...VECTOR FUNCTIONS VECTOR FUNCTIONS Motion in Space: Velocity and Acceleration In this section, we will learn about: The motion of an object using tangent and normal vectors. MOTION IN SPACE: VELOCITY AND ACCELERATION Here, we show how the ideas of tangent and normal vectors and curvature can be used in physics to study: The motion of an object, including its velocity and acceleration, along a space curve. VELOCITY AND ACCELERATION In particular, we follow in the footsteps of Newton by using these methods to derive Kepler’s First Law of planetary motion. VELOCITY Suppose a particle moves through space so that its position vector at time t is r(t). VELOCITY Vector 1 Notice from the figure that, for small values of h, the vector r(t h) r(t ) h approximates the direction of the particle moving along the curve r(t). VELOCITY Its magnitude measures the size of the displacement vector per unit time. VELOCITY The vector 1 gives the average velocity over a time interval of length h. VELOCITY VECTOR Equation 2 Its limit is the velocity vector v(t) at time t : r(t h) r(t ) v(t ) lim h 0 h r '(t ) VELOCITY VECTOR Thus, the velocity vector is also the tangent vector and points in the direction of the tangent line. SPEED The speed of the particle at time t is the magnitude of the velocity vector, that is, |v(t)|. SPEED This is appropriate because, from Equation 2 ...
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...Tutorial 1 – Vector Calculus 1. Find the magnitude of the vector PQ with P (−1,2) and Q (5,5) . 2. Find the length of the vector v = 2,3,−7 . 3. Given the points in 3-dimensional space, P ( 2,1,5), Q (3,5,7), R (1,−3,−2) and S ( 2,1,0) . Does r PQ = RS ? ˆ ˆ 4. Find a vector of magnitude 5 in the direction of v = 3i + 5 ˆ − 2k . j r r ˆ ˆ ˆ j ˆ 5. Given u = 3i − ˆ − 6k and v = −i + 12k , find (a) u • v , r r (b) the angle between vectors u and v , r (c) the vector proju v , r r r r (d) the scalar component of v in the direction of u . 6. Given P (1,−1,3), Q ( 2,0,1) and R (0,2,−1) , find (a) the area of the triangle determined by the points P, Q and R. (b) the unit vector perpendicular to the plane PQR. 7. Find the volume of the parallelepiped determined by the vectors u = 4,1,0 , v = 2,−2,3 and r r r r r w = 0,2,5 . 8. Find the area of the parallelogram whose vertices are given by the points A (0, 0, 0), B (3, 2, 4), C (5, 1, 4) and D (2, -1, 0). ˆ j 9. Find the equation of the line through (2, 1, 0) and perpendicular to both i + ˆ and ˆ + k . j ˆ 10. Find the parametric equation of the line through the point (1, 0, 6) and perpendicular to the plane x+3y+z=5. 11. Determine whether the given lines are skew, parallel or intersecting. If the lines are intersecting, what is the angle between them? L1: x −1 y −3 z−2 = = 2 2 −1 x−2 y−6 z+3 L2 : = = 1 −1 3 12. Find the point in which the line x = 1 –t, y = 3t, z = 1 + t meets...
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...Excerpt More information 1 Vector and tensor analysis Vectors and scalars Vector methods have become standard tools for the physicists. In this chapter we discuss the properties of the vectors and vector ®elds that occur in classical physics. We will do so in a way, and in a notation, that leads to the formation of abstract linear vector spaces in Chapter 5. A physical quantity that is completely speci®ed, in appropriate units, by a single number (called its magnitude) such as volume, mass, and temperature is called a scalar. Scalar quantities are treated as ordinary real numbers. They obey all the regular rules of algebraic addition, subtraction, multiplication, division, and so on. There are also physical quantities which require a magnitude and a direction for their complete speci®cation. These are called vectors if their combination with each other is commutative (that is the order of addition may be changed without aecting the result). Thus not all quantities possessing magnitude and direction are vectors. Angular displacement, for example, may be characterised by magnitude and direction but is not a vector, for the addition of two or more angular displacements is not, in general, commutative (Fig. 1.1). In print, we shall denote vectors by boldface letters (such as A) and use ordinary italic letters (such as A) for their magnitudes; in writing, vectors are usually ~ ~ represented by a letter with an arrow above it such as A. A given vector A (or A) can be written as ...
