Syllabus Objectives: 5.1 The student will eplore methods of vector addition and subtraction. 5. The student will develop strategies for computing a vector s direction angle and magnitude given its coordinates. 5.4 The student will resolve vectors into unit vectors. 5.7 The student will solve real-world application problems using vectors in two and three dimensions. Directed Line Segment: a segment with direction and distance B AB A A: Initial Point (start); B: Terminal Point (end) Coordinates of A: A, A y Coordinates of B: B, B y Magnitude (length) of a Directed Line Segment AB Note: This is the distance formula! AB B A B A : y y Vector (v): the set of all directed line segments that are equivalent to a given directed line segment Note: Equivalent means same magnitude and direction. Component Form of a Vector: B A, B A y y E: Graph the vector AB 3, One possible graph: y and find the magnitude. Note: AB 3, could be placed anywhere on the coordinate grid. We have placed it in standard position, which is with the initial point at the origin. AB 30 0 94 13 Magnitude: Note: If a vector u is written in component form, u u1, u, then the magnitude of u is u u u. This is because the initial point is the origin, 1 0,0. Vector Addition: Let u u1, u and v v1, v. Then uv u1v1, u v. Scalar Multiplication: Let u u1, u and k be any constant. Then ku ku1, ku. Note: If k 0, then ku is in the opposite direction. Page 1 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
E: Use the graph of the vectors to complete each eample below. w u v Show that 1. Show that u v. u v. u 61 41 59 34 v 4 1 1 4 59 34 Show that the direction of u is the same as the direction of v. Use slope: Direction of u = 4 1 3 ; direction of v = 61 5 The direction and magnitude are the same, so u v. 1 4 3 4 1 5. Find the component form and the magnitude of u and w. Component form of u: u 16,14 5, 3 u 34 (see above) Component form of w: w 4,5 3,8 w 8 464 68 17 3. Find the component form of u 3w. u3w 5, 3 3,8 10, 6 6, 4 10 6, 6 4 16, 30 Unit Vector: a vector with a magnitude of 1 A unit vector in the direction of a vector v can be found by dividing v by the magnitude of v. v Unit Vector in the Direction of v: v Standard Unit Vectors: unit vectors i and j in standard position along the positive - and y-aes i 1,0 & j 0,1 Any vector can be written in terms of the standard unit vectors. E: Write the vector v,5 in terms of the standard unit vectors. v,5 1,0 5 0,1 i5j Page of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
E: Find a unit vector in the direction of the given vector. Verify your answer is a unit vector and give your answer in component form and standard unit vector form. i 4j Find the magnitude: i4j 4 0 5 Divide the original vector by its magnitude: i 4j i 4j i j 5 5 i j (SUV) 5 5 5 5 5 5 5 Component Form: 5 5, 5 5 Verify magnitude of unit vector: 5 5 5 5 5 0 5, 1 5 5 5 5 5 5 5 Recall: In the unit circle, cos, y sin. This leads into another way of epressing a vector, in terms of its direction angle, θ. Direction Angle: in standard position, the angle the vector makes with the positive -ais (counterclockwise) Resolving a Vector: in terms of its direction angle, θ, a vector can be written as u cos,sin u cosi u sinj E: Find the magnitude and direction angle of v i6j. v 6 40 10 Magnitude: Direction angle: v cosi v sinj v cos OR 6 v sin 10cos 1 cos 10 1 1 cos 108.43 10 6 10sin 3 sin 10 1 3 sin 71.57 10 but since we know v i6j is in Quadrant II, 180 71.57 108.43 Page 3 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
E: Find the component form of v given its magnitude and its direction angle. v 5, 30 5 3 5 v v cosi v sinj v 5cos30 i5sin30 j v i j Application: Resultant Force E: Two forces act on an object: u 3, 45 and v 4, 30. Find the direction and magnitude of the resultant force. Write each vector in component form: 3 3 u u cosui u sinuj u 3cos 45 i 3sin 45 j u i j u v v cos i v sin j v 4cos 30 i4sin 30 j v 3ij v v u The resultant force is the sum u v: 3 3 3, Application: Bearing E: A plane flies due east at 500 km/h and there is a 60 km/h with a bearing of 45. Find the ground speed and the actual bearing of the plane. 60 km/hr 45 θ 500 km/hr Sketch a diagram: w p v Find the vectors p and w: p 500cos,500sin w 60cos45,60sin 45 Note: The 45 is the direction angle, not the bearing. Vector v is the sum p + w: v 60cos 45 500cos,60sin 45 500sin The second component of vector v must equal zero, because the plane is headed due east. 60sin 45 1 60sin 45 60sin 45 500sin 0 sin sin 4.868 500 500 Bearing of the plane: 90 94.868 Ground speed of the plane: v 60 cos 45 500 cos 0 60 cos 45 500 cos 4.868 540.63km/hr Page 4 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
