Download Velocity and Acceleration

13
VECTOR FUNCTIONS
VECTOR FUNCTIONS
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.
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.
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.
VELOCITY
Suppose a particle moves through
space so that its position vector at
time t is r(t).
VELOCITY
Vector 1
Notice from the figure that, for small values
of h, the vector
r (t  h)  r (t )
h
approximates
the direction of the
particle moving along
the curve r(t).
VELOCITY
Its magnitude measures the size
of the displacement vector per unit
time.
VELOCITY
The vector 1 gives the average
velocity over a time interval of
length h.
VELOCITY VECTOR
Equation 2
Its limit is the velocity vector v(t)
at time t :
r (t  h)  r(t )
v(t )  lim
h 0
h
 r '(t )
VELOCITY VECTOR
Thus, the velocity vector is also
the tangent vector and points in
the direction of the tangent line.
SPEED
The speed of the particle at time t
is the magnitude of the velocity vector,
that is, |v(t)|.
SPEED
This is appropriate because, from Equation 2
and from Equation 7 in Section 13.3,
we have:
ds
| v(t ) || r '(t ) |
dt
= rate of change
of distance with
respect to time
ACCELERATION
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)
VELOCITY & ACCELERATION
Example 1
The position vector of an object moving
in a plane is given by:
r(t) = t3 i + t2 j
 Find its velocity, speed, and acceleration
when t = 1 and illustrate geometrically.
VELOCITY & ACCELERATION
Example 1
The velocity and acceleration at time t
are:
v(t) = r’(t) = 3t2 i + 2t j
a(t) = r”(t) = 6t I + 2 j
VELOCITY & ACCELERATION
Example 1
The speed at t is:
| v(t ) | (3t )  (2t )
2 2
 9t  4t
4
2
2
VELOCITY & ACCELERATION
Example 1
When t = 1, we have:
v(1) = 3 i + 2 j
a(1) = 6 i + 2 j
|v(1)| = 13
VELOCITY & ACCELERATION
Example 1
These velocity and acceleration vectors
are shown here.
VELOCITY & ACCELERATION
Example 2
Find the velocity, acceleration, and speed
of a particle with position vector
r(t) = ‹t2, et, tet›
Example 2
VELOCITY & ACCELERATION
v(t )  r '(t )  2t , e , (1  t )e
t
a(t )  v '(t )  2, e , (2  t )e
t
t
t
| v(t ) | 4t  e  (1  t ) e
2
2t
2
2t
VELOCITY & ACCELERATION
The figure shows the path of the particle in
Example 2 with the velocity and acceleration
vectors when t = 1.
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.
VELOCITY & ACCELERATION
Example 3
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
Since a(t) = v’(t), we have:
v(t) = ∫ a(t) dt
= ∫ (4t i + 6t j + k) dt
=2t2 i + 3t2 j + t k + C
VELOCITY & ACCELERATION
Example 3
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
It follows:
v(t) = 2t2 i + 3t2 j + t k + i – j + k
= (2t2 + 1) i + (3t2 – 1) j + (t + 1) k
VELOCITY & ACCELERATION
Example 3
Since 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
VELOCITY & ACCELERATION
Example 3
Putting 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
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.
VELOCITY & ACCELERATION
In general, vector integrals allow us
to recover:
 Velocity, when acceleration is known
t
v(t )  v(t0 )   a(u) du
t0
 Position, when velocity is known
t
r (t )  r (t0 )   v(u ) du
t0
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.
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),
then
F(t) = ma(t)
VELOCITY & ACCELERATION
Example 4
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
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) = –aω2 cos ωt i – aω2 sin ωt j
VELOCITY & ACCELERATION
Example 4
Therefore, Newton’s Second Law gives
the force as:
F(t) = ma(t)
= –mω2 (a cos ωt i + a sin ωt j)
VELOCITY & ACCELERATION
Example 4
Notice that:
F(t) = –mω2r(t)
 This shows that the force acts in the direction
opposite to the radius vector r(t).
VELOCITY & ACCELERATION
Example 4
Therefore, it points toward
the origin.
CENTRIPETAL FORCE
Example 4
Such a force is called a centripetal
(center-seeking) force.
VELOCITY & ACCELERATION
Example 5
A projectile is fired with:
 Angle of elevation α
 Initial velocity v0
VELOCITY & ACCELERATION
Example 5
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
What value of α maximizes the range
(the horizontal distance traveled)?
VELOCITY & ACCELERATION
Example 5
We set up the axes so that the projectile
starts at the origin.
VELOCITY & ACCELERATION
Example 5
As the force due to gravity acts downward,
we have:
F = ma = –mg j
where g = |a| ≈ 9.8 m/s2.
 Therefore, a = –g j
VELOCITY & ACCELERATION
Example 5
Since v(t) = a, we have:
v(t) = –gt j + C
where C = v(0) = v0.
 Therefore,
r’(t) = v(t) = –gt j + v0
VELOCITY & ACCELERATION
Example 5
Integrating again, we obtain:
r(t) = –½ gt2 j + t v0 + D
 However,
D = r(0) = 0
VELOCITY & ACCELERATION
E. g. 5—Equation 3
So, the position vector of the projectile
is given by:
r(t) = –½gt2 j + t v0
VELOCITY & ACCELERATION
Example 5
If we write |v0| = v0 (the initial speed
of the projectile), then
 v0 = v0 cos α i + v0 sin α j
 Equation 3 becomes:
r(t) = (v0 cos α)t i + [(v0 sin α)t – ½gt2] j
VELOCITY & ACCELERATION
E. g. 5—Equations 4
Therefore, the parametric equations
of the trajectory are:
x = (v0 cos α)t
y = (v0 sin α)t – ½gt2
VELOCITY & ACCELERATION
Example 5
If you eliminate t from Equations 4,
you will see that y is a quadratic function
of x.
VELOCITY & ACCELERATION
Example 5
So, the path of the projectile is part
of a parabola.
VELOCITY & ACCELERATION
Example 5
The horizontal distance d is the value
of x when y = 0.
 Setting y = 0, we obtain:
t = 0 or t = (2v0 sin α)/g
VELOCITY & ACCELERATION
Example 5
That second value of t then gives:
2v0 sin 
d  x  (v0 cos  )
g
v (2sin  cos  ) v sin 2


