this line since substituting these coordinates into the equation creates a false statement. Now instead of the equation "y equals x" let's see what graph is created by the equation "y equals two x". In this equation, the variable x is multiplied by the constant, "two". A constant multiplier of a variable is called the "coefficient" of the variable. In this equation "two" is the coefficient of the variable x. This equation states that the y-coordinate of each point in the graph has a value twice that of the x-coordinate. For example, if the x-coordinate of a point is three then the y-coordinate of that point must be six. Or if x is two then y will be four. If we continue to calculate more points for this equation we will see that these points all fall on a line running through the origin. We can confirm that this line passes through the origin since substituting the coordinates (0,0) into this equation, results in a true statement. Therefore, the point (0,0) lies on this line. Let's see what happens if we make the x-coefficient even larger say, three. We can see that if x equals two y will equal three times that value, or six. So the point (2,6) lies on this graph. Calculating y for several more values of x we see that once again, the collection of all the points which satisfy this equation form a line passing through the origin. The graphs of all of the equations which we have seen are lines which pass through the origin and slope upwards as we move towards the right. The only difference between these equations is the coefficient of x. Notice that the larger the x-coefficient, the greater the slope of the line. But what if the x-coefficient is negative? For example, let's say we pick an x-coefficient of negative one. Now if x is equal to three y must be negative three. Likewise, if x is negative three then y will be positive three. This equation also describes a line which passes through the origin but slopes in the opposite direction from the lines formed by equations with positive coefficients. This line slopes downwards as we move to the right. Increasing the magnitude of this negative coefficient produces lines with a greater downward slope. So the sign of the multiplier determines whether the line slopes up or down and the magnitude of the multiplier determines the magnitude of the slope. Lines which slope up as we move to the right are said to have a positive slope and lines which slope down are said to have a negative slope. So what happens if we make the x coefficient zero? Since zero times x is always zero regardless of the value of x this equation can be written as "y equals zero". This equation states that regardless of the value of the x-coordinate every point's y-coordinate will be zero. So this equation describes a horizontal line running through the origin. This fits the relationship we have seen between the x-coefficient and the slope since an x-coefficient of zero produces a graph with zero slope. So for linear functions, the coefficient of the independent variable x determines the slope of the graph. The coefficient's sign determines the sign of the slope and the coefficient's magnitude determines the magnitude of the slope. The smaller the magnitude of the x-coefficient, the smaller the slope. And when the x-coefficient is zero, the slope will be zero and the graph is a horizontal line. But is it possible for an equation to describe a vertical line? We have seen that as the x-coefficient gets larger the slope of the graph becomes closer to vertical. However, no matter how large the coefficient becomes the line will never be exactly vertical. It would require an infinitely large x-coefficient to produce a vertical line. But infinity is not a number and therefore cannot be used as a coefficient. However, a vertical line can be described by the equation "x equals zero". But this equation does not describe a function of x since it associates multiple values of y with a single x value. So with an equation of the form "y = mx" where "m" is some constant we can describe any non-vertical line passing through the origin. By choosing the value of "m" we can create a line with any slope we please. In the next lecture we will see how to create functions whose graphs are lines which do not pass through the origin.
188092	https://mathbitsnotebook.com/Geometry/Similarity/SMSplitter.html	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| "Side Splitter" Theorem MathBitsNotebook.com Topical Outline \| Geometry Outline \| MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts \| In the previous lesson, we saw the Proportional Line Segment Theorem that stated that "three or more parallel lines intercept proportional segments on two or more transversals". On this page, we will examine an application of that theorem associated with similar triangles. The "Side Splitter" Theorem says that if a line intersects two sides of a triangle and is parallel to the third side of the triangle, it divides those two sides proportionally. \| \| \| --- \| \| Find x. \| Apply the Side Splitter Theorem: (form a proportion using the segment lengths) Solve the proportion for x: 4x = (2)(7) 4x = 14 x = 3.5 (Answer) \| \| Notice in this example, that two of the four labeled "segments" are NOT sides of a triangle. This is an important observation when working with this theorem. \| In the diagram above, while the ratio of the segments shown is 2:4, or 1:2, the ratio of the small triangle to the large triangle is NOT 1:2. To find the ratio of the small triangle to the large triangle requires using the FULL SIDES of each triangle, which gives 2:6 or 1:3. In addition, once we know the ratio of the triangles' sides (1:3), we also know the ratio of their areas is 1:9. NGMS Quote (pg.132): "When students use area as a device to establish results about proportions, such as the important theorem (Side-Splitter Theorem), they are making use of structure. Proof of Side Splitter Theorem and its Converse \| \| \| --- \| \| \| \| \| (Side Splitter Theorem): If a line is parallel to a side of a triangle and intersects the other two sides, then this line divides those two sides proportionally. (proof below) \| \| \| \| \| While this theorem may look somewhat like the "mid-segment" theorem, the segment in this theorem does not necessarily connect the "midpoints" of the sides. Proof: \| \| \| --- \| \| Statements \| Reasons \| \| 1. Given \| \| 2. If 2 \|\| lines are cut by a transversal, the corresponding angles are congruent. \| \| 3. (AA) If two ∠s of one Δ are congruent to the corresponding ∠s of another Δ, the Δs are similar. \| \| 4. Corresponding sides of similar triangles are in proportion. \| \| 5. Segment Addition Postulate (or whole quantity equals the sum of its parts) \| \| 6. Substitution \| \| 7. In a proportion, the product of the means = the product of the extremes. \| \| 8. Distributive property \| \| 9. Subtraction \| \| 10. In a proportion, the product of the means = the product of the extremes. \| \| \| \| \| --- \| \| \| \| --- \| \| Converse \| (Side Splitter Theorem): If a line intersects two sides of a triangle and divides the sides proportionally, the line is parallel to the third side of the triangle. \| \| \| \| \| Proof: \| \| \| --- \| \| Statements \| Reasons \| \| 1. Given \| \| 2. In a proportion, the product of the means = the product of the extremes. \| \| 3. Reflexive (Identity) \| \| 4. Addition \| \| 5. Distributive property \| \| 6. In a proportion, the product of the means = the product of the extremes. \| \| 7. Segment Addition Postulate (or whole quantity equals the sum of its parts) \| \| 8. Substitution \| \| 9. Reflexive (Identity) \| \| 10. (SAS for Similarity). In two triangles, if two sets of corresponding sides are proportional and the included angle is congruent, the triangles are similar. \| \| 11. Corresponding angles of similar triangles are congruent. \| \| 12. If 2 lines are cut by a transversal and the corresponding angles are congruent, the lines are parallel. \| \| \| \| NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". \| Topical Outline \| Geometry Outline \| MathBitsNotebook.com \| MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts \| \|
188093	https://www.johndcook.com/blog/2017/01/09/three-proofs-that-2017-is-prime/	Three proofs that 2017 is prime Aaron Meurer asked on Twitter whether there’s a proof that 2017 is prime that would fit into 140 characters. My first reply was this: sqrt(2017) < 45. 2017 not divisible by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, or 43. Ergo prime. I’m not sure that’s what he had in mind. There’s some implied calculation (which I didn’t actually do), so it’s kinda cheating. It would be interesting if there were something special about 2017 that would allow a more transparent proof. (Implicit in the proof above is the theorem that if a number has a prime factor, it has a prime factor less than it’s square root. If you multiply together numbers bigger than the square root of n, you get a product bigger than n.) Then for fun I gave two more proofs that are more sophisticated but would require far too much work to actually carry out by hand. The first uses Fermat’s little theorem: For 0 < a < 2017, a2016 − 1 is divisible by 2017. 2017 is not one of the three Carmichael numbers < 2465. Ergo prime. Fermat’s little theorem says that if p is prime, then for any 0 < a < p, ap − 1 − 1 is divisible by p. This is actually an efficient way to prove that a number is not prime. Find a number a such that the result doesn’t hold, and you’ve proved that p isn’t prime. For small numbers, the easiest way to show a number is not prime is to show its factors. But for very large numbers, such as those used in cryptography, it’s efficient to have a way to prove that a number has factors without having to actually produce those factors. Unfortunately, Fermat’s little theorem gives a necessary condition for a number to be prime, but not a sufficient condition. It can appear to be prime for every witness (the bases a are called witnesses) and still not be a prime. The Carmichael numbers pass the Fermat primality test without being prime. The first four are 561, 1105, 1729, and 2465. For more on using Fermat’s little theorem to test for primality, see Probability that a number is prime. The final proof was this: 2016! + 1 is divisible by 2017, and so by Wilson’s theorem 2017 is prime. Unlike Fermat’s little theorem, Wilson’s theorem gives necessary and sufficient conditions for a number to be prime. A number n is prime if and only if (n − 1)! + 1 is divisible by n. In theory you could use Wilson’s theorem to test whether a number is prime, but this would be horrendously inefficient. 2016! has 5,789 digits. (You can find out how many digits n! has without computing it using a trick described here.) Despite its inefficiency, you can actually use Wilson’s theorem and SymPy to prove that 2017 is prime. ``` >>> from sympy import factorial >>> x = factorial(2016) + 1 >>> x % 2017 0 ``` 9 thoughts on “Three proofs that 2017 is prime” Aaron Meurer I actually like the first reply best. It’s the only one you can verify with just mental calculations. It’s easy to forget that you only have to test primes p < sqrt(n) to check if n is prime. 2. Peter Norvig No need for sympy: import math (math.factorial(2016) + 1) % 2017 3. John Thanks! I didn’t think of that. 4. Jan Van lent The following is a bit elliptic, but it provides easy to check certificates that show that 2017 is not divisible by any of the primes to be tested. test primes < sqrt(2017): 2 3 5 7 11 13 17 19 23 29 31 37 41 43 2017=21759+11=32329+2^4=251331+2=23^337+19=417^2+2^3=22343+313 5. vonjd In R you could use the Rmpfr package: library(Rmpfr) (factorial(as.bigz(2016)) + 1) %% 2017 ## Big Integer (‘bigz’) : ## 0 6. Benoît R. Kloeckner You do not need to actually compute 2016! to apply Wilson’s theorem: you can do the computation modulo 2017, saving time and memory space. This is actually doable by hand, though still tedious. 7. Timothy Gowers How about this? 2, 3 and 5 are not factors; 711131719232931374143=436092044389001; hcf(2017,436092044389001) = 1. Again there is some implied computation, but the Euclidean algorithm part would be fairly quick. 8. Tom Spencer A variation on 1: for 1 < a < 2016, a^1008-1 or a^1008+1 is divisible by 2017 ergo prime If n is a Carmichael number, then x^2 = 1 (mod n) has at least 8 solutions. 9. Tom Spencer I goofed. I should be for 1 <= a <= 2016, a^1008-1 or a^1008+1 is divisible by 2017 and for 1 < a < 2016, a^2-1 is not divisible by 2017, ergo prime.
