Precalculus Notes: Unit 6 Vectors, Parametrics, Polars, & Complex Numbers. A: Initial Point (start); B: Terminal Point (end) : ( ) ( )

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1 Syllabus Objectives: 5.1 The student will explore 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: ( Ax, A y Coordinates of B: ( Bx, B y Magnitude (length of a Directed Line Segment AB Note: This is the distance formula! AB = B A + B A : ( ( x x 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 x x y y Ex: Graph the vector AB = 3, One possible graph: y and find the magnitude. x 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. Magnitude: AB = u Note: If a vector u is written in component form, u = u1, u, then the magnitude of u is ( 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 u+ v = u1+ v1, u + v. Scalar Multiplication: Let u = u1, u and k be any constant. Then ku = k u1, ku. Note: If k < 0, then k u is in the opposite direction. Page 1 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

2 Ex: Use the graph of the vectors to complete each example below. w u v 1. Show that u= v. Show that u = v. u = v = Show that the direction of u is the same as the direction of v. Use slope: Direction of u = ; direction of v = The direction and magnitude are the same, so u= v.. Find the component form and the magnitude of u and w. Component form of u: u = u = (see above Component form of w: w = w = u 3w = 3. Find the component form of u 3w. 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 x- and y-axes i = 1,0 & j = 0,1 Any vector can be written in terms of the standard unit vectors. Ex: Write the vector v =,5 in terms of the standard unit vectors. v = Ex: 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: i 4j = Page of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

3 Divide the original vector by its magnitude: i 4j = 5 Component Form: Verify magnitude of unit vector: 5 5, 5 5 = Recall: In the unit circle, x = cos θ, y = sinθ. This leads into another way of expressing a vector, in terms of its direction angle, θ. Direction Angle: in standard position, the angle the vector makes with the positive x-axis (counterclockwise Resolving a Vector: in terms of its direction angle, θ, a vector can be written as u cos θ,sinθ = u cosθi+ u sinθj Ex: Find the magnitude and direction angle of v = i+ 6j. Magnitude: v = Direction angle: v cosθi+ v sinθj but since we know v = i+ 6j is in Quadrant II, θ = = Page 3 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

4 Ex: Find the component form of v given its magnitude and its direction angle. v = 5, θ = 30 v = v cosθi+ v sinθj Application: Resultant Force Ex: Two forces act on an object: u = 3, θ u = 45 and v = 4, θ u = 30. Find the direction and magnitude of the resultant force. Write each vector in component form: u = v = v cosθvi+ v sinθvj The resultant force is the sum u + v: Application: Bearing Ex: 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 = w = Note: The 45 is the direction angle, not the bearing. Vector v is the sum p + w: v = The second component of vector v must equal zero, because the plane is headed due east. 60sin sinθ = 0 Bearing of the plane: 90 + θ Ground speed of the plane: v = You Try: Find the component form of v given its magnitude and the angle it makes with the positive x- axis. v =, direction: i+ 3j OD: In the examples in your notes, we used sine or cosine to find the direction angle of a vector. Explain how you could use tangent to find the direction angle. Page 4 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

5 Syllabus Objective: 5.3 The student will explore 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 i = uv 1 1+ uv. Note: The dot product is a scalar. Ex: Evaluate 5, i 3, 4. Properties of the Dot Product: 1. uv i = vu i. uu= i u 3. 0u i = 0 4. ui( v+w =uiv+ ui w 5. ( cu iv = ui( cv = c( ui v Ex: Evaluate the following given u= 3,6; v = 1,0; w = 5, (a ww i (b w (c ( v+ w i u (d vu i + wu i Angle Between Two Vectors: uv i 1 cosθ = θ = cos uv i u v u v Proof: Use the triangle. Law of Cosines: v θ Property of Dot Product: ( i ( Expand: Property of Dot Product: u u v v u = u + v u v cosθ v-u v-u = u + v u v cosθ v v viu uiv+ ui u= u + v u v cosθ v ui v+ u = u + v u v cosθ Page 5 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

