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8 Pages

### tcu11_13_02

Course: MATH 1240, Spring 2010
School: Westminster UT
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Word Count: 2434

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AWL/Thomas_ch13p906-964 4100 8/25/04 2:48 PM Page 920 920 Chapter 13: Vector-Valued Functions and Motion in Space 13.2 Modeling Projectile Motion When we shoot a projectile into the air we usually want to know beforehand how far it will go (will it reach the target?), how high it will rise (will it clear the hill?), and when it will land (when do we get results?). We get this information from the direction and...

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AWL/Thomas_ch13p906-964 4100 8/25/04 2:48 PM Page 920 920 Chapter 13: Vector-Valued Functions and Motion in Space 13.2 Modeling Projectile Motion When we shoot a projectile into the air we usually want to know beforehand how far it will go (will it reach the target?), how high it will rise (will it clear the hill?), and when it will land (when do we get results?). We get this information from the direction and magnitude of the projectiles initial velocity vector, using Newtons second law of motion. The Vector and Parametric Equations for Ideal Projectile Motion To derive equations for projectile motion, we assume that the projectile behaves like a particle moving in a vertical coordinate plane and that the only force acting on the projectile during its flight is the constant force of gravity, which always points straight down. In practice, none of these assumptions really holds. The ground moves beneath the projectile as the earth turns, the air creates a frictional force that varies with the projectiles speed and altitude, and the force of gravity changes as the projectile moves along. All this must be taken into account by applying corrections to the predictions of the ideal equations we are about to derive. The corrections, however, are not the subject of this section. We assume that the projectile is launched from the origin at time t = 0 into the first quadrant with an initial velocity v0 (Figure 13.9). If v0 makes an angle a with the horizontal, then v0 = s v0 cos a di + s v0 sin a dj. If we use the simpler notation y0 for the initial speed v0 , then v0 = s y0 cos a di + s y0 sin a dj. The projectiles initial position is r0 = 0i + 0j = 0. (2) (1) Copyright 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch13p906-964 8/25/04 2:48 PM Page 921 13.2 y Modeling Projectile Motion 921 Newtons second law of motion says that the force acting on the projectile is equal to the projectiles mass m times its acceleration, or ms d 2r> dt 2 d if r is the projectiles position vector and t is time. If the force is solely the gravitational force - mg j, then m v0 v0 sin j x d 2r = - mg j dt 2 and d 2r = - g j. dt 2 We find r as a function of t by solving the following initial value problem. Differential equation: Initial conditions: The first integration gives d 2r = -gj dt 2 r = r0 and dr = v0 dt when t = 0 r 0 at time t 0 v0 cos i a g j (a) y (x, y) v g j dr = - s gt dj + v0 . dt A second integration gives r=12 gt j + v0 t + r0 . 2 a r 0 xi yj Substituting the values of v0 and r0 from Equations (1) and (2) gives x R Horizontal range (b) r=- 12 gt j + s y0 cos a dt i + s y0 sin a dt j + 0 2 ('''''')''''''* v0 t FIGURE 13.9 (a) Position, velocity, acceleration, and launch angle at t = 0 . (b) Position, velocity, and acceleration at a later time t. Collecting terms, we have Ideal Projectile Motion Equation r = s y0 cos a dt i + as y0 sin a dt 12 gt b j. 2 (3) Equation (3) is the vector equation for ideal projectile motion. The angle a is the projectiles launch angle (firing angle, angle of elevation), and y0 , as we said before, is the projectiles initial speed. The components of r give the parametric equations x = s y0 cos a dt and y = s y0 sin a dt 12 gt , 2 (4) where x is the distance downrange and y is the height of the projectile at time t 0. EXAMPLE 1 Firing an Ideal Projectile A projectile is fired from the origin over horizontal ground at an initial speed of 500 m> sec and a launch angle of 60. Where will the projectile be 10 sec later? Copyright 