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Vector-Valued Functions and
Motion in Space
Dr. Ching I Chen
12.1 Vector-Valued Functions and Space Curves (1)
Space Curve
When a particle moves through space during a time interval  ,
one thinks of the particle's coordinates as functions defined on  :
x  f (t ), y  g (t ), z  h (t ) t  
P (f (t ), g (t ), h (t ))
z
parametric vector :
r(t)
y
O
x
curve
P : position
r (t ) : position vector
r (t )  OP  f (t )i  g (t ) j  h (t )k
12.1 Vector-Valued Functions and Space Curves (2)
Space Curve (Example 1)
Example 1 GRAPHING A HELIX
Graph the function r (t )  cos t i  sin t j  t k
40
30
20
10
0
1
1
0
0
-1 -1
12.1 Vector-Valued Functions and Space Curves (3)
Space Curve
Figure 12.4 GRAPHING A HELIX
Graph the function r (t )  2 cos 3t i  e t j  2 sin3t k
2
0
-2
5
10
15
20
-2
0
2
12.1 Vector-Valued Functions and Space Curves (4)
Space Curve (Exploration 1-1~4)
Exploration 1 How Vector - Valued Function in Space Ought
to Work
1. If r (t )  (t 2  3)i   cos t  j   sin t / t
r(t )?
 k, What is lim
t 0
2. Is g t   e 3t 1  i   t  j   tan2 t  k continuous at t  0?
Is it differentiable there?
3. Find dr/dt if r t   t 3  i  ln t  3   j   tan t  k.
4. Find

4
0
r t  dt if r t   t 2  i   sin t  j 
 t  k.
12.1 Vector-Valued Functions and Space Curves (5)
Space Curve (Exploration 1-5~7)
Exploration 1 How Vector - Valued Function in Space Ought
to Work
5. A particle moves through space with position vector at time
t given by r (t )  (2t 3 )i   sint  j  e 2t  k, Find the velocity
vector?
6. Find the acceleration vector for the partical in part  5  .
7. Find the speed and direction of motion of the partical in  5 
at time t  0.
12.1 Vector-Valued Functions and Space Curves (6)
Space Curve (Exploration 1-8~10)
Exploration 1 How Vector - Valued Function in Space Ought
to Work
8. If u and v are vectors in three dimensional space, what is the
rule for finding (d / dt )(u v )?
9. If r is a differentiable function of t and t is a differentiable
function of s, what would the Chain Rule say about dr/ds?
10.We did not consider the cross product of two-dimensional
vectors, so we have no two dimensional formula for
 d/dt  u  v . Nonetheless, what do you suppose the
three-dimentional formula for the derivative of a cross
product would be?
12.1 Vector-Valued Functions and Space Curves (7)
Limit and Continuity
Definition Limit
Let r t   f t  i  g t  j  h t  k be a vector function and L a vector.
We say that r has limit L as t approaches t 0 and write
lim r t   L
t t 0
if  for every number   0there exists a corresponding   0 such
that for all t
0  t - t 0    r t   L  .
12.1 Vector-Valued Functions and Space Curves (8)
Limit and Continuity (Example 2)
Example 2 FINDING LIMITS OF VECTOR FUNCTIONS
If r (t )  cos t i  sint j  t k, then

