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VECTOR FUNCTIONS 13. 13.4 Motion in Space: Velocity and Acceleration In this section, we will learn about: The motion of an object using tangent and normal.

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Presentation on theme: "VECTOR FUNCTIONS 13. 13.4 Motion in Space: Velocity and Acceleration In this section, we will learn about: The motion of an object using tangent and normal."— Presentation transcript:


2 13.4 Motion in Space: Velocity and Acceleration In this section, we will learn about: The motion of an object using tangent and normal vectors. VECTOR FUNCTIONS

3 Here, we show how the ideas of tangent and normal vectors and curvature can be used in physics to study:  The motion of an object, including its velocity and acceleration, along a space curve. MOTION IN SPACE: VELOCITY AND ACCELERATION

4 In particular, we follow in the footsteps of Newton by using these methods to derive Kepler’s First Law of planetary motion. VELOCITY AND ACCELERATION

5 Suppose a particle moves through space so that its position vector at time t is r(t). VELOCITY

6 Notice from the figure that, for small values of h, the vector approximates the direction of the particle moving along the curve r(t). VELOCITY Vector 1

7 Its magnitude measures the size of the displacement vector per unit time. VELOCITY

8 The vector 1 gives the average velocity over a time interval of length h. VELOCITY

9 Its limit is the velocity vector v(t) at time t : VELOCITY VECTOR Equation 2

10 Thus, the velocity vector is also the tangent vector and points in the direction of the tangent line. VELOCITY VECTOR

11 The speed of the particle at time t is the magnitude of the velocity vector, that is, |v(t)|. SPEED

12 This is appropriate because, from Equation 2 and from Equation 7 in Section 13.3, we have: = rate of change of distance with respect to time SPEED

13 As in the case of one-dimensional motion, the acceleration of the particle is defined as the derivative of the velocity: a(t) = v’(t) = r”(t) ACCELERATION

14 The position vector of an object moving in a plane is given by: r(t) = t 3 i + t 2 j  Find its velocity, speed, and acceleration when t = 1 and illustrate geometrically. VELOCITY & ACCELERATION Example 1

15 The velocity and acceleration at time t are: v(t) = r’(t) = 3t 2 i + 2t j a(t) = r”(t) = 6t I + 2 j VELOCITY & ACCELERATION Example 1

16 The speed at t is: VELOCITY & ACCELERATION Example 1

17 When t = 1, we have: v(1) = 3 i + 2 j a(1) = 6 i + 2 j |v(1)| = VELOCITY & ACCELERATION Example 1

18 These velocity and acceleration vectors are shown here. VELOCITY & ACCELERATION Example 1

19 Find the velocity, acceleration, and speed of a particle with position vector r(t) = ‹t 2, e t, te t › VELOCITY & ACCELERATION Example 2


21 The figure shows the path of the particle in Example 2 with the velocity and acceleration vectors when t = 1. VELOCITY & ACCELERATION

22 The vector integrals that were introduced in Section 13.2 can be used to find position vectors when velocity or acceleration vectors are known, as in the next example. VELOCITY & ACCELERATION

23 A moving particle starts at an initial position r(0) = ‹1, 0, 0› with initial velocity v(0) = i – j + k Its acceleration is a(t) = 4t i + 6t j + k  Find its velocity and position at time t. VELOCITY & ACCELERATION Example 3

24 Since a(t) = v’(t), we have: v(t) = ∫ a(t) dt = ∫ (4t i + 6t j + k) dt =2t 2 i + 3t 2 j + t k + C VELOCITY & ACCELERATION Example 3

25 To determine the value of the constant vector C, we use the fact that v(0) = i – j + k  The preceding equation gives v(0) = C.  So, C = i – j + k VELOCITY & ACCELERATION Example 3

26 It follows: v(t) = 2t 2 i + 3t 2 j + t k + i – j + k = (2t 2 + 1) i + (3t 2 – 1) j + (t + 1) k VELOCITY & ACCELERATION Example 3

27 Since v(t) = r’(t), we have: r(t) = ∫ v(t) dt = ∫ [(2t 2 + 1) i + (3t 2 – 1) j + (t + 1) k] dt = (⅔t 3 + t) i + (t 3 – t) j + (½t 2 + t) k + D VELOCITY & ACCELERATION Example 3

28 Putting t = 0, we find that D = r(0) = i. So, the position at time t is given by: r(t) = (⅔t 3 + t + 1) i + (t 3 – t) j + (½t 2 + t) k VELOCITY & ACCELERATION Example 3