6 Property of Equality: uv= u vcosθ cosθ = uv i i u v cosθ = Ex: Find uv i, where θ is the angle between u and v. uv i u v 5π u = 6, v = 8, θ = 6 Application Ex: Find the interior angles of the triangle with vertices ( 3, 0, ( 4,, ( 5,1. y B A C x AB = cos A = BA = cos B = C Angles: Orthogonal Vectors: two vectors whose dot product is equal to 0 What is the angle between two non-zero orthogonal vectors? uv i 0 cosθ = cosθ = cosθ = 0 θ = 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 i uv i uv i uv i cosθ = cos180 = 1 = uv i or cosθ = cos360 = 1 = uv i u v u v u v u v Page 6 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

7 Parallel Vectors: two vectors whose dot product is equal to 1 or 1 Ex: Are the vectors orthogonal, parallel, or neither? v = 3i j, w = 3i+ 4j Find vw i : The vectors are. uv i Vector Projection: the projection of u onto v is denoted by: projvu= v v u u = u u 1 u 1 = proj v u v Ex: 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 proj w vw i v = w w v proj v = w v = Application: Force Ex: Find the force required to keep a 00-lb cart from rolling down a 30 incline. f Draw a diagram and label: The force due to gravity: g = 00j (gravity acts vertically downward 3 1 v = i+ j= i+ j Incline vector: ( cos30 ( sin 30 gv i Force vector required to keep the cart from rolling: f = projvg = v v Page 7 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

8 gv i f = v = v Magnitude of Force: f = Application: Work W = cosθ ( force( distance Ex: 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? 40 ( ( w = cos40 15 lbs 500 ft 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 = i 3 j; u= i j QOD: If u is a unit vector, what is uu? i Explain why. Page 8 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

9 Syllabus Objective: 1.10 The student will solve problems using parametric equations. Parametric Curve: the set of all points ( x, y, where x = f ( t and y g( t on an interval I (called the parameter interval Parameter: the variable t Parametric Equations: x = f ( t and y = g( t = are continuous functions of t Orientation: the directions that results from plotting the points as the values of t increase Graphing Parametric Equations Ex: Graph x =, t y = t 1, 1 t Note: Choose appropriate values for t first. Then substitute in and find x and y. Make t x y a table: 0 1 Plot points: 1 y x 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. Ex: Write the parametric equation as a function of y in terms of x. a x =, t y = t 1 x Solve for t in the x-equation (easier to solve for: x = t t = Substitute into the y-equation: 1 t b x =, y t = + t + Solve for t in the x-equation (easier to solve for: Substitute into the y-equation: t y = t + Page 9 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

10 Ex: Write the parametric equation as a function of y and graph. π x = tan θ, y = tanθ 1, 0 θ < Note: A parametric equation can be written in terms of θ instead of t. The x-equation is already solved for the trig function: x = tanθ Substitute into the y-equation: y = x 1 Graph: Using a Trig Identity Ex: Eliminate the parameter. x = 6cos t, y = 6sin t, 0 t π Solving for a trig function won t help, so we need to use the identity sin t+ cos t = 1. Square both equations: Add the equations: Trig identity: Note: The graph is a circle. The parameter interval lets us know that it would go around 1 time. Writing a Parameterization Ex: Find the parameterization of the line segment through the points A = ( 1, and B = (, 3. Sketch a graph: y O x A B P OP = OA + AP ; AP is a scalar multiple of AB, so OP = OA + t AB OP = OA + t AB x, y = x, y = 1+ 3 t, t x =, y = These equations define the LINE. Page 10 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

11 Find the parameter interval for the line segment: We want 1 x. x = 1+ 3t : So, 0 t 1 Solution: Simulating Horizontal Motion Ex: A dog is running on a horizontal path with the coordinates of his position (in meters given s = 0. t 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: ( x = 0. t 19t + 100t 70, 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: x = ( v θ t height: ( sinθ 0 cos Note: On Earth, g = 3 ft/sec or g = 9.8 m/sec. 1 y = gt + v t + h 0 0 Height: Ex: A golf ball is hit at 150 ft/sec at a 30 angle to the horizontal. a When does it reach its maximum height? Simplify: ( h 0 = 0 because a golf ball is hit from the ground Maximum height is at vertex: b t = t = a Page 11 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

