2 Motion in Space: Velocity and Acceleration VECTOR FUNCTIONS13.4Motion in Space: Velocity and AccelerationIn this section, we will learn about:The motion of an objectusing tangent and normal vectors.
3 MOTION IN SPACE: VELOCITY AND ACCELERATION 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.
4 VELOCITY AND ACCELERATION In particular, we follow in the footsteps of Newton by using these methods to derive Kepler’s First Law of planetary motion.
5 VELOCITYSuppose a particle moves through space so that its position vector at time t is r(t).
6 VELOCITYVector 1Notice from the figure that, for small values of h, the vector approximates the direction of the particle moving along the curve r(t).
7 VELOCITYIts magnitude measures the size of the displacement vector per unit time.IS THE PREVIOUS IMAGE NEEDED HERE TOO?
8 VELOCITYThe vector 1 gives the average velocity over a time interval of length h.
9 Its limit is the velocity vector v(t) at time t : Equation 2Its limit is the velocity vector v(t) at time t :
10 VELOCITY VECTORThus, the velocity vector is also the tangent vector and points in the direction of the tangent line.
11 SPEEDThe speed of the particle at time t is the magnitude of the velocity vector, that is, |v(t)|.
12 = rate of change of distance with respect to time SPEEDThis 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
13 ACCELERATIONAs 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)
14 VELOCITY & ACCELERATION Example 1The position vector of an object moving in a plane is given by: r(t) = t3 i + t2 jFind its velocity, speed, and acceleration when t = 1 and illustrate geometrically.
15 VELOCITY & ACCELERATION Example 1The velocity and acceleration at time t are:v(t) = r’(t) = 3t2 i + 2t ja(t) = r”(t) = 6t I + 2 j
16 VELOCITY & ACCELERATION Example 1The speed at t is:
17 VELOCITY & ACCELERATION Example 1When t = 1, we have: v(1) = 3 i + 2 j a(1) = 6 i + 2 j |v(1)| =
18 VELOCITY & ACCELERATION Example 1These velocity and acceleration vectors are shown here.
19 VELOCITY & ACCELERATION Example 2Find the velocity, acceleration, and speed of a particle with position vector r(t) = ‹t2, et, tet›
21 VELOCITY & ACCELERATION The figure shows the path of the particle in Example 2 with the velocity and acceleration vectors when t = 1.
22 VELOCITY & ACCELERATION 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.
23 VELOCITY & ACCELERATION Example 3A moving particle starts at an initial position r(0) = ‹1, 0, 0› with initial velocity v(0) = i – j + kIts acceleration is a(t) = 4t i + 6t j + kFind its velocity and position at time t.
24 VELOCITY & ACCELERATION Example 3Since a(t) = v’(t), we have:v(t) = ∫ a(t) dt = ∫ (4t i + 6t j + k) dt=2t2 i + 3t2 j + t k + C
25 VELOCITY & ACCELERATION Example 3To determine the value of the constant vector C, we use the fact that v(0) = i – j + kThe preceding equation gives v(0) = C.So, C = i – j + k
26 VELOCITY & ACCELERATION Example 3It follows:v(t) = 2t2 i + 3t2 j + t k + i – j + k= (2t2 + 1) i + (3t2 – 1) j + (t + 1) k
27 VELOCITY & ACCELERATION Example 3Since v(t) = r’(t), we have:r(t) = ∫ v(t) dt= ∫ [(2t2 + 1) i + (3t2 – 1) j + (t + 1) k] dt= (⅔t3 + t) i + (t3 – t) j + (½t2 + t) k + D
28 VELOCITY & ACCELERATION Example 3Putting t = 0, we find that D = r(0) = i.So, the position at time t is given by:r(t) = (⅔t3 + t + 1) i + (t3 – t) j + (½t2 + t) k
29 VELOCITY & ACCELERATION 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.
30 VELOCITY & ACCELERATION In general, vector integrals allow us to recover:Velocity, when acceleration is knownPosition, when velocity is known
31 VELOCITY & ACCELERATION If the force that acts on a particle is known, then the acceleration can be found from Newton’s Second Law of Motion.
32 VELOCITY & ACCELERATION 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), thenF(t) = ma(t)
33 VELOCITY & ACCELERATION Example 4An 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 jFind the force acting on the object and show that it is directed toward the origin.