188094	https://www.collinsdictionary.com/us/sentences/english/last	English French German Italian Spanish Portuguese Hindi Chinese Korean Japanese More English Italiano Português 한국어 简体中文 Deutsch Español हिंदी 日本語 English French German Italian Spanish Portuguese Hindi Chinese Korean Japanese Definitions Summary Synonyms Sentences Pronunciation Collocations Conjugations Grammar Examples of 'last' in a sentence Examples from Collins dictionaries I got married last July. He never made it home at all last night. It is not surprising they did so badly in last year's elections. Much has changed since my last visit. At the last count inflation was 10.9 per cent. I split up with my last boyfriend three years ago. The last few weeks have been hectic. The next tide, it was announced, would be even higher than the last. When were you there last? The house is a little more dilapidated than when I last saw it. Hunting on the trust's 625,000 acres was last debated two years ago. This is his last chance as prime minister. She said it was the very last house on the road. They didn't come last in their league. I'm not the first employee she has done this to and I probably won't be the last. The trickiest bits are the last on the list. I was always picked last for the football team at school. The foreground, nearest the viewer, is painted last. She was the last to go to bed. Riccardo and I are always the last to know what's going on. Jed nodded, finishing off the last piece of pizza. He finished off the last of the coffee. The last of the ten inmates gave themselves up after twenty eight hours. He reached the last four at Wimbledon. The last thing I wanted to do was teach. He would be the last person who would do such a thing. I would be the last to say that science has explained everything. She disappeared shouting, 'To the river, to the river!' And that was the last we saw of her. I had a feeling it would be the last I heard of him. The marriage had lasted for less than two years. The games lasted only half the normal time. Enjoy it because it won't last. You only need a very small blob of glue, so one tube lasts for ages. The repaired sail lasted less than 24 hours. The implication is that this battery lasts twice as long as other batteries. If you build more plastics into cars, the car lasts longer. They wouldn't have lasted the full game. I almost lasted the two weeks. I only had a couple of days to do. It'll be a miracle if the band lasts out the tour. A breakfast will be served to those who last out till dawn! Examples from the Collins Corpus These examples have been automatically selected and may contain sensitive content that does not reflect the opinions or policies of Collins, or its parent company HarperCollins. We welcome feedback: report an example sentence to the Collins team. Read more… Their marriage had lasted less than a year. The Sun (2016) We should have had one last week. Times, Sunday Times (2016) She disappeared a couple of hours after she was last seen alive. Times, Sunday Times (2017) When we played there last year we had opportunities. The Sun (2016) The last four home games we had two wins and two draws. Times, Sunday Times (2016) Yet for him this may be his last chance of affection. Times, Sunday Times (2016) In its simplest form this may just be by finding a play to use the last of your letters. Times, Sunday Times (2016) They moved to a house in Ealing last spring. Times, Sunday Times (2016) If design was dead last week, is this the revival? Times, Sunday Times (2016) Once again last night, they were unconvincing. The Sun (2016) We went into the last week knowing less than nine points gave us no chance. Times, Sunday Times (2006) They rounded the corner of the last house in the street and emerged into a small square. Iain Gale Man of Honour (2007) Then came the invitation to play golf one last time. Christianity Today (2000) It was so funny to see her twirling again last night. The Sun (2010) We were used to it last year. The Sun (2014) You mean something that lasts a long time can be interesting? Times, Sunday Times (2006) He was also the last person to say goodnight to me yesterday. Times, Sunday Times (2010) Last time we were a very happy team. Times, Sunday Times (2012) This is probably your last chance at courtship. Christina Dodd SOMEDAY MY PRINCE (1999) When did you last hear of an unemployed solicitor? Times, Sunday Times (2009) Why is the ministry so often the last one to know what its personnel are doing? Times, Sunday Times (2006) He saw part of his wish come true last week. Times, Sunday Times (2006) He is the last man to concede that it is flawed or failing as a concept. Times, Sunday Times (2010) Twenty years ago last week the educational landscape changed. Times, Sunday Times (2008) City have lost three of their last four at home and fail too often against the top clubs. The Sun (2006) What did you earn last year? Times, Sunday Times (2015) There were two summits taking place in Europe at the end of last week. Times, Sunday Times (2007) We got a heart-warming letter from our electricity provider last week. Times, Sunday Times (2014) He was jeered by fans as recently as last month after a 1-0 defeat at home by Wigan. The Sun (2012) The vote last night showed that is needed for a deal. The Guardian (2019) Last week he said he would consider whether to wear it given the controversy it has caused. The Guardian (2018) So this is what the last roll of the dice looks like. The Guardian (2016) Some embrace knowing that they have a finite amount of time left and grab at their last years with gusto. The Guardian (2017) We saw that in the last series. The Guardian (2020) How long did the police interview last? The Guardian (2016) His personality inspired the upturn last time. The Guardian (2015) This is the last unknown sport. The Guardian (2015) She got lit up early last time but expect a bold show here. The Guardian (2016) We have very large facilities here that were built last year. The Guardian (2019) What has he been up to since last summer? The Guardian (2019) The interest in the last three or four years has increased significantly. The Guardian (2022) That is the last of the expected resignations. The Guardian (2022) There is more to write a book about this year than there is about the last eight years. The Guardian (2022) As this blog reported last week, some of these powers are conditional. The Guardian (2015) We have to convince those we didn't convince last year. The Guardian (2015) The meeting lasted little more than 15 minutes. The Guardian (2016) It was at just 3.05% early last week. The Guardian (2018) She's always fighting until the last point. The Guardian (2018) What happened last week didn't involve tanks or live rounds. The Guardian (2018) He 'll need a miracle to finish this race anything but last. The Guardian (2018) The third and last phase is unfolding. Royal Society Interface (2023) The last two datasets show different patterns. Royal Society Interface focus (2022) This conformity has existed during the last forty centuries. 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188095	https://math.stackexchange.com/questions/186940/mathematics-behind-intersection-points-of-two-lines-using-quadratic-equation	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Mathematics behind intersection points of two lines using quadratic equation Ask Question Asked Modified 12 years, 7 months ago Viewed 2k times 2 $\begingroup$ This is the question I am trying to solve. I do not need any code examples just help on mathematics. Suppose two line segments intersect. The two endpoints for the first line segment are $(x_1, y_1)$ and $(x_2, y_2)$ and for the second line segment are $(x_3, y_3)$ and $(x_4, y_4)$. Write a program that prompts the user to enter these four endpoints and displays the intersecting point. (Hint: Use the LinearEquation class from the preceding exercise.) For the program to work I have to convert the endpoints of the lines into $a$, $b$, $c$, $d$, $e$, and $f$ so I can use the quadratic equation: $$x = \frac{(ed - bf)}{(ad - bc)} \ \text{ } \ y = \frac{(af - ec)}{(ad - bc)}$$ I cannot for the life of me through use of Google and such figure out how to convert the endpoints of each line into the variables I need for that equation. I have the class LinearEquation to use I just need to figure out how to solve for the variables in it. Any help would be greatly appreciated. This is not homework by the way, I have been self teaching myself Java so I don't have many resources. Thanks ahead of time! P.S. I do not need any code examples. I am just really looking for someone to point me in the right direction on the mathematics behind how to do this. I have been looking for too long with no real answer. algebra-precalculus analytic-geometry quadratics Share edited Dec 16, 2012 at 3:40 Martin Argerami 219k1717 gold badges161161 silver badges299299 bronze badges asked Aug 26, 2012 at 2:20 Alec HorneAlec Horne 2122 bronze badges $\endgroup$ 1 $\begingroup$ I've changed algebra tag to algebra-precalculus, since we don't use algebra tag anymore, see meta for details. $\endgroup$ Martin Sleziak – Martin Sleziak 2012-08-26 03:21:29 +00:00 Commented Aug 26, 2012 at 3:21 Add a comment \| 2 Answers 2 Reset to default 2 $\begingroup$ Since you have two points for each line, we can find the formula for the unique line (assuming the points are distinct). For the first line, defined by $(x_1,y_1)$ and $(x_2,y_2)$, the slope is "rise-over-run", i.e. $\frac {y_2 - y_1} {x_2 - x_1}$, thus any point $x$ and $y$ on this line must have the same slope, so $$\frac {y - y_1} {x - x_1}= \frac {y_2 - y_1} {x_2 - x_1}\implies y=\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x - \left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1+y_1\,.$$For the second line, we get the analogous result: $$y=\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\,.$$ Now we just have to solve these two equations, since we are looking for the point of intersection where the $x$ and $y$ variables for each line are equivalent, we set the $y's$ equal to each other and arrive at the equation: $$\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x - \left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1+y_1=\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\,,$$ which we can simplify to $$\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)\right]x= \left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1-y_1 - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\,.$$ Now if $\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)\right]\ne 0$ we can divide to find a specific formula for $x$: $$x=\frac 1 {\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)}\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1-y_1 - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\right]\,.