lim r (t )  lim cos t
t  / 4
t  / 4
 i   lim sint  j   lim t  k
t  / 4
 2   2 


i  
 j k
4
 2   2 
t  / 4
12.1 Vector-Valued Functions and Space Curves (9)
Limit and Continuity
Definition continuity at a point
A vector function r t  is continuity at a poi nt t  t 0 in its
domain if
lim r t   r t 0  .
t →t 0
12.1 Vector-Valued Functions and Space Curves (10)
Limit and Continuity
Component Test for continuity at a point
The vector function r t   f t  i  g t  j  h t  k is continuous
at t  t 0 if and only if f g  and h are continuous at t 0
12.1 Vector-Valued Functions and Space Curves (11)
Limit and Continuity (Example 3)
Example 3 FINDING POINTS OF CONTINUITY
AND DISCONTINUITY
(a) The function
r (t )  cos t i  sint j  t k
is continuous because cos t ,sin t ,and t are continuity.
(b) The function
g(t )  cos t i  sint j   int t  k
is discontinuous at every integer.
12.1 Vector-Valued Functions and Space Curves (12)
Derivatives and Motion on Smooth Curves
Suppose that r(t) = f(t) i + g(t) j + h(t) k is the position of a
particle moving along a curve in the plane and that f(t), g(t)
and h(t) are differentiable functions of t. Then the difference
between the particle’s positions at time t+Dt and the time t is
Dr (t )  r (t  Dt )  r (t )
 f (t  Dt ) i  g (t  Dt ) j  h (t  Dt ) k   f (t ) i  g (t ) j  h (t ) k 
 f (t  Dt )  f (t ) i   g (t  Dt )  g (t ) j  h (t  Dt )  h (t ) k
z
y
r(t+Dt)
r(t)
x
Q
Drt)
P
12.1 Vector-Valued Functions and Space Curves (13)
Derivatives and Motion on Smooth Curves
Definition Derivative at a point
The vector function r t   f t  i  g t  j  h t  k is differentiable
at t  t 0 if f , g ,and h are differentiable at t 0 . The derivative is
the vector
r t  Dt   r t  df
dr
dg dh
 lim

i
j
k
dt Dt 0
Dt
dt
dt
dt
12.1 Vector-Valued Functions and Space Curves (14)
Derivatives and Motion on Smooth Curves
Definitions Velocity, Speed, Acceleration, Direction of Motion
If r is the position vector of a particle moving along a smooth curve in
space, then at any time t .
1. v t  =
dr
 the derivative of position is the particles velocity vector
dt
and is tangent to the curve.
2. v t   the magnitude of v is the particles speed.
dv d 2r
3. a t  
 2  the derivative of velocity and the second derivative.
dt dt
of positionis the particles acceleration vector.
v
4.    a unit vector  is the direction of motion.
v
12.1 Vector-Valued Functions and Space Curves (15)
Derivatives and Motion on Smooth Curves
v
Velocity  v     speed  direction 
v
12.1 Vector-Valued Functions and Space Curves (16)
Derivatives and Motion on Smooth Curves (Example 4)
Example 4 STUDYING MOTION
The vector r (t )  cos t i  sin t j  t 2 k gives the position of a
moving body at time t . Find
(a) the velocity and acceleration vectors.
(b) the body's speed and direction at t  2.
(c) at what times the body's velocity and acceleration are
orthogonal.
12.1 Vector-Valued Functions and Space Curves (17)
Derivatives and Motion on Smooth Curves
Differention Rules for Vector Function
Let u(t ) and v(t ) be differentiable vector functions of t ,
C any constant vector, c any scalar, and f any differentiable
scalar function of t .
d
1. Constant Function Rule :
C0
dt
2. Scalar Multiple Rule :
d
d
c u   c u 
dt
dt
d
df
d
f u  u  f
u

dt
dt
dt
12.1 Vector-Valued Functions and Space Curves (18)
Differentiation Rules
Differention Rules for Vector Function
Let u(t ) and v(t ) be differentiable vector functions of t ,
C any constant vector, c any scalar, and f any differentiable
scalar function of t .
d
d
d
3. Sum Rule :
u  v   u  v
dt
dt
dt
d
d
d
4. Difference Rule :
u  v   u  v
dt
dt
dt
d
d u
d v
5. Dot Product Rule :
v u
u v  
dt
dt
dt
d
d u
d v
6. Cross Product Rule :
u

v


v

u



dt
dt
dt
12.1 Vector-Valued Functions and Space Curves (19)
Differentiation Rules
Differention Rules for Vector Function
Let u(t ) and v(t ) be differentiable vector functions of t ,
C any constant vector, c any scalar, and f any differentiable
scalar function of t .
7. Chain Rule : If r is a differentiable function of t and t is a
differentiable function of s . then
d u d u dt