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch13p906-964 8/25/04 2:48 PM Page 922 922 Chapter 13: Vector-Valued Functions and Motion in Space Solution We use Equation (3) with y0 = 500, a = 60, g = 9.8, and t = 10 to find the projectiles components 10 sec after firing. r = s y0 cos a dt i + as y0 sin a dt 23 1 1 = s 500 d a b s 10 di + as 500 d a b 10 - a b s 9.8 ds 100 d b j 2 2 2 12 gt b j 2 L 2500i + 3840j. Ten seconds after firing, the projectile is about 3840 m in the air and 2500 m downrange. Height, Flight Time, and Range Equation (3) enables us to answer most questions about the ideal motion for a projectile fired from the origin. The projectile reaches its highest point when its vertical velocity component is zero, that is, when dy = y0 sin a - gt = 0, dt For this value of t, the value of y is ymax = s y0 sin a d a s y0 sin a d2 y0 sin a y0 sin a 2 1 b - ga g b = . g 2 2g or t= y0 sin a g. To find when the projectile lands when fired over horizontal ground, we set the vertical component equal to zero in Equation (3) and solve for t. s y0 sin a dt t ay0 sin a 12 gt = 0 2 1 gt b = 0 2 t = 0, t= 2y0 sin a g Since 0 is the time the projectile is fired, s 2y0 sin a d> g must be the time when the projectile strikes the ground. To find the projectiles range R, the distance from the origin to the point of impact on horizontal ground, we find the value of the horizontal component when t = s 2y0 sin a d> g . x = s y0 cos a dt R = s y0 cos a d a 2y0 sin a y02 y02 b = g s 2 sin a cos a d = g sin 2a g The range is largest when sin 2a = 1 or a = 45. Copyright 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch13p906-964 8/25/04 2:48 PM Page 923 13.2 Modeling Projectile Motion 923 Height, Flight Time, and Range for Ideal Projectile Motion For ideal projectile motion when an object is launched from the origin over a horizontal surface with initial speed y0 and launch angle a : Maximum height: Flight time: Range: ymax = t= s y0 sin a d2 2g 2y0 sin a g y02 R = g sin 2a . EXAMPLE 2 Investigating Ideal Projectile Motion Find the maximum height, flight time, and range of a projectile fired from the origin over horizontal ground at an initial speed of 500 m> sec and a launch angle of 60 (same projectile as Example 1). Solution Maximum height: ymax = = Flight time: t s y0 sin a d2 2g s 500 sin 60 d2 L 9566 m 2s 9.8 d 2y0 sin a = g = 2s 500 d sin 60 L 88.4 sec 9.8 Range: y 10,000 8000 6000 4000 2000 0 2000 5000 10,000 15,000 20,000 x R y02 = g sin 2a = s 500 d2 sin 120 L 22,092 m 9.8 From Equation (3), the position vector of the projectile is r = s y0 cos a dt i + as y0 sin a dt 12 gt b j 2 1 s 9.8 dt 2 b j 2 = s 500 cos 60 dt i + as 500 sin 60 dt = 250t i + FIGURE 13.10 The graph of the projectile described in Example 2. A A 250 23 B t - 4.9t 2 B j. A graph of the projectiles path is shown in Figure 13.10. Ideal Trajectories Are Parabolic It is often claimed that water from a hose traces a parabola in the air, but anyone who closely looks enough will see this is not so. The air slows the water down, and its forward progress is too slow at the end to keep pace with the rate at which it falls. Copyright 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch13p906-964 8/25/04 2:48 PM Page 924 924 y Chapter 13: Vector-Valued Functions and Motion in Space What is really being claimed is that ideal projectiles move along parabolas, and this we can see from Equations (4). If we substitute t = x> s y0 cos a d from the first equation into the second, we obtain the Cartesian-coordinate equation v0 y = -a g 2y02 cos2 a b x 2 + s tan a dx . (x 0 , y0 ) x This equation has the form y = ax 2 + bx , so its graph is a parabola. 0 Firing from s x0, y0 d If we fire our ideal projectile from the point s x0, y0 d instead of the origin (Figure 13.11), the position vector for the path of motion is r = s x0 + s y0 cos a dt di + ay0 + s y0 sin a dt as you are asked to show in Exercise 19. 