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12 Vector-Valued Functions Copyright © Cengage Learning. All rights reserved. HWQ 3/19/15 Find v(t) given the following conditions: v t 4cos ti 3j v 0 3j v 0 4i 3t 2 v t 4cos t i + 3t j 2 12.3 Day 1 Velocity and Acceleration Copyright © Cengage Learning. All rights reserved. Objectives Describe the velocity and acceleration associated with a vector-valued function. Use a vector-valued function to analyze projectile motion. 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. Velocity and Acceleration Velocity and Acceleration The velocity vector is the tangent vector to the curve at point P Speed v t “Absolute value” means “distance from the origin” so we must use the Pythagorean theorem. The magnitude of the velocity vector r'(t) gives the speed of the object at time t. velocity vector Direction speed v t v t Velocity and Acceleration 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 = Example 1 – Finding Velocity and Acceleration Along a Plane Curve 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 – Solution The acceleration vector is Note that the velocity and acceleration vectors are orthogonal at any point in time. This is characteristic of motion at a constant speed. 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.. Example 2 (you try): r t 3cos t i 3sin t j a) Find the velocity and acceleration vectors. dr v 3sin t i 3cos t j dt dv a 3cos t i 3sin t j dt b) Find the velocity, acceleration, speed and direction of motion at t / 4 . Example 2: r t 3cos t i 3sin t j dr v 3sin t i 3cos t j dt dv a 3cos t i 3sin t j dt b) Find the velocity, acceleration, speed and direction of motion at t / 4 . 3 3 i j velocity: v 3sin i 3cos j 4 4 2 2 4 3 3 i j acceleration: a 3cos i 3sin j 4 4 4 2 2 Example 2: r t 3cos t i 3sin t j dr v 3sin t i 3cos t j dt dv a 3cos t i 3sin t j dt b) Find the velocity, acceleration, speed and direction of motion at t / 4 . 3 3 v i j 2 2 4 3 3 a i j 2 2 4 2 speed: direction: 3 3 v 2 2 4 v / 4 v / 4 2 9 9 3 2 2 3 / 2 3/ 2 1 i 1 j i j 2 2 3 3 Example 3: r t 2t 3 3t 2 i t 3 12t j a) Write the equation of the tangent where t 1. dr v t 6t 2 6t i 3t 2 12 j dt At t 1 : position: tangent: v 1 12i 9j r 1 5i 11j 5,11 slope: y y1 m x x1 3 y 11 x 5 4 9 3 12 4 3 29 y x 4 4 Example 3: r t 2t 3 3t 2 i t 3 12t j dr v t 6t 2 6t i 3t 2 12 j dt b) Find the coordinates of each point on the path where the horizontal component of the velocity is 0. 2 6 t 6t . The horizontal component of the velocity is 6t 6t 0 2 t t 0 r 0 0i 0j 0, 0 2 t t 1 0 t 0, 1 r 1 2 3 i 1 12 j r 1 1i 11j 1, 11 0, 0 1, 11 Example 4: An object starts from rest at the point (1,2,0) and moves with an acceleration of a t j + 2k , where a t is measured in feet per second per second. Find the location of the object after 2 seconds. t2 r t i 2 j + t 2k 2 r 2 1 4 j + 4k Example 5 (you try) : Use the given acceleration function to find the velocity and position vectors. Then find the position at time t = 2. a t cos t i - sint j v 0 j + k r 0 i r 2 cos2 i sin 2 j + 2k Homework: Section 12.3 Day 1: pg. 854 # 1-23 odd Day 2: MMM pg. 181 + 12.3, pg. 854 #25, 34, 37, 38 Day 2: Projectile Motion Fort Pulaski, GA Photo by Vickie Kelly, 2002 Greg Kelly, Hanford High School, Richland, Washington 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) vo We assume that the projectile is launched from the origin at time t =0 with initial velocity vo. Let vo vo then vo vo