14.3 Arc Length and Curvature
Arc Length
Theorem (Arc length) Let r(t) =< x(t), y(t), z(t) > be a differentiable vector function. Let [a, b] be an interval in the
domain of r. Then the length L of the arc r from r(a) to r(b) can be computed by
Z
b
kr0 (t)kdt
L=
a
(Explanation) Suppose the interval [a, b] is cut into n subintervals of equal length ;
a = to < t1 < t2 < · · · < tn−1 < tn = b
If we let ∆xi = x(ti ) − x(ti−1 ), ∆yi = y(ti ) − y(ti−1 ), and ∆zi = z(ti ) − z(ti−1 ),
the arc length on the curve r from r(ti ) to r(ti−1 ) is, by Pythagorean Theorem,
∆Li =
From this, L ≈
p
(∆xi )2 + (∆yi )2 + (∆zi )2
n q
X
(∆xi )2 + (∆yi )2 + (∆zi )2 , and by assuming the interval [a, b] is divided into sufficiently many subintervals, we
i=1
have
L = lim
n q
X
n→∞
(∆xi )2 + (∆yi )2 + (∆zi )2
i=1
We conveniently use the differential notation dL and write its formula by
dL
=
p
(dx)2 + (dy)2 + (dz)2
s
=
=
Z
dx
dt
2
+
dy
dt
2
+
dz
dt
2
dt
kr0 (t)kdt
b
Notice L =
dL.
a
Example Find the length of the arc of the circular helix r(t) = cos(πt)i + sin(πt)j + tk when t changes from 0 to 3.
√
(Answer) L = 3 π 2 + 1
Definition (Arc length function s(t)) Let r(t) be a vector function who is differentiable in an interval [a, b]. Then we call
Z
s(t) =
a
t
kr0 (u)kdu =
Z tp
x0 (u)2 + y 0 (u)2 + z 0 (u)2 du
(a ≤ t ≤ b)
a
the arc-length function of r. The relation between s(t) and r(t) can be written by
ds
= kr0 (t)k
dt
Technique (Parametrization of a curve with respect to arc length) Let C be the curve drawn by a vector function
Z t
r(t) =< x(t), y(t), z(t) > which is differentiable in an interval [a, b], and let s = s(t) =
kr0 (u)kdu be
a
the arc length function of r. If we solve the equation s = s(t) for t to write t = t(s) and substitute it to
r(t), we get
r(s) =< x(t(s)), y(t(s)), z(t(s)) >
1
Above vector function draws the same curve C but uses s as its parameter. As a result, r(s) at s = c(c is a
constant) is the arc length of C from r(0) to r(c).
Example Reparametrize the curve
r(t) = e2t cos(2t)i + 2j + e2t sin(2t)k
with respect to arc length measured from the point where t = 0 in the direction of increasing t.
(Answer) We want to find an equation relating s and t, and use it to describe t in terms of s. By substituting the result t = t(s), we will get r(s).
t
Z
s(t) =
q
(2e2u cos 2t − 2 sin(2u)e2u )2 + 0 + (2e2u sin 2u + 2 cos(2u)e2u )2 du =
0
s+
√
2=
r(s) =
√
√
2e2t
√
s+ 2
cos
√
2
;
t
Z
√
8e4u du =
√
2e
2t
−
√
2
0
√
s+ 2
1
s+ 2
2t
=e
; t = ln √
√
2
2
2
√ !
√
√ !
s+ 2
s+ 2
s+ 2
ln √
i + 2j + √
sin ln √
k
2
2
2
Example Suppose you start at the point (0, 0, 3) and move 5 units along the curve < 3 sin t, 4t, 3 cos t > in the positive direction. Where are you now?
t
Z
(Answer) s(t) =
q
(3 cos u)2 + 42 + (3 sin u)2 du = 5t
;
t=
0
s
5
;
r(s) =< 3 sin
s 4
s
, s, 3 cos >
5 5
5
The answer is r(5) =< 3 sin 1, 4, 3 cos 1 >.
Curvature
Remark (Smooth parametrization and smooth curve) A parametrization(or a vector function) r(t) is said to be smooth
on an interval I if r0 is continuous and r0 6= 0 on the interval I. A curve is called smooth if it has a smooth
parametrization.
Definition Let r(t) be a vector function whose graph is a smooth curve. Let T =
of r. We define
dT κ=
ds the curvature of the curve.
T changes direction rapidly
T changes direction slowly
(Curvature is large)
(Curvature is small)
2
r(t)
be the unit tangent vector
kr0 (t)k
Formulas Let r(t) be a vector function whose graph is a smooth curve.
(a) Without re-parameterizing with respect to