12 gt b j, 2 (5) FIGURE 13.11 The path of a projectile fired from s x0 , y0 d with an initial velocity v0 at an angle of a degrees with the horizontal. EXAMPLE 3 Firing a Flaming Arrow To open the 1992 Summer Olympics in Barcelona, bronze medalist archer Antonio Rebollo lit the Olympic torch with a flaming arrow (Figure 13.12). Suppose that Rebollo shot the arrow at a height of 6 ft above ground level 90 ft from the 70-ft-high cauldron, and he wanted the arrow to reach maximum height exactly 4 ft above the center of the cauldron (Figure 13.12). Photograph is not available. FIGURE 13.12 Spanish archer Antonio Rebollo lights the Olympic torch in Barcelona with a flaming arrow. (a) Express ymax in terms of the initial speed y0 and firing angle a . (b) Use ymax = 74 ft (Figure 13.13) and the result from part (a) to find the value of y0 sin a . (c) Find the value of y0 cos a . (d) Find the initial firing angle of the arrow. Copyright 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch13p906-964 8/25/04 2:48 PM Page 925 13.2 Solution v0 y Modeling Projectile Motion 925 ymax 74' (a) We use a coordinate system in which the positive x-axis lies along the ground toward the left (to match the second photograph in Figure 13.12) and the coordinates of the flaming arrow at t = 0 are x0 = 0 and y0 = 6 (Figure 13.13). We have 6' x 90' NOT TO SCALE 0 y = y0 + s y0 sin a dt = 6 + s y0 sin a dt - 12 gt 2 12 gt . 2 Equation (5), j-component y0 = 6 FIGURE 13.13 Ideal path of the arrow that lit the Olympic torch (Example 3). We find the time when the arrow reaches its highest point by setting dy> dt = 0 and solving for t, obtaining t= For this value of t, the value of y is ymax = 6 + s y0 sin a d a =6+ y0 sin a y0 sin a 2 1 g b - 2 ga g b y0 sin a g. s y0 sin a d2 . 2g (b) Using ymax = 74 and g = 32, we see from the preceeding equation in part (a) that y0 sin a = 2s 68 ds 64 d . 74 = 6 + s y0 sin a d2 2s 32 d or (c) When the arrow reaches ymax , the horizontal distance traveled to the center of the cauldron is x = 90 ft. We substitute the time to reach ymax from part (a) and the horizontal distance x = 90 ft into the i-component of Equation (5) to obtain x = x0 + s y0 cos a dt 90 = 0 + s y0 cos a dt = s y0 cos a d a y0 sin a g b. Equation (5), i-component x = 90, x0 = 0 t = s y0 sin a d> g Solving this equation for y0 cos a and using g = 32 and the result from part (b), we have y0 cos a = (d) Parts (b) and (c) together tell us that y0 sin a tan a = y0 cos a = 90g s 90 ds 32 d = . y0 sin a 2s 68 ds 64 d A 2s 68 ds 64 d B 2 s 90 ds 32 d = 68 45 Copyright 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch13p906-964 8/25/04 2:48 PM Page 926 926 Chapter 13: Vector-Valued Functions and Motion in Space or a = tan-1 a This is Rebollos firing angle. 68 b L 56.5. 45 Projectile Motion with Wind Gusts The next example shows how to account for another force acting on a projectile. We also assume that the path of the baseball in Example 4 lies in a vertical plane. EXAMPLE 4 Hitting a Baseball A baseball is hit when it is 3 ft above the ground. It leaves the bat with initial speed of 152 ft> sec, making an angle of 20 with the horizontal. At the instant the ball is hit, an instantaneous gust of wind blows in the horizontal direction directly opposite the direction the ball is taking toward the outfield, adding a component of - 8.8i s ft> sec d to the balls initial velocity s 8.8 ft> sec = 6 mph d . (a) Find a vector equation (position vector) for the path of the baseball. (b) How high does the baseball go, and when does it reach maximum height? (c) Assuming that the ball is not caught, find its range and flight time. Solution (a) Using Equation (1) and accounting for the gust of wind, the initial velocity of the baseball is v0 = s y0 cos a di + s y0 sin a dj - 8.8i = s 152 cos 20 di + s 152 sin 20 dj - s 8.8 di = s 152 cos 20 - 8.8 di + s 152 sin 20 dj. The initial position is r0 = 0i + 3j. Integration of d 2r> dt 2 = - g j gives dr = - s gt dj + v0 . dt A second integration gives r=12 gt j + v0 t + r0 . 