g
g
2
0
2
0
 Clearly, d has its maximum value when
sin 2α = 1, that is, α = π/4.
VELOCITY & ACCELERATION
Example 6
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
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.
Example 6
VELOCITY & ACCELERATION
With v0 = 150 m/s, α = 45°, and g = 9.8 m/s2,
we have:
x  150 cos( / 4)t  75 2 t
y  10  150sin( / 4)t  (9.8)t
1
2
 10  75 2 t  4.9t
2
2
VELOCITY & ACCELERATION
Example 6
Impact occurs when y = 0, that is,
4.9t2 – 75 2 t – 10 = 0
 Solving this quadratic equation (and using
only the positive value of t), we get:
75 2  11, 250  196
t
 21.74
9.8
VELOCITY & ACCELERATION
Example 6
Then,
x ≈ 75 2 (21.74)
≈ 2306
 So, the projectile hits the ground
about 2,306 m away.
VELOCITY & ACCELERATION
Example 6
The velocity of the projectile is:
v(t )  r '(t )
 75 2 i  (75 2  9.8t ) j
VELOCITY & ACCELERATION
Example 6
So, its speed at impact is:
| v(21.74) | (75 2)  (75 2  9.8  21.74)
2
 151m/s
2
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)
ACCELERATION—COMPONENTS
If we write v = |v| for the speed of the particle,
then
r '(t )
v(t )
v
T(t ) 


| r '(t ) | | v(t ) | v
 Thus,
v = vT
ACCELERATION—COMPONENTS
Equation 5
If we differentiate both sides of that
equation with respect to t, we get:
a  v '  v ' T  vT '
ACCELERATION—COMPONENTS
Equation 6
If we use the expression for the curvature
given by Equation 9 in Section 13.3,
we have:
| T' | | T' |


so | T ' |  v
|r'|
v
ACCELERATION—COMPONENTS
The unit normal vector was defined
in Section 13.4 as N = T’/ |T’|
 So, Equation 6 gives:
T ' | T ' | N   vN
ACCELERATION—COMPONENTS
Formula/Equation 7
Then, Equation 5 becomes:
a  v 'T  v N
2
ACCELERATION—COMPONENTS
Equations 8
Writing aT and aN for the tangential and
normal components of acceleration,
we have
a = aTT + aNN
where
aT = v’
and
aN = Kv2
ACCELERATION—COMPONENTS
This resolution is illustrated
here.
ACCELERATION—COMPONENTS
Let’s look at what Formula 7
says.
a  v 'T  v N
2
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.
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.
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.
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.
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”.
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)
ACCELERATION—COMPONENTS
Equation 9
Therefore,
v a
aT  v ' 
v
r '(t )  r "(t )

| r '(t ) |
ACCELERATION—COMPONENTS
Equation 10
Using the formula for curvature given by
Theorem 10 in Section 13.3, we have:
| r '(t )  r "(t ) |
2
aN   v 
| r '(t ) |
3
| r '(t ) |
| r '(t )  r "(t ) |

| r '(t ) |
2
ACCELERATION—COMPONENTS
Example 7
A particle moves with position function
r(t) = ‹t2, t2, t3›
 Find the tangential and normal
components of acceleration.
ACCELERATION—COMPONENTS
2
2
3
r (t ) = t i + t j + t k
2
r '(t ) = 2t i + 2t j + 3t k
r "(t ) = 2 i + 2 j + 6t k
| r'(t ) |
8t  9t
2
4
Example 7
Example 7
ACCELERATION—COMPONENTS
Therefore, Equation 9 gives the tangential
component as:
r '(t )  r "(t )
aT 
| r '(t ) |

