Chap13_Sec3

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13
VECTOR FUNCTIONS
VECTOR FUNCTIONS
13.3
Arc Length
and Curvature
In this section, we will learn how to find:
The arc length of a curve and its curvature.
PLANE CURVE LENGTH
In Section 10.2, we defined the length
of a plane curve with parametric equations
x = f(t),
y = g(t),
a≤t≤b
as the limit of lengths of inscribed polygons.
PLANE CURVE LENGTH
Formula 1
For the case where f’ and g’ are continuous,
we arrived at the following formula:
L
b
 f
a
'( t )    g '( t )  dt
2
2


b
a
2
2
 dx 
 dy 

 
 dt
 dt 
 dt 
SPACE CURVE LENGTH
The length of a space curve is defined
in exactly the same way.
SPACE CURVE LENGTH
Suppose that the curve has the vector
equation
r(t) = <f(t), g(t), h(t)>,
a≤t≤b
 Equivalently, it could have the parametric
equations
x = f(t), y = g(t), z = h(t)
where f’, g’ and h’ are continuous.
SPACE CURVE LENGTH
Formula 2
If the curve is traversed exactly once as t
increases from a to b, then it can be shown
that its length is:
L
b
 f
a
'( t )    g '( t )    h '( t )  dt
2
2


b
a
2
2
2
2
 dx 
 dy 
 dz 

 
 
 dt
 dt 
 dt 
 dt 
ARC LENGTH
Formula 3
Notice that both the arc length formulas
1 and 2 can be put into the more compact
form
L 

b
r '( t ) d t
a
ARC LENGTH
That is because:
 For plane curves r(t) = f(t) i + g(t) j
r '( t )  f '( t ) i  g '( t ) j 
f
'( t )    g '( t ) 
2
 For space curves r(t) = f(t) i + g(t) j + h(t) k
r '( t )  f '( t ) i  g '( t ) j  h '( t ) k

f
'( t )    g '( t )    h '( t ) 
2
2
2
2
ARC LENGTH
Example 1
Find the length of the arc of the circular helix
with vector equation
r(t) = cos t i + sin t j + t k
from the point (1, 0, 0) to the point (1, 0, 2π).
ARC LENGTH
Example 1
Since r’(t) = -sin t i + cos t j + k,
we have:
r '( t ) 

(  sin t )  cos t  1
2
2
2
ARC LENGTH
Example 1
The arc from (1, 0, 0) to (1, 0, 2π)
is described by the parameter interval
0 ≤ t ≤ 2π.
 So, from Formula 3, we have: L 



2
r '( t ) dt
0
2
2 dt
0
 2 2
ARC LENGTH
A single curve C can be
represented by more than
one vector function.
ARC LENGTH
Equations 4 & 5
For instance, the twisted cubic
r1(t) = <t, t 2, t 3>
1≤t≤2
could also be represented by the function
r2(u) = <eu, e2u, e3u>
0 ≤ u ≤ ln 2
 The connection between the parameters
t and u is given by t = eu.
PARAMETRIZATION
We say that Equations 4 and 5 are
parametrizations of the curve C.
PARAMETRIZATION
If we were to use Equation 3 to compute
the length of C using Equations 4 and 5,
we would get the same answer.
 In general, it can be shown that, when Equation 3
is used to compute arc length, the answer is
independent of the parametrization that is used.
ARC LENGTH
Now, we suppose that C is a curve given by
a vector function
r(t) = f(t) i + g(t) j + h(t) k
a≤t≤b
where:
 r’ is continuous.
 C is traversed exactly once as t increases
from a to b.
ARC LENGTH FUNCTION
Equation 6
We define its arc length function s
by:
s (t ) 

t
r '( u ) du
a
2


t
a
2
2
 dx 
 dy 
 dz 

 
 