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...6302 Vector Aeromotive Corporation Vector sold exotic sports cars and was the only US based manufacturer. Their major competition, Ferrari and Lamborghini, took up 75% of the market share. Gerry’s idea was to make a car based off of aerospace technology. They created the V8 twin turbo which was highly advanced and priced. After selling a total of 13 cars, 45 people were employed. Vector built two other models to increase sell volume and decrease losses. Vector’s Board of Directors composed of 3 individuals, Berry with a background in real estate; John, a financial consultant, and Vector’s CFO; and Gerry, the president of Vector. Berry and John grew to disagree with Gerry more often; therefore, Gerry brought in Dan, an attorney and associate of Gerry, to help shift the balance in his favor. In June 1990, vector hired a vice president; this V.P did not agree with Gerry’s management style and reported it to the board. Dan Harnett even began to see the malevolent behaviors of Gerry, and called for his removal. The other board members did not agree to this because there was no one to take his place and Gerry was allowed to be a bad manager, according to his employment contract, he just could not do anything illegal. Instead the board decided to require Gerry to submit formal expense reports in order to get reimbursed for expenses. Dan later resigned from the board and was replaced by George, another associate of Gerry. June 1992, 2 million dollars was given to Vector by Setiwon...
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...Support Vector Machines Operations Management Project Report by- Suryansh Kapoor PGPM (2011 – 2013) 11P171 Section – ‘C’ Supervised by- Prof. Manoj Srivastava Abstract In today’s highly competitive world markets, high reliability plays increasingly important role in the modern manufacturing industry. Accurate reliability predictions enable companies to make informed decisions when choosing among competing designs or architecture proposals. This is all the more important in case of specialized fields where operations management is a necessary requirement. Therefore, predicting machine reliability is necessary in order to execute predictive maintenance, which has reported benefits include reduced downtime, lower maintenance costs, and reduction of unexpected catastrophic failures. Here, the role of Support Vector Machines or SVMs comes in to predict the reliability of the necessary equipment. SVMs are cited by various sources in the field of medical researches⁶ and other non-mining fields¹ to be better than other classifying methods like Monte-Carlo simulation etc. because SVM models have nonlinear mapping capabilities, and so can more easily capture reliability data patterns than can other models. The SVM model minimizes structural risk rather than minimizing training errors improves the generalization ability of the models. Contents 1. Objective 2. Literature Review * Introduction of Reliability * Introduction...
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...Offline handwritten Arabic character recognizer based on Feature extraction and Support vector machine Thahira banu, Assistant professor in MCA department Sankara College of Science and comerce, Coimbatore-35. thahirshanth@gmail.com. ABSTRACT: Since the problem of Arabic text recognition is a large and complex one, it makes sense to try a simple method to see what performance can be achieved. The characters are written by many people using a great variety of sizes, writing styles, instruments, and with a widely varying amount of care. Some of the characters or words are poorly formed and are hard to classify, even for a human. Of the 280 sample characters used for training, 280 have been used for test purposes. The captured image of a character is normalized and set to eight feature values as parameter values of a vector. Training has given for a character by SVM (Support Vector machine) algorithm. It attempts to work with a subset of the features in a character that a human would typically see for the identification of Arabic characters. 1. Introduction One of the most classical applications of the Artificial Neural Network is the Character Recognition System. Cost effective and less time consuming, businesses, post offices, banks, security systems, and even the field of robotics employ this system as the base of their operations. Handwriting recognition can be defined as the task of transforming text represented in the spatial form...