12 b How far does it go before it hits the ground? Hits the ground when y = 0 : Note: We could have doubled the time it took for the ball to reach its highest point! Distance: x = ( v 0 cosθ t = c Does the ball hit a 6 ft tall golfer, standing directly in the path of the ball 580 feet away? Find the time it takes for the ball to be 580 ft away: ( θ x = v 0 cos t Find the height of the ball at this time: Application: Ferris Wheel Ex: 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. Time to complete one revolution = 4 sec: π π = 4 1 When t = 0, x = 0 & y = 30 : 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 1 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

13 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, θ ; r: the directed distance from the pole (origin; θ: the directed angle from the polar axis (x-axis Plotting Points on a Polar Graph Ex: Plot the points 3 π A 3,, B( 8, 40, & C, 5 π. 6 Point A: Start at the polar axis and go counter-clockwise 3 π (70. Place the point 3 units from the pole (origin. Point B: Start at the polar axis 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 π 6 (150 from the polar axis. 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 Ex: Find four different polar coordinates of P. 6 4 P Starting at the polar axis and going counter-clockwise: ( 6,30 Page 13 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

14 Starting at the polar axis and going clockwise: ( 6,330 Going counter-clockwise more than one revolution: ( 6,390 Using r < 0 and rotating counter-clockwise: ( 6, 10 Note: There are infinitely many correct answers! Polar Conversions Polar to Rectangular: Rectangular to Polar: rcosθ = x rsinθ = y 1 y tan = θ x x + y = r r = x + y Converting from Polar to Rectangular Coordinates Ex: Convert to rectangular coordinates. a ( 5,45 rcosθ = x rsinθ = y rcosθ = x π b 3, 3 rsinθ = y Converting from Rectangular to Polar Coordinates (Note: Be careful with the quadrant! Ex: Convert to polar coordinates. a ( 1, 3 1 y tan = θ x x + y = r r = x + y 1 ( θ = tan 3 = 60 r = 1+ 3 = ( 1, 3 is in Quadrant II, so the polar coordinates are. b ( 0, 4 This point is on the negative y-axis, so we know θ = 70. Note: There are other possible answers to these! Page 14 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

15 Converting from Polar to Rectangular Equations Ex: Convert the equations and sketch the graph. a r = y Graph is a circle with center at origin & radius. x π b θ = 4 π x = r cosθ x = r cos x 4 = r π y = r sin θ y = r sin y = r 4 y x r = secθ c r = secθ 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 π = 45 4 for all values of r. y = x represents the angle Page 15 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

16 Converting from Rectangular to Polar Equations Ex: Convert the equations. a x = 4 x = rcosθ : b 3x 6y+ = 0 c ( x+ + ( y 1 = 5 Expand: Substitute: r + 4rcosθ rsinθ = 0 So r = 0 or r + 4cosθ sinθ = 0. But r = 0 is a single point. So Application: Finding Distance Ex: 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 = 70. Using the Law of Cosines: d = You Try: Convert the coordinates. Polar: (,π ; Rectangular: ( 7,7 QOD: How could you write an expression for all of the possible polar coordinates of a point? Page 16 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

17 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 x-axis: ( r, θ is equivalent to ( r, θ. Symmetry about y-axis: ( r, θ is equivalent to ( r, θ 3. Symmetry about origin: ( r, θ is equivalent to ( r, θ Graphing Polar Curves Ex: Graph and find the domain, range, symmetry, and maximum r-value. a r = + 4cosθ θ 0 π π π π 3 3 r Domain: Range: Max r-value: Symmetry: Substitute θ. Symmetric about x-axis b r = 6sin( 3θ θ 0 This curve is called a limaçon. π π 5π π 7π π r Domain: Range: Max r-value: Symmetry: Substitute r & θ. r = Symmetric about y-axis c r = 4cos( θ θ 0 π 6 π 4 This curve is called a rose. r ± ± 0 Domain: Range: Max r-value: Symmetry: Substitute θ. r = 4cos( θ r = 4cos( θ Symmetric about x-axis Page 17 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