ds
dt ds
12.1 Vector-Valued Functions and Space Curves (20)
Vector Functions of Constant Length
If u is a differentiable vector function of t of constant
length then
du
u
0
dt
12.1 Vector-Valued Functions and Space Curves (21)
Vector Functions of Constant Length (Example 5)
Example 5 SUPPORTING EQUATION (3)
Show that u(t )  cos t i  sint j  3 k has constant length
and is orthogonal to its derivative.
12.1 Vector-Valued Functions and Space Curves (22)
Integrals of Vector Functions
Definition Indefinite Integral
The indefinite integral of r with respect to t is the set of all
antiderivatives or r  denoted by  r t dt. If R is any
antiderivative of r then
 r t dt  R t   C .
12.1 Vector-Valued Functions and Space Curves (23)
Integrals of Vector Functions (Example 6)
Example 6 FINDING ANTIDERIVATI VES
   cos t  i  j  2t k dt    cos t
   dt  j    2t dt  k
dt i 
  sint  C1  i  t  C 2  j  t 2  C 3  k
  sint  i  t j  t 2k  C
C  C1i  C 2 j  C 3k
12.1 Vector-Valued Functions and Space Curves (20)
Integrals of Vector Functions
Definition Indefinite Integral
If the components of r t   f t  i  g t  j  h t  k are
integrable over a ,b  then so is r and the definite integral
of r from a to b is
a r t dt   a f t dt  i   a g t dt  j   a h t dt  k
b
b
b
b
12.1 Vector-Valued Functions and Space Curves (24)
Integrals of Vector Functions (Example 7)
Example 7 EVALUATING DEFINITE INTEGRALS
 (cost )i  j  2t k dt   


0
0

   dt  j    2t dt  k
cos t dt i 


0
0
   
 sint 0 i  t 0 j  t 2


0  k

  0  0  i     0  j   2  0 2  k
 j  2k
12.1 Vector-Valued Functions and Space Curves (25)
Integrals of Vector Functions (Example 8)
Example 8 FINDING A PARTICLE'S POSITION FUNCTION
The velocity of a partical moving in space is
d r
  cos t  i   sint  j  k
dt
Find the particle's position function as a function of t
if r  2i  k when t  0
12.2 Arc Length and the Unit Tangent Vector T(1)
Arc length
Definition Arc Length : Length of a Smooth Curve
The length of a smooth curve r t   f t  i  g t  j  h t  k 
a  t  b that is traced exactly once as t increases form
t  a to t  b is
L
b

b
a
a
b
2
2
2
2
2
 df   dg   dh 
  
 
 dt
 dt   dt   dt 
 dx   dy   dz 

 
 
 dt
 dt   dt   dt 
  v dt
a
2
12.2 Arc Length and the Unit Tangent Vector T(2)
Arc length (Example 1)
Example 1 APPLYING THE DEFINITION OF LENGTH
Find the length of one turn of the helix r (t )  cos t i  sint j  t k
Solution
The length of this portion of the curve is
b
L   v dt  
2
a
0

2
0
  sint 2   cos t 2  1 dt
2 dt  2 2 units
8
6
4
2
0
1
1
0
0
-1 -1
12.2 Arc Length and the Unit Tangent Vector T(3)
Arc length
Arc Length Parameter with Base Point P t 0 
on a Smooth curve
s t   
t
t0
2
2
2
t
 x        y        z      d    v   d 
t0
12.2 Arc Length and the Unit Tangent Vector T(4)
Arc length (Example 2)
Example 2 FINDING AN ARC LENGTH PARAMETE R
If t 0  0, the arc length parameter along the helix
r (t )  cos t i  sint j  t k
from t 0 to t is
s t   
t
t0
v   d 

t
0
2 d   2t .
Thus, s  2   2 2,
s  2   2 2, and so on.
12.2 Arc Length and the Unit Tangent Vector T(5)
The Unit Tangent Vector T
Definition Unit Tangent Vector
The unit tangent vector of a smooth curve r t  is
dr dr / dt
v
T


ds ds / dt v
12.2 Arc Length and the Unit Tangent Vector T(6)
The Unit Tangent Vector T (Example 4)
Example 4 FINDING THE UNIT TANGENT VECTOR
Find the unit vector of the helix r(t )  cos t i  sint j  t k
Solution
v(t )   sint i  cos t j  k
v(t ) 
  sint 2   cos t 2  1  2
v (t )  sint
cos t
1
T