$$ Now, finding $y$ will be easy - we just "back substitute" $x$ into either equation for $y$, say the first one: $$y=\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)\left (\frac 1 {\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)}\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1-y_1 - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\right] \right) - \left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1+y_1\,.$$ This will always work, unless we have that $\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)\right]=0$. In that case, you either have coincident lines or parallel lines - to determine which, just check the y-intercept of each line (we already know the slopes are equal because of the above condition). If the y-intercepts of each line are equal to each other, then the lines are coincident (so there are infinitely many "points of intersection"). If not, they are parallel (and there are no points of intersection). Share answered Aug 26, 2012 at 2:43 mboratkomboratko 4,6032828 silver badges3636 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ You can use the two-point equation for a line to determine the equations for two lines. $$ y = y_1 + \frac{y_2-y_1}{x_2-x_1}(x-x_1) $$ and $$ y = y_3 + \frac{y_4-y_3}{x_4-x_3}(x-x_3)$$ Rearrange these to give $ax+by=c_1$ and $cx+dy=c_2$ and solve these by some usual technique for solving two-equations and two unknowns. I assume you have software for this. You'll have to isolate special cases for this to work. If either of the line segments are vertical then the formulas break down. Also, if some sarcastic prof. puts the same pair of points in then you might want to account for the case of infinitely many points of intersection. Good luck. Share answered Aug 26, 2012 at 2:34 James S. CookJames S. Cook 17.3k33 gold badges5252 silver badges131131 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algebra-precalculus analytic-geometry quadratics See similar questions with these tags. 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188096	https://docs.data.world/documentation/sql/concepts/intermediate/aggregations.html	Concepts Basic Learn SQL Basics The SELECT and FROM Clauses The SELECT EXCLUDE and SELECT RENAME Clauses LIMIT WHERE and Comparison Operators ORDER BY AND, OR, and NOT Arithmetic Operations IS NULL and IS NOT NULL LIKE and NOT LIKE BETWEEN OFFSET Intermediate Working With Joins, Aggregations, and Built-In Functions Joins CAST Aggregations GROUP BY and FILTER HAVING UNION, INTERSECT, and MINUS POSITION CASE IN and NOT IN Working With Strings Working With Dates Working With Arrays Advanced WITH Inline Subqueries Value blocks Special Uses of Inline Subqueries Advanced Aggregate Functions Windowed Aggregations ROLLUP and CUBE PIVOT UNPIVOT TABLESAMPLE Metadata Tables Lexical Issues data.world Specific Multi-Tables Federated Queries Parameterized Queries Special Columns Query past instances of data Text formatting for query outputs Reference Boolean Aggregations Regression Aggregations Standard Aggregations Statistical Aggregations Other Aggregations PRAGMA Statements All Functions Color Functions Array Functions Date Functions Formatting Functions General Functions Linked Data Functions Hashing and Related Functions Math Functions Security Functions String Functions URL Extraction Functions Aggregations APPROX_MEDIAN APPROX_DISTINCT ARBITRARY APPROX_PERCENTILE ARRAY_AGG AVG BOOL_AND BOOL_OR CHECKSUM COUNT CORRELATION COUNT_IF COVAR_POP COVAR_SAMP CUME_DIST DENSE_RANK FIRST_VALUE GROUP_CONCAT LAG KURTOSIS GROUPING LEAD LAST_VALUE MAX MIN MIN_BY MAX_BY NTH_VALUE PERCENT_RANK NTILE RANK REGR_AVGY REGR_COUNT REGR_AVGX REGR_R2 REGR_INTERCEPT REGR_SXX REGR_SLOPE REGR_SXY REGR_SYY ROW_NUMBER STD_SAMP SKEWNESS STD_POP STDEV SUM VAR_SAMP VAR_POP VARIANCE Functions ABS ARRAY ACOS ARRAY_CONCAT ARRAY_CONTAINS ARRAY_LOWER ARRAY_JOIN ARRAY_APPEND ARRAY_LENGTH ARRAY_PREPEND ARRAY_UPPER ATAN ASIN AT_TIME_ZONE ATAN2 ATTR_OF BASE64_TO_HEX BOLD CHAR CONCAT CEILING COLOR COALESCE COS CURRENT_USER COSH CURRENT_USER_IN_GROUP DATE_ADD DATE_COMPARE DATE_FORMAT DATE_DIFF DATE_PART DAY DATE_PARSE DATE_SUB DATE_TRUNC DEGREES ELEMENT_AT ENDSWITH EXP EXP10 FLOOR GENERATE_SERIES GREATEST IF HEX_TO_BASE64 IRI_OF IRI_TO_RESOURCE_HASH ITALIC HOURS JARO_WINKLER_DISTANCE LEAST LABEL_OF LEFT LENGTH LEVENSHTEIN_DISTANCE LOG LOG10 LOWER LPAD LTRIM MD5 MID MINUTES MOD NOW MONTH NOW_UTC POSITION PI POW RADIANS RAND RANDOM REGEXP_EXTRACT REGEX REPLACE REVERSE RIGHT RGB ROUND RTRIM RPAD SECONDS SHA1 SHA256 SHA384 SHA512 SIGN SIN SINH SOUNDEX SOUNDEX_DISTANCE SQRT STARTSWITH STATUS STR_AFTER STR_BEFORE STR_CONTAINS TAN STRING_SPLIT SUBSTRING TANH TRIM TRUNC UPPER URL_EXTRACT_FRAGMENT URL_EXTRACT_HOST URL_EXTRACT_PARAMETER URL_EXTRACT_PATH URL_EXTRACT_PORT URL_EXTRACT_PROTOCOL URL_EXTRACT_QUERY WITH_BACKGROUND_COLOR WITH_COLOR WITH_LINK WITH_STATUS YEAR Concepts/Intermediate/ Aggregations An introduction to aggregations, aggregation functions, and aggregation modifiers. An aggregation is the result returned from running an aggregate function in a query and is displayed in a new column in the results table. An aggregate function is a function that results of a set of the data queried being condensed into an aggregation. The most common aggregate functions are SUM, COUNT, AVG, MIN, MAX, and GROUP_CONCAT. SUM returns the sum of all numeric value in a specified column. COUNT preceding a column name returns the number of rows for which the value is not NULL and can be used with both numeric and non-numeric field-types. COUNT() returns the number of rows in the table specified in the FROM clause. AVG returns the average of all the entries in a column which are not NULL. AVG can only be used with numeric data. MIN returns the lowest value in a specified column and can be used with numeric and non-numeric data. MAX returns the greatest value in a specified column and can be used with both numeric and non-numeric data. GROUP_CONCAT is the only aggregate function designed to work with strings. It takes string values and aggregates them into one value for each row with an optional separator. SUM If we wanted to know the total value of all the deals closed we would write: SELECT SUM FROM siyeh/sql-crm-example-data Run query Copy code and the resulting table would be: \| sum \| \| 10,005,534 \| COUNT Similarly, if we wanted to see the number of deals we had won in our sales pipeline we could write a query like this: SELECT COUNT FROM WHERE "Won" siyeh/sql-crm-example-data Run query Copy code and the result returned would look like: \| count \| \| 4,238 \| AVG Likewise we could show the average deal won: SELECT AVG FROM WHERE "Won" siyeh/sql-crm-example-data Run query Copy code \| avg \| \| 2,360.91 \| Our pipeline data has a stored value of ‘0’ instead of a blank field for close_value of lost deals. If we run an average on the close_value column without specifying we want to only include the deals won we’ll get an average of all the completed deals regardless of whether they were won and lost. While an average of all finished deals is potentially interesting data, it isn’t the same as an average value of the actual sales. MAXand MIN For the largest deal we would query: SELECT MAX FROM siyeh/sql-crm-example-data Run query Copy code \| max \| \| 30,288 \| And for the smallest deal: SELECT MIN FROM WHERE "Won" siyeh/sql-crm-example-data Run query Copy code \| min \| \| 38 \| As for AVG, we had to include the qualification to only use data from deals that were won because otherwise the smallest deal size would have been ‘0’–the value for lost sales. GROUP CONCAT Until we look at the GROUP BY clause in the next section, the only use we have for GROUP_CONCAT at this point is to get a list of the values from one column in a table and to return it as a string. For example, if we wanted a list of all of the managers we could run the query: SELECT GROUP_CONCAT FROM siyeh/sql-crm-example-data Run query Copy code The result would be: \| Managers \| \| Rocco Neubert Celia Rouche Summer Sewald Summer Sewald Celia Rouche Melvin Marxen Melvin Marxen Summer Sewald Dustin Brinkmann Melvin Marxen Melvin Marxen Rocco Neubert Summer Sewald Melvin Marxen Rocco Neubert Cara Losch Celia Rouche Cara Losch Celia Rouche Rocco Neubert Dustin Brinkmann Cara Losch Rocco Neubert Cara Losch Dustin Brinkmann Cara Losch Summer Sewald Celia Rouche Dustin Brinkmann Dustin Brinkmann \| There are two problems with that result. First, it is not a list of the managers, it’s a list of all the instances of manager (one for each sale agent in the table). It we want to return each manager’s name only once we need to use the DISTINCT modifier. Second, the names all run together, and it’s likely that if we wanted a list we’d want the entries separated. Fortunately, GROUP_CONCAT has the modifier SEPARATOR which inserts the specified string between each value returned by GROUP_CONCAT. The amended search would look like: SELECT GROUP_CONCAT DISTINCT", " AS FROM siyeh/sql-crm-example-data Run query Copy code And the result would be: \| Managers \| \| Rocco Neubert, Celia Rouche, Summer Sewald, Melvin Marxen, Dustin Brinkmann, Cara Losch \| The keyword ASwas used in the preceding query because without it the column name used in the table would have been undefined. Exercises (Continued from previous section) There are two ways to do these exercises. The first is to use the “Try query” links to test your queries without saving them. The second is to create a data.world project and save your queries to it. If you are reading this documentation and completing the exercises as a tutorial, you will need to create your own project to save your work. Details and instructions are in the SQL tutorial which has instructions for setting up your project and links to all the current exercises. Exercise 24 Write a query that returns the description and the number of patients who have an allergy to fish from the allergies table. You should get a warning that the “Field of aggregated query neither grouped no aggregated”. Grouping is not necessary for this query and will be discussed in the next section. Try query Hint SELECT the column description, and a COUNT of all the records () FROM the allergies table in the case WHERE the description column = “Allergy to fish” See solution Exercise 25 As we saw in exercise 23, real-world data is not always nice and tidy and sometimes we have to do a little clean-up to get it in the format we need to analyze it. While the CAST expression is very good at turning strings into integers or decimals, it’s a pain to have to include it in every aggregation clause of your queries to clean up your data whenever there is a datatype mismatch. Fortunately we’re doing these exercises on data.world, and we have a little shortcut to take the sting out of data clean-up. For the next exercise, take the query result from exercise 23 and modify it to include the descriptions “Body Height” and “Body Mass Index”. Save your query as “Observations Cleaned”. Then go to the Download button underneath the query text and select “Save to dataset or project”. If you need help with the “Save to dataset or project option, see the help article Why “Save to dataset or project” is important and how to use it You’ll be prompted to save the table with the name of the query to your current project. Select Save. Once you have saved this table you can use it to do aggregations and other numerical calculations on the data. Now write a query against this new table that returns the description column, the average body weight of all the patients, and the units from the observations table. You should get a warning that the “Field of aggregated query neither grouped no aggregated”. Grouping is not necessary for this query and will be discussed in the next section. Try query Hint SELECT the column description, the AVG of the column (value), and the column units FROM the observations_cleaned table for the records WHERE the description column = “Body Weight” See solution Exercise 26 Write a query that returns the description column, the maximum body height of all the patients, and the units from the observations table. You should get a warning that the “Field of aggregated query neither grouped no aggregated”. Grouping is not necessary for this query and will be discussed in the next section. Try query Hint SELECT the column description, the MAX of the column (value), column, and the units column FROM the observations_cleaned table for the records WHERE the description column = “Body Height” See solution Next up: GROUP BY and FILTER An introduction to the GROUP BY clause and FILTER modifier.