 du
 du 
 du 
 du 
ARC LENGTH FUNCTION
Thus, s(t) is the length of the part of C
between r(a) and r(t).
ARC LENGTH FUNCTION
Equation 7
If we differentiate both sides of Equation 6
using Part 1 of the Fundamental Theorem of
Calculus (FTC1), we obtain:
ds
dt
 r '( t )
PARAMETRIZATION
It is often useful to parametrize a curve
with respect to arc length.
 This is because arc length arises naturally from
the shape of the curve and does not depend on
a particular coordinate system.
PARAMETRIZATION
If a curve r(t) is already given in terms of
a parameter t and s(t) is the arc length
function given by Equation 6, then we may
be able to solve for t as a function of s:
t = t(s)
REPARAMETRIZATION
Then, the curve can be reparametrized
in terms of s by substituting for t:
r = r(t(s))
REPARAMETRIZATION
Thus, if s = 3 for instance, r(t(3)) is
the position vector of the point 3 units
of length along the curve from its starting
point.
REPARAMETRIZATION
Example 2
Reparametrize the helix
r(t) = cos t i + sin t j + t k
with respect to arc length measured from
(1, 0, 0) in the direction of increasing t.
REPARAMETRIZATION
Example 2
The initial point (1, 0, 0) corresponds to
the parameter value t = 0.
ds
From Example 1, we have:
 r '( t ) 
dt
 So,
s  s (t ) 

t
0
r '( u ) du 

t
0
2 du 
2t
2
REPARAMETRIZATION
Example 2
Therefore, t  s / 2 and the required
reparametrization is obtained by substituting
for t:
r ( t ( s ))

 cos s /


2 i  sin s /
 
2 j s /

2 k
SMOOTH PARAMETRIZATION
A parametrization r(t) is called smooth
on an interval I if:
 r’ is continuous.
 r’(t) ≠ 0 on I.
SMOOTH CURVE
A curve is called smooth if it has
a smooth parametrization.
 A smooth curve has no sharp corners or cusps.
 When the tangent vector turns, it does so
continuously.
SMOOTH CURVES
If C is a smooth curve defined by the vector
function r, recall that the unit tangent vector
T(t) is given by:
T (t ) 
r '( t )
r '( t )
 This indicates the direction of the curve.
SMOOTH CURVES
You can see that T(t) changes direction:
 Very slowly when C is fairly straight.
 More quickly when C bends or twists more sharply.
CURVATURE
The curvature of C at a given point
is a measure of how quickly the curve
changes direction at that point.
CURVATURE
Specifically, we define it to be the magnitude
of the rate of change of the unit tangent vector
with respect to arc length.
 We use arc length so that the curvature will
be independent of the parametrization.
CURVATURE—DEFINITION
Definition 8
The curvature of a curve is:
 
dT
ds
where T is the unit tangent vector.
CURVATURE
The curvature is easier to compute if
it is expressed in terms of the parameter
t instead of s.
CURVATURE
So, we use the Chain Rule (Theorem 3
in Section 13.2, Formula 6) to write:
dT
dt

d T ds
ds dt
and
 
dT
ds

d T / dt
ds / dt
CURVATURE
Equation/Formula 9
However, ds/dt = |r’(t)| from Equation 7.
So,
 (t ) 
T '( t )
r '( t )
CURVATURE
Example 3
Show that the curvature of a circle
of radius a is 1/a.
 We can take the circle to have center the origin.
 Then, a parametrization is:
r(t) = a cos t i + a sin t j
CURVATURE
Example 3
 Therefore, r’(t) = –a sin t i + a cos t j
and
 So,
|r’(t)| = a
T (t ) 
r '( t )
  sin t i  cos t j
r '( t )
and
T '( t )   cos t i  sin t j
CURVATURE
Example 3
 This gives |T’(t)| = 1.
 So, using Equation 9,
we have:
 (t ) 
T '( t )
r '( t )

1
a
CURVATURE
The result of Example 3 shows—in
accordance with our intuition—that:
 Small circles have large curvature.
 Large circles have small curvature.
CURVATURE
We can see directly from the definition of
curvature that the curvature of a straight line
is always 0—because the tangent vector is
constant.
CURVATURE
Formula 9 can be used in all cases
to compute the curvature.
Nevertheless, the formula given by
the following theorem is often more
convenient to apply.
CURVATURE
Theorem 10
The curvature of the curve given by
the vector function r is:
 (t ) 
r '( t )  r ''( t )
r '( t )
3
CURVATURE
Proof
T = r’/|r’| and |r’| = ds/dt.
So, we have:
r' r'T 
ds
dt
T
CURVATURE
Proof
Hence, the Product Rule (Theorem 3
in Section 13.2, Formula 3) gives:
2
r '' 
d s
dt
2
T
ds
dt
T'
CURVATURE
Proof
Using the fact that T x T = 0 (Example 2
in Section 12.4), we have:
 ds 
r ' r ''  

 dt 
2
 T  T '
CURVATURE
Proof
Now, |T(t)| = 1 for all t.
So, T and T’ are orthogonal
by Example 4 in Section 13.2
CURVATURE
Proof
Hence, by Theorem 6 in Section 12.4,
2
 ds 
r ' r ''  
 TT'
 dt 
2
 ds 