34 VELOCITY & ACCELERATION Example 4To find the force, we first need to know the acceleration:v(t) = r’(t) = –aω sin ωt i + aω cos ωt ja(t) = v’(t) = –aω2 cos ωt i – aω2 sin ωt j
35 VELOCITY & ACCELERATION Example 4Therefore, Newton’s Second Law gives the force as:F(t) = ma(t)= –mω2 (a cos ωt i + a sin ωt j)
36 VELOCITY & ACCELERATION Example 4Notice that: F(t) = –mω2r(t)This shows that the force acts in the direction opposite to the radius vector r(t).
37 VELOCITY & ACCELERATION Example 4Therefore, it points toward the origin.
38 Such a force is called a centripetal (center-seeking) force. CENTRIPETAL FORCEExample 4Such a force is called a centripetal (center-seeking) force.
39 VELOCITY & ACCELERATION Example 5A projectile is fired with:Angle of elevation αInitial velocity v0
40 VELOCITY & ACCELERATION Example 5Assuming that air resistance is negligible and the only external force is due to gravity, find the position function r(t) of the projectile.
41 VELOCITY & ACCELERATION Example 5What value of α maximizes the range (the horizontal distance traveled)?
42 VELOCITY & ACCELERATION Example 5We set up the axes so that the projectile starts at the origin.
43 VELOCITY & ACCELERATION Example 5As the force due to gravity acts downward, we have: F = ma = –mg j where g = |a| ≈ 9.8 m/s2.Therefore, a = –g j
44 VELOCITY & ACCELERATION Example 5Since v(t) = a, we have:v(t) = –gt j + Cwhere C = v(0) = v0.Therefore, r’(t) = v(t) = –gt j + v0
45 VELOCITY & ACCELERATION Example 5Integrating again, we obtain:r(t) = –½ gt2 j + t v0 + DHowever, D = r(0) = 0
46 VELOCITY & ACCELERATION E. g. 5—Equation 3So, the position vector of the projectile is given by: r(t) = –½gt2 j + t v0
47 VELOCITY & ACCELERATION Example 5If we write |v0| = v0 (the initial speed of the projectile), thenv0 = v0 cos α i + v0 sin α jEquation 3 becomes: r(t) = (v0 cos α)t i + [(v0 sin α)t – ½gt2] j
48 VELOCITY & ACCELERATION E. g. 5—Equations 4Therefore, the parametric equations of the trajectory are: x = (v0 cos α)t y = (v0 sin α)t – ½gt2
49 VELOCITY & ACCELERATION Example 5If you eliminate t from Equations 4, you will see that y is a quadratic function of x.
50 VELOCITY & ACCELERATION Example 5So, the path of the projectile is part of a parabola.
51 VELOCITY & ACCELERATION Example 5The horizontal distance d is the value of x when y = 0.Setting y = 0, we obtain: t = 0 or t = (2v0 sin α)/g
52 VELOCITY & ACCELERATION Example 5That second value of t then gives:Clearly, d has its maximum value when sin 2α = 1, that is, α = π/4.
53 VELOCITY & ACCELERATION Example 6A 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?
54 VELOCITY & ACCELERATION Example 6If 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.
55 VELOCITY & ACCELERATION Example 6With v0 = 150 m/s, α = 45°, and g = 9.8 m/s2, we have:
56 VELOCITY & ACCELERATION Example 6Impact occurs when y = 0, that is, t2 – t – 10 = 0Solving this quadratic equation (and using only the positive value of t), we get:
57 VELOCITY & ACCELERATION Example 6Then, x ≈ (21.74) ≈ 2306So, the projectile hits the ground about 2,306 m away.
58 VELOCITY & ACCELERATION Example 6The velocity of the projectile is:
59 VELOCITY & ACCELERATION Example 6So, its speed at impact is:
60 ACCELERATION—COMPONENTS 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)
61 ACCELERATION—COMPONENTS If we write v = |v| for the speed of the particle, thenThus, v = vT
62 ACCELERATION—COMPONENTS Equation 5If we differentiate both sides of that equation with respect to t, we get:
63 ACCELERATION—COMPONENTS Equation 6If we use the expression for the curvature given by Equation 9 in Section 13.3, we have:
64 ACCELERATION—COMPONENTS The unit normal vector was defined in Section 13.4 as N = T’/ |T’|So, Equation 6 gives:
66 ACCELERATION—COMPONENTS Equations 8Writing aT and aN for the tangential and normal components of acceleration, we havea = aTT + aNNwhereaT = v’ and aN = Kv2PUT THE CORRECT FONT FOR ‘K.’