188097	https://www.researchgate.net/publication/1920193_A_Structural_Approach_to_Subset-Sum_Problems	Published Time: 2008-05-20 (PDF) A Structural Approach to Subset-Sum Problems Article PDF Available A Structural Approach to Subset-Sum Problems May 2008 DOI:10.1007/978-3-540-85221-6_19 Authors: Van Vu Yale University Download full-text PDFRead full-text Download full-text PDF Read full-text Download citation Copy link Link copied Read full-textDownload citation Copy link Link copied Citations (6)References (87) Abstract We discuss a structural approach to subset-sum problems in additive combinatorics. The core of this approach are Freiman-type structural theorems, many of which will be presented through the paper. These results have applications in various areas, such as number theory, combinatorics and mathematical physics. Discover the world's research 25+ million members 160+ million publication pages 2.3+ billion citations Join for free Public Full-text 1 Content uploaded by Van Vu Author content All content in this area was uploaded by Van Vu on Aug 14, 2015 Content may be subject to copyright. arXiv:0804.3211v1 [math.CO] 20 Apr 2008 A STR UCTURAL APPRO A CH TO SUBSET-SUM PROBL EMS V AN VU Abstract.W e discuss a structural approach to subset-sum problems in ad- ditiv e combinat orics.The core of this approach are F reiman-t ype structural theorems,many of which w ill b e presen ted through the paper.These results ha ve applications in v arious areas,suc h as nu mber theory,combinat orics and mathematical physics. 1.Introduction Let A={a 1 ,a 2 ,...}be a subset of an a dditiv e group G(all groups discussed in this pap er will b e ab elian).Let S A be the collection of subset sums of A S A := { X x∈B x\|B⊂A,\|B\|<∞}. Two rela ted notions that a re frequently considered are l A:= {a 1 +···+a l \|a i ∈A} l ∗ A:= {a 1 +···+a l \|a i ∈A,i 6= j}. W e hav e the trivial relations l ∗ A⊂lA a nd∪ l l ∗ A= S A . One can hav e similar deﬁnitions for A b eing a sequence(rep etitions allow ed). Example. A= {0,1,4}, G= Z,2 A= {0,1,2,4,5,8}, 2 ∗ A= {1,4,5}, S A = {0,1,4,5}. A= {0,1,4},G= Z 5 ,2 A= G,2 ∗ A= {0,1,4}= S A . V.V u is supp orted by NSF Career Gran t 0635606. 1 2 V AN VU Now let A b e a sequence: A= {1,1,9}, G= Z,3 A= {3,11,1 9,27}, 3 ∗ A= {11},S A = {1,1,2,9,1 0,11}. Notice that for a large l,l A can be signiﬁca n tly diﬀerent from S A and l ∗ A.In general,it is eas ier to handle than the later tw o. Many basic problems in additive combinatorics ha ve the following form: If A is suﬃciently dense in G,then S A (or l ∗ A or lA)c ontains a sp e cial element (such as 0 or a squar e),or a lar ge stru ctur e(such as a long arithmetic pr o gr ession G itself). The main question is to ﬁnd the threshold for“dense”.As examples,we presen t below a few w ell-known r esults/problems in the area.In the whole pap er,w e a re going to focus mostly on t wo special cases:(1) G=Z p , where Z p denotes the cyclic group of residues mo dulo a large prime p;(2)G= Z,the set o f integers. F ollowing the literature,we say that A is zer o-sum-fr e e if 0/∈S A .F urthermore,A is c omplete if S A =G and inc omplete otherwis e.The a symptotic notation is used under the assumption that\|A\|→∞. A basic result concerning zero-sum-free sets is the following theorem of Olso n and Szemer´edifrom the late 1960 s,addressing a problem of Er d˝os and Heil- bronn. Theorem 1.1.(Olson-Szemer´edi)L et A b e a subset of Z p with c ar dinality C √ p, for a suﬃciently lar ge c onstant C.Then S A c ontains zer o. T o see that o rder √ p is necessar y,consider A:={1,2,...,n},where n≈ √ 2 p is the lar gest integer suc h that 1+···+n<p. Concerning completeness,Olson[5 2],proved the following r esult Theorem 1.2.(Olson)Any subset A of Z p with c ar dinality at le ast √ 4 p−3 +1 is c omplete. T o see that the b ound is clo se to o ptimal,take A:={−m,...,−1,0,1,...,m} where m is the lar gest integer such that 1+···+m<⌊p/2⌋. Another class ical result concerning z ero sums is that o f Erd˝o s-Ginburg-Ziv, again from the 1 960s. Theorem 1.3.(Er d˝os-Ginbur g-Ziv)If A is a se quen c e of 2 p−1 elements in Z p , then p ∗ A c ontains zer o. This theorem is shar p b y the following example:A= {0 [p−1] ,1 [p−1] }F urthermor e, instead o f 0 and 1,one ca n use any t wo diﬀerent elemen ts of Z p .(Here and la ter x [k] means x app ears with multiplicit y k in A.) A STRUC TURAL APPR OA CH TO SUBSET-SUM PROBLEMS 3 Now we discuss tw o problems in volving in tegers.Set[n]:={1,2,...,n}.An old and p opular c onjecture concer ning subset sums of integers is F o lkman’s co njecture, made in 1966.F olkman’s conjecture is a strengthening of a conjecture b y Erd˝os ab out ﬁnding a necessar y and suﬃcien t condition for a sequence A such that S A contains all but ﬁnite exception of the p ositive integers. Conjecture 1.4.(F olkman’s c onje cture)The fo l lowing holds for a ny suﬃciently lar ge c onstant C.L et A b e an st rictly incr e asing se quen c e of p ositive inte gers with (asymptotic)density at le ast C √ n(namely,\|A∩[n]\|≥C √ n for al l suﬃciently lar ge n ) .Then S A c ontains an inﬁnite arithmetic pr o gr ession. Casselsand E rd˝osshow e d that density √ n is indeed needed;th us,F olk- man’s conjecture is sharp up to the v alue of C.F or mor e discussion ab out F olk- man’s co njecture and its relation with Erd˝os’conjecture,w e refer toand the monogra ph [2 1]by Er d˝os and Graham. Finally,a problem inv olving a no n-linear rela tion,p osed b y Erd˝os in 1986[1 9]. Problem 1 .5.(Er d˝os’squar e-sum-fr e e pr oblem)A set A of inte gers is squar e- sum-fr e e if S A do es not c ontain a squar e.Find the lar gest size of a squar e-su m-fr e e subset of[n]. Erd˝os o bserved that one can construct suc h a square-sum-free subset of[n]with at least Ω(n 1/3 )element s.T o s ee this,co nsider A:={q,2 q,...,k q}with q prime, (k+1)k<2 q,k q≤n.Since the sum of all elements of A is less than q 2 ,S A do es not contain a squa re.Erd˝o sc onjectured that the truth is close to this low er bo und. Problems inv olving subset sums such a s the above(and many others)ha v e b een attack ed,with considerable succe ss,using v a rious techniques:co mbinatorial,har- monic analysis,algebraic etc.The reader who is interested in these techniques ma y wan t to lo ok at[3,5 7,64,48]a nd the references ther ein. The goal of this paper is to intro duce the so-called“structural approach”,which has b een developed s ystematically in recent years.T his appr oach is based on the following simple plan Step I:F or c e a structur e on A.I n this step,one tr ies to show the following:If A is relatively dense(close to the c onjectured threshold but not yet there)and S A do es not contain the des ired ob ject,then A has a very sp ecial structure. Alternately,o ne can can try to Step I’:Find a structur e in S A .If A is r elatively dense(aga in close to the conjec- tured threshold but not yet there)then S A contains a sp ecial structure. Step II:Completion.Since\|A\|is s till below the threshold,we can add(usually many)new element s to A.Using these elements together with the existing struc- ture, one can, in most cases, obtain the desired ob ject in a relativ ely simple manner. 4 V AN VU The success of the method dep ends on the quality of the information we can obtain on the structure of A(or S A )in Step I (or I’). In several recent studies, it has turned out that o ne c an frequently o btain something clos e to a c omplete char acterization of these sets.Thanks to these r esults,one is able to make considerable progres ses on man y old pro blems and also reprove and strengthen sev eral existing ones (with a better understa nding and a complete classiﬁcatio n of the extremal cons tructions). The res t of this pa per is dev oted to the pr esentation of these structural theorems and their representative applications. 2.Freiman’s structural theorem A corner stone in additive combinatorics is the structura l theorem o f F reiman(some- time referred to as F r eiman’s inv erse theorem),which writes down the structure of sets with small doubling. A gener alize d ari thmetic pr o gr ession(GAP)of rank d in a group G is a set of the form {a 0 +a 1 x 1 +···+a d x d \|M i ≤x i ≤N i }, where a i are e lemen ts o f G and M i ≤N i are integers.It is in tuitiv e to view a GAP Q as the image of the d-dimensional in teg ral box B:= {(x 1 ,...,x d )\|M i ≤x i ≤N i } under the linear ma p Φ(x 1 ,...,x d ) = a 0 +a 1 x 1 +···+a d x d . W e say that Q is pr op er if Φ is one-to-one.It is easy to s ee that if Q is a prop er GAP o f rank d and A is a subset of density δ o f Q ,then\|2 A\|≤C(d,δ)\|A\|.Indeed, \|2 A\|≤\|2 Q\|≤\|2 B\|= 2 d \|B\|= 2 d \|Q\|≤ 2 d δ \|A\| since the volume of a b ox increases by a factor 2 d if its sizes are do ubled. F reiman’s theorem shows that this is the only construction of sets with co nstant doubling. Theorem 2.1.(F r eiman’s the or em)F or any p ositive c onstant C,ther e ar e p ositive c onstants d=d(C)ad δ=δ(C)such that the fol lowing holds.L et A b e a ﬁnite su bset of a t orsion-fr e e gr oup G such that\|2 A\|≤C\|A\|.Then ther e is a pr op er GAP Q of dimension d such that A⊂Q and\|A\|≥δ \|Q\|. A STRUC TURAL APPR OA CH TO SUBSET-SUM PROBLEMS 5 F reiman theorem has b een extended recently to the tor sion case by Gr een and Ruzsa.[64,Chapter 5]con tains a detailed discussion of b oth theorems and related re sults. One can use F reiman’s theo rem iteratively to treat the sumset l A,for l>2.F or simplicit y,assume that l=2 s is a pow er o f 2.Thus,the se t A s :=l A=2 s A can be viewed as 2 A s−1 where A s−1 :=2 s−1 A.Using a m ulti-scale analys is combined with F remain’s theorem,one can o btain use ful structural information ab o ut l A o r A itself.F or an example of this technique,w e refer toor[6 4,Chapter 12]. The tr eatment of l ∗ A and S A is more diﬃcult.How ever,one ca n still develop structural theorems in these cases.While the conten t of most theorems in this direction are quite diﬀ erent from that of F reiman’s,they do bear a similar spirit that someho w the most natural cons truction happens to b e(essentially)the only one. 3.Structure of zer o-sum-free sets Let A be a zero-sum-fre e subset of Z p .W e reca ll the exa mple following Theo rem 1.1. Let A:= {1,2,...,n}.If 1+··· +n<p, then obviously S A do es not contain 0.This shows that a zero-sum-free set can hav e close to √ 2 p elements.In,Szemer´edi and V u show ed that having elements with smal l sum is essential ly the only r e ason for a set to b e zer o-su m-fr e e.More quan titativ e versions of this s tatement were work ed o ut inand.F or example,we have [50,Theorem 2.2] Theorem 3.1.After a pr op er dilation(by some n on-zer o element),any zer o-sum- fr e e subset A of Z p has the form A= A ′ ∪A ′′ wher e t he elements of A ′ (viewe d as inte gers b etwe en 0 and p−1)ar e smal l, P x∈A ′ x<p,and A ′′ is ne gligible,\|A ′′ \|≤p 6/13+o(1) . One ca n perha ps improve the constant 6/13 b y tigh tening the ana lysis in.It is not clear,howev er,what would b e the b est constant here.In mo st applications, it suﬃces to have a ny constant strictly less than 1/2. The dilation is neces sary.Notice that if A is zero-sum-free(incomplete),then the set A x := {xa\|a∈A}is also zero-sum-free(incomplete)for a ny 0 6= x∈Z p . W e can a lso prove similar results for lA a nd l ∗ A,and for A b eing a sequence(see for details).In the rest of this section,we presen t few applications of these results. 