VECTOR FUNCTIONS

VECTOR FUNCTIONS

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:
2 2 2

| v(t ) |

(3t ) 9t
4

(2t )
2

4t

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›

VELOCITY & ACCELERATION

Example 2 t v(t ) r '(t )

2t , e , (1 t )e t t

a(t )

v '(t )
2

2, e , (2 t )e
2t 2

t

| v(t ) |

4t

e

(1 t ) e

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

v(t )

v(t0 )

t t0

a(u ) du

 Position, when velocity is known

r (t ) r (t0 )

t t0

v(u ) du

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:

d

2v0 sin x (v0 cos ) g v (2sin cos ) g
2 0

v sin 2 g

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.

VELOCITY & ACCELERATION

Example 6

With v0 = 150 m/s, α = 45 , and g = 9.8 m/s2,

we have:

x 150 cos( / 4)t

75 2 t
1 2

y 10 150sin( / 4)t 10 75 2 t 4.9t
2

(9.8)t

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:

t

75 2

11, 250 196 9.8

21.74

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) 151m/s

2

(75 2 9.8 21.74)

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

T(t )

r '(t ) | r '(t ) |

v(t ) | v(t ) |

v 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' | |r'|

| T' | so | T ' | v

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

vN

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

vN

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,

aT

v a 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:

aN

v

2

| r '(t ) r "(t ) | 2 | r '(t ) | 3 | r '(t ) | | r '(t ) r "(t ) | | r '(t ) |

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

Example 7

r (t ) = t i + t j + t k r '(t ) = 2t i + 2t j + 3t k r "(t ) = 2 i + 2 j + 6t k | r'(t ) | 8t
2 2

2

2

3

9t

4

ACCELERATION—COMPONENTS

Example 7

Therefore, Equation 9 gives the tangential

component as:

aT

r '(t ) r "(t ) | r '(t ) | 8t 18t 8t
2 3 4

9t

ACCELERATION—COMPONENTS

Example 7

r '(t ) r "(t )

i j k 2 2t 2t 3t 2 2 6t 6t i 6t j
2 2

ACCELERATION—COMPONENTS

Example 7

Hence, Equation 10 gives the normal

component as:

aN

r '(t ) r "(t ) | r '(t ) | 6 2t 8t
2 2 4

9t

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:

F

GMm r 3 r GMm u 2 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:

a
 So, a is parallel to r.  It follows that r x a = 0.

GM r 3 r

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) r (u u ')
2 2

KEPLER’S FIRST LAW—PROOF

Then, a h GM 2 u (r u u ') 2 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| h where h = |h|.
2 2

KEPLER’S FIRST LAW—PROOF

Thus,

r

h /(GM ) 1 e cos 2 eh / c 1 e cos

2

KEPLER’S FIRST LAW—PROOF

Equation 12

Writing d = h2/c, we obtain:

r

ed 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.