18 Substitute r. ( r = 4cos( θ r = Symmetric about origin Substitute ( ( r & θ. r = 4cos θ Symmetric about y-axis This curve is called a lemniscate. Classifications of Polar Curves Limaçon Curves: r = a± bsinθ and r = a± bcosθ Rose Curves: r = acos( nθ and r = asin ( nθ Petals: odd = n and even = n Lemniscate Curves: r = a cos( θ and r = a sin ( θ You Try: Use your graphing calculator to explore variations of r = a sin ( nθ. Describe the effects of changing the window, the θ-step, a, n, and changing sinθ to cosθ. QOD: Are all polar curves bounded? Explain. Page 18 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

19 Syllabus Objectives: 7.1 The student will graph a complex number on the complex/argand plane. 7. The student will represent a complex number in trigonometric (polar form. 7.3 The student will simplify expressions involving complex numbers in trigonometric (polar form. 7.4 The student will compute the powers of complex numbers using DeMoivre s Theorem and find the nth roots of a complex number. Complex Number Plane (Argand Plane: horizontal axis real axis; vertical axis imaginary axis Plotting Points in the Complex Plane Ex: Plot the points A( 3+ 4 i, B( 1+ 3 i, & ( i in the complex plane. Imaginary B A C Real Absolute Value (Modulus of a Complex Number: the distance a complex number is from the origin on the complex plane a+ bi = a + b (This can be shown using the Pythagorean Theorem. Ex: Evaluate 3 i. Recall: Trigonometric form of a vector: u cos θ,sinθ Trigonometric Form of a Complex Number z = a + bi: z = r( cosθ + isinθ Note: This can also be written as z = rcisθ. b a = rcos θ, b= rsin θ, r = a + b tanθ = a r = modulus; θ = argument Writing a Complex Number in Trig Form Ex: Find the trigonometric form of 3 i. Find r: r = a + b = b Find θ: tanθ = tanθ = 3 i = a Page 19 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

20 Writing a Complex Number in Standard Form (a + bi Ex: Write 9cisπ in standard form. Expand: 9cisπ 9( cosπ i sinπ = + = Multiplying and Dividing Complex Numbers Let z = r ( cosθ + isinθ and z = r ( cosθ + isinθ Multiplication: z z = r r cos( θ + θ + isin( θ + θ z z r r 1 1 Division: = cos( θ θ + i sin( θ θ 1 1 Ex: Express the product of and z in standard form. z1 n Powers of a Complex Number: De Moivre s Theorem ( ( Ex: Evaluate ( i 5 +. z n = r cosθ + isinθ = r n cos nθ + isin nθ Rewrite in trig form: n th Roots of a Complex Number: n n θ πk θ πk n θ + π z = r cos + isin rcis n n + + = n n n Note: Every complex number has a total of n n th roots. Ex: Find the cube roots of 8i. Write in trig form: Evaluate the roots: k, k = 0,1,,... ( n 1 Page 0 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6

21 Roots of Unity: the n th roots of 1 Ex: Express the fifth roots of unity in standard form and graph them in the complex plane. 5 th Roots of Unity: i r = 1, θ = 0 1cis0 + πk 1cis0 = 1cis = cis πk You Try: 1. Write each complex number in trigonometric form. Then find the product and the quotient. 1+ 3, i 3i. Solve the equation 4 x + 1= 0. (You should have 4 solutions! QOD: Is the trigonometric form of a complex number unique? Explain. Page 1 of 1 Precalculus Graphical, Numerical, Algebraic: Pearson Chapter 6