1. In chapter 1, we talked about parametric equations. Parametric equations can be used to describe motion that is not a function. If f and g have derivatives at t, then the parametrized curve also has a derivative at t. 10.1 Parametric Functions

2. The formula for finding the slope of a parametrized curve is: This makes sense if we think about canceling dt. 10.1 Parametric Functions

3. The formula for finding the slope of a parametrized curve is: We assume that the denominator is not zero. 10.1 Parametric Functions

4. To find the second derivative of a parametrized curve, we find the derivative of the first derivative: 1.Find the first derivative (dy/dx). 2. Find the derivative of dy/dx with respect to t. 3. Divide by dx/dt. 10.1 Parametric Functions

6. 1.Find the first derivative ( dy/dx ). 10.1 Parametric Functions

7. 2. Find the derivative of dy/dx with respect to t. Quotient Rule 10.1 Parametric Functions

8. 3. Divide by dx/dt. 10.1 Parametric Functions

9. The equation for the length of a parametrized curve is similar to our previous “length of curve” equation: 10.1 Parametric Functions

11. This curve is: 10.1 Parametric Functions

12. Quantities that we measure that have magnitude but not direction are called scalars. Quantities such as force, displacement or velocity that have direction as well as magnitude are represented by directed line segments. A B initial point terminal point The length is 10.2 Vectors in the Plane

13. A B initial point terminal point A vector is represented by a directed line segment. Vectors are equal if they have the same length and direction (same slope). 10.2 Vectors in the Plane

14. A vector is in standard position if the initial point is at the origin. x y The component form of this vector is: 10.2 Vectors in the Plane

15. A vector is in standard position if the initial point is at the origin. x y The component form of this vector is: The magnitude (length) ofis: 10.2 Vectors in the Plane

16. P Q (-3,4) (-5,2) The component form of is: v (-2,-2) 10.2 Vectors in the Plane

17. If Then v is a unit vector. is the zero vector and has no direction. 10.2 Vectors in the Plane

18. Vector Operations: (Add the components.) (Subtract the components.) 10.2 Vectors in the Plane

19. Vector Operations: Scalar Multiplication: Negative (opposite): 10.2 Vectors in the Plane

23. v v u u u+v u + v is the resultant vector. (Parallelogram law of addition) 10.2 Vectors in the Plane

24. The angle between two vectors is given by: This comes from the law of cosines. See page 524 for the proof if you are interested. 10.2 Vectors in the Plane

25. The dot product (also called inner product) is defined as: Read “u dot v” This could be substituted in the formula for the angle between vectors to give: 10.2 Vectors in the Plane

28. Application: Example 7 A Boeing 727 airplane, flying due east at 500 mph in still air, encounters a 70-mph tail wind acting in the direction of 60 o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they? 10.2 Vectors in the Plane

29. N E u

30. N E v u 60 o 10.2 Vectors in the Plane

31. N E v u We need to find the magnitude and direction of the resultant vector u + v. u+v 10.2 Vectors in the Plane

32. N E v u The component forms of u and v are: u+v 500 70 Therefore: 10.2 Vectors in the Plane

33. N E The new ground speed of the airplane is about 538.4 mph, and its new direction is about 6.5 o north of east. 538.4 6.5 o 10.2 Vectors in the Plane

34. Any vector can be written as a linear combination of two standard unit vectors. The vector v is a linear combination of the vectors i and j. The scalar a is the horizontal component of v and the scalar b is the vertical component of v. 10.3 Vector-valued Functions

35. If we separate r ( t ) into horizontal and vertical components, we can express r ( t ) as a linear combination of standard unit vectors i and j. 10.3 Vector-valued Functions

36. Let A = (-2,3) and B = (4,6) Find AB in terms of i and j AB = = 6i + 3j

37. In three dimensions the component form becomes: 10.3 Vector-valued Functions

38. Most of the rules for the calculus of vectors are the same as we have used, except: “Absolute value” means “distance from the origin” so we must use the Pythagorean theorem. 10.3 Vector-valued Functions

39. a) Find the velocity and acceleration vectors. b) Find the velocity, acceleration, speed and direction of motion at. 10.3 Vector-valued Functions

40. b) Find the velocity, acceleration, speed and direction of motion at. velocity: acceleration: 10.3 Vector-valued Functions

41. b) Find the velocity, acceleration, speed and direction of motion at. speed: direction: 10.3 Vector-valued Functions

42. a) Write the equation of the tangent where. At : position:slope: tangent: 10.3 Vector-valued Functions

43. The horizontal component of the velocity is. b) Find the coordinates of each point on the path where the horizontal component of the velocity is 0. 10.3 Vector-valued Functions

44. a) Write the equation of the tangent where t = 1. position:slope: tangent: 10.3 Vector-valued Functions r(t) = (2t 2 - 3t)i + (3t 3 - 2t)jv(t) = (4t - 3)i + (9t 2 - 2)j At t = 1 r(1) = -1i + 1j v(1) = 1i + 7j (-1,1)

45. The horizontal component of the velocity is 4t - 3. b) Find the coordinates of each point on the path where the horizontal component of the velocity is 0. 10.3 Vector-valued Functions 4t – 3 = 0

46. 10.3 Vector-valued Functions Find r

47. One early use of calculus was to study projectile motion. In this section we assume ideal projectile motion: Constant force of gravity in a downward direction Flat surface No air resistance (usually) 10.4 Projectile Motion

48. We assume that the projectile is launched from the origin at time t =0 with initial velocity v o. The initial position is: 10.4 Projectile Motion