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...Calculus and Vectors – How to get an A+ 7.4 Dot Product of Algebraic Vectors A Dot Product for Standard Unit Vectors The dot product of the standard unit vectors is given by: r r r r r r i ⋅ i =1 j ⋅ j =1 k ⋅ k =1 r r r r r r i ⋅ j =0 j ⋅k =0 k ⋅i = 0 B Dot Product for two Algebraic Vectors The dot product of two algebraic vectors r r r r a = (a x , a y , a z ) = a x i + a y j + a z k and r r r r b = (b x , b y , b z ) = b x i + b y j + b z k is given by: r r a ⋅ b = a x bx + a y b y + a z bz Proof: r r r r r r r r a ⋅ b = (a x i + a y j + a z k ) ⋅ (bx i + b y j + bz k ) r r r r r r = (a x bx )(i ⋅ i ) + (a x b y (i ⋅ j ) + (a x bz )(i ⋅ k ) + r r r r r r + (a y bx )( j ⋅ i ) + (a y b y ( j ⋅ j ) + (a y bz )( j ⋅ k ) + r r r r r r + (a z bx )(k ⋅ i ) + (a z b y (k ⋅ j ) + (a z bz )(k ⋅ k ) = a x bx + a y b y + a z bz Proof: r r r r i ⋅ i =|| i || || i || cos 0° = (1)(1)(1) = 1 r r r r i ⋅ j =|| i || || j || cos 90° = (1)(1)(0) = 0 r Ex 1. For each case, find the dot product of the vectors a r and b . r r a) a = (1,−2,0) , b = (0,−1,2) r r a ⋅ b = (1)(0) + (−2)(−1) + (0)(2) = 2 r r r r r r r b) a = −i + 2 j , b = i − 2 j − k r r a ⋅ b = (−1)(1) + (2)(−2) + (0)(−1) = −1 − 4 = −5 r r r r r c) a = (−1,1,−1) , b = −i + 2 j − 2k r r a ⋅ b = (−1)(−1) + (1)(2) + (−1)(−2) = 1 + 2 + 2 = 5 C Angle between two Vectors r r r The angle θ = ∠(a , b ) between two vectors a and r b (when positioned tail to tail) is given by: r r a x bx + a y b y + a z bz a ⋅b cosθ = r r = | a ||...
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...corrected/improved versions. Problem 1(a) - Fall 2008 Find parametric equations for the line L which contains A(1, 2, 3) and B(4, 6, 5). Solution: To get the parametric equations of L you need a point through which the line passes and a vector parallel to the line. −→ Take the point to be A and the vector to be the AB. The vector equation of L is −→ −→ r(t) = OA + t AB = 1, 2, 3 + t 3, 4, 2 = 1 + 3t, 2 + 4t, 3 + 2t , where O is the origin. The parametric equations are: x = 1 + 3t y = 2 + 4t, z = 3 + 2t t ∈ R. Problem 1(b) - Fall 2008 Find parametric equations for the line L of intersection of the planes x − 2y + z = 10 and 2x + y − z = 0. Solution: The vector part v of the line L of intersection is orthogonal to the normal vectors 1, −2, 1 and 2, 1, −1 . Hence v can be taken to be: i j k v = 1, −2, 1 × 2, 1, −1 = 1 −2 1 = 1i + 3j + 5k. 2 1 −1 Choose P ∈ L so the z-coordinate of P is zero. Setting z = 0, we obtain: x − 2y = 10 2x + y = 0. Solving, we find that x = 2 and y = −4. Hence, P = 2, −4, 0 lies on the line L. The parametric equations are: x =2+t y = −4 + 3t z = 0 + 5t = 5t. Problem 2(a) - Fall 2008 Find an equation of the plane which contains the points P(−1, 0, 1), Q(1, −2, 1) and R(2, 0, −1). Solution: Method 1 −→ −→ Consider the vectors...
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...International Journal of Engineering Trends and Technology- Volume3Issue3- 2012 Hiding Messages Using Motion Vector Technique In Video Steganography P.Paulpandi1, Dr.T.Meyyappan,M.sc.,M.Phil.,M.BA.,Ph.D2 Research Scholar1, Associate professor2 Department of Computer Science & Engineering, Alagappa University,Karaikudi. Tamil Nadu,India. Abstract- Steganography is the art of hiding information in ways that avert the revealing of hiding messages.Video files are generally a collection of images. so most of the presented techniques on images and audio can be applied to video files too. The great advantages of video are the large amount of data that can be hidden inside and the fact that it is a moving stream of image. In this paper, we proposed a new technique using the motion vector, to hide the data in the moving objects. Moreover, to enhance the security of the data, the data is encrypted by using the AES algorithm and then hided. The data is hided in the horizontal and the vertical components of the moving objects. The PSNR value is calculated so that the quality of the video after the data hiding is evaluated. Keywords- Data hiding, Video Steganography,PSNR, Moving objects, AES Algorithm. I. INTRODUCTION Since the rise of the Internet one of the most important factors of information technology and communication has been the security of information. Steganography is a technology that hides a user defined information within an object, a text...
