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1 © 2011 Pearson Education, Inc. All rights reserved 1 © 2010 Pearson Education, Inc. All rights reserved © 2011 Pearson Education, Inc. All rights reserved.

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Presentation on theme: "1 © 2011 Pearson Education, Inc. All rights reserved 1 © 2010 Pearson Education, Inc. All rights reserved © 2011 Pearson Education, Inc. All rights reserved."— Presentation transcript:

1 1 © 2011 Pearson Education, Inc. All rights reserved 1 © 2010 Pearson Education, Inc. All rights reserved © 2011 Pearson Education, Inc. All rights reserved Chapter 7 Applications of Trigonometric Functions

2 OBJECTIVES Vectors SECTION 7.5 1 2 Represent vectors geometrically. Represent vectors algebraically. Find a unit vector in the direction of v. Write a vector in terms of its magnitude and direction. Use vectors in applications. 3 4 5

3 3 © 2011 Pearson Education, Inc. All rights reserved Many physical quantities such as length, area, volume, mass, and temperature are completely described by their magnitudes in appropriate units. Such quantities are called scalar quantities. Other physical quantities such as velocity, acceleration, and force are completely described only if both a magnitude (size) and a direction are specified. Such quantities are called vector quantities. VECTORS

4 4 © 2011 Pearson Education, Inc. All rights reserved GEOMETRIC VECTORS A vector can be represented geometrically by a directed line segment with an arrow-head. The arrow specifies the direction of the vector, and its length describes the magnitude. The tail of the arrow is the vector’s initial point, and the tip of the arrow is its terminal point. Vectors are denoted by lowercase boldfaced type. With vectors, real number are scalars. Scalars are denoted by lowercase italic type.

5 5 © 2011 Pearson Education, Inc. All rights reserved If the initial point of a vector v is P and the terminal point is Q, we write The magnitude (or norm) of a vector denoted by is the length of the vector v and is a scalar quantity. GEOMETRIC VECTORS

6 6 © 2011 Pearson Education, Inc. All rights reserved EQUIVALENT VECTORS Two vectors having the same length and same direction are called equivalent vectors.

7 7 © 2011 Pearson Education, Inc. All rights reserved EQUIVALENT VECTORS Equivalent vectors are regarded as equal even though they may be located in different positions.

8 8 © 2011 Pearson Education, Inc. All rights reserved ZERO VECTOR The vector of length zero is called the zero vector and is denoted by 0. The zero vector has zero magnitude and arbitrary direction. If vectors v and a, as in the figure to the right, have the same length and opposite direction, then a is the opposite vector of v and we write a = –v.

9 9 © 2011 Pearson Education, Inc. All rights reserved Let v and w be any two vectors. Place the vector w so that its initial point coincides with the terminal point of v. The sum v + w is the resultant vector whose initial point coincides with the initial point of v, and whose terminal point coincides with the terminal point of w. v w v + w GEOMETRIC VECTOR ADDITION

10 10 © 2011 Pearson Education, Inc. All rights reserved w v + w w v v w + v As shown in the figure, v + w = w + v. The sum coincides with the diagonal of the parallelogram determined by v and w when v and w have the same initial point. GEOMETRIC VECTOR ADDITION

11 11 © 2011 Pearson Education, Inc. All rights reserved VECTOR SUBTRACTION For any two vectors v and w, v – w = v + (–w), where –w is the opposite of w. w v – w v –w w v – w v

12 12 © 2011 Pearson Education, Inc. All rights reserved SCALAR MULTIPLES OF VECTORS Let v be a vector and c a scalar (a real number). The vector cv is called the scalar multiple of v. If c > 0, cv has the same direction as v and magnitude c||v||. If c < 0, cv has the opposite direction as v and magnitude |c| ||v||. If c = 0, cv = 0v = 0.

13 13 © 2011 Pearson Education, Inc. All rights reserved EXAMPLE 1 Geometric Vectors Use the vectors u, v, and w in the figure to the right to graph each vector. a. u – 2w b. 2v – u + w Solution a. u – 2w

14 14 © 2011 Pearson Education, Inc. All rights reserved EXAMPLE 1 Geometric Vectors Solution continued b. 2v – u + w

15 15 © 2011 Pearson Education, Inc. All rights reserved ALGEBRAIC VECTORS Specifying the terminal point of the vector will completely determine the vector. For the position vector v with initial point at the origin O and terminal point at P(v 1, v 2 ), we denote the vector by A vector drawn with its initial point at the origin is called a position vector.