49. Newton’s second law of motion: Vertical acceleration 10.4 Projectile Motion

50. Newton’s second law of motion: The force of gravity is: Force is in the downward direction 10.4 Projectile Motion

51. Newton’s second law of motion: The force of gravity is: 10.4 Projectile Motion

52. Newton’s second law of motion: The force of gravity is: 10.4 Projectile Motion

53. Initial conditions: 10.4 Projectile Motion

54. Vector equation for ideal projectile motion: 10.4 Projectile Motion

55. Vector equation for ideal projectile motion: Parametric equations for ideal projectile motion: 10.4 Projectile Motion

56. Example 1: A projectile is fired at 60 o and 500 m/sec. Where will it be 10 seconds later? Note: The speed of sound is 331.29 meters/sec Or 741.1 miles/hr at sea level. The projectile will be 2.5 kilometers downrange and at an altitude of 3.84 kilometers. 10.4 Projectile Motion

57. The maximum height of a projectile occurs when the vertical velocity equals zero. time at maximum height 10.4 Projectile Motion

58. The maximum height of a projectile occurs when the vertical velocity equals zero. We can substitute this expression into the formula for height to get the maximum height. 10.4 Projectile Motion

59. maximum height 10.4 Projectile Motion

60. When the height is zero: time at launch: 10.4 Projectile Motion

61. When the height is zero: time at launch: time at impact (flight time) 10.4 Projectile Motion

62. If we take the expression for flight time and substitute it into the equation for x, we can find the range. 10.4 Projectile Motion

63. If we take the expression for flight time and substitute it into the equation for x, we can find the range. Range 10.4 Projectile Motion

64. The range is maximum when Range is maximum. Range is maximum when the launch angle is 45 o. 10.4 Projectile Motion

65. If we start with the parametric equations for projectile motion, we can eliminate t to get y as a function of x. 10.4 Projectile Motion

66. If we start with the parametric equations for projectile motion, we can eliminate t to get y as a function of x. This simplifies to: which is the equation of a parabola. 10.4 Projectile Motion

67. If we start somewhere besides the origin, the equations become: 10.4 Projectile Motion

68. A baseball is hit from 3 feet above the ground with an initial velocity of 152 ft/sec at an angle of 20 o from the horizontal. A gust of wind adds a component of -8.8 ft/sec in the horizontal direction to the initial velocity. The parametric equations become:

69. These equations can be graphed on the TI-83 to model the path of the ball: the calculator must be in degrees.

71. Max height about 45 ft Distance traveled about 442 ft Time about 3.3 sec Using the trace function:

72. In real life, there are other forces on the object. The most obvious is air resistance. If the drag due to air resistance is proportional to the velocity: (Drag is in the opposite direction as velocity.) Equations for the motion of a projectile with linear drag force are given on page 546. 10.4 Projectile Motion

73. Polar graphing is like the second method of giving directions. Each point is determined by a distance and an angle. Initial ray A polar coordinate pair determines the location of a point. 10.5 Polar Graphing

74. (Circle centered at the origin) (Line through the origin) 10.5 Polar Graphing

75. More than one coordinate pair can refer to the same point. All of the polar coordinates of this point are: 10.5 Polar Graphing

76. Equations Relating Polar and Cartesian Coordinates θ x y r Polar Rectangular Rectangular Polar

77. 10.5 Polar Graphing Change from rectangular to polar.

78. 10.5 Polar Graphing Change from polar to rectangular.

79. 10.5 Polar Graphing Write as a Cartesian Equation r = 2 cos θ +1

80. 10.5 Polar Graphing Write as a Cartesian Equation r = 4 tan θ sec θ

81. 10.5 Polar Graphing Write as a Polar Equation xy = 2

82. 10.5 Polar Graphing Write as a Polar Equation x - y = 6

83. Tests for Symmetry: x-axis: If (r, q) is on the graph, so is (r, -q). 10.5 Polar Graphing

84. y-axis: If (r, q) is on the graph, so is (r, p-q) or (-r, -q). 10.5 Polar Graphing

85. origin: If (r, q) is on the graph, so is (-r, q) or (r, q+p). 10.5 Polar Graphing

86. If a graph has two symmetries, then it has all three: 10.5 Polar Graphing

87. Try graphing this on the TI-83. 10.6 Calculus of Polar Curves

88. To find the slope of a polar curve: We use the product rule here. 10.6 Calculus of Polar Curves

89. To find the slope of a polar curve: 10.6 Calculus of Polar Curves

90. Example: 10.6 Calculus of Polar Curves

91. The length of an arc (in a circle) is given by r. q when q is given in radians. For a very small q, the curve could be approximated by a straight line and the area could be found using the triangle formula: 10.6 Calculus of Polar Curves

92. We can use this to find the area inside a polar graph. 10.6 Calculus of Polar Curves

93. Example: Find the area enclosed by: 10.6 Calculus of Polar Curves

95. Notes: To find the area between curves, subtract: Just like finding the areas between Cartesian curves, establish limits of integration where the curves cross. 10.6 Calculus of Polar Curves

96. When finding area, negative values of r cancel out: Area of one leaf times 4:Area of four leaves: 10.6 Calculus of Polar Curves

97. To find the length of a curve: Remember: For polar graphs: So: 10.6 Calculus of Polar Curves

98. There is also a surface area equation similar to the others we are already familiar with: When rotated about the x-axis: 10.6 Calculus of Polar Curves