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Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-2 Vectors, Operations, and the Dot Product 7.4 Basic Terminology ▪ Algebraic Interpretation of Vectors ▪ Operations with Vectors ▪ Dot Product and the Angle Between Vectors

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-4 Vector v has magnitude 14.5 and direction angle 220°. Find the horizontal and vertical components. 7.4 Example 2 Finding Horizontal and Vertical Components Horizontal component: –11.1 Vertical component: –9.3

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-5 Write each vector in the form  a, b . 7.4 Example 3 Writing Vectors in the Form  a, b  u: magnitude 8, direction angle 135° v: magnitude 4, direction angle 270° w: magnitude 10, direction angle 340°

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-6 Two forces of 32 and 48 newtons act on a point in the plane. If the angle between the forces is 76°, find the magnitude of the resultant vector. 7.4 Example 4 Finding the Magnitude of a Resultant because the adjacent angles of a parallelogram are supplementary. Law of cosines Find square root.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-8 Find the angle θ between the two vectors u =  5, –12  and v =  4, 3 . 7.4 Example 7 Finding the Angle Between Two Vectors

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-10 Find the force required to keep a 2500-lb car parked on a hill that makes a 12° angle with the horizontal. 7.5 Example 2 Finding a Required Force The vertical force BA represents the force of gravity. BA = BC + (–AC)

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-11 7.5 Example 2 Finding a Required Force (cont.) Vector BC represents the force with which the weight pushes against the hill. Vector BF represents the force that would pull the car up the hill. Since vectors BF and AC are equal, gives the magnitude of the required force.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-12 7.5 Example 2 Finding a Required Force (cont.) Vectors BF and AC are parallel, so the measure of angle EBD equals the measure of angle A. Since angle BDE and angle C are right angles, triangles CBA and DEB have two corresponding angles that are equal and, thus, are similar triangles. Therefore, the measure of angle ABC equals the measure of angle E, which is 12°.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-13 7.5 Example 2 Finding a Required Force (cont.) From right triangle ABC, A force of approximately 520 lb will keep the car parked on the hill.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-14 A force of 18.0 lb is required to hold a 74.0-lb crate on a ramp. What angle does the ramp make with the horizontal? 7.5 Example 3 Finding an Incline Angle Vector BF represents the force required to hold the crate on the incline. In right triangle ABC, the measure of angle B equals θ, the magnitude of vector BA represents the weight of the crate, and vector AC equals vector BF.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-16 A ship leaves port on a bearing of 25.0° and travels 61.4 km. The ship then turns due east and travels 84.6 km. How far is the ship from port? What is its bearing from port? 7.5 Example 4 Applying Vectors to a Navigation Problem Vectors PA and AE represent the ship’s path. We are seeking the magnitude and bearing of PE.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-17 Triangle PNA is a right triangle, so the measure of angle NAP = 90° − 25.0° = 65.0°. 7.5 Example 4 Applying Vectors to a Navigation Problem (cont.) The measure of angle PAE = 180° − 65.0 = 115.0°. Law of cosines The ship is about 123.8 km from port.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-18 To find the bearing of the ship from port, first find the measure of angle APE. 7.5 Example 4 Applying Vectors to a Navigation Problem (cont.) Law of sines Now add 38.3° to 25.0° to find that the bearing is 63.3°.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-19 A plane with an airspeed of 355 mph is headed on a bearing of 62°. A west wind is blowing (from west to east) at 28.5 mph. Find the groundspeed and the actual bearing of the plane. 7.5 Example 5 Applying Vectors to a Navigation Problem The groundspeed is represented by |x|.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-22 8.5 (part I) Polar Coordinates 8.1 Complex Numbers 8.2 Trigonometric (Polar) Form of Complex Numbers 8.3 The Product and Quotient Theorems 8.4 De Moivre’s Theorem; Powers and Roots of Complex Numbers 8.5 (part II) Polar Equations and Graphs Complex Numbers, Polar Equations, and Parametric Equations 8

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-24 Plot each point by hand in the polar coordinate system. Then, determine the rectangular coordinates of each point. 8.5 Example 1 Plotting Points With Polar Coordinates The rectangular coordinates of P(4, 135°) are

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-27 Give three other pairs of polar coordinates for the point P(5, –110°). 8.5 Example 2(a) Giving Alternative Forms for Coordinates of a Point Three pairs of polar coordinates for the point P(5, −110º) are (5, 250º), (−5, 70º), and (−5, −290º). Other answers are possible.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-28 Give two pairs of polar coordinates for the point with the rectangular coordinates 8.5 Example 2(b) Giving Alternative Forms for Coordinates of a Point The point lies in quadrant II. Since, one possible value for θ is 300°. Other answers are possible. Two pairs of polar coordinates are (12, 300°) and (−12, 120°).