67 ACCELERATION—COMPONENTS This resolution is illustrated here.
68 ACCELERATION—COMPONENTS Let’s look at what Formula 7 says.
69 ACCELERATION—COMPONENTS 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.
70 ACCELERATION—COMPONENTS Next, we notice that:The tangential component of acceleration is v’, the rate of change of speed.The normal component of acceleration is ĸv2, the curvature times the square of the speed.PUT THE CORRECT FONT FOR ‘K.’
71 ACCELERATION—COMPONENTS 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.PUT THE CORRECT FONT FOR ‘K.’
72 ACCELERATION—COMPONENTS High speed around the turn has the same effect.In fact, if you double your speed, aN is increased by a factor of 4.
73 ACCELERATION—COMPONENTS 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”.
74 ACCELERATION—COMPONENTS Thus, we take the dot product of v = vT with a as given by Equation 7: v · a = vT · (v’ T + ĸv2N) = vv’ T · T + ĸv3T · N = vv’ (Since T · T = 1 and T · N = 0)PUT THE CORRECT FONT FOR ‘K.’
81 ACCELERATION—COMPONENTS Example 7Hence, Equation 10 gives the normal component as:
82 KEPLER’S LAWS OF PLANETARY MOTION 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.
83 KEPLER’S LAWS OF PLANETARY MOTION 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.
84 KEPLER’S FIRST LAWA planet revolves around the sun in an elliptical orbit with the sun at one focus.
85 KEPLER’S SECOND LAWThe line joining the sun to a planet sweeps out equal areas in equal times.
86 KEPLER’S THIRD LAWThe square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit.
87 KEPLER’S LAWSIn 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 MotionLaw of Universal Gravitation
88 In what follows, we prove Kepler’s First Law. The remaining laws are proved as exercises (with hints).
89 KEPLER’S FIRST LAW—PROOF 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.
90 KEPLER’S FIRST LAW—PROOF We use a coordinate system with the sun at the origin.We let r = r(t) be the position vector of the planet.
91 KEPLER’S FIRST LAW—PROOF Equally well, r could be the position vector of any of:The moonA satellite moving around the earthA comet moving around a star
92 KEPLER’S FIRST LAW—PROOF The velocity vector is: v = r’The acceleration vector is: a = r”
93 KEPLER’S FIRST LAW—PROOF We use the following laws of Newton.Second Law of Motion: F = maLaw of Gravitation:
94 KEPLER’S FIRST LAW—PROOF In the two laws,F is the gravitational force on the planetm and M are the masses of the planet and the sunG is the gravitational constantr = |r|u = (1/r)r is the unit vector in the direction of r
95 KEPLER’S FIRST LAW—PROOF First, we show that the planet moves in one plane.
96 KEPLER’S FIRST LAW—PROOF 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.
97 KEPLER’S FIRST LAW—PROOF We use Formula 5 in Theorem 3 in Section 13.2 to write:
98 KEPLER’S FIRST LAW—PROOF Therefore, r x v = hwhere h is a constant vector.We may assume that h ≠ 0; that is, r and v are not parallel.
99 KEPLER’S FIRST LAW—PROOF 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.
100 KEPLER’S FIRST LAW—PROOF Thus, the orbit of the planet is a plane curve.
101 KEPLER’S FIRST LAW—PROOF To prove Kepler’s First Law, we rewrite the vector h as follows:
102 KEPLER’S FIRST LAW—PROOF Then,(Property 6, Th. 8, Sec. 12.4)
103 KEPLER’S FIRST LAW—PROOF However, u · u = |u|2 = 1Also, |u(t)| = 1It follows from Example 4 in Section that: u · u’ = 0
105 KEPLER’S FIRST LAW—PROOF Equation 11Integrating both sides of that equation, we get: where c is a constant vector.
106 KEPLER’S FIRST LAW—PROOF 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.
107 KEPLER’S FIRST LAW—PROOF As both v x h and u are perpendicular to h, Equation 11 shows that c lies in the xy-plane.
108 KEPLER’S FIRST LAW—PROOF This means that we can choose the x- and y-axes so that the vector i lies in the direction of c.
109 KEPLER’S FIRST LAW—PROOF If θ is the angle between c and r, then (r, θ) are polar coordinates of the planet.
110 KEPLER’S FIRST LAW—PROOF From Equation 11 we have: where c = |c|.
111 KEPLER’S FIRST LAW—PROOF Then, where e = c/(GM).
112 KEPLER’S FIRST LAW—PROOF However,where h = |h|.