vector functions 13. 13.3 arc length and curvature in this section, we will learn how to find: the...
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13.3Arc Length
and Curvature
In this section, we will learn how to find:
The arc length of a curve and its curvature.
VECTOR FUNCTIONS
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.
[ ] [ ]2 2
2 2
'( ) '( )b
a
b
a
L f t g t dt
dx dydt
dt dt
= +
⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
∫
∫
Formula 1 PLANE CURVE LENGTH
For the case where f’ and g’ are continuous,
we arrived at the following formula:
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
If the curve is traversed exactly once as t
increases from a to b, then it can be shown
that its length is:
[ ] [ ] [ ]2 2 2
2 2 2
'( ) '( ) '( )b
a
b
a
L f t g t h t dt
dx dy dzdt
dt dt dt
= + +
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
∫
∫
Formula 2 SPACE CURVE LENGTH
ARC LENGTH
Notice that both the arc length formulas
1 and 2 can be put into the more compact
form
'( )b
aL t dt=∫ r
Formula 3
That is because:
For plane curves r(t) = f(t) i + g(t) j
For space curves r(t) = f(t) i + g(t) j + h(t) k
[ ] [ ]2 2'( ) '( ) '( ) '( ) '( )t f t g t f t g t= + = +r i j
ARC LENGTH
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).
Example 1 ARC LENGTH
Since r’(t) = -sin t i + cos t j + k,
we have:
2 2'( ) ( sin ) cos 1
2
t t t= − + +
=
r
Example 1 ARC LENGTH
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:2
0
2
0
'( )
2
2 2
L t dt
dt
π
π
π
=
=
=
∫∫
r
Example 1 ARC LENGTH
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.
Equations 4 & 5 ARC LENGTH
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.
PARAMETRIZATION
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
ARC LENGTH FUNCTION
We define its arc length function s
by:
2 2 2
( ) '( )t
a
t
a
s t u du
dx dy dzdu
du du du
=
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
∫
∫
r
Equation 6
If we differentiate both sides of Equation 6
using Part 1 of the Fundamental Theorem of
Calculus (FTC1), we obtain:
'( )ds
tdt
=r
Equation 7 ARC LENGTH FUNCTION
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)
PARAMETRIZATION
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
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.
Example 2 REPARAMETRIZATION
The initial point (1, 0, 0) corresponds to
the parameter value t = 0.
From Example 1, we have:
So,
'( ) 2ds
tdt
= =r
0 0( ) '( ) 2 2
t ts s t u du du t= = = =∫ ∫r
Example 2 REPARAMETRIZATION
Therefore, and the required
reparametrization is obtained by substituting
for t:
/ 2t s=
( ) ( ) ( )( ( ))
cos / 2 sin / 2 / 2
t s
s s s= + +
r
i j k
Example 2 REPARAMETRIZATION
A parametrization r(t) is called smooth
on an interval I if:
r’ is continuous.
r’(t) ≠ 0 on I.
SMOOTH PARAMETRIZATION
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 CURVE
If C is a smooth curve defined by the vector
function r, recall that the unit tangent vector
T(t) is given by:
This indicates the direction of the curve.
'( )( )
'( )
tt
t=r
Tr
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.
SMOOTH CURVES
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
The curvature of a curve is:
where T is the unit tangent vector.
d
dsκ =
T
Definition 8 CURVATURE—DEFINITION
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:
/and
/
d d ds d d dt
dt ds dt ds ds dtκ= = =
T T T T
CURVATURE
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
Example 3 CURVATURE
Therefore, r’(t) = –a sin t i + a cos t j
and |r’(t)| = a
So,
and
Example 3 CURVATURE
'( )( ) sin cos
'( )
'( ) cos sin
tt t t
t
t t t
= =− +
=− −
rT i j
r
T i j
This gives |T’(t)| = 1.