i
j
k
v (t )
2
2
2
T
dr dr / dt
v


ds ds / dt
v
12.2 Arc Length and the Unit Tangent Vector T(7)
The Unit Tangent Vector T (Example 5)
Example 5 FINDING THE UNIT TANGENT VECTOR T
For the countclockwise motion r(t )  cos t i  sint j
around the unit circle v(t )  (  sin t ) i  (cos t ) j is already
a unit vector, so
Tv
y
Tv
r
t
O
P(x,y)
x
T
dr dr / dt
v


ds ds / dt
v
12.3 Curvature, Torsion, and the TNB Frame (1)
Curvature, Torsion, and TNB Frame
As a partical moves along a smooth curve in the plane,T  dr/ds
turns as the curve bends. Since T is a unit vector, its lengh
remains constant and only its direction changes per unit of lengh
along the curve. The rate at which the direction changes per unit
of lengh along the curve is the curveature. The traditional symbol
for the curvature is the Greek letter   "kappa"  .
y
P
P0
O
T
x
12.3 Curvature, Torsion, and the TNB Frame (2)
Curvature, Torsion, and TNB Frame
Definition Curvature
If T is the unit tangent vector of a smooth curve, the curvature
dT
function of the curve is  
ds
The curvature measures how sharply a curve bends at a point. If
dT/ds is large, T turns sharply as the particle passes through P
and the curvature at P is large. If dT/ ds is close to zero, T turns
more slowly and the curvature at P is smaller.
12.3 Curvature, Torsion, and the TNB Frame (3)
Curvature, Torsion, and TNB Frame (Example 1)
Example 1 FINDING THE CURVATURE OF A STRAIGHT LINE
Show that the curvature of a straight line is zero.
Solution
On a straignt line, the unit tangent vector T always points in the
same direction, so its components are constants.Therefore,
dT/ds  0  0.
T
12.3 Curvature, Torsion, and the TNB Frame (4)
Curvature, Torsion, and TNB Frame (Example 2)
Example 2 FINDING THE CURVATURE OF A CIRCLE
Show that the curvature of a circle of radius a is 1/ a .
12.3 Curvature, Torsion, and the TNB Frame (5)
The Principal Unit Normal Vector for Plane Curves
Since T has constant length, the vector dT / ds is orthogonal to T.
Therefore, if at a point where dT / ds  0, we divide dT / ds by its
length , we obtain a unit vector orthogonal to T.
Unlike the curvature , which is never negative, the torsion 
12.3
Curvature,
Torsion,
and the TNB Frame (6)
may
be positive,
negative,or
zero.
Thethree
Principal
Unit Normal
forBPlane
Curves
The
planedetermined
byVector
T, N and
are shown
in
Figure 12.25. The curvature   dT / ds can be thought
of as the rate at which the normal plane turns as the
point P moves along the curve.Similarly, the torsion
   dB / ds  N is the rate at which the osculating plane
The vector dT/ds points in the direction in which T turns as the
turns
about
T Therefore,
as P moves
curve
bends.
if wealong
face inthe
thecurve.
directionTo
ofrsion
increasing
measure
how the
curve
twists.
arc length,the
vector
dT / ds
points to the right if T turns clockwise
and toward the left it T turns counterclockwise. In other words,the
principle normal vector N will point toward the concave side of the
curve.
12.3 Curvature, Torsion, and the TNB Frame (7)
The Principal Unit Normal Vector for Plane Curves
Because the arc length parameter for a smooth curve
r (t )  f (t ) i  g (t ) j is defined with ds / dt positive,
ds / dt  ds / dt , and the Chain rule gives
dT / ds
N
dT / ds
(dT / dt )(dt / ds )

dT / dt ds / ds
(dT / dt )