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...5 6 1 5 2 0 <= Available Input a text file that includes the number of processes, resources, and the matrixes for allocations, max, and available. Output System Safe or Unsafe 1) Read the # of processes and the # of resources 2) Read allocation, max and available for each process and each resource 3) Print whether this system is safe or not. Data Structures for the Banker’s Algorithm Let n = number of processes, and m = number of resources types. Available: Vector of length m. If available [j] = k, there are k instances of resource type Rj available Input.txt Input.txt Max: n x m matrix. If Max [i,j] = k, then process Pi may request at most k instances of resource type Rj Allocation: n x m matrix. If Allocation[i,j] = k then Pi is currently allocated k instances of Rj Need: n x m matrix. If Need[i,j] = k, then Pi may need k more instances of Rj to complete its task Need [i,j] = Max[i,j] – Allocation [i,j] Safety Algorithm 1. Let Work and Finish be vectors of length m and n, respectively. Initialize: Work = Available Finish [i] = false for i = 0, 1, …, n- 1 2. Find an i such that both: (a) Finish [i] = false (b) Needi ≤ Work If no such i exists, go...
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...[pic] (____)2. El valor de la pendiente de la recta tangente a la curva [pic] ; [pic] , en [pic] es 75. (____) 3. La integral que representa la longitud de arco de la curva [pic] ; [pic] , en el intervalo [pic] es [pic] (____) 4. La representación polar de la ecuación [pic] es [pic] (____) 5. El producto vectorial de dos vectores u [pic] v = 0 , si y solo si u y v son múltiplos escalares uno del otro. (____) 6. El dominio de la función [pic] es [pic] (____) 7. Considera la función [pic], el resultado de [pic] es[pic]. (____) 8. [pic] (____) 9. Si [pic], entonces [pic] (____) 10. La derivada direccional de la función [pic], en el punto P(1,[pic] y en dirección del vector v = -i es [pic] (____) 11. Multiplicación de Números Complejos: ( 2 + 3i)(4 - 5i) = 23 + 2i (____) 12. El módulo [pic] y el argumento [pic]del número z = -4+i son: [pic] , [pic] Instrucciones. En esta sección del examen se te pide que incluyas todos los procedimientos que realizas para llegar a la solución de los ejercicios, deben estar escritos en forma limpia, clara y ordenada. En caso de que los procedimientos no avalen tu...
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...by beginners, just starting to code using Java. The Array and ArrayList are both used to store elements, which can be a primitive or an object in the case of ArrayList in Java. A main difference between the ArrayList and an Array in Java would be the static nature of the Array, but the ArrayList has a dynamic nature. Once an Array is created, programmers cannot change the size of it, but an ArrayList will be able to re-size itself at any time. There is one more notable difference between ArrayList and an Array (Paul, 2012). The Array is a core part of Java programming that has a special syntax and a semantics support within Java. An ArrayList is a part of the collection framework of popular classes, such as HashMap, Hashtable, and Vector. There are six more differences between Array and ArrayList which will be listed in numeral order: 1. First and Major difference between Array and ArrayList in Java would be that Array is a fixed length data structure, while ArrayList is a variable length collection class. 2. Another difference is that an Array cannot use Generics, due to it cannot store files, unlike the ArrayList that allows users to use Generics to ensure storage. 3. Programmers can compare the Array vs. ArrayList on how to calculate length of Array or size of an ArrayList. 4. One more major difference within an ArrayList and Array is that programmers cannot store primitives in Arraylist, which can only store objects. Array can contain both an object...
