Section 11.7: Arc Length and Curvature
Theorem: (Arc Length)
~
Consider the curve C defined by R(t)
= hx(t), y(t), z(t)i, a ≤ t ≤ b. If C is traversed exactly
once as t increases from a to b, then its length is
s
Z b
Z b 2 2 2
dx
dy
dz
~ 0 (t)||dt =
L=
||R
+
+
dt.
dt
dt
dt
a
a
√
~
Example: Find the length of the curve defined by R(t)
= h6t, 3 2t2 , 2t3 i, 0 ≤ t ≤ 1.
√
~ 0 (t) = h6, 6 2t, 6t2 i, we have
Since R
√
36 + 72t2 + 36t4
p
=
36(1 + 2t2 + t4 )
p
=
36(1 + t2 )2
= 6(1 + t2 ).
~ 0 (t)|| =
||R
Then
Z
L=
0
1
1
6(1 + t2 )dt = (6t + 2t3 )0 = 4.
~
Definition: Suppose C is a space curve defined by R(t)
= hx(t), y(t), z(t)i, a ≤ t ≤ b. Then
its arc length function is defined by
s
Z t
Z t 2 2 2
dx
dy
dz
~ 0 (u)||du =
s(t) =
||R
+
+
du.
du
du
du
a
a
~
~
Thus, s(t) is the length of the part of C between R(a)
and R(t).
Note: A single curve C can be represented by more than one vector function. Different
representations of curves are called parameterizations of the curve C. It is often useful to
parameterize a curve with respect to arc length.
~
Example: Reparameterize the helix R(t)
= h3 sin t, 4t, 3 cos ti with respect to arc length
measured from the point where t = 0 in the direction of increasing t.
~ 0 (t) = h3 cos t, 4, −3 sin ti, we have
Since R
q
p
√
0
2
2
~
||R (t)|| = 9 cos t + 16 + 9 sin t = 9(sin2 t + cos2 t) + 16 = 25 = 5.
The arc length function is
Z
t
5du = 5t.
s = s(t) =
0
Thus, t = s/5 and the reparameterization is given by
s 4s
s ~
R(s) = 3 sin
, , 3 cos
.
5
5
5
~
Definition: The curvature of a curve defined by R(t)
is
κ(t) =
~ 0 (t) × R
~ 00 (t)||
||T~ 0 (t)||
||R
=
,
~ 0 (t)||
~ 0 (t)||3
||R
||R
where T~ is the unit tangent vector.
Note: The curvature of a cuve C at a given point is a measure of how quickly the curve
changes direction at that point. Specifically, it measures the rate of change of the unit
tangent vector (which indicates the direction of C) with respect to arc length.
Figure 1: Unit tangent vectors for a space curve C.
~
Example: Find the curvature of the twisted cubic R(t)
= ht, t2 , t3 i at a general point and at
(0, 0, 0).
First,
~ 0 (t) = h1, 2t, 3t2 i
R
~ 00 (t) = h0, 2, 6ti
R
√
~ 0 (t)|| = 1 + 4t2 + 9t4
||R
~i ~j ~k ~ 0 (t) × R
~ 00 (t) = 1 2t 3t2 = h6t2 , −6t, 2i.
R
0 2 6t Then
~ 0 (t) × R
~ 00 (t)|| =
||R
√
√
36t4 + 36t2 + 4 = 2 9t4 + 9t2 + 1.
Thus, the curvature is
√
~ 0 (t) × R
~ 00 (t)||
||R
2 1 + 9t2 + 9t4
κ(t) =
=
.
~ 0 (t)||3
(1 + 4t2 + 9t4 )3/2
||R
At the origin the curvature is κ(0) = 2.
Theorem: (Curvature of a Plane Curve)
The curvature of the plane curve y = f (x) is given by
κ(x) =
|f 00 (x)|
.
[1 + (f 0 (x))2 ]3/2
Example: Find the curvature of y = sin x at a general point and at
π
2
,0 .
Since y 0 = cos x and y 00 = − sin x,
|y 00 |
| sin x|
κ(x) =
=
.
0
2
3/2
[1 + (y ) ]
(1 + cos2 x)3/2
At
π
2
, 0 the curvature is κ(1) = 1.
~
Definition: Consider the curve defined by R(t).
The principal unit normal vector is given
by
~0
~ (t) = T (t) ,
N
||T~ 0 (t)||
and the binormal vector is given by
~
~ (t).
B(t)
= T~ (t) × N
~ (t), and B(t)
~
Note: The vectors T~ (t), N
form a set of orthogonal vectors, called a frame, that
moves along the curve as t varies. This frame has applications in differential geometry and
motion of spacecraft.
Figure 2: Illustration of the frame formed by the unit tangent, unit normal, and binormal
vectors.
~
= hsin t, cos t, ti.
Example: Find the unit normal and binormal vectors for the helix R(t)
~ 0 (t) = hcos t, − sin t, 1i,
Since R
~ 0 (t)|| =
||R
p
√
cos2 t + sin2 t + 1 = 2.
Thus, the unit tangent vector is
~ 0 (t)
R
1
T~ (t) =
= √ hcos t, − sin t, 1i.
0
~ (t)||
2
||R
Therefore,
1
T~ 0 (t) = √ h− sin t, − cos t, 0i,
2
The unit normal vector is
and
1
||T~ 0 (t)|| = √ .
2
~0
~ (t) = T (t) = h− sin t, − cos t, 0i.
N
||T~ 0 (t)||
Now the binormal vector is
~i
~j
1
~
~ (t) = √ cos t − sin t
B(t)
= T~ (t) × N
2 − sin t − cos t
~k
1
0
= √1 hcos t, − sin t, −1i.
2
~ and B
~ at a
Definition: The plane determined by the unit normal and binormal vectors N
point P on the curve C is called the normal plane of C at P . The unit tangent vector is
orthogonal to the normal plane. Similarly, the plane determined by the unit tangent and
~ is called the osculating plane of C at P . The binormal vector
unit normal vectors T~ and N
is orthogonal to the osculating plane.
~
= hsin t, cos t, ti
Example: Find equations for the normal and osculating planes of the helix R(t)
at the point P = (0, −1, π).
~ 0 (t) = hcos t, − sin t, 1i is normal to the normal plane. At the point P ,
The tangent vector R
~ 0 (π) = h−1, 0, 1i. Thus, an equation of the normal plane is
R
−1(x − 0) + 0(y + 1) + 1(z − π) = 0
−x + z = π.
The binormal vector is normal to the osculating plane. From the previous example,
1
~
B(t)
= √ hcos t, − sin t, −1i
2
and so
1
~
B(π)
= √ h−1, 0, −1i.
2
An easier normal vector is h−1, 0, 1i. So an equation of the osculating plane is
−1(x − 0) + 0(y + 1) − 1(z − π) = 0
z − x = −π
x − z = π.