29 The expression for r(t) that we obtained in Example 3 was used to plot the path of the particle here for 0 ≤ t ≤ 3. VELOCITY & ACCELERATION

30 In general, vector integrals allow us to recover:  Velocity, when acceleration is known  Position, when velocity is known VELOCITY & ACCELERATION

31 If the force that acts on a particle is known, then the acceleration can be found from Newton’s Second Law of Motion. VELOCITY & ACCELERATION

32 The vector version of this law states that if, at any time t, a force F(t) acts on an object of mass m producing an acceleration a(t), then F(t) = ma(t) VELOCITY & ACCELERATION

33 An object with mass m that moves in a circular path with constant angular speed ω has position vector r(t) = a cos ωt i + a sin ωt j  Find the force acting on the object and show that it is directed toward the origin. VELOCITY & ACCELERATION Example 4

34 To find the force, we first need to know the acceleration: v(t) = r’(t) = –aω sin ωt i + aω cos ωt j a(t) = v’(t) = –aω 2 cos ωt i – aω 2 sin ωt j VELOCITY & ACCELERATION Example 4

35 Therefore, Newton’s Second Law gives the force as: F(t) = ma(t) = –mω 2 (a cos ωt i + a sin ωt j) VELOCITY & ACCELERATION Example 4

36 Notice that: F(t) = –mω 2 r(t)  This shows that the force acts in the direction opposite to the radius vector r(t). VELOCITY & ACCELERATION Example 4

37 Therefore, it points toward the origin. VELOCITY & ACCELERATION Example 4

38 Such a force is called a centripetal (center-seeking) force. CENTRIPETAL FORCE Example 4

39 A projectile is fired with:  Angle of elevation α  Initial velocity v 0 VELOCITY & ACCELERATION Example 5

40 Assuming that air resistance is negligible and the only external force is due to gravity, find the position function r(t) of the projectile. VELOCITY & ACCELERATION Example 5

41 What value of α maximizes the range (the horizontal distance traveled)? VELOCITY & ACCELERATION Example 5

42 We set up the axes so that the projectile starts at the origin. VELOCITY & ACCELERATION Example 5

43 As the force due to gravity acts downward, we have: F = ma = –mg j where g = |a| ≈ 9.8 m/s 2.  Therefore, a = –g j VELOCITY & ACCELERATION Example 5

44 Since v(t) = a, we have: v(t) = –gt j + C where C = v(0) = v 0.  Therefore, r’(t) = v(t) = –gt j + v 0 VELOCITY & ACCELERATION Example 5

45 Integrating again, we obtain: r(t) = –½ gt 2 j + t v 0 + D  However, D = r(0) = 0 VELOCITY & ACCELERATION Example 5

46 So, the position vector of the projectile is given by: r(t) = –½gt 2 j + t v 0 VELOCITY & ACCELERATION E. g. 5—Equation 3

47 If we write |v 0 | = v 0 (the initial speed of the projectile), then  v 0 = v 0 cos α i + v 0 sin α j  Equation 3 becomes: r(t) = (v 0 cos α)t i + [(v 0 sin α)t – ½gt 2 ] j VELOCITY & ACCELERATION Example 5

48 Therefore, the parametric equations of the trajectory are: x = (v 0 cos α)t y = (v 0 sin α)t – ½gt 2 VELOCITY & ACCELERATION E. g. 5—Equations 4

49 If you eliminate t from Equations 4, you will see that y is a quadratic function of x. VELOCITY & ACCELERATION Example 5

50 So, the path of the projectile is part of a parabola. VELOCITY & ACCELERATION Example 5

51 The horizontal distance d is the value of x when y = 0.  Setting y = 0, we obtain: t = 0 or t = (2v 0 sin α)/g VELOCITY & ACCELERATION Example 5

52 That second value of t then gives:  Clearly, d has its maximum value when sin 2α = 1, that is, α = π/4. VELOCITY & ACCELERATION Example 5

53 A projectile is fired with muzzle speed 150 m/s and angle of elevation 45° from a position 10 m above ground level.  Where does the projectile hit the ground?  With what speed does it do so? VELOCITY & ACCELERATION Example 6

54 If we place the origin at ground level, the initial position of the projectile is (0, 10).  So, we need to adjust Equations 4 by adding 10 to the expression for y. VELOCITY & ACCELERATION Example 6