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...Gravedad (m/s) | Mercurio | 3.3x10 23| 4870| 3.72 | Venus | 4.87x1024 | 12100 | 8.82 | Tierra | 5.98x1024 | 12756 | 9.8 | Marte | 6.4x1023 | 6670 | 3.72 | Júpiter | 1.9x1027 | 143760| 23.13 | Saturno | 5.68x1026 | 120420| 9.01 | Urano | 8.7x1025 | 51300 | 8.72 | Neptuno| 1x1022 | 40500 | 10.97 | LAS LEYES DE KEPLER Las leyes de Kepler fueron enunciadas por Johannes Kepler para describir matemáticamente el movimiento de los planetas en sus órbitas alrededor del Sol. Aunque él no las describió así, en la actualidad se enuncian como sigue: Primera ley (1609): Todos los planetas se desplazan alrededor del Sol describiendo órbitas elípticas. El Sol se encuentra en uno de los focos de la elipse. Segunda ley (1609): el radio vector que une un planeta y el Sol barre áreas iguales en tiempos iguales. La ley de las áreas es equivalente a la constancia del momento angular, es decir, cuando el planeta está más alejado del Sol (afelio) su velocidad es menor que cuando está más cercano al Sol (perihelio). En el afelio y en el perihelio, el momento angular es el producto de la masa del planeta, su velocidad y su distancia al centro del Sol. Tercera ley (1618): para cualquier planeta, el cuadrado de su período orbital es directamente proporcional al cubo de la longitud del semieje de su órbita elíptica. Donde, T es el periodo orbital (tiempo que tarda en dar una vuelta alrededor del Sol), R la distancia media del planeta con el Sol y C la constante de proporcionalidad...
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...UNIDAD IV:ESPACIOS CON PRODUCTO INTERNO Objetivos particulares: Identificar los espacios con producto interno Comprobar la ortogonalidad entre vectores Definir bases ortonormales Verificar la ortogonalidad en una transformación lineal Contenido programático: Espacio con producto interno Producto interno. Definición y ejemplo El producto interno estándar en dimensión tres (3) Ortogonalidad Conjuntos ortonormales. Definición Bases ortonormales Proceso de ortonormalización de Gram- Schmidt Matrices ortogonales Complemento transformaciones ortogonales Producto interno Los elementos fundamentales que constituyen un espacio vectorial son: Un conjunto de vectores, y conjunto de escalares y dos operaciones: Adición y producto por un escalar. Tanto la definición de espacio vectorial como sus consecuencias inmediatas (teoremas) se refieren a propiedades que podríamos denominar algebraicas. Existen diversas formas de introducir en un espacio vectorial dichos conceptos. Uno de ellos consiste en definirlos a partir de una operación conocida como producto interno Producto interno Un producto interior dentro de un espacio vectorial V es una función que asocia a cada par ordenado de vectores u y v en V, un número real único < u,v >, llamado producto interno de u y v, y que satisface los siguientes axiomas para todos u, v y w en V y para todo escalar c : 1. 2. 3. 4. 5. < u,v > = < v,u > < u + v , w > = < u, w > + < v , w > < cu, v >...
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...Los Nanotubos de Carbono Galo Fabara Alex Ronquillo Jon Yugcha galoartur_123@hotmail.com alexhino.sonic@hotmail.com jonyugcha@hotmail.com Universidad de las Fuerzas Armadas ESPE Carrera de Ingeniería Electromecánica Latacunga - Ecuador Resumen Un nanotubo de carbono es una muy pequeña configuración de átomos de este elemento en forma cilíndrica. En este documento se hace una revisión de la estructura y principales características de estos elementos, así como de su descubrimiento y las distintas aplicaciones, actuales y futuras, en las que pueden ser utilizados, además se hace mención de los distintos inconvenientes actuales y de cómo el avance de esta tecnología podría cambiar drásticamente el futuro. __ Palabras Clave -- Configuración, Nanotubos, Características, Aplicaciones, Futuro. __ Abstract -- A carbon nanotube is a very small set of atoms of this element in a cylindrical shape. In this paper we review the structure and main characteristics of these elements, their discovery and the current and future applications of this elements, also makes mention of the various current problems and how this technology will be able to change drastically the future. . Keywords -- Configuration, Nanotubes, Features, Applications, Future. 1. Introducción La nanotecnología promete ser la revolución tecnológica de los años venideros por ello es de vital importancia el notar como se han mejorado las distintas características de los materiales con el uso de la...
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