1. Vector-Valued Functions 12 Copyright © Cengage Learning. All rights reserved.

2. HWQ 3/19/15 Find v(t) given the following conditions:

3. Day 1 Velocity and Acceleration Copyright © Cengage Learning. All rights reserved. 12.3

4. Describe the velocity and acceleration associated with a vector-valued function. Use a vector-valued function to analyze projectile motion. Objectives

5. Velocity and Acceleration As an object moves along a curve in the plane, the coordinates x and y of its center of mass are each functions of time t. Rather than using the letters f and g to represent these two functions, it is convenient to write x = x(t) and y = y(t). So, the position vector r(t) takes the form r(t) = x(t)i + y(t)j.

6. Velocity and Acceleration

7. The velocity vector is the tangent vector to the curve at point P The magnitude of the velocity vector r'(t) gives the speed of the object at time t. “Absolute value” means “distance from the origin” so we must use the Pythagorean theorem.

8. For motion along a space curve, the definitions are similar. That is, if r(t) = x(t)i + y(t)j + z(t)k, you have Velocity = v(t) = r'(t) = x'(t)i + y'(t)j + z'(t)k Acceleration = a(t) = r''(t) = x''(t)i + y''(t)j + z''(t)k Speed = Velocity and Acceleration

9. Find the velocity vector, speed, and acceleration vector of a particle that moves along the plane curve C described by Solution: The velocity vector is The speed (at any time) is Example 1 – Finding Velocity and Acceleration Along a Plane Curve

10. The acceleration vector is Example 1 – Solution Note that the velocity and acceleration vectors are orthogonal at any point in time. This is characteristic of motion at a constant speed.

11. Example 1, con’t Write the parametric and rectangular equations for r(t). Because the velocity vector has a constant magnitude, but a changing direction as t increases, the particle moves around the circle at a constant speed..

12. Example 2 (you try): a) Find the velocity and acceleration vectors. b) Find the velocity, acceleration, speed and direction of motion at.

13. Example 2: b) Find the velocity, acceleration, speed and direction of motion at. velocity: acceleration:

14. Example 2: b) Find the velocity, acceleration, speed and direction of motion at. speed: direction:

15. Example 3: a) Write the equation of the tangent where. At : position: slope: tangent:

16. The horizontal component of the velocity is. Example 3: b) Find the coordinates of each point on the path where the horizontal component of the velocity is 0.

18. Example 4: An object starts from rest at the point (1,2,0) and moves with an acceleration of, where is measured in feet per second per second. Find the location of the object after 2 seconds.

19. Example 5 (you try) : Use the given acceleration function to find the velocity and position vectors. Then find the position at time t = 2.

20. Homework: Section 12.3 Day 1: pg. 854 # 1-23 odd Day 2: MMM pg. 181 + 12.3, pg. 854 #25, 34, 37, 38

21. Day 2: Projectile Motion Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 2002 Fort Pulaski, GA

22. 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)

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

24. Newton’s second law of motion: Vertical acceleration

25. Newton’s second law of motion: The force of gravity is: Force is in the downward direction

26. Newton’s second law of motion: The force of gravity is:

27. Newton’s second law of motion: The force of gravity is:

28. Initial conditions:

29. Vector equation for ideal projectile motion:

30. Parametric equations for ideal projectile motion:

31. 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.

32. The maximum height of a projectile occurs when the vertical velocity equals zero. time at maximum height

33. 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.

35. maximum height (not including initial height)

36. When the height is zero: time at launch:

37. When the height is zero: time at launch: time at impact (flight time)

38. If we take the expression for flight time and substitute it into the equation for x, we can find the range.

39. Range

40. The range is maximum when Range is maximum. Range is maximum when the launch angle is 45 o.

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

42. This simplifies to: which is the equation of a parabola.

43. If we start somewhere besides the origin, the equations become:

44. Example 4: 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:

45. These equations can be graphed on the TI-89 to model the path of the ball: Note that the calculator is in degrees. t2t2

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

48. A baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45° with respect to the ground, as shown in Figure 12.19. Find the maximum height reached by the baseball. Will it clear a 10-foot-high fence located 300 feet from home plate? Example 6 – Describing the Path of a Baseball

49. The maximum height occurs when which implies that So, the maximum height reached by the ball is Example 6 – Solution

50. The ball is 300 feet from where it was hit when Solving this equation for t produces At this time, the height of the ball is = 303 – 288 = 15 feet. Therefore, the ball clears the 10-foot fence for a home run. Example 6 – Solution

51. 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. You are not responsible for memorizing these formulas. 

52. Homework: Section 12.3 Day 1: pg. 854 # 1-23 odd Day 2: MMM pg. 181 + 12.3, pg. 854 #25, 34, 37, 38