cos i vo sin j The initial position is: ro 0i 0 j 0 vo Newton’s second law of motion: f ma 2 d r f m 2 dt Vertical acceleration vo Newton’s second law of motion: f ma 2 d r f m 2 dt The force of gravity is: f mg j Force is in the downward direction vo Newton’s second law of motion: f ma 2 d r f m 2 dt The force of gravity is: f mg j 2 d r m 2 mg j dt vo Newton’s second law of motion: f ma 2 d r f m 2 dt The force of gravity is: f mg j 2 d r m 2 mg j dt 2 d r g j 2 dt Initial conditions: r ro dr vo dt when t o dr gt j v o dt 1 2 r gt j vot ro 2 1 2 r gt j vo cos t i vo sin t j 0 2 1 2 r gt j vo cos t i vo sin t j 0 2 Vector equation for ideal projectile motion: 1 2 r vo cos t i vo sin t gt j 2 1 2 r gt j vo cos t i vo sin t j 0 2 Vector equation for ideal projectile motion: 1 2 r vo cos t i vo sin t gt j 2 Parametric equations for ideal projectile motion: x vo cos t 1 2 y vo sin t gt 2 Example 1: A projectile is fired at 60o and 500 m/sec. Where will it be 10 seconds later? x 500 cos6010 x 2500 1 y 500sin 60 10 9.8 102 2 y 3840.13 The projectile will be 2.5 kilometers downrange and at an altitude of 3.84 kilometers. Note: The speed of sound is 331.29 meters/sec Or 741.1 miles/hr at sea level. The maximum height of a projectile occurs when the vertical velocity equals zero. dy vo sin gt 0 dt vo sin gt vo sin t g time at maximum height The maximum height of a projectile occurs when the vertical velocity equals zero. dy vo sin gt 0 dt vo sin gt vo sin t g We can substitute this expression into the formula for height to get the maximum height. 1 2 y vo sin t gt 2 ymax vo sin 1 vo sin vo sin g g 2 g ymax vo sin g 2 vo sin 2 2 2g 1 2 y vo sin t gt 2 ymax vo sin 1 vo sin vo sin g g 2 g 2 vo sin vo sin 2g 2g 2 ymax ymax vo sin 2g 2 2 2 maximum height (not including initial height) When the height is zero: time at launch: 1 2 0 vo sin t gt 2 1 0 t vo sin gt 2 t 0 When the height is zero: time at launch: t 0 1 2 0 vo sin t gt 2 1 0 t vo sin gt 2 1 vo sin gt 0 2 1 vo sin gt 2 time at impact (flight time) 2vo sin t g If we take the expression for flight time and substitute it into the equation for x, we can find the range. x vo cos t 2vo sin x vo cos g If we take the expression for flight time and substitute it into the equation for x, we can find the range. x vo cos t 2vo sin x vo cos g 2 vo x 2cos sin g 2 vo x sin 2 g Range The range is maximum when sin 2 is maximum. sin 2 1 2 90o 45 o Range is maximum when the launch angle is 45o. 2 vo x sin 2 g Range If we start somewhere besides the origin, the equations become: x xo vo cos t 1 2 y yo vo sin t gt 2 Example 4: A baseball is hit from 3 feet above the ground with an initial velocity of 152 ft/sec at an angle of 20o 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: x 152 cos 20 8.8 t o vo y 3 152sin 20o t 16t 2 yo 1 g 2 These equations can be graphed on the TI-89 to model the path of the ball: t2 Note that the calculator is in degrees. Using the trace function: Max height about 45 ft Time about 3.3 sec Distance traveled about 442 ft Example 6 – Describing the Path of a Baseball 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? x vo cos t 1 2 y vo sin t gt 2 Example 6 – Solution The maximum height occurs when which implies that So, the maximum height reached by the ball is Example 6 – Solution 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. Homework: Section 12.3 Day 1: pg. 854 # 1-23 odd Day 2: MMM pg. 181 + 12.3, pg. 854 #25, 34, 37, 38