8t  18t
3
8t  9t
2
4
ACCELERATION—COMPONENTS
j
i
r '(t )  r "(t )  2t
2
k
2t 3t
2
6t
 6t i  6t j
2
2
2
Example 7
Example 7
ACCELERATION—COMPONENTS
Hence, Equation 10 gives the normal
component as:
r '(t )  r "(t )
aN 
| r '(t ) |

6 2t
2
8t  9t
2
4
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.
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.
KEPLER’S FIRST LAW
A planet revolves around the sun
in an elliptical orbit with the sun at
one focus.
KEPLER’S SECOND LAW
The line joining the sun to
a planet sweeps out equal areas
in equal times.
KEPLER’S THIRD LAW
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 LAWS
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 FIRST LAW
In what follows, we prove Kepler’s
First Law.
 The remaining laws are proved
as exercises (with hints).
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.
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.
KEPLER’S FIRST LAW—PROOF
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
The velocity vector is:
v = r’
The acceleration vector is:
a = r”
KEPLER’S FIRST LAW—PROOF
We use the following laws of Newton.
Second Law of Motion:
F = ma
Law of Gravitation:
GMm
F 3 r
r
GMm
 2 u
r
KEPLER’S FIRST LAW—PROOF
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
First, we show that
the planet moves in
one plane.
KEPLER’S FIRST LAW—PROOF
By equating the expressions for F in
Newton’s two laws, we find that:
GM
a 3 r
r
 So, a is parallel to r.
 It follows that r x a = 0.
KEPLER’S FIRST LAW—PROOF
We use Formula 5 in Theorem 3 in
Section 13.2 to write:
d
(r  v)  r ' v  r  v '
dt
 v v  r a
 00
0
KEPLER’S FIRST LAW—PROOF
Therefore,
rxv=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
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
Thus, the orbit of the planet is
a plane curve.
KEPLER’S FIRST LAW—PROOF
To prove Kepler’s First Law, we rewrite
the vector h as follows:
h  r v  rr '
 r u  ( r u) '
 r u  (ru ' r ' u)
 r (u  u ')  rr '(u  u)
2
 r (u  u ')
2
KEPLER’S FIRST LAW—PROOF
Then,
GM
a  h   2 u  (r 2u  u ')
r
 GM u  (u  u ')
 GM  (u  u ')u  (u  u)u '
(Property 6,
Th. 8, Sec. 12.4)
KEPLER’S FIRST LAW—PROOF
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
Therefore,
a  h  GM u '
Thus,
( v  h) '  v ' h  a  h  GM u '
KEPLER’S FIRST LAW—PROOF
Equation 11
Integrating both sides of that equation,
we get:
v  h  GM u  c
where c is a constant vector.
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.
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.
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.
KEPLER’S FIRST LAW—PROOF
If θ is the angle between c and r,
then (r, θ) are polar coordinates of
the planet.
KEPLER’S FIRST LAW—PROOF
From Equation 11 we have:
r  ( v  h)  r  (GM u  c)
 GM r  u  r  c
 GMr u  u  | r || c | cos 
 GMr  rc cos 
where c = |c|.
KEPLER’S FIRST LAW—PROOF
Then,
r  ( v  h)
r
GM  c cos 
1 r  ( v  h)

GM 1  e cos 
where e = c/(GM).
KEPLER’S FIRST LAW—PROOF
However,
r  ( v  h)  (r  v)  h
 h h
| h |
2
h
where h = |h|.
2
KEPLER’S FIRST LAW—PROOF
Thus,
2
h /(GM )
r
1  e cos 
2
eh / c

1  e cos 
KEPLER’S FIRST LAW—PROOF
Equation 12
Writing d = h2/c, we obtain:
ed
r
1  e cos 
KEPLER’S FIRST LAW—PROOF
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
We know that the orbit of a planet is
a closed curve.
 Hence, the conic must be an ellipse.
KEPLER’S FIRST LAW—PROOF
This completes the derivation
of Kepler’s First Law.
KEPLER’S LAWS
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.