 T T'
 dt 
2
 ds 

 T'
 dt 
CURVATURE
Proof
Thus,
T' 
r ' r ''
 ds / dt 
and
 
T'
r'

2

r ' r ''
r'
r ' r ''
r'
3
2
CURVATURE
Example 4
Find the curvature of the twisted cubic
r(t) = <t, t2, t3>
at:
 A general point
 (0, 0, 0)
CURVATURE
Example 4
First, we compute the required
ingredients:
r '( t )  1, 2 t , 3 t
r '( t ) 
r ''( t )  0, 2, 6 t
2
1  4t  9t
2
4
CURVATURE
Example 4
i
j
k
r '( t )  r ''( t )  1
2t
3t
0
2
6t
2
 6t i  6t j  2 k
2
r '( t )  r ''( t ) 
36 t  36 t  4
4
2
 2 9t  9t  1
4
2
CURVATURE
Example 4
Then, Theorem 10 gives:
 (t ) 
r '( t )  r ''( t )
r '( t )
3
2 1  9t  9t
2

1  4 t
2
 9t
4

 At the origin, where t = 0, the curvature is:
ĸ(0) = 2
4
3/2
CURVATURE
For the special case of a plane curve
with equation y = f(x), we choose x as
the parameter and write:
r(x) = x i + f(x) j
CURVATURE
Then,
r’(x) = i + f’(x) j
and
r’’(x) = f’’(x) j
CURVATURE
Since i x j = k and j x j = 0,
we have:
r’(x) x r’’(x) = f’’(x) k
CURVATURE
We also have:
r '( x ) 
1   f '( x ) 
2
CURVATURE
Formula 11
So, by Theorem 10,
 ( x) 
f ''( x )
1   f '( x )  


2
3/2
CURVATURE
Example 5
Find the curvature of the parabola y = x2
at the points
(0, 0), (1, 1), (2, 4)
CURVATURE
Example 5
Since y’ = 2x and y’’ = 2, Formula 11
gives:
 ( x) 
y ''
1  ( y ') 


2

3/2
2
1  4 x 
2
3/2
CURVATURE
Example 5
At (0, 0), the curvature is κ(0) = 2.
At (1, 1), it is κ(1) = 2/53/2 ≈ 0.18
At (2, 4), it is κ(2) = 2/173/2 ≈ 0.03
CURVATURE
Example 5
Observe from the expression for κ(x)
or the graph of κ here that:
κ(x) → 0 as x → ±∞
 This corresponds
to the fact that
the parabola appears
to become flatter
as x → ±∞
NORMAL AND BINORMAL VECTORS
At a given point on a smooth space curve
r(t), there are many vectors that are
orthogonal to the unit tangent vector T(t).
NORMAL VECTORS
We single out one by observing that,
because |T(t)| = 1 for all t, we have T(t) · T’(t)
by Example 4 in Section 13.2.
So, T’(t) is orthogonal to T(t).
 Note that T’(t) is itself not a unit vector.
NORMAL VECTOR
However, if r’ is also smooth, we can
define the principal unit normal vector N(t)
(simply unit normal) as:
N (t ) 
T '( t )
T '( t )
NORMAL VECTORS
We can think of the normal vector as
indicating the direction in which the curve
is turning at each point.
BINORMAL VECTOR
The vector
B(t) = T(t) x N(t)
is called the binormal vector.
BINORMAL VECTORS
It is perpendicular to both T and N
and is also a unit vector.
NORMAL & BINORMAL VECTORS
Example 6
Find the unit normal and binormal vectors
for the circular helix
r(t) = cost i + sin t j + t k
NORMAL & BINORMAL VECTORS
Example 6
First, we compute the ingredients
needed for the unit normal vector:
r '( t )   sin t i  cos t j  k
T (t ) 
r '( t )
r '( t )