3.2.The size of the l argest zero-sum-free set in Z p .Let m p denote the size of t he largest zero-sum-free set in Z p .The pro blem of determining m p was p o sed b y 6 V AN VU Erd˝os and Heilbronn and has a long history.Szemer´edi prov ed tha t m p ≤C √ p, for some suﬃcien tly lar ge C indep endent of p.Olson show ed that C= 2 suﬃces .Muc h later,Hamidoune and Z´emo rshowed that m p ≤ √ 2 p+5 log p, which is a symptotically sha rp.Using a n ea rlier v ersion o f Theor em 3.1,Szemer´edi, Nguyen a nd V urecently obtained the exact v alue o f m p . Theorem 3.3.L et n p b e the lar gest inte ger so that 1+···+(n p −1) <p. •If p 6= n p (n p +1) 2 −1,then m p = n p −1. •If p= n p (n p +1) 2 −1,t hen m p = n p .F u rthermor e,up to a dilation,the only zer o-sum-fr e e set with n p elements is{−2,1,3,4,...,n p }. The same re sult was o btained by Deshouiller s and Pr ak a sh(p ers onal communica- tion by Deshouillers)at ab out the same time. 3.4.The structure of relativ ely l arge zero-sum-free s e ts.Let us now con- sider the structure of zero-sum-free sets of size close to √ 2 p.Let k x k denote the in teger nor m of x.In,Desho uillers proved Theorem 3.5.L et A b e a zer o-sum-fr e e subset of Z p of size at le ast √ p.T hen (after a pr op er dilation) X x∈A,x<p/2 k x/p k≤1+O(p −1/4 log p) X x∈A,x>p/2 k x/p k≤O(p −1/4 log p). Deshouillers showed(by a construction)that the error t erm p −1/4 cannot be replace b y o(p −1/2 ).Using a n ear lier version of Theorem 3.1,Nguyen,Sze mer´edi and V u improved Theor em 4.4 to obtain the b est p ossible erro r term O(p −1/2 ),under a stro nger assumption on the size o f\|A\|. Theorem 3.6.L et A b e a zer o-sum-fr e e subset of Z p of size at le ast.99 √ 2 p.Then (after a pr op er dilation) X x∈A,x<p/2 k x/p k≤1+O(p −1/2 ) X x∈A,x>p/2 k x/p k≤O(p −1/2 ). The co nstant .99 is,of course,ad-ho c and can b e improved by redoing the analysis carefully.On the other hand, it is not clear what the b est assumption on \|A\|should be. A STRUC TURAL APPR OA CH TO SUBSET-SUM PROBLEMS 7 3.7.Erd˝os-Ginbur g-Ziv revisited.Using a version of Theorem 3.1 for sequences, Nguyen and V uobtained the following characterization for a sequence o f size slightly mor e than p and does not con tain a subse quence of p elemen ts summing up to 0. Corollary 3.8.[50,Theor em 6.2]L et ε b e an arbitr ary p ositive c onst ant.Assume that A is a p-zer o-sum-fr e e se quenc e and p+ p 12/13+ǫ ≤\|A\|≤2 p−2.T hen A c ontains t wo elements a and b with m u ltiplicities m a ,m b satisfying m a m b ≥ 2(\|A\|−p−p 12/13+ǫ ). The interesting p oint here is that the s tructure kicks in v ery soo n, when A has just slightly more than p elements.F ew years ago, Gao, Panigrahi, and Thangdurai prov ed a similar statement under the stronger ass umption that\|A\|≥3 p/2. It is easy to deduce Er d˝os-Ginburg-Ziv theorem from Corollar y 3.8,together with a complete c haracter ization of the extrema l s ets.The reader ma y wan t to co nsider as a n exercise o r chec k[5 0]for details. 4.Incomplete Sets Now we turn o ur atten tion to incomplete sets,namely sets A where S A 6= Z p .The situation here is very s imilar to that with zero-sum-free sets.Szemer´edi and V u showed that having elements with smal l sum is essential ly the only r e ason for a set to b e inc omplete.More quantitativ e versions of this statemen t were w orked out inand.F o r example,in,the following analogue of Theorem 3.1 was proved Theorem 4.1 .After a pr op er dilation(by some non-zer o element),any inc omplete subset A of Z p has the form A= A ′ ∪A ′′ wher e the el ements of A ′ ar e smal l(in the inte ger norm), P x∈A ′ k x/p k<1 and A ′′ is n e gligible,\|A ′′ \|≤p 6/13+o(1) . The reader can ﬁnd similar results for l A and l ∗ A and for A being a sequence in .W e next discuss so me applications o f these results. 4.2.The structur e of relatively large i ncomplete sets.Theor em 4.1 enables us to prov e results similar to those in the last section for incomplete sets.The problem of determining the la rgest s ize of an incomplete set in Z p was ﬁrst consid- ered b y Erd˝os and Heilbronnand essenti ally solved by Olson(Theorem 1.2). da Silv a and Hamidounetigh tened the bound to √ 4 p−7 + 1,whic h is bes t po ssible.W e ar e not going to g o in to these results here,but note that one can per haps obtain a new pro of(with a c haracteriza tion of the extremal sets)using Theorem 4.1. 8 V AN VU Concerning the structure of rela tiv ely lar ge incomplete s ets,Deshouillers and F r eiman prov ed Theorem 4.3.L et A b e an inc omplete su bset of Z p of size at le ast √ 2 p.Then (after a pr op er dilation) X x∈A k x/p k≤1+O(p −1/4 log p). They conjectured that the err or term may b e repla ced b y O( √ p),which would be bes t p os sible due to a later c onstruction o f Deshouiller s. Using Theo rem 4.1,Nguyen a nd V uco nﬁrmed this conjecture,provided that A is suﬃcientl y close to 2 √ p. Theorem 4.4.L et A b e an inc omplete subset of Z p of size at le ast 1.99 √ p.Then (after a pr op er dilation) X x∈A k x/p k≤1+O( √ p). Similar to the constant.99(in the previous section),the constant 1.99 is a d-ho c and can be improv ed by redoing the analysis car efully.On the o ther hand,it is not clear what the best assumption on\|A\|is. 4.5.The struc ture of incompl ete sequences.Let us now dis cuss(ra ther brieﬂy) the situation w ith seq uences.The main diﬀ erence betw een sets and sequences is that a sequence can have elemen ts with high multiplicit ies.It has turned o ut that when the maximum multiplicit y of incomplete sequence A is deter mined,o ne ca n obtain stro ng structural information ab o ut A. Let 1≤m≤p be a p ositive in teger and A b e an incomplete sequence of Z p with maximum multipli city m.T ry ing to make A as larg e as po ssible,we come up with the following example, B m = {−n [k] ,(n−1) [m] ,...,−1 [m] ,0 [m] ,1 [m] ,...,(n−1) [m] ,n [k] } where 1 ≤k≤m a nd n are the unique in tegers satisfying 2 m(1+2+···+n−1)+2 k n<p≤2 m(1+2+···+n−1)+2(k+1)n. It is clea r that any subsequence of B m is incomplete a nd has multiplicit y at mo st m.In[5 0],we prov e d that any incomplete sequence A with max im um multiplicit y m and cardinality close to\|B m \|is esse nt ially a subset o f B m . A STRUC TURAL APPR OA CH TO SUBSET-SUM PROBLEMS 9 Theorem 4.6.L et 6/13<α<1/2 b e a c onst ant.Assume that A is an inc om- plete se quenc e of Z p with maximum multipli city m and c ar dinality\|A\|=\|B m \|− O((pm) α ).Then after a pr op er dilation,we c an have A= A ′ ∪A ′′ wher e A ′ ⊂B m and\|A ′′ \|= O((pm) (α+1/2)/2 ). 4.7.Coun ting problems.Sometime one would lik e to c ount the num b er of sets which forbid certain additive c onﬁgurations.A well-known example of problems of this type is the Camer on-Erd˝o s pro blem,which asked f or the nu mber of subsets o f[n] = {1,2,...,n}which do es not co nt ain three diﬀerent elements x,y,z such that x+ y−z=0.Camer on an E rd˝os noticed that any s et of o dd num b er s has this prop er ty.Th us,in[n]there are at lea st Ω(2 n/2 )subsets with the required prop erty.They conjectured that 2 n/2 is the right order of magnitude.There were several par tial results[2,9,22]b efore Green settled the co njecture. Using structural theorems suc h as Theorem 3.1,one ca n obtain results of similar spirit for the n umber o f ze ro-sum-free o r incomplete sets and sequences.F or e xam- ple,using an earlier version of Theorem 3.1 and standard fa cts from the theory of partitions,Szemer´edi and V uproved Corollary 4.8.The numb er of inc omplete subsets of Z p is exp(( q 2 3 π+o(1)) √ p). Using Theor em 4.6,one obtains the following gener alizations. Corollary 4.9.The numb er of inc omplete se quenc es A with highest mu ltiplicity m in Z p is exp(( q (1− 1 m+1 ) 4 3 π+o(1)) √ p). It is an interesting question to deter mine the erro r term o(1). 5.Incomplete sets in a general abelian group Let us now consider the problem of ﬁnding the largest size of an incomplete set in a general a b elian g roup G,whic h w e denote b y In(G)in the rest of this section. The situation with a g eneral gr oup G is quite diﬀerent from that with Z p ,due to the existence o f no n-trivial subgro ups.It is clea r that any such subgroup is incomplete.Th us,In(G) ≥h,wher e h is the larges t no n-trivial divisor of\|G\|.The in tuition behind the discussion in this section is that a lar ge inc omplete set should b e essent ial ly c ontaine d in a pr op er sub gr oup. In 197 5,Diderrichconjectured that if\|G\|=ph,wher e p≥3 is the smallest prime divisor of\|G\|and h is comp osite,then c(G) =h+ p−2.(The cases where p= 2 o r h is a pr ime is simpler and w ere treated ea rlier,some by Diderrich himself 10 V AN VU [13,47,14].)Didderich’s conjecture was settled b y Gao and Hamidoune in 1999 . The following simple fact explains the app ea rance of the term p−2. F act.If S A∩H =H for some maximal subgroup H of(prime)index q,then \|A\|≤\|H\|+q−2. T o verify this fact,notice that A/H is a sequence in the group Z q .It is easy to show(exercis e)that if B is a sequence o f q−1 non-zero element s in Z q ,then S B ∪{0}= Z q . W e say that subset A of G is sub-c omplete if there is a subgroup H of prime index such tha t S A∩H = H. Once w e know that an inco mplete set A is sub-c omplete,we can write down it s structure completely.There is a subgr oup H with prime index q suc h that \|A\H\|≤ q−2, and the sequence A/H is incomplete in Z q .(The structure of suc h a sequence was discuss ed in the pr evious s ection.)It is natural to p ose the fo llowing Problem 5.1.Find the thr eshold for sub-c ompleteness. Recent ly, Gao, Hamidoune, Llad´o and Serra showed(under some weak assump- tion)that a ny subset of at least p p+2 h+p elements is sub-co mplete.F urthermor e, one can choose H to hav e index p,wher e p is the smallest prime divisor of\|G\|.V u showed (aga in under some w eak assumption)that 5 6 h is suﬃcient to guarantee sub-completeness.It is not clear,how ever,that what the sha rp b ound is. The situation is muc h b etter if w e assume that \|G\|is suﬃciently compos ite.In par- ticular,if the pro duct of the tw o smallest prime divisors of\|G\|is signiﬁcantly smaller than p \|G\|,then one can determine the sharp threshold for sub-co mpleteness. Theorem 5.2.F or any p ositive c onst ant δ ther e i s a p ositive c onst ant D(δ) such that the fol lowing ho lds.Assume that\|G\|=p 1 ...p t ,wher e t≥3 and p 1 ≤ p 2 ···≤p t ar e primes such that p 1 p 2 ≤ 1 D(δ) p \|G\|/log\|G\|.Then any inc omplete subset A of G with c ar dinality at le ast(1+δ) n p 1 p 2 is sub c omplete.F urthermor e,the lower b oun d(1+δ) n p 1 p 2 c annot b e r eplac e d by n p 1 p 2 +n 1/4−α ,for any c onstant α. 6.Structures in S A As menti oned in the intro duction,an alternativ e wa y to implemen t our plan is to ﬁnd a structure in S A rather than in A(Step I’).A w ell-known result concerning the structure of S A is the following theorem,pr ov ed by F reimanand S´ark¨ozy indep endent ly. Theorem 6.1.Ther e ar e p ositive c onstants C and c such that t he fol lowing hold s for al l su ﬃciently lar ge n.L et A b e a su bset of[n]:={1,...,n}with at le ast C √ n log n elements.Then S A c ontains an arithmetic pr o gr ession of length c\|A\| 2 . A STRUC TURAL APPR OA CH TO SUBSET-SUM PROBLEMS 11 It is clea r that the b ound on the length of the arithmetic prog ression(AP)is optimal,as one can take A to b e an interv al.The lo w er b ound on\|A\|,how ever, can b e improv ed to C √ n,a s showed by Szemer´edi and V u. Theorem 6.2.Ther e ar e p ositive c onstants C and c such that t he fol lowing hold s for al l suﬃciently lar ge n.L et A b e a subset of[n] := {1,...,n}with at le ast C √ n elements.Then S A c ontains an arithmetic pr o gr ession of length c\|A\| 2 . The assumption\|A\|≥C √ n is optimal,up to the v alue of C,as one can construct a set A⊂[n]of ǫ √ n elemen ts,for some small constan t ε,such that S A do es not contain any arithmetic prog ression o f length larger than n 3/4 (seeor[63, Section 3.4]). Theorem 6.2 can b e extended cons iderably.Szemer´edi and V ushow ed that for any set A⊂[n]and an y in teg er l suc h that l d \|A\|≥n for some constan t d,the sumset l ∗ A contains a lar ge pr op er genera lized ar ithmetic progression(GAP).The parameters o f this GAP is o ptimal,up to a constant factor(see[63,Section 3]fo r more details). Theorem 6.3.