55 With v 0 = 150 m/s, α = 45°, and g = 9.8 m/s 2, we have: VELOCITY & ACCELERATION Example 6

56 Impact occurs when y = 0, that is, 4.9t 2 – 75 t – 10 = 0  Solving this quadratic equation (and using only the positive value of t), we get: VELOCITY & ACCELERATION Example 6

57 Then, x ≈ 75 (21.74) ≈ 2306  So, the projectile hits the ground about 2,306 m away. VELOCITY & ACCELERATION Example 6

58 The velocity of the projectile is: VELOCITY & ACCELERATION Example 6

59 So, its speed at impact is: VELOCITY & ACCELERATION Example 6

60 When we study the motion of a particle, it is often useful to resolve the acceleration into two components:  Tangential (in the direction of the tangent)  Normal (in the direction of the normal) ACCELERATION—COMPONENTS

61 If we write v = |v| for the speed of the particle, then  Thus, v = vT ACCELERATION—COMPONENTS

62 If we differentiate both sides of that equation with respect to t, we get: ACCELERATION—COMPONENTS Equation 5

63 If we use the expression for the curvature given by Equation 9 in Section 13.3, we have: ACCELERATION—COMPONENTS Equation 6

64 The unit normal vector was defined in Section 13.4 as N = T’/ |T’|  So, Equation 6 gives: ACCELERATION—COMPONENTS

65 Then, Equation 5 becomes: ACCELERATION—COMPONENTS Formula/Equation 7

66 Writing a T and a N for the tangential and normal components of acceleration, we have a = a T T + a N N where a T = v’ and a N = Kv 2 ACCELERATION—COMPONENTS Equations 8

67 This resolution is illustrated here. ACCELERATION—COMPONENTS

68 Let’s look at what Formula 7 says. ACCELERATION—COMPONENTS

69 The first thing to notice is that the binormal vector B is absent.  No matter how an object moves through space, its acceleration always lies in the plane of T and N (the osculating plane).  Recall that T gives the direction of motion and N points in the direction the curve is turning. ACCELERATION—COMPONENTS

70 Next, we notice that:  The tangential component of acceleration is v’, the rate of change of speed.  The normal component of acceleration is ĸ v 2, the curvature times the square of the speed. ACCELERATION—COMPONENTS

71 This makes sense if we think of a passenger in a car.  A sharp turn in a road means a large value of the curvature ĸ.  So, the component of the acceleration perpendicular to the motion is large and the passenger is thrown against a car door. ACCELERATION—COMPONENTS

72 High speed around the turn has the same effect.  In fact, if you double your speed, a N is increased by a factor of 4. ACCELERATION—COMPONENTS

73 We have expressions for the tangential and normal components of acceleration in Equations 8. However, it’s desirable to have expressions that depend only on r, r’, and r”. ACCELERATION—COMPONENTS

74 Thus, we take the dot product of v = vT with a as given by Equation 7: v · a = vT · (v’ T + ĸ v 2 N) = vv’ T · T + ĸ v 3 T · N = vv’ (Since T · T = 1 and T · N = 0) ACCELERATION—COMPONENTS

75 Therefore, ACCELERATION—COMPONENTS Equation 9

76 Using the formula for curvature given by Theorem 10 in Section 13.3, we have: ACCELERATION—COMPONENTS Equation 10

77 A particle moves with position function r(t) = ‹t 2, t 2, t 3 ›  Find the tangential and normal components of acceleration. ACCELERATION—COMPONENTS Example 7


79 Therefore, Equation 9 gives the tangential component as: ACCELERATION—COMPONENTS Example 7


81 Hence, Equation 10 gives the normal component as: ACCELERATION—COMPONENTS Example 7

82 We now describe one of the great accomplishments of calculus by showing how the material of this chapter can be used to prove Kepler’s laws of planetary motion. KEPLER’S LAWS OF PLANETARY MOTION

83 After 20 years of studying the astronomical observations of the Danish astronomer Tycho Brahe, the German mathematician and astronomer Johannes Kepler (1571–1630) formulated the following three laws. KEPLER’S LAWS OF PLANETARY MOTION

84 A planet revolves around the sun in an elliptical orbit with the sun at one focus. KEPLER’S FIRST LAW

85 The line joining the sun to a planet sweeps out equal areas in equal times. KEPLER’S SECOND LAW

86 The square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit. KEPLER’S THIRD LAW

87 In his book Principia Mathematica of 1687, Sir Isaac Newton was able to show that these three laws are consequences of two of his own laws:  Second Law of Motion  Law of Universal Gravitation KEPLER’S LAWS