2 Substituting the values of v0 and r0 into the last equation gives the position vector of the baseball. r=12 gt j + v0 t + r0 2 = s 152 cos 20 - 8.8 dt i + A 3 + (152 sin 20 dt - 16t 2 B j. = - 16t 2j + s 152 cos 20 - 8.8 dt i + s 152 sin 20 dt j + 3j Copyright 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch13p906-964 8/25/04 2:48 PM Page 927 13.2 Modeling Projectile Motion 927 (b) The baseball reaches its highest point when the vertical component of velocity is zero, or dy = 152 sin 20 - 32t = 0. dt Solving for t we find t= 152 sin 20 L 1.62 sec. 32 Substituting this time into the vertical component for r gives the maximum height ymax = 3 + s 152 sin 20 ds 1.62 d - 16s 1.62 d2 L 45.2 ft. That is, the maximum height of the baseball is about 45.2 ft, reached about 1.6 sec after leaving the bat. (c) To find when the baseball lands, we set the vertical component for r equal to 0 and solve for t: 3 + s 152 sin 20 dt - 16t 2 = 0 3 + s 51.99 dt - 16t 2 = 0. The solution values are about t = 3.3 sec and t = - 0.06 sec. Substituting the positive time into the horizontal component for r, we find the range R = s 152 cos 20 - 8.8 ds 3.3 d L 442 ft. Thus, the horizontal range is about 442 ft, and the flight time is about 3.3 sec. In Exercises 29 through 31, we consider projectile motion when there is air resistance slowing down the flight. Copyright 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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4100 AWL/Thomas_ch13p906-964 8/25/04 2:48 PM Page 92713.2 Modeling Projectile Motion927EXERCISES 13.2Projectile flights in the following exercises are to be treated as ideal unless stated otherwise. All launch angles are assumed to be measured from th
Westminster UT - MATH - 1240
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Westminster UT - MATH - 1240
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Westminster UT - MATH - 1240
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Westminster UT - MATH - 1240
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Westminster UT - MATH - 1240
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Westminster UT - MATH - 1240
4100 AWL/Thomas_ch15p1067-1142 8/25/04 2:57 PM Page 107915.1 Double Integrals1079EXERCISES 15.1Finding Regions of Integration and Double IntegralsIn Exercises 110, sketch the region of integration and evaluate the integral.3 2 3 0 1 4 - 2x21. 23. 2
Westminster UT - MATH - 1240
4100 AWL/Thomas_ch15p1067-1142 8/25/04 2:57 PM Page 108115.2 Area, Moments, and Centers of Mass108115.2Area, Moments, and Centers of MassIn this section, we show how to use double integrals to calculate the areas of bounded regions in the plane and t
Westminster UT - MATH - 1240
4100 AWL/Thomas_ch15p1067-1142 8/25/04 2:57 PM Page 108915.2 Area, Moments, and Centers of Mass1089EXERCISES 15.2Area by Double IntegrationIn Exercises 18, sketch the region bounded by the given lines and curves. Then express the regions area as an i
Westminster UT - MATH - 1240
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Westminster UT - MATH - 1240
4100 AWL/Thomas_ch15p1067-1142 8/25/04 2:57 PM Page 109715.3 Double Integrals in Polar Form1097EXERCISES 15.3Evaluating Polar Integrals1 21 - y 21 - x 2 1In Exercises 116, change the Cartesian integral into an equivalent polar integral. Then evaluat
Westminster UT - MATH - 1240
4100 AWL/Thomas_ch15p1067-1142 8/25/04 2:57 PM Page 10981098Chapter 15: Multiple Integrals15.4Triple Integrals in Rectangular CoordinatesJust as double integrals allow us to deal with more general situations than could be handled by single integrals,
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4100 AWL/Thomas_ch15p1067-1142 8/25/04 2:57 PM Page 11061106Chapter 15: Multiple IntegralsEXERCISES 15.4Evaluating Triple Integrals in Different Iterations1. Evaluate the integral in Example 2 taking Fs x, y, z d = 1 to find the volume of the tetrahe
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4100 AWL/Thomas_ch15p1067-1142 8/25/04 2:57 PM Page 110915.5 Masses and Moments in Three Dimensions110915.5Masses and Moments in Three DimensionsThis section shows how to calculate the masses and moments of three-dimensional objects in Cartesian coor
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4100 AWL/Thomas_ch15p1067-1142 8/25/04 2:57 PM Page 11121112Chapter 15: Multiple IntegralsEXERCISES 15.5Constant DensityThe solids in Exercises 112 all have constant density d = 1. 1. (Example 1 Revisited.) Evaluate the integral for Ix in Table 15.3
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4100 AWL/Thomas_ch15p1067-1142 8/25/04 2:58 PM Page 11141114Chapter 15: Multiple Integrals15.6Triple Integrals in Cylindrical and Spherical CoordinatesWhen a calculation in physics, engineering, or geometry involves a cylinder, cone, or sphere, we ca