16 16 © 2011 Pearson Education, Inc. All rights reserved The magnitude of position vector follows directly from the Pythagorean Theorem. Notice the difference between the notations for the point (v 1, v 2 ) and the position vector We call v 1 and v 2 the components of the vector v; v 1 is the first component, and v 2 is the second component. ALGEBRAIC VECTORS

17 17 © 2011 Pearson Education, Inc. All rights reserved ALGEBRAIC VECTORS If equivalent vectors, v and w, are located so that their initial points are at the origin, then their terminal points must coincide. Thus, for the vectors v = w if and only if v 1 = w 1 and v 2 = w 2. R

18 18 © 2011 Pearson Education, Inc. All rights reserved REPRESENTING A VECTOR AS A POSITION VECTOR The vector with initial point P(x 1, y 1 ) and terminal point Q(x 2, y 2 ) is equal to the position vector R

19 19 © 2011 Pearson Education, Inc. All rights reserved EXAMPLE 2 Representing a Vector in the Cartesian Plane Let v be the vector with initial point P(4, –2) and terminal point Q(–1, 3). Write v as a position vector. Solution v has initial point P(4, –2), so x 1 = 4 and y 1 = –2. terminal point Q(–1, 3), so x 2 = –1 and y 2 = 3. So,

20 20 © 2011 Pearson Education, Inc. All rights reserved ARITHMETIC OPERATIONS AND PROPERTIES OF VECTORS If are vectors and c and d are any scalars, then

21 21 © 2011 Pearson Education, Inc. All rights reserved EXAMPLE 3 Operations on Vectors Find each expression. a. v + w b. –2v c. 2v – w d. ||2v – w|| Solution

22 22 © 2011 Pearson Education, Inc. All rights reserved EXAMPLE 3 Operations on Vectors Solution continued

23 23 © 2011 Pearson Education, Inc. All rights reserved UNIT VECTORS A vector of length 1 is a unit vector. The unit vector in the same direction as v is given by

24 24 © 2011 Pearson Education, Inc. All rights reserved UNIT VECTORS IN i, j FORM In a Cartesian coordinate plane, two important unit vectors lie along the positive coordinate axes: The unit vectors i and j are called standard unit vectors.

25 25 © 2011 Pearson Education, Inc. All rights reserved UNIT VECTORS IN i, j FORM Every vector can be expressed in terms of i and j as follows: A vector v from (0,0) to (v 1,v 2 ) can therefore be represented in the form v = v 1 i + v 2 j with

26 26 © 2011 Pearson Education, Inc. All rights reserved VECTORS IN TERMS OF MAGNITUDE AND DIRECTION Let  is the smallest nonnegative angle that v makes with the positive x-axis. The angle  is called the direction angle of v. The formula be a position vector and suppose expresses a vector v in terms of its magnitude ||v|| and its direction angle .

27 27 © 2011 Pearson Education, Inc. All rights reserved EXAMPLE 6 Writing a Vector with Given Length and Direction Angle Write the vector of magnitude 3 that makes an angle of with the positive x-axis. Solution

28 28 © 2011 Pearson Education, Inc. All rights reserved APPLICATIONS OF VECTORS If a system of forces acts on a particle, that particle will move as though it were acted on by a single force equal to the vector sum of the forces. This single force is called the resultant of the system of forces. If the particle does not move, the resultant is zero and we say that the particle is in equilibrium.

29 29 © 2011 Pearson Education, Inc. All rights reserved EXAMPLE 8 Finding the Resultant Solution Find the magnitude and bearing of the resultant R of two forces F 1 and F 2, where F 1 is a 50 lb force acting northward and F 2 is a 40 lb force acting eastward. First, set up a coordinate system.

30 30 © 2011 Pearson Education, Inc. All rights reserved EXAMPLE 8 Finding the Resultant Solution continued

31 31 © 2011 Pearson Education, Inc. All rights reserved EXAMPLE 8 Finding the Resultant Solution continued The angle between R and the y-axis (north) is 90º −51.3º = 38.7º. Therefore, the resultant R is a force of approximately 64.0 lbs in the direction N 38.7º E.


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