So, using Equation 9, we have:
'( ) 1( )
'( )
tt
t aκ = =
T
r
Example 3 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
The curvature of the curve given by
the vector function r is:
3
'( ) ''( )( )
'( )
t tt
tκ
×=r r
r
Theorem 10 CURVATURE
Hence, the Product Rule (Theorem 3
in Section 13.2, Formula 3) gives:
2
2'' 'd s ds
dt dt= +r T T
Proof CURVATURE
Using the fact that T x T = 0 (Example 2
in Section 12.4), we have:
( )2
' '' 'ds
dt⎛ ⎞× = ×⎜ ⎟⎝ ⎠
r r T T
Proof CURVATURE
Thus,
and
( )2 2
3
' '' ' '''
/ '
' ' ''
' '
ds dt
κ
× ×= =
×= =
r r r rT
r
T r r
r r
Proof CURVATURE
Find the curvature of the twisted cubic
r(t) = <t, t2, t3>
at: A general point
(0, 0, 0)
Example 4 CURVATURE
Example 4 CURVATURE
2
2
4 2
4 2
'( ) ''( ) 1 2 3
0 2 6
6 6 2
'( ) ''( ) 36 36 4
2 9 9 1
t t t t
t
t t
t t t t
t t
× =
= − +
× = + +
= + +
i j k
r r
i j k
r r
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
Since y’ = 2x and y’’ = 2, Formula 11
gives:
( )
3/ 22
3/ 22
''( )
1 ( ')
2
1 4
yx
y
x
κ =⎡ ⎤+⎣ ⎦
=+
Example 5 CURVATURE
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
Example 5 CURVATURE
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 → ±∞
Example 5 CURVATURE
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).
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 VECTORS
However, if r’ is also smooth, we can
define the principal unit normal vector N(t)
(simply unit normal) as:
'( )( )
'( )
tt
t=T
NT
NORMAL VECTOR
NORMAL VECTORS
We can think of the normal vector as
indicating the direction in which the curve
is turning at each point.
NORMAL & BINORMAL VECTORS
Find the unit normal and binormal vectors
for the circular helix
r(t) = cost i + sin t j + t k
Example 6
First, we compute the ingredients
needed for the unit normal vector:
( )
'( ) sin cos '( ) 2
'( ) 1( ) sin cos
'( ) 2
t t t t
tt t t
t
= − + + =
= = − + +
r i j k r
rT i j k
r
Example 6 NORMAL & BINORMAL VECTORS
( )1 1'( ) cos sin '( )
2 2
'( )( ) cos sin
'( )
cos , sin ,0
t t t t
tt t t
t
t t
= − − =
= = − −
= − −
T i j T
TN i j
T
Example 6 NORMAL & BINORMAL VECTORS
This shows that the normal vector
at a point on the helix is horizontal and
points toward the z-axis.
Example 6 NORMAL & BINORMAL VECTORS
The binormal vector is:
1( ) ( ) ( ) sin cos 1
2cos sin 0
1sin , cos ,1
2
t t t t t
t t
t t
⎡ ⎤⎢ ⎥= × = −⎢ ⎥⎢ ⎥− −⎣ ⎦
= −
i j k
B T N
Example 6 NORMAL & BINORMAL VECTORS
The figure illustrates
Example 6 by showing
the vectors T, N, and B
at two locations on the helix.
NORMAL & BINORMAL VECTORS
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.
TNB FRAME
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.’
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 PLANE
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).
It is the circle that best describes how
C behaves near P.
It shares the same tangent, normal, and curvature at P.
OSCULATING CIRCLE
NORMAL & OSCULATING PLANES
Find the equations of the normal
plane and osculating plane of the helix
in Example 6 at the point
P(0, 1, π/2)
Example 7
The normal plane at P has normal
vector r’(π/2) = <–1, 0, 1>.
So, an equation is:
or
( ) ( )1 0 0 1 1 02
2
x y z
z x
π
π
⎛ ⎞− − + − + − =⎜ ⎟⎝ ⎠
= +
Example 7 NORMAL & OSCULATING PLANES
The osculating plane at P contains
the vectors T and N.
So, its normal vector is:
T x N = B
Example 7 NORMAL & OSCULATING PLANES
From Example 6, we have:
1( ) sin , cos ,1
2
1 1,0,
2 2 2
t t t
π
= −
⎛ ⎞=⎜ ⎟⎝ ⎠
B
B
Example 7 NORMAL & OSCULATING PLANES
A simpler normal vector is <1, 0, 1>.
So, an equation of the osculating plane is:
or
( ) ( )1 0 0 1 1 02
2
x y z
z x
π
π
⎛ ⎞− + − + − =⎜ ⎟⎝ ⎠
= − +
Example 7 NORMAL & OSCULATING PLANES
OSCULATING CIRCLES
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).
Example 8
For the graph, we use parametric
equations of this circle:
x = ½ cos t y = ½ + ½ sin t
Example 8 OSCULATING CIRCLES
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