[6 3, Theor em 7.1]F or any ﬁxe d p ositive inte ger d ther e ar e p ositive c onstant s C and c dep en ding on d such that t he fol lowing h olds.F or any p ositive inte gers n and l and any set A⊂[n]satisfying l d \|A\|≥C n,l ∗ A c ontains a pr op er GAP of r ank d ′ and volume at le ast cl d ′ \|A\|,for some inte ger 1 ≤d ′ ≤d. There are v ar iants of Theorem 6.3 f or ﬁnite ﬁelds, and also f or sums of diﬀeren t sets (see[63,Section 5]and[63,Section 10]).In the follo wing subsections,we discuss few a pplications of Theorems 6.2 and 6.3. 6.4.F olkman conjectures on inﬁnite arithmeti c progressi ons.Let us recall to the conjecture of F o lkman,mentioned in the introduction. Conjecture 6.5.(F olkman’s c onje cture)The fo l lowing holds for a ny suﬃciently lar ge c onstant C.L et A b e an st rictly incr e asing se quen c e of p ositive inte gers with (asymptotic)density A(n)at le ast C √ n(namely A(n):=\|A∩[n]\|≥C √ n for al l suﬃciently lar ge n).Then S A c ontains an inﬁ nite arithmetic pr o gr ession. F olkmanshow ed that the conjecture holds u nder a stronger assumption that A(n) ≥n 1/2+ǫ ,wher e ǫ is a n ar bitrarily small po sitive c onstant.(A n ear lier result of E rd˝os o n a closely related problem ca n p er haps be ada pted to give a w eaker bo und n ( √ 5−1)/2 .)Hegyv´ar iand Luczak a nd Schoen,independentl y,re- duced the densit y n 1+ǫ to C √ n log n,using Theorem 6.1. Using the stronger Theorem 6.2,together with some additional arguments,Sze- mer´edi a nd V uprov ed the full conjecture. Theorem 6.6.Conje ctur e 6.5 hold s. 12 V AN VU In the same pap er,F olkman also made a r elated conjecture for increas ing,but not strictly increasing sequences.Let A(n)no w b e t he n um b er o f elements of A (count ing multiplicities)at most n. Conjecture 6.7.(F olkman’s se c ond c onje ctur e)The following holds for any suﬃ- ciently lar ge c onstant C.L et A b e an incr e asing se quenc e of p ositive inte gers with such t hat A(n)≥C n for al l su ﬃciently lar ge n.Then S A c ontains an inﬁnite arithmetic pr o gr ession. Despite the huge change f rom √ n to n in the density bound,this conjecture is also sharp,and(for some time)app ear ed more subtle than the ﬁrst one(see a discussion in[21,Chapter 6]).F olkmanproved t he co njecture under the stronger ass umption that A(n)≥n 1+ε .It do es no t seem that one can obtain the analogue of Heg yv´ari a nd Luczak-Schoen results due to the lac k of a”sequence” v aria nt of Theorem 6.1.How ever, t he metho d in is suﬃcien tly r obust to enable one to obtain such a v aria n t for the stro nger Theo rem 6.2.With the help of this result,one can settle Conjecture 6.7 Theorem 6.8.[63,Sectio n 6]Conje ct u r e 6.7 holds. The s trategy for the pro ofs of Theorems 6.6 a nd 6.8 is the following.W e ﬁrst ﬁnd a suﬃcient condition for a sequence A such that S A contains an inﬁnite AP. W e say that a n inﬁnite sequence A admits a go o d partition if it can b e partitioned in to tw o subsequences A ′ and A ′′ with the following tw o prop erties •There is a n um b er d suc h that S A ′ contains an arbitrar y lo ng ar ithmetic progres sion with diﬀerence d. •Let A ′′ =b 1 ≤b 2 ≤b 3 ≤....F or an y n um b er K,there is an index i(K) such that P i−1 j=1 b j ≥b i +K for all i≥i(K). Lemma 6.9.If a se quenc e A admits a go o d p artition t hen S A c ontains an inﬁnite AP. The s econd as sumption is eas y to satisﬁed given that A has prop er density.Thus, the key is the ﬁr st assumption.The main feature here is that in this assumption, we o nly need to guara nt ee the existence of long(but ﬁnite)APs.So,Theorem 6.2 and its v ar iants can b e us ed with full power to achiev e this goal. 6.10.Erd˝os conjecture on square-sum-free sets.In this section,we return to Erd˝os conjecture on squar e-sum-free sets,mentioned in the in tr o duction.L et S F(n)denote the s ize of the largest subset A of[n]such that S A do es not contain a square(or A is s quare-sum-free).Erd˝osobser ved that S F(n) = Ω(n 1/3 )and conjectured that the truth is close to this lower bound.Since then,there hav e been several attempts on his conjecture.Alon proved that S F(n) = O( n log n ).In Lipkin improv ed the bound to S F(n)=O(n 3/4+ε ).L ater,Alon and F reima n obtained another improv ement S F(n)=O(n 2/3+ε ).Ab out ﬁfteen years ago, S´ark¨o zyshow ed S F(n) = O( √ n log n). A STRUC TURAL APPR OA CH TO SUBSET-SUM PROBLEMS 13 Let us now address the pr oblem from our structura l appro ach p oint o f v iew.The- orem 6.2 is no longer useful,as w e are dealing with sets of s ize around n 1/3 ,wa y below the low er b o und √ n required in this theor em.F ortunately,w e hav e a more general result,Theorem 6.3,which enables us to ﬁnd structures in S A for any set of s ize n δ ,for any co nstant δ.In particular,we can deduce fro m this theorem the following co rollary. Corollary 6.11.Ther e ar e p ositive c onstants C and c su ch that the fol lowing holds for al l suﬃciently lar ge n.L et A b e a subset of[n]with c ar dinality at le ast C n 1/3 . Then S A c ontains either an A P of length c\|A\| 2 or a pr op er GAP of r ank 2 and volume c\|A\| 3 . Combinin g this corollar y with some num b er theor etic a rguments,Nguyen and the authorcan get close to the conjectured b o und. Theorem 6.12.Ther e is a c onstant C such that S F(n) ≤n 1/3 log C n. W e strongly b elieve that the log term can b e removed.Details will app ear else where. 7.Inverse Littlew ood-Offord theorems and random ma trices In this ﬁnal section,we discuss a pro blem with a slig ht ly diﬀeren t nature.Let A be a sequence of non-zero int egers.Now we are g oing to view S A as a multi-set of 2 n element s.W e denote by M A be the largest m ultiplicit y in S A .F or example,if A= {1,...,1},then M A =  n ⌊n/2⌋  = Θ(2 n / √ n). The problem of b ounding M A originated from Littlewoo d and O ﬀord’s work on random polynomia ls.In particular,they prov ed that M A =O(2 n log n/ √ n). The log term w as remo ved by E rd˝os,who obtained a sharp b ound for M A . Many ex tensions of this r esult were obtained b y v a rious resear cher:Erd˝o s-Moser ,S´ark¨ozy-Szemer´edi,K atonaKleitman,Hal´asz,Grigg s et.a l. ,F rankl-F¨uredi,Stanley[5 9]etc.Among others,it w as sho wed that the bo und on M A keeps improving,if o ne forbids more and mo re additiv e structures in A.F or example,Er d˝os and Mosershow ed that if the elements of A are diﬀerent(i.e.,A is a set),then M A =O(2 n log n/n 3/2 ).In ge neral,the following can b e deduced from results of44 Theorem 7.1.F or any ﬁxe d inte ger k ther e is a c onstant C such that the fol lowing holds.Le t A= {a 1 ,...,a n }and R k b e the numb er of ro ots of the e quation ε 1 a i 1 +···+ε 2 k a i 2 k = 0 with ε i = ±1 and i 1 ,...,i 2 k ∈[n].Then M A ≤C n −2 k−1/2 R k . In,T ao and V u in tro duced the notion of In verse Littlewoo d-O ﬀord theorems. The intuiti on here is that if M A is larg e(of order 2 n /n C for any constant C,say), then A should have a very strong structure. 14 V AN VU The most gener al example we found with larg e M A is the fo llowing.Let Q be a prop er GAP of co nstant rank d and v olume V.If A is a subset of Q,it is eas y to show that M A = Ω( 1 n d/2 V ) (in o rder to see this, view the element s o f S A as random sums P n i=1 ξ i a i where a i are elements of A and ξ i are iid random v ariables taking v alues 0 and 1 with probability 1/2).Thus,if the volume o f Q is small,then M A is larg e. In,T ao and V u prov ed the in verse statemen t,asserting that having A as a subset of a smal l GAP is essent ial ly t he only way to guar ante e make M A lar ge. Theorem 7.2.F or any c onst ant C and ǫ ther e ar e c onstant s B and d such that the fol lowing holds.L et A b e a se qu enc e of n elements in a torsion-fr e e gr oup G.If M A ≥2 n /n C for some c onstant C,then al l but at most n 1−ǫ elements of A is c ontaine d in a pr op er GAP Q of r ank d and c ar dinality n B . In a mor e recent pap er ,the same authors obtained a(nea r)optimal re lationship betw een the pa rameters C,ǫ,d a nd B.As a corolla ry,one can deduce(a symptotic versions of)many earlier results,such as Theorem 7.1.(In spirit,this pro ces s is somewhat s imilar to the pro cess of using Theorem 3.1 to reprov e,s ay,Erd˝os- Gin burg-Ziv theo rem.) W e would lik e to conclude this survey with a r ather unexp ected applica tion.Let us leav e combinatorial num ber theor y and jump to the(fairy remote)area of mathe- matical ph ysics.In the 1 950s,Wigner o bserved and prov ed his famous semi-circle law concerning the limiting distribution of eigenv alues in a s ymmet ric random ma- trix .A brother of this law, t he so-called circular law for non-symmetric ra ndom matrices,has b een conjectured,but remains o p en since that time. Conjecture 7.3.(Cir cular L aw Conje ctur e)L et ξ b e a r andom variable with me an 0 and varia nc e 1 and M n b e the r andom matrix whose entries ar e iid c opies of ξ. Then t he limiting distribution of the eigenvalues of 1 √ n M n c onver ges to the uniform distribution on t he unit disk. Girko[3 1] and Ba i obtained important partial results concerning this conjecture. These results and many r elated results are ca refully discussed in the b o ok .There has b een a series of rapid developments recently by G¨otze-Tikhomirov[32, 33], P an- Zhou , and T ao-V u .In par ticular, T ao and V u conﬁrmed the conjecture under the slightly stronge r a ssumption that the(2 +η)-moment o f ξ is bo unded, for any η>0. Theorem 7.4.The Cir cu lar L aw holds(with str ong c onver genc e)under an addi- tional assumption t hat E(\|ξ\| 2+η ) <∞ for some ﬁx e d η>0. The k ey elemen t of this pro of is a v ariant of Theorem 7.2, whic h enables u s to coun t the num ber of sequences A with b ounded elements such that M A (more precisely a contin uous version of it)is large.F or details,we refer to. A STRUC TURAL APPR OA CH TO SUBSET-SUM PROBLEMS 15 Reference s G.E.Andrews,The theory of partitions.Cambridge univ ersity press,1998. N.Alon,Independent sets in regular graphs and sum-free subsets of abelian groups,Isr ael Journal Math.73(1991),247-256. 