dT / dt
12.3 Curvature, Torsion, and the TNB Frame (8)
The Principal Unit Normal Vector for Plane Curves (EX.3)
Example 3 FINDING T AND N
Find T and N for the circlular motion r (t )  cos 2t i  sin2t j
Solution
v  2 sin2t i  2 cos 2t j
v  4 sin2 2t  4 cos 2 2t  2
T  v / v   sin2t i  cos 2t j
from this we find
dT / dt  2 cos 2t i  2 sin2t j
dT / dt  4 cos 2 2t  4 sin2 2t  2
and
N  dT / dt  /  dT / dt    cos 2t i  sin2t j
12.3 Curvature, Torsion, and the TNB Frame (9)
Circle of Curvature and Radius of Curvature
The circle of curvature or osculating circle at a point P on
a plane curve where   0 is the circle in the plane of the
curve that
1. is tangent to the curve at P ,
2. has the same curvature the curve has at P ,
3. lies toward the concave or inner side of the curve.
12.3 Curvature, Torsion, and the TNB Frame (10)
Circle of Curvature and Radius of Curvature
The radius of curvature of the curve at P is the radius of
the radius of the circle of curvature, which is
Radius of curvature    1/ 
The center of curvature of the curva at P is the center of the
circle of curvature.
12.3 Curvature, Torsion, and the TNB Frame (11)
Curvature and Normal Vectors for Space Curves (Ex. 4-1)
Example 4 FINDING CURVATURE ON A HELIX
Find the curvature of the helix
r (t )  a cos t i  a sint j  bt k,
Solution
v  a sint i  a cos t j  b k
a ,b  0, a 2  b 2  0
v  a 2 sin2 t  a 2 cos2 t  b 2  a 2  b 2
v  a sint i  a cos t j  b k 
T

v
a2 b2
12.3 Curvature, Torsion, and the TNB Frame (12)
Curvature and Normal Vectors for Space Curves (Ex. 4-2)
Solution
then we find dT/ds as
dT dT dt dT 1


ds dt ds dt v
a cos t


i  a sint j
a2 b2
a   cos t i  sint j

a2 b2
1
a2 b2
dT
a cos 2 t + sin2 t
a


 2
2
2
ds
a b
a b2
12.3 Curvature, Torsion, and the TNB Frame (13)
Curvature and Normal Vectors for Space Curves (Example 5)
Example 5 FINDING THE PRINCIPAL UNIT NORMAL
VECTOR N
Find N for the helix r (t )  a cos t i  a sint j  bt k, a , b  0,
a2  b2  0
Solution
then we find dT/ds as
dT  a cos t i  a sint j

,
2
2
dt
a b
dT a cos2 t + sin2 t
a

 2
2
2
2
dt
a

b
a b
a cos t i  a sint j a 2  b 2
dT / dt

N

dT / dt
a
a2 b2
  cos t i  sin t j
12.3 Curvature, Torsion, and the TNB Frame (14)
Torsion and the Binormal Vector
The binormal vector of a curve in space is B = T  N, a unit
vector orthogonal to both T and N. Together define a moving
right-handed vector frame that always travel with a body
moving along a curve in space. It is the Frenet (“fre-nay”)
frame, or the TNB frame. This vector frame plays a
significant role in calculating the flight paths of space vehicles.
B
N
T
12.3 Curvature, Torsion, and the TNB Frame (15)
Torsion and the Binormal Vector
How does dB / ds behavior in relation to T, N and B?
dB dT
dN

N  T 
B
ds
ds
ds
dN
(since N is the direction of dT / ds )
ds
T
we know dB / ds is orthogonal to T.
 T
N
since dB / ds is also orthogonal to B (due to constant vector),
it follows that dB / ds is orthogonal to the plane B and T.
In other word, dB / ds is parallel to N, so dB / ds is a scalar
multiple of N. In symbols,
dB
 N
ds
12.3 Curvature, Torsion, and the TNB Frame (16)
Torsion and the Binormal Vector
dB
 N
ds
The minus sign is traditional. The scalar  is the torsion along
the curve. Notice that
so that
dB
N  N N  
ds

dB
N
ds
B
N
T
12.3 Curvature, Torsion, and the TNB Frame (17)
Torsion and the Binormal Vector
Definition Torsion
Let B  Τ  Ν. The torsion function of a smooth curve is
dB