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-35 8.1 Example 6 Adding and Subtracting Complex Numbers Find each sum or difference. (a) (4 – 5i) + (–5 + 8i) = –1 + 3i = [4 + (–5)] + (–5i + 8i) (b) (–6 + 3i) + (12 – 9i)= 6 – 6i (c) (–10 + 7i) – (5 – 3i) = –15 + 10i (d) (15 – 8i) – (–10 + 4i) + (–25 + 12i) = 0 + 0i = (–10 – 5) + [7i + (3i)] = [15 – (–10) + (–25)] + [–8i – 4i + 12i]

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-36 8.1 Example 7 Multiplying Complex Numbers Find each product. (a) (5 + 3i)(2 – 7i) (b) (4 – 5i) 2 = 5(2) + (5)(–7i) + (3i)(2) + (3i)(–7i) = 10 – 35i + 6i – 21i 2 = 10 – 29i – 21(–1) = 31 – 29i = 4 2 – 2(4)(5i) + (5i) 2 = 16 – 40i + 25i 2 = 16 – 40i + 25(–1) = –9 – 40i

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-37 8.1 Example 7 Multiplying Complex Numbers (cont.) (d) (9 – 8i)(9 + 8i)= 9 2 – (8i) 2 = 81 – 64i 2 = 81 – 64(–1) = 81 + 64 = 145 or 145 + 0i (c) (3 – i)(–3 + i)= –9 + 3i + 3i – i 2 = –9 + 6i – (–1) = –9 + 6i + 1 = –8 + 6i

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-40 8.1 Example 9(a) Dividing Complex Numbers Write the quotient in standard form a + bi. Multiply the numerator and denominator by the complex conjugate of the denominator. i 2 = –1 Combine terms. Lowest terms; standard form Multiply.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-41 8.1 Example 9(b) Dividing Complex Numbers Write the quotient in standard form a + bi. Multiply the numerator and denominator by the complex conjugate of the denominator. –i 2 = 1 Lowest terms; standard form Multiply.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-42 Trigonometric (Polar) Form of Complex Numbers 8.2 The Complex Plane and Vector Representation ▪ Trigonometric (Polar) Form ▪ Converting Between Rectangular and Trigonometric (Polar) Forms ▪ An Application of Complex Numbers to Fractals

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-43 Find the sum of 2 + 3i and –4 + 2i. Graph both complex numbers and their resultant. 8.2 Example 1 Expressing the Sum of Complex Numbers Graphically (2 + 3i) + (–4 + 2i) = –2 + 5i

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-45 Write 8 – 8i in trigonometric form. 8.2 Example 3(a) Converting From Rectangular Form to Trigonometric Form The reference angle for θ is 45°. The graph shows that θ is in quadrant IV, so θ = 315°.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-46 Write –15 in trigonometric form. 8.2 Example 3(b) Converting From Rectangular Form to Trigonometric Form –15 + 0i is on the negative x-axis, so θ = 180°. –15 = –15 + 0i

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-47 Write each complex number in its alternative form, using calculator approximations as necessary. 8.2 Example 4 Converting Between Trigonometric and Rectangular Forms Using Calculator Approximations (a)7(cos 205° + i sin 205°)≈ –6.3442 – 2.9583i

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-48 8.2 Example 4 Converting Between Trigonometric and Rectangular Forms Using Calculator Approximations (cont.) (b)–7 + 2i x = −7 and y = 2 The reference angle for θ is approximately 15.95°. The graph shows that θ is in quadrant II, so θ = 180° – 15.95° = 164.05°.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-51 Find the product of 4(cos 120° + i sin 120°) and 5(cos 30° + i sin 30°). Write the result in rectangular form. 8.3 Example 1 Using the Product Theorem

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-54 Use a calculator to find the following. Write the results in rectangular form. 8.3 Example 3 Using the Product and Quotient Theorems With a Calculator

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-57 Find and express the result in rectangular form. 8.4 Example 1 Finding a Power of a Complex Number First write in trigonometric form. and Because x and y are both positive, θ is in quadrant I, so θ = 45°.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-59 Find the three cube roots of 8 (cos 180 o + i sin 180 o ). Write the roots in rectangular form. 8.4 Example 2 Finding Complex Roots Note that [2 (cos 60 o + i sin 60 o )] 3 = 8 (cos 180 o + i sin 180 o ). So one cube root is 2 (cos 60 o + i sin 60 o ) The other 2 are apart 2 (cos 180 o + i sin 180 o ) and 2 (cos 300 o + i sin 300 o ) If cubed all 3 will yield 8 in rectangular form

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-60 Find all fourth roots of Write the roots in rectangular form. 8.4 Example 3 Finding Complex Roots First write in trigonometric form. Because x and y are both negative, θ is in quadrant III, so θ = 240°.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-63 8.5 Example 3 Examining Polar and Rectangular Equations of Lines and Circles For each rectangular equation, give the equivalent polar equation and sketch its graph. (a)y = 2x – 4 In standard form, the equation is 2x – y = 4, so a = 2, b = –1, and c = 4. The general form for the polar equation of a line is y = 2x – 4 is equivalent to

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-65 8.5 Example 3 Examining Polar and Rectangular Equations of Lines and Circles (cont.) This is the graph of a circle with center at the origin and radius 5. Note that in polar coordinates it is possible for r < 0.

Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 8-69 8.5 Example 4 Graphing a Polar Equation (Cardioid) (cont.) Connect the points in order from (1, 0°) to (.5, 30°) to (.1, 60°) and so on.