88 In what follows, we prove Kepler’s First Law.  The remaining laws are proved as exercises (with hints). KEPLER’S FIRST LAW

89 The gravitational force of the sun on a planet is so much larger than the forces exerted by other celestial bodies.  Thus, we can safely ignore all bodies in the universe except the sun and one planet revolving about it. KEPLER’S FIRST LAW—PROOF

90 We use a coordinate system with the sun at the origin. We let r = r(t) be the position vector of the planet. KEPLER’S FIRST LAW—PROOF

91 Equally well, r could be the position vector of any of:  The moon  A satellite moving around the earth  A comet moving around a star KEPLER’S FIRST LAW—PROOF

92 The velocity vector is: v = r’ The acceleration vector is: a = r” KEPLER’S FIRST LAW—PROOF

93 We use the following laws of Newton. Second Law of Motion: F = ma Law of Gravitation: KEPLER’S FIRST LAW—PROOF

94 In the two laws,  F is the gravitational force on the planet  m and M are the masses of the planet and the sun  G is the gravitational constant  r = |r|  u = (1/r)r is the unit vector in the direction of r KEPLER’S FIRST LAW—PROOF

95 First, we show that the planet moves in one plane. KEPLER’S FIRST LAW—PROOF

96 By equating the expressions for F in Newton’s two laws, we find that:  So, a is parallel to r.  It follows that r x a = 0. KEPLER’S FIRST LAW—PROOF

97 We use Formula 5 in Theorem 3 in Section 13.2 to write: KEPLER’S FIRST LAW—PROOF

98 Therefore, r x v = h where h is a constant vector.  We may assume that h ≠ 0; that is, r and v are not parallel. KEPLER’S FIRST LAW—PROOF

99 This means that the vector r = r(t) is perpendicular to h for all values of t.  So, the planet always lies in the plane through the origin perpendicular to h. KEPLER’S FIRST LAW—PROOF

100 Thus, the orbit of the planet is a plane curve. KEPLER’S FIRST LAW—PROOF

101 To prove Kepler’s First Law, we rewrite the vector h as follows: KEPLER’S FIRST LAW—PROOF

102 Then, (Property 6, Th. 8, Sec. 12.4) KEPLER’S FIRST LAW—PROOF

103 However, u · u = |u| 2 = 1 Also, |u(t)| = 1  It follows from Example 4 in Section 13.2 that: u · u’ = 0 KEPLER’S FIRST LAW—PROOF

104 Therefore, Thus, KEPLER’S FIRST LAW—PROOF

105 Integrating both sides of that equation, we get: where c is a constant vector. KEPLER’S FIRST LAW—PROOF Equation 11

106 At this point, it is convenient to choose the coordinate axes so that the standard basis vector k points in the direction of the vector h.  Then, the planet moves in the xy-plane. KEPLER’S FIRST LAW—PROOF

107 As both v x h and u are perpendicular to h, Equation 11 shows that c lies in the xy-plane. KEPLER’S FIRST LAW—PROOF

108 This means that we can choose the x- and y-axes so that the vector i lies in the direction of c. KEPLER’S FIRST LAW—PROOF

109 If θ is the angle between c and r, then (r, θ) are polar coordinates of the planet. KEPLER’S FIRST LAW—PROOF

110 From Equation 11 we have: where c = |c|. KEPLER’S FIRST LAW—PROOF

111 Then, where e = c/(GM). KEPLER’S FIRST LAW—PROOF

112 However, where h = |h|. KEPLER’S FIRST LAW—PROOF


114 Writing d = h 2 /c, we obtain: KEPLER’S FIRST LAW—PROOF Equation 12

115 Comparing with Theorem 6 in Section 10.6, we see that Equation 12 is the polar equation of a conic section with:  Focus at the origin  Eccentricity e KEPLER’S FIRST LAW—PROOF

116 We know that the orbit of a planet is a closed curve.  Hence, the conic must be an ellipse. KEPLER’S FIRST LAW—PROOF

117 This completes the derivation of Kepler’s First Law. KEPLER’S FIRST LAW—PROOF

118 The proofs of the three laws show that the methods of this chapter provide a powerful tool for describing some of the laws of nature. KEPLER’S LAWS

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