N
ds
12.3 Curvature, Torsion, and the TNB Frame (18)
Torsion and the Binormal Vector
Unlike the curvature , which is never negative, the torsion  may be positive,
negative,or zero. The three plane determined by T,N and B are shown in
Figure 12.25. The curvature   dT / ds can be thought of as the rate at
which the normal plane turns as the point P moves along the curve.
Similarly, the torsion    dB / ds  N is the rate at which the osculating
plane turns about T as P moves along the cu rve. Torsion measure how
the curve twists.
12.3 Curvature, Torsion, and the TNB Frame (19)
Tangential and Normal Components of Acceleration
When a body move in a path, the velovity is
d r d r ds
ds
v

T
(in the dirction of T)
ds ds dt
dt
the acceleration is
d v d
a

dt
dt
2
ds d T
 ds  d s
T   2 T 
dt dt
 dt  dt
2
d s
 ds 
 2 T    N
dt
 dt 
2
 ds 
aN    
 dt 
s
a
2
N
T
d 2s
aT  2
dt
12.3 Curvature, Torsion, and the TNB Frame (20)
Tangential and Normal Components of Acceleration
Definition Tangential and Normal Components of
Acceleration
a  aT T  aN N
where
d 2s d
v
aT  2 
dt
dt
2
 ds 
and aN       v 2
 dt 
are the tangential and normal scalar components of
acceleration.
12.3 Curvature, Torsion, and the TNB Frame (21)
Tangential and Normal Components of Acceleration
1. If a body moves in a circle at a
d 2s
T
2
dt
a
constant speed d2s / dt 2 is zero
P
2
v N
C
v
2

N
and all the acceleration points
along N toward the circle's
center.
2. If the body is speeding up or
slowing down, a has a nonzero
tangential component.
aN 
2
a  a T2
12.3 Curvature, Torsion, and the TNB Frame (22)
Tangential and Normal Components of Acceleration
(Ex. 6-1)
Example 6 FINDING AN ACCELERATION VECTOR
Without finding T and N, write the acceleration of the motion
r (t )  (cos 2t  t sint )i  (sint  t cos t ) j, t  0, in the form of
a  a T T  aNN.
T
P (x , y )
0
1
v  (t cos t )2  (t sint )2  t
d
aT 
v 1
dt
r
-1
v  d r / dt  (t cos t )i  (t sint ) j
2
12.3 Curvature, Torsion, and the TNB Frame (23)
Tangential and Normal Components of Acceleration
(Ex. 6-2)
Example 6 FINDING AN ACCELERATION VECTOR
Without finding T and N, write the acceleration of the motion
r (t )  (cos 2t  t sint )i  (sint  t cos t ) j, t  0, in the form of
a  a T T  aNN.
a  d 2r / dt 2
T
 (cos t  t sint )i  (sint  t cos t ) j
t N
P (x , y )
aN 
r
0
2
a  a T2 
t
2
 1  1  t
a  a T T  aNN  T  t N
t
-1
2
a  t 2 1
1
2
12.3 Curvature, Torsion, and the TNB Frame (24)
Formulas for Computing Curvature and Torsion
2
2

ds
d
s


 ds  
va  
T    2 T     N
 dt   dt
 dt  
3
3
ds
ds
 
 
     T  N     B
 dt 
 dt 
3
va   v B   v
3
A Vector Formula for Curvature

va
v
3
12.3 Curvature, Torsion, and the TNB Frame (24)
Formulas for Computing Curvature and Torsion
Formula for curves in Space
Unit tangent vector 
v
T
v
dT / dt
Principal unit normal vector N 
dT / dt
Binormal vector B  T  N
dT v  a
Curvature  

3
ds
v
12.3 Curvature, Torsion, and the TNB Frame (24)
Formulas for Computing Curvature and Torsion
Formula for curves in Space
dB
Torsion T  
N
ds
Tangential and normal
scalar components
of acceleration 

x
x
x
y
y
y
va
a  a T T  aNN
d
aT 
v
dt

aN  k v 
2
a  a T 2 
z
z
z
2
12.4 Planetary Motion and Satellites
omitted