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A STRUC TURAL APPR OA CH TO SUBSET-SUM PROBLEMS 17 T.T ao and V.V u,p ap er in pr ep ar ation. V.V u,Structure of lar ge incomplete sets in ab elian groups,to app ea r in Combinatoric a. Wigner,O n the distribution of the roots of certain symmetric matrices,A nnals of Mathe- matics(2)67 1958 325–327. Dep ar tment of Ma thema tics, R utgers, Pisca t a w a y, NJ 08854 E-mail addr ess:vanvu@math.r utgers.edu Citations (6) References (87) ... When is A m-incomplete? There is a number of results concerning these questions (see for instance for surveys) including classical results such as Olson's theorem and the Erd˝ os-Ginzburg-Ziv theorem. Our goal is to study the above problems for the basic group F d p , as d is fixed and p is a large prime. ... ... Our understanding in the case d = 1 is more or less satisfying, due to the results from (see also for a survey). However, the proofs of these results do not extend to higher dimensions. ... A characterization of incomplete sequences in vector spaces Article Jan 2012 Hoi H. Nguyen Van Vu A sequence A of elements an additive group G is incomplete if there exists a group element that cannot be expressed as a sum of elements from A. The study of incomplete sequences is a popular topic in combinatorial number theory. However, the structure of incomplete sequences is still far from being understood, even in basic groups.The main goal of this paper is to give a characterization of incomplete sequences in the vector space Fpd, where d is a fixed integer and p is a large prime. As an application, we give a new proof for a recent result by Gao, Ruzsa and Thangadurai on Olsonʼs constant of Fp2 and partially answer their conjecture concerning Fp3. View Show abstract ... The following questions are among the most popular in classical combinatorial number theory. There is a number of results concerning these questions (see for instance [3, 5, 11, 12, 14] for surveys) including classical results such as Olson's theorem and the Erd˝ os-Ginzburg-Ziv theorem. Our goal is to study the above problems for the basic group F d p , as d is fixed and p is a large prime. ... ... (Here and later F p denote the finite field with p elements.) Our understanding in the case d = 1 is more or less satisfying, due to the results from [7, 8] (see also for a survey). However, the proofs of these results do not extend to higher dimensions. ... A characterization of incomplete sequences in F p d F_p^d Article Full-text available Dec 2011 Hoi H. Nguyen Van Vu A sequence A of elements an additive group G is {\it incomplete} if there exists a group element that {\it can not} be expressed as a sum of elements from A. The study of incomplete sequences is a popular topic in combinatorial number theory. However, the structure of incomplete sequences is still far from being understood, even in basic groups. The main goal of this paper is to give a characterization of incomplete sequences in the vector space F p d F_p^d, where d is a fixed integer and p is a large prime. As an application, we give a new proof for a recent result by Gao-Ruzsa-Thangadurai on the Olson's constant of \F_p^2 and partially answer their conjecture concerning F p 3 F_p^3. View Show abstract ... , w n , see e.g. [Vu08,TV08]. Can corresponding extensions of Lemma 26 be established, proving that every threshold function admits a representation with weights that have the required structure? ... ... Roughly speaking, if one forbids more and more additive structure in the w i 's, then one gets better and better anti-concentration; see e.g.[Vu08,TV08] and Chapter 7 of[TV06]. ... Improved Approximation of Linear Threshold Functions Article Full-text available Oct 2009 Ilias Diakonikolas Rocco A. Servedio We prove two main results on how arbitrary linear threshold functions f(x) = \sign(w\cdot x - \theta) over the n-dimensional Boolean hypercube can be approximated by simple threshold functions. Our first result shows that every n-variable threshold function f is \eps-close to a threshold function depending only on \Inf(f)^2 \cdot \poly(1/\eps) many variables, where \Inf(f) denotes the total influence or average sensitivity of f. This is an exponential sharpening of Friedgut's well-known theorem \cite{Friedgut:98}, which states that every Boolean function f is \eps-close to a function depending only on 2^{O(\Inf(f)/\eps)} many variables, for the case of threshold functions. We complement this upper bound by showing that \Omega(\Inf(f)^2 + 1/\epsilon^2) many variables are required for ϵ\epsilon-approximating threshold functions. Our second result is a proof that every n-variable threshold function is \eps-close to a threshold function with integer weights at most \poly(n) \cdot 2^{\tilde{O}(1/\eps^{2/3})}. This is a significant improvement, in the dependence on the error parameter \eps, on an earlier result of \cite{Servedio:07cc} which gave a \poly(n) \cdot 2^{\tilde{O}(1/\eps^{2})} bound. Our improvement is obtained via a new proof technique that uses strong anti-concentration bounds from probability theory. The new technique also gives a simple and modular proof of the original \cite{Servedio:07cc} result, and extends to give low-weight approximators for threshold functions under a range of probability distributions beyond just the uniform distribution. Comment: full version of CCC'09 paper View Show abstract ... An important generalization of Subset Sum is the Modular Subset Sum problem, in which sums are taken over the finite cyclic group Z m for some given integer m. This problem and its structural properties has been studied extensively in Additive Combinatorics [EGZ61,Ols68,Sze70,Ols75,Alo87,HLS08,Vu08]. The trivial algorithm for deciding whether a given target is achievable modulo m runs in time O(nm). ... Fast Modular Subset Sum using Linear Sketching Preprint Jul 2018 Kyriakos Axiotis Arturs Backurs Christos Tzamos Given n positive integers, the Modular Subset Sum problem asks if a subset adds up to a given target t modulo a given integer m. This is a natural generalization of the Subset Sum problem (where m=+\infty) with ties to additive combinatorics and cryptography. Recently, in [Bringmann, SODA'17] and [Koiliaris and Xu, SODA'17], efficient algorithms have been developed for the non-modular case, running in near-linear pseudo-polynomial time. For the modular case, however, the best known algorithm by Koiliaris and Xu [Koiliaris and Xu, SODA'17] runs in time O~(m^{5/4}). In this paper, we present an algorithm running in time O~(m), which matches a recent conditional lower bound of [Abboud et al.'17] based on the Strong Exponential Time Hypothesis. Interestingly, in contrast to most previous results on Subset Sum, our algorithm does not use the Fast Fourier Transform. Instead, it is able to simulate the "textbook" Dynamic Programming algorithm much faster, using ideas from linear sketching. This is one of the first applications of sketching-based techniques to obtain fast algorithms for combinatorial problems in an offline setting. View Show abstract ... Subset sums are helpful in the design of algorithms that prune the search space depending on the computation table, for instance, rectangle packing . Subset sums over Z u are also widely studied in additive combinatorics [30,3,17], however, there is little study around computing them. ... A Faster Pseudopolynomial Time Algorithm for Subset Sum Article Full-text available Jul 2015 Konstantinos Koiliaris Chao Xu Given a (multi) set S of n positive integers and a target integer u , the subset sum problem is to decide if there is a subset of S that sums up to u . We present a series of new algorithms that compute and return all the realizable subset sums up to the integer u in Õ(min { √ n u , u 5/4 ,σ }), where σ is the sum of all elements of S and Õ hides polylogarithmic factors. We also present a modified algorithm for integers modulo m , which computes all the realizable subset sums modulo m in Õ(min { √ n m , m 5/4 }) time. Our contributions improve upon the standard dynamic programming algorithm that runs in O ( nu ) time. To the best of our knowledge, the new algorithms are the fastest deterministic algorithms for this problem. The new results can be employed in various algorithmic problems, from graph bipartition to computational social choice. Finally, we also improve a result on covering Z m , which might be of independent interest. View Show abstract On a conjecture of Alon Article Nov 2009 Linh Tran Van Vu Philip Matchett Wood Text: Let f (n, m) be the cardinality of largest subset of {1, 2, ..., n} which does not contain a subset whose elements sum to m. In this note, we show thatf (n, m) = (1 + o (1)) frac(n, snd (m)) for all n (log n)1 + ε{lunate} ≤ m ≤ frac(n2, 9 log2 n), where snd (m) is the smallest integer that does not divide m. This proves a conjecture of Alon posed in [N. Alon, Subset sums, J. Number Theory 27 (2) (1987) 196-205]. Video: For a video summary of this paper, please visit View Show abstract On a conjecture of Erdös and Heilbronn Article Jan 1970 E. Szemerédi View On the addition of residue classes mod p Article Jan 1964 P. Erdös Hans Heilbronn View On the Number of Sets of Integers With Various Properties: Proceedings of the First Conference of the Canadian Number Theory Association held at the Banff Center, Banff, Alberta, April 17–27, 1988 Chapter Jan 1990 Peter J. Cameron P. Erdös View On a conjecture of Erdos and a stronger form of Sperner's theorem Article Jan 1966 Gyula Katona View On a lemma of Littlewood and Offord Article BULL AMER MATH SOC P. Erdos View The circle law Article Jan 1983 V.L. Girko The following assertion is proved with the help of the V-transform ∫∫e ipx+iqy dv n (x,y)=(q 2 +p 2 )(2ipπ) -1 ∫∫d ∂t[∫ 0 ∞ lnxdμ n (x,z)]e itp+isq dtds, p≠0, where v n (x,y) is the normalized spectral function of a random complex matrix Ξ n , μ n (x,z) is the normalized spectral function of the Hermitian matrix (Iz-Ξ n )(Iz-Ξ n ) , and z is a complex number. If for each n the random elements ξ pl (n) (p,l=1,···,n) of the complex matrix Ξ n =(ξ pl (n) ) are independent, Eξ pl (n) =0, E\|ξ pl (n) \| 2 =σ 2 n -1 , 0<σ 2 <∞, lim n→∞ n -1 ∑ p,l=1 n E\|ξ pl (n) \| 2 χ(/ξ pl (n) \|>r)=0 for any τ>0, and sup n ∫ 0 ∞ ln 1+δ xdμ n <∞, then plim n→∞ v n (B)=π -1 σ -2 ∫ B∩(z:σ -1 \|z\|<1) dxdy for any measurable subset B of the plane, where v n (B)=n -1 ∑ k=1 n χ B (λ k ), χ B (λ k )=1 if λ k ∈B and χ B (λ k )=0 if λ k ∉B, and the λ k are the eigenvalues of the matrix Ξ n . View Show abstract Über ein Problem von Erdős und Moser Article Jan 1965 András Sárközy Endre Szemerédi View Combinatorial Problems in Finite Abelian Groups Article Dec 1973 George T Diderrich H. B. Mann This chapter presents combinatorial problems in finite Abelian groups. An Abelian group, also called a commutative group, is a group in which the result of applying the group operation to two group elements does not depend on their order. Abelian groups generalize the arithmetic of addition of integers. The concept of an Abelian group is one of the first concepts encountered in undergraduate abstract algebra, with many other basic objects, such as a module and a vector space, being its refinements. The theory of abelian groups is generally simpler than that of their non-Abelian counterparts, and finite Abelian groups are also well understood. Every finite abelian group G can be expressed as the direct sum of cyclic subgroups of prime-power order. This is a special case of the fundamental theorem of finitely generated Abelian groups when G has zero rank. View Show abstract Foundations of a structural theory of set addition. Translation from Russian Article G.A. Freiman View On the representation of integers as the sums of distinct summands taken from a fixed set Article Jan 1960 J.W.S. Cassels View Show more Recommended publications Discover more Article Full-text available Mellin transforms with only critical zeros: Chebyshev and Gegenbauer functions June 2013 Mark W. Coffey Matthew Lettington We consider the Mellin transforms of certain Chebyshev functions based upon the Chebyshev polynomials. We show that the transforms have polynomial factors whose zeros lie all on the critical line or on the real line. The polynomials with zeros only on the critical line are identified in terms of certain 3 F 2(1)_3F_2(1) hypergeometric functions. Furthermore, we extend this result to a 1-parameter ... [Show full abstract] family of polynomials with zeros only on the critical line. These polynomials possess the functional equation p n(s;β)=(−1)⌊n/2⌋p n(1−s;β)p_n(s;\beta)=(-1)^{\lfloor n/2 \rfloor} p_n(1-s;\beta). We then present the generalization to the Mellin transform of certain Gegenbauer functions. The results should be of interest to special function theory, combinatorics, and analytic number theory. View full-text Article Full-text available