You Try: Find the component form of v given its magnitude and the angle it makes with the positive - ais. v, direction: i 3j QOD: In the eamples in your notes, we used sine or cosine to find the direction angle of a vector. Eplain how you could use tangent to find the direction angle. Page 5 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
Syllabus Objective: 5.3 The student will eplore methods of vector multiplication. 5.5 The student will determine if two vectors are parallel or perpendicular (orthogonal). 5.6 The student will derive an equation of a line or plane by using vector operations. 5.7 The student will solve real-world application problems using vectors in two and three dimensions. Dot Product: Let u u1, u and v v1, v. The dot product is uv uv 1 1 uv. Note: The dot product is a scalar. E: Evaluate 5, 3, 4. 5, 3, 4 5 3 4 15 8 7 Properties of the Dot Product: 1. uv vu. uu= u 3. 0u 0 4. uv+w =uv u w 5. cuv ucv cu v E: Evaluate the following given u 3,6; v 1,0; w 5, (a) ww ww 55 9 (b) w w 5 54 9 (c) v w u v wu u (d) vu wu vu wu 15,0 6, 3,6 6 3 6 30 1 3 0 6 5 3 6 3 7 30 Angle Between Two Vectors: uv 1 cos cos uv u v u v v u v θ Proof: Use the triangle. Law of Cosines: u vu u v u v cos Property of Dot Product: v-u v-u u v u v cos Epand: vvvuuvu u u v u v cos Property of Dot Product: v u v u u v u v cos Property of Equality: uv u vcos cos uv u v Page 6 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
E: Find uv, where θ is the angle between u and v. uv 5 uv 3 cos cos uv 48 uv 4 3 u v 6 68 Application E: Find the interior angles of the triangle with vertices. y 5 u 6, v 8, 6 B A C AB 43,0 1, ; AC 53,10,1 BA 34,0 1, ; BC 54,1 1, 1 11 1 1 1 cos B B cos B 71.565 5 10 C 180 36.87 71.565 71.565 11 1 4 cos A Acos A36.87 5 5 5 Angles: 36.87, 71.565, 71.565 Orthogonal Vectors: two vectors whose dot product is equal to 0 What is the angle between two non-zero orthogonal vectors? uv 0 cos cos cos 0 90 u v u v Note: If the angle between the vectors is 90, we may also say they are perpendicular. The word orthogonal is used instead for vectors because the zero vector is orthogonal to any other vector, but is not perpendicular. What is the dot product of two vectors that are parallel? The angle between them would have to be either 180 or 360. uv uv uv uv cos cos180 1 uv or cos cos360 1 uv u v u v u v u v Parallel Vectors: two vectors whose dot product is equal to 1 or 1 E: Are the vectors orthogonal, parallel, or neither? v 3i j, w 3i4j Find vw : vw 33 4 1 The vectors are parallel. Page 7 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
uv Vector Projection: the projection of u onto v is denoted by: projvu v v u u 1 = proj v u v u = u u 1 Note: The projection of u onto v is the shadow formed by vector u onto v as light comes straight down. E: Find the projection of v onto w. Then write v as the sum of two orthogonal vectors, with one the proj w v. v 1, 3 ; w 1,1 vw 11 31 proj w v w proj 1,1 proj 1,1 proj, w v w v w v w 1 1 vproj v 1,3, 1,1 v, 1,1 w Application: Force E: Find the force required to keep a 00-lb cart from rolling down a 30 incline. f 00 v 30 g Draw a diagram and label: The force due to gravity: g 00j (gravity acts vertically downward) 3 1 Incline vector: v cos30 isin 30 j i j gv Force vector required to keep the cart from rolling: f projvg v v gv 100 3 100 3 1 f v v 100 3 100, 150 50 3, 50 3 50 3 1 v 4 4 Magnitude of Force: f 150 50 3, 50 3 50 73.05 pounds Application: Work W cos forcedistance E: A person pulls a wagon with a constant force of 15 lbs at a constant angle of 40 for 500 ft. What is the person s work? Page 8 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