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4100 AWL/Thomas_ch15p1067-1142 8/25/04 2:58 PM Page 11241124Chapter 15: Multiple Integrals 12. Let D be the region bounded below by the cone z = 2x 2 + y 2 and above by the paraboloid z = 2 - x 2 - y 2. Set up the triple integrals in cylindrical coordin
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4100 AWL/Thomas_ch15p1067-1142 8/25/04 2:58 PM Page 11281128Chapter 15: Multiple Integrals15.7Substitutions in Multiple IntegralsThis section shows how to evaluate multiple integrals by substitution. As in single integration, the goal of substitution
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4100 AWL/Thomas_ch15p1067-1142 8/25/04 2:58 PM Page 113515.7 Substitutions in Multiple Integrals1135EXERCISES 15.7Finding Jacobians and Transformed Regions for Two Variables1. a. Solve the system u = x - y, y = 2x + y u = x + 2y, for x and y in terms
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4100 AWL/Thomas_ch15p1067-1142 08/27/04 12:43 PM Page 11401140Chapter 15: Multiple IntegralsChapter 15VolumesAdditional and Advanced Exercisesz 2 sin1. Sand pile: double and triple integrals The base of a sand pile covers the region in the xy-plane
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4100 AWL/Thomas_ch15p1067-1142 08/27/04 12:43 PM Page 11381138Chapter 15: Multiple IntegralsChapter 15Practice ExercisesMasses and Moments19. Centroid Find the centroid of the triangular region bounded by the lines x = 2, y = 2 and the hyperbola xy
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4100 AWL/Thomas_ch15p1067-1142 08/27/04 12:42 PM Page 1137Chapter 15 Questions to Guide Your Review1137Chapter 15Questions to Guide Your Review7. How are triple integrals in rectangular coordinates used to calculate volumes, average values, masses, m
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4100 AWL/Thomas_ch15p1067-1142 08/27/04 12:43 PM Page 11421142Chapter 15: Multiple IntegralsChapter 15Technology Application ProjectsMathematica / Maple ModuleTake Your Chances: Try the Monte Carlo Technique for Numerical Integration in Three Dimens
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4100 AWL/Thomas_ch16p1143-12288/27/047:26 AMPage 1143Chapter1616.1Line IntegralsINTEGRATION IN VECTOR FIELDSOVERVIEW This chapter treats integration in vector fields. It is the mathematics that engineers and physicists use to describe fluid flow,
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4100 AWL/Thomas_ch16p1143-12288/27/047:26 AMPage 114716.1 Line Integrals1147EXERCISES 16.1Graphs of Vector EquationsMatch the vector equations in Exercises 18 with the graphs (a)(h) given here. a. b.z 1 1 y x 1 x y x 2 1 1 zc.zd.z (2, 2, 2) 2
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4100 AWL/Thomas_ch16p1143-12288/27/047:26 AMPage 114916.2 Vector Fields, Work, Circulation, and Flux114916.2Vector Fields, Work, Circulation, and FluxWhen we study physical phenomena that are represented by vectors, we replace integrals over close
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4100 AWL/Thomas_ch16p1143-12289/2/0411:18 AMPage 11581158Chapter 16: Integration in Vector FieldsEXERCISES 16.2Vector and Gradient FieldsFind the gradient fields of the functions in Exercises 14. 1. s x, y, z d = s x 2 + y 2 + z 2 d-1&gt;22 2 22. s
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4100 AWL/Thomas_ch16p1143-12289/2/0411:18 AMPage 11601160Chapter 16: Integration in Vector Fields16.3Path Independence, Potential Functions, and Conservative FieldsIn gravitational and electric fields, the amount of work it takes to move a mass or
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4100 AWL/Thomas_ch16p1143-12288/27/047:26 AMPage 11681168Chapter 16: Integration in Vector FieldsEXERCISES 16.3Testing for Conservative FieldsWhich fields in Exercises 16 are conservative, and which are not? 1. F = yz i + xz j + xyk 2. F = s y sin
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4100 AWL/Thomas_ch16p1143-12288/27/047:26 AMPage 116916.4 Greens Theorem in the Plane116916.4Greens Theorem in the PlaneFrom Table 16.2 in Section 16.2, we know that every line integral 1C M dx + N dy can be b written as a flow integral 1a F # T d