From the Littlewood-Offord problem to the Circular Law: universality of the spectral distribution of... November 2008 · Bulletin of the American Mathematical Society Van Vu Terence Tao The famous \emph{circular law} asserts that if M n M_n is an n×n n \times n matrix with iid complex entries of mean zero and unit variance, then the empirical spectral distribution (ESD) of the normalized matrix 1 n M n\frac{1}{\sqrt{n}} M_n converges almost surely to the uniform distribution on the unit disk {z \in \C: \|z\| \leq 1 }. After a long sequence of partial results that verified this law ... [Show full abstract] under additional assumptions on the distribution of the entries, the full circular law was recently established in \cite{TVcir2}. In this survey we describe some of the key ingredients used in the establishment of the circular law, in particular recent advances in understanding the Littlewood-Offord problem and its inverse. View full-text Article Full-text available Random Matrices: The circular Law September 2007 · Communications in Contemporary Mathematics Van Vu Terence Tao Let x be a complex random variable with mean zero and bounded variance σ ² . Let N n be a random matrix of order n with entries being i.i.d. copies of x. Let λ 1 , …, λ n be the eigenvalues of [Formula: see text]. Define the empirical spectral distributionμ n of N n by the formula [Formula: see text] The following well-known conjecture has been open since the 1950's: Circular Law Conjecture: μ n ... [Show full abstract] converges to the uniform distribution μ ∞ over the unit disk as n tends to infinity. We prove this conjecture, with strong convergence, under the slightly stronger assumption that the (2 + η)th-moment of x is bounded, for any η > 0. Our method builds and improves upon earlier work of Girko, Bai, Götze–Tikhomirov, and Pan–Zhou, and also applies for sparse random matrices. The new key ingredient in the paper is a general result about the least singular value of random matrices, which was obtained using tools and ideas from additive combinatorics. View full-text Article Full-text available Random matrices: Universality of ESDs and the circular law August 2008 · The Annals of Probability Van Vu Terence Tao Manjunath Krishnapur Given an n×n n \times n complex matrix A, let μ A(x,y):=1 n∣{1≤i≤n,ℜ λ i≤x,ℑ λ i≤y}∣\mu_{A}(x,y):= \frac{1}{n} \|{1\le i \le n, \Re \lambda_i \le x, \Im \lambda_i \le y}\| be the empirical spectral distribution (ESD) of its eigenvalues \lambda_i \in \BBC, i=1, ... n. We consider the limiting distribution (both in probability and in the almost sure convergence sense) of the normalized ESD μ 1 n A n\mu_{\frac{1}{\sqrt{n}} A_n} of a ... [Show full abstract] random matrix A n=(a i j)1≤i,j≤n A_n = (a_{ij}){1 \leq i,j \leq n} where the random variables a{ij} - \E(a_{ij}) are iid copies of a fixed random variable x with unit variance. We prove a \emph{universality principle} for such ensembles, namely that the limit distribution in question is {\it independent} of the actual choice of x. In particular, in order to compute this distribution, one can assume that x is real of complex gaussian. As a related result, we show how laws for this ESD follow from laws for the \emph{singular} value distribution of 1 n A n−z I\frac{1}{\sqrt{n}} A_n - zI for complex z. As a corollary we establish the Circular Law conjecture (in both strong and weak forms), that asserts that μ 1 n A n\mu_{\frac{1}{\sqrt{n}} A_n} converges to the uniform measure on the unit disk when the a i j a_{ij} have zero mean. View full-text Article Optimal Inverse Littlewood-Offord theorems April 2010 · Advances in Mathematics Hoi H. Nguyen Van Vu Let eta_i be iid Bernoulli random variables, taking values -1,1 with probability 1/2. Given a multiset V of n integers v_1,..., v_n, we define the concentration probability as rho(V) := sup_{x} Pr(v_1 eta_1+...+ v_n eta_n=x). A classical result of Littlewood-Offord and Erdos from the 1940s asserts that if the v_i are non-zero, then rho(V) is O(n^{-1/2}). Since then, many researchers obtained ... [Show full abstract] improved bounds by assuming various extra restrictions on V. About 5 years ago, motivated by problems concerning random matrices, Tao and Vu introduced the Inverse Littlewood-Offord problem. In the inverse problem, one would like to give a characterization of the set V, given that rho(V) is relatively large. In this paper, we introduce a new method to attack the inverse problem. As an application, we strengthen a previous result of Tao and Vu, obtaining an optimal characterization for V. This immediately implies several classical theorems, such as those of Sarkozy-Szemeredi and Halasz. The method also applies in the continuous setting and leads to a simple proof for the beta-net theorem of Tao and Vu, which plays a key role in their recent studies of random matrices. All results extend to the general case when V is a subset of an abelian torsion-free group and eta_i are independent variables satisfying some weak conditions. Read more Discover the world's research Join ResearchGate to find the people and research you need to help your work. Join for free ResearchGate iOS App Get it from the App Store now. 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188098	https://brilliant.org/wiki/circumscribed-squares/	Circumscribed Squares \| Brilliant Math & Science Wiki HomeCourses Sign upLog in The best way to learn math and computer science. Log in with GoogleLog in with FacebookLog in with email Join using GoogleJoin using email Reset password New user? Sign up Existing user? Log in Circumscribed Squares Sign up with FacebookorSign up manually Already have an account? Log in here. A Former Brilliant Member contributed This wiki is incomplete. A circumscribed square of a circle is a square surrounding a circle such that the circumference of the circle touches the midpoints of the four sides of the square. The diameter of the circle is equal to the side length of the square. Cite as: Circumscribed Squares. Brilliant.org. Retrieved 03:49, September 29, 2025, from Join Brilliant The best way to learn math and computer science.Sign up Sign up to read all wikis and quizzes in math, science, and engineering topics. Log in with GoogleLog in with FacebookLog in with email Join using GoogleJoin using email Reset password New user? Sign up Existing user? Log in
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Long-term ocular outcome in congenital toxoplasmosis: a prospective cohort of treated children B Faucher1,P Garcia-Meric,J Franck,P Minodier,P Francois,S Gonnet,C L'ollivier,R Piarroux Affiliations Expand Affiliation 1 Laboratoire de Parasitologie et Mycologie, Hôpital de la Timone, Centre Hospitalo-Universitaire de la Timone, Assistance Publique-Hôpitaux de Marseille (AP-HM), 264 rue St-Pierre, 13385 Marseille cedex 05, France. benoit.faucher@ap-hm.fr PMID: 22051915 DOI: 10.1016/j.jinf.2011.10.008 Item in Clipboard Long-term ocular outcome in congenital toxoplasmosis: a prospective cohort of treated children B Faucher et al. J Infect.2012 Jan. Show details Display options Display options Format J Infect Actions Search in PubMed Search in NLM Catalog Add to Search . 2012 Jan;64(1):104-9. doi: 10.1016/j.jinf.2011.10.008. Epub 2011 Oct 24. Authors B Faucher1,P Garcia-Meric,J Franck,P Minodier,P Francois,S Gonnet,C L'ollivier,R Piarroux Affiliation 1 Laboratoire de Parasitologie et Mycologie, Hôpital de la Timone, Centre Hospitalo-Universitaire de la Timone, Assistance Publique-Hôpitaux de Marseille (AP-HM), 264 rue St-Pierre, 13385 Marseille cedex 05, France. benoit.faucher@ap-hm.fr PMID: 22051915 DOI: 10.1016/j.jinf.2011.10.008 Item in Clipboard Cite Display options Display options Format Abstract Objectives: Congenital toxoplasmosis remains a public health problem throughout the world. Long-term longitudinal studies are still needed to argument controversial screening and treatment strategies and to enable to accurately counsel parents. Methods: We conducted a prospective cohort study over 16 years in Marseilles, France. Seronegative pregnant women underwent monthly serological testing. Children were treated antenatally with rovamycine as soon as maternal infection was detected and with pyrimethamine and sulfadoxine in case of positive Toxoplasma PCR on amniotic fluid. Postnatal treatment with pyrimethamine and sulfadoxine was systematically prescribed for one year and possibly continued at the physician discretion. Results: 127 children were included. 24 children (18.9%) presented ocular lesions causing visual impairment in eight cases. Eleven children (8.7%) presented with ocular lesions at birth, mostly macular. Sixteen children (12.6%) developed ocular lesions during follow-up, mostly peripheral. The first ocular lesion could occur as late as 12 years after birth. No significant risk factor of chorioretinitis was identified including gestational age at infection, type of antenatal treatment and shorter postnatal treatment. Conclusions: These results confirm the overall good prognosis of congenital toxoplasmosis in Europe but highlight though a low risk of late ocular manifestation. Chorioretinitis affected 18.9% of children suffering from congenital toxoplasmosis despite antenatal and neonatal screening associated with early treatment. Long-standing follow-up is needed because first lesion can occur as late as 12 years after birth. Late lesions were less often macular but nevertheless caused sometimes visual impairment. Copyright © 2011 The British Infection Association. Published by Elsevier Ltd. All rights reserved. PubMed Disclaimer Similar articles Longitudinal study of new eye lesions in treated congenital toxoplasmosis.Phan L, Kasza K, Jalbrzikowski J, Noble AG, Latkany P, Kuo A, Mieler W, Meyers S, Rabiah P, Boyer K, Swisher C, Mets M, Roizen N, Cezar S, Remington J, Meier P, McLeod R; Toxoplasmosis Study Group.Phan L, et al.Ophthalmology. 2008 Mar;115(3):553-559.e8. doi: 10.1016/j.ophtha.2007.06.022. Epub 2007 Sep 6.Ophthalmology. 2008.PMID: 17825418 Long-term outcome of children with congenital toxoplasmosis.Berrébi A, Assouline C, Bessières MH, Lathière M, Cassaing S, Minville V, Ayoubi JM.Berrébi A, et al.Am J Obstet Gynecol. 2010 Dec;203(6):552.e1-6. doi: 10.1016/j.ajog.2010.06.002. Epub 2010 Jul 15.Am J Obstet Gynecol. 2010.PMID: 20633868 Impact of visual impairment on measures of cognitive function for children with congenital toxoplasmosis: implications for compensatory intervention strategies.Roizen N, Kasza K, Karrison T, Mets M, Noble AG, Boyer K, Swisher C, Meier P, Remington J, Jalbrzikowski J, McLeod R, Kipp M, Rabiah P, Chamot D, Estes R, Cezar S, Mack D, Pfiffner L, Stein M, Danis B, Patel D, Hopkins J, Holfels E, Stein L, Withers S, Cameron A, Perkins J, Heydemann P.Roizen N, et al.Pediatrics. 2006 Aug;118(2):e379-90. doi: 10.1542/peds.2005-1530. Epub 2006 Jul 24.Pediatrics. 2006.PMID: 16864640 [Treatment of ocular toxoplasmosis. Part 1: Basic principles and diagnosis].Mittelviefhaus H.Mittelviefhaus H.Kinderarztl Prax. 1993 May;61(3):90-6.Kinderarztl Prax. 1993.PMID: 8326704 Review.German. [Management of congenital toxoplasmosis in France: current data].Garcia-Méric P, Franck J, Dumon H, Piarroux R.Garcia-Méric P, et al.Presse Med. 2010 May;39(5):530-8. doi: 10.1016/j.lpm.2009.09.016. Epub 2009 Nov 18.Presse Med. 2010.PMID: 19926248 Review.French. See all similar articles Cited by Immunological Molecular Responses of Human Retinal Pigment Epithelial Cells to Infection With Toxoplasma gondii.Lie S, Rochet E, Segerdell E, Ma Y, Ashander LM, Shadforth AMA, Blenkinsop TA, Michael MZ, Appukuttan B, Wilmot B, Smith JR.Lie S, et al.Front Immunol. 2019 May 1;10:708. doi: 10.3389/fimmu.2019.00708. eCollection 2019.Front Immunol. 2019.PMID: 31118929 Free PMC article. 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