40 w cos40 15 lbs 500 ft 5745.33 foot lbs You Try: Find the projection of v onto u. Then write v as the sum of two orthogonal vectors, with one the proj u v. v i3 j; uij QOD: If u is a unit vector, what is uu? Eplain why. Page 9 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
Syllabus Objective: 1.10 The student will solve problems using parametric equations. Parametric Curve: the set of all points, y, where on an interval I (called the parameter interval) f t and y gt are continuous functions of t Parameter: the variable t Parametric Equations: f t and y gt Orientation: the directions that results from plotting the points as the values of t increase Graphing Parametric Equations E: Graph, t y t 1, 1t Note: Choose appropriate values for t first. Then substitute in and find and y. Make a table: t y 1 1 1 1 Plot points: 0 0 1 y 4 7 Eliminating the Parameter 1. Solve one of the equations for t. (Or if a trig function, isolate the trig function.). Substitute for t in the other equation. (Or use an identity if a trig function.) E: Write the parametric equation as a function of y in terms of. a), t y t 1 Solve for t in the -equation (easier to solve for): t t 1 Substitute into the y-equation: y t 1 y 1 y 1 1 t b), y t t Solve for t in the -equation (easier to solve for): 1 1 1 1 1 t t t t t Page 10 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
Substitute into the y-equation: 1 1 t y y y y 1 t 1 1 E: Write the parametric equation as a function of y and graph. tan, y tan 1, 0 Note: A parametric equation can be written in terms of θ instead of t. The -equation is already solved for the trig function: tan Substitute into the y-equation: y 1 Graph: y Using a Trig Identity E: Eliminate the parameter. 6cos t, y 6sin t, 0t Solving for a trig function won t help, so we need to use the identity sin t cos t 1. Square both equations: 36cos t, y 36sin t Add the equations: y 36cos t36sin t y 36cos t sin t Trig identity: y 36cos tsin t y 36 Note: The graph is a circle. The parameter interval lets us know that it would go around 1 time. Writing a Parameterization E: Find the parameterization of the line segment through the points A 1, and B, 3. Sketch a graph: y O A B P OP OA AP ; AP is a scalar multiple of AB, so OP OA t AB Page 11 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
OP OA t AB, y 1, t 1, 3, y 1, t 3, 1, y 13 t, t 13 t, y t These equations define the LINE. Find the parameter interval for the line segment: We want 1. 1 3t : 113t t 0 13t t 1 So, 0 t 1 Solution: 13 t, y t, 0t 1 Simulating Horizontal Motion E: A dog is running on a horizontal path with the coordinates of his position (in meters) given s 0. t 3 19t 100t 70 where 0 t 15. Use parametric equations and a graphing by calculator to simulate the dog s motion. Choose any horizontal line to simulate the motion: We will choose y 3. Parametric Equations: 0. t 19t 100t70, y 3, 0 t 15 Graph (Calculator must be in Parametric mode): Note: To see the motion, change the type of line to a bubble. If you would like the bubble to move slower, make the Tstep smaller. Parametric Equations for Projectile Motion distance: v 0 cos t height: 1 y gt v sin t h Note: On Earth, g 3 ft/sec or 0 0 g 9.8 m/sec. v 0 is initial velocity; h 0 is initial height. E: A golf ball is hit at 150 ft/sec at a 30 angle to the horizontal. a) When does it reach its maimum height? 1 Height: y gt v0 sin th0 y 16t 150sin 30 t0 ( h 0 0 because a golf ball is hit from the ground) Simplify: y 16t 150sin 30 t0 y 16t 75t b 75 75 11 Maimum height is at verte: t t sec a 16 3 3 b) How far does it go before it hits the ground? 11 Hits the ground when y 0 : y 16t 75t 0 t16t750 16t75 0 t 4 sec 16 Note: We could have doubled the time it took for the ball to reach its highest point! Page 1 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