1
2
r '( t ) 
  sin t i  cos t j  k 
2
NORMAL & BINORMAL VECTORS Example 6
T '( t ) 
1
2
N (t ) 
  cos t i  sin t j 
T '( t )
  cos t i  sin t j
T '( t )
  cos t ,  sin t , 0
T '( t ) 
1
2
NORMAL & BINORMAL VECTORS Example 6
This shows that the normal vector
at a point on the helix is horizontal and
points toward the z-axis.
NORMAL & BINORMAL VECTORS
Example 6
The binormal vector is:
 i
1 
B (t )  T (t )  N (t ) 
 sin t
2 
  cos t

1
2
j
cos t
 sin t
sin t ,  cos t ,1
k

1

0 
NORMAL & BINORMAL VECTORS
The figure illustrates
Example 6 by showing
the vectors T, N, and B
at two locations on the helix.
TNB FRAME
In general, the vectors T, N, and B, starting
at the various points on a curve, form a set
of orthogonal vectors—called the TNB
frame—that moves along the curve as
t varies.
TNB FRAME
This TNB frame plays an important
role in:
 The branch of mathematics known as
differential geometry.
 Its applications to the motion of spacecraft.
NORMAL PLANE
The plane determined by the normal and
binormal vectors N and B at a point P on a
curve C is called the normal plane of C at P.
 It consists of all lines that are orthogonal
to the tangent vector T.
OSCULATING PLANE
The plane determined by the vectors
T and N is called the osculating plane
of C at P.
 The name comes from the Latin osculum,
meaning ‘kiss.’
OSCULATING PLANE
It is the plane that comes closest to
containing the part of the curve near P.
 For a plane curve, the osculating plane is
simply the plane that contains the curve.
OSCULATING CIRCLE
The osculating circle (the circle of
curvature) of C at P is the circle that:
 Lies in the osculating plane of C at P.
 Has the same tangent as C at P.
 Lies on the concave side of C (toward which N points).
 Has radius ρ = 1/ĸ (the reciprocal of the curvature).
OSCULATING CIRCLE
It is the circle that best describes how
C behaves near P.
 It shares the same tangent, normal,
and curvature at P.
NORMAL & OSCULATING PLANES Example 7
Find the equations of the normal
plane and osculating plane of the helix
in Example 6 at the point
P(0, 1, π/2)
NORMAL & OSCULATING PLANES Example 7
The normal plane at P has normal
vector r’(π/2) = <–1, 0, 1>.
 So, an equation is:
 

 1  x  0   0  y  1  1  z    0
2

or
z  x

2
NORMAL & OSCULATING PLANES Example 7
The osculating plane at P contains
the vectors T and N.
 So, its normal vector is:
TxN=B
NORMAL & OSCULATING PLANES Example 7
From Example 6, we have:
B (t ) 
1
sin t ,  cos t ,1
2
 
B 
 2
1
2
, 0,
1
2
NORMAL & OSCULATING PLANES Example 7
A simpler normal vector is <1, 0, 1>.
 So, an equation of the osculating plane
is:
 

1  x  0   0  y  1  1  z    0
2

or
z  x 

2
NORMAL & OSCULATING PLANES
The figure shows the helix and
the osculating plane in Example 7.
OSCULATING CIRCLES
Example 8
Find and graph the osculating circle
of the parabola y = x2 at the origin.
 From Example 5, the curvature of the parabola
at the origin is ĸ(0) = 2.
 So, the radius of the osculating circle at
the origin is 1/ĸ = ½ and its center is (0, ½).
OSCULATING CIRCLES
Example 8
Therefore, its equation is:
x y
2
1
2

2

1
4
OSCULATING CIRCLES
Example 8
For the graph, we use parametric
equations of this circle:
x = ½ cos t
y = ½ + ½ sin t
OSCULATING CIRCLES
Example 8
The graph is displayed.
SUMMARY
We summarize the formulas for unit tangent,
unit normal and binormal vectors, and
curvature.
T (t ) 
r '( t )
N (t ) 
r '( t )
 
T '( t )
B (t )  T (t )  N (t )
T '( t )
dT
ds

T '( t )
r '( t )

r '( t )  r ''( t )
r '( t )
3

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