11 v0 cos t 150cos 30 4 608.84 ft 16 c) Does the ball hit a 6 ft tall golfer, standing directly in the path of the ball 580 feet away? v cos t 580 150cos30 t t 4.46sec Distance: Find the time it takes for the ball to be 580 ft away: 0 Find the height of the ball at this time: y t t No the ball misses him by about 10 feet. 16 75 16 4.46 75 4.46 16.3ft Application: Ferris Wheel E: Zac is on a Ferris wheel of radius 0 ft that turns counterclockwise at a rate of one revolution every 4 sec. The lowest point of the Ferris wheel (6 o clock) is 10 ft above ground level at the point (0, 10) on a rectangular coordinate system. Find the parametric equations for the position of Zac as a function of time t (in seconds) if the Ferris wheel starts (t = 0) with Zac at the point (0, 30). Remember: rcos, y rsin Time to complete one revolution = 4 sec: 4 1 So, in one second, the ferris wheel travels through an angle of. 1 When t 0, 0 & y 30 : 0cos t, y 30 0sin t, t 0 1 1 You Try: A baseball is hit at 3 ft above the ground with an initial speed of 160 ft/sec at an angle of 17 with the horizontal. Will the ball clear a 0-ft wall that is 400 ft away? QOD: How would you write a parametrization for a semicircle? Page 13 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
Syllabus Objectives: 3.3 The student will differentiate between polar and Cartesian (rectangular) coordinates. 6. The student will transform functions between Cartesian and polar form. 6.4 The student will solve real-world application problems using polar coordinates. Polar Coordinate: r, ; r: the directed distance from the pole (origin); θ: the directed angle from the polar ais (-ais) Plotting Points on a Polar Graph E: Plot the points 3, 3 A, B 8, 40, & C, 5. 6 Point A: Start at the polar ais and go counter-clockwise 3 (70 ). Place the point 3 units from the pole (origin). Point B: Start at the polar ais and go clockwise 40. Place the point 8 units from the pole. (Note: Each radius drawn in the grid is 15.) Point C: Start by going counter-clockwise 5 3 (300 ) from the polar ais. Place a point units from the pole. Because r, you must place the point on the opposite side of the pole. B 6 4 C 5 5 A 4 6 Writing the Polar Coordinates of a Point E: Find four different polar coordinates of P. 6 4 P 5 5 4 6 Starting at the polar ais and going counter-clockwise: 6,30 Page 14 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
Starting at the polar ais and going clockwise: 6, 330 Going counter-clockwise more than one revolution: 6,390 Using 0 r and rotating counter-clockwise: 6,10 Note: There are infinitely many correct answers! Polar Conversions Polar to Rectangular: Rectangular to Polar: rcos rsin y 1 y tan y r r y Converting from Polar to Rectangular Coordinates E: Convert to rectangular coordinates. a) 5,45 rcos rsin y rcos rsin y 1 5 cos45 5 5 1 y 5 sin45 5 5 b) 3, 3 1 3 3cos 3 3 3 3 3 y 3sin 3 3 5,5 3 3 3, Converting from Rectangular to Polar Coordinates (Note: Be careful with the quadrant!) E: Convert to polar coordinates. a) 1, 3 1 y tan y r r y 1, 3 1 tan 3 60 r 13 is in Quadrant II, so the polar coordinates are b) 0, 4 This point is on the negative y-ais, so we know 70. 4,70 Note: There are other possible answers to these!,10. Page 15 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
Converting from Polar to Rectangular Equations E: Convert the equations and sketch the graph. a) r 4 4 Graph is a circle with center at origin & radius. r r y b) 4 rcos rcos r 4 y rsin y rsin y r 4 y y c) r sec 1 r sec r rcos 1 1 cos y Graphing in Polar Coordinates on the Calculator We will check our graphs above. Calculator must be in Polar mode. a) c) Note: We cannot check the graph of b) on the calculator, but the line y 45 4 for all values of r. represents the angle Page 16 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
Converting from Rectangular to Polar Equations E: Convert the equations. a) 4 rcos : rcos 4 r 4sec b) 36y 0 36y0 3rcos 6rsin 0 r3cos 6sin r or r 3cos 6sin 6sin 3cos c) y1 5 Epand: 44 y y15 y 4y 0 Substitute: r 4rcos rsin 0 rr4cos sin 0 So r 0 or r 4cos sin 0. But r 0 is a single point. So r 4cos sin Application: Finding Distance E: The location of two ships from the shore patrol station, given in polar coordinates, are mi, 150 & 3mi, 80. Find the distance between the ships. Sketch a diagram: Note: The angle between the ships (from the patrol station) is 150 80 70 Using the Law of Cosines: d. 3 3 cos70 d 131cos70.983mi You Try: Convert the coordinates. Polar:, ; Rectangular: 7,7 QOD: How could you write an epression for all of the possible polar coordinates of a point? Page 17 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
Syllabus Objectives: 6.3 The student will sketch the graph of a polar function and analyze it. Tests for Symmetry of Polar Curves 1. Symmetry about -ais: r, is equivalent to r,. Symmetry about y-ais: r, is equivalent to r, 3. Symmetry about origin: r, is equivalent to r, Graphing Polar Curves E: Graph and find the domain, range, symmetry, and maimum r-value. a) r 4cos θ 0 3 3 r 6 4 0 Domain: All real numbers Range: r 6 Ma r-value: r 6 Symmetry: Substitute. r 4cos 4cos Symmetric about -ais b) r 6sin3 This curve is called a limaçon. θ 0 5 7 18 6 18 3 18 r 0 3 6 3 0 3 6 Domain: All real numbers Range: 6 r 6 Ma r-value: r 6 Symmetry: Substitute r &. r 6sin3 r 6sin3 r 6sin3 Symmetric about y-ais c) r 4cos θ 0 6 4 r 0 This curve is called a rose. Domain: All real numbers Range: r Ma r-value: r Symmetry: Substitute. r 4cos r 4cos Symmetric about -ais Page 18 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
Substitute r. r 4cos r 4cos Symmetric about origin Substitute r &. r 4cos r 4cos Symmetric about y-ais This curve is called a lemniscate. Classifications of Polar Curves Limaçon Curves: r a bsin and r a bcos Rose Curves: r acosn and r asin n Petals: odd = n and even = n Lemniscate Curves: r a cos and r a sin You Try: Use your graphing calculator to eplore variations of r a sin n. Describe the effects of changing the window, the θ-step, a, n, and changing sin to cos. QOD: Are all polar curves bounded? Eplain. Page 19 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
Syllabus Objectives: 7.1 The student will graph a comple number on the comple/argand plane. 7. The student will represent a comple number in trigonometric (polar) form. 7.3 The student will simplify epressions involving comple numbers in trigonometric (polar) form. 7.4 The student will compute the powers of comple numbers using DeMoivre s Theorem and find the nth roots of a comple number. Comple Number Plane (Argand Plane): horizontal ais: real ais; vertical ais: imaginary ais Plotting Points in the Comple Plane E: Plot the points A34 i, B13 i, & i in the comple plane. Imaginary B A C Real Absolute Value (Modulus) of a Comple Number: the distance a comple number is from the origin on the comple plane abi a b (This can be shown using the Pythagorean Theorem.) E: Evaluate 3 i. 3i 3 1 10 Recall: Trigonometric form of a vector: u cos,sin Trigonometric Form of a Comple Number z = a + bi: z rcos isin Note: This can also be written as z rcis. b a rcos, b rsin, r a b tan a r = modulus; θ = argument Writing a Comple Number in Trig Form E: Find the trigonometric form of 3 i. Find r: r a b 3 1 4 Find θ: b 1 11 tan tan a 3 6 6 11 11 11 3 i cos isin or cis 6 6 6 Page 0 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
Writing a Comple Number in Standard Form (a + bi) E: Write 9cis in standard form. Epand: 9cis 9cos isin 9 1i0 9 0i 9 Multiplying and Dividing Comple Numbers Let z r cos isin and z r cos isin. 1 1 1 1 Multiplication: z z r r cos isin z z 1 1 1 1 r r 1 1 Division: cos i sin 1 1 E: Epress the product of z 1 and z in standard form. z1 4 cos isin, z cos isin 4 4 6 6 5 5 z1 z 4 cos isin 4 cos isin 1.464 5.464i 4 6 4 6 1 1 n Powers of a Comple Number: De Moivre s Theorem cos sin n n z r i r cos n isin n E: Evaluate i 5 Rewrite in trig form:. 1 tan 45 135 5 5 cis135 cis5135 3cis675 3cis315 r 4 n th Roots of a Comple Number: n n k k n k z r cos isin rcis n n n n n Note: Every comple number has a total of n n th roots. E: Find the cube roots of 8i. Write in trig form: r 8, 90 8cis90 3 3 90 360 k Evaluate the roots: 8cis90 8cis cis30 10 k k 0: cis 30 0 cis 30 cos30 isin30 3i k 1: cis 30 10 cis 150 cos150 isin150 3 i k : cis 30 40 cis 70 cos 70 i sin 70 i Roots of Unity: the n th roots of 1 3, k 0,1,,... n 1 Page 1 of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6
E: Epress the fifth roots of unity in standard form and graph them in the comple plane. 5 th Roots of Unity: 5 1 0i r 1, 0 1cis0 k 0:cis0 1 k k 1cis0 1cis cis 5 5 5 5 0 k 1: cis 0.310.95i 5 4 k : cis 0.810.59i 5 6 k 3: cis 0.810.59i 5 8 k 4 : cis 0.310.95i 5 You Try: 1. Write each comple number in trigonometric form. Then find the product and the quotient. 1 3, i 3i 4. Solve the equation 1 0. (You should have 4 solutions!) QOD: Is the trigonometric form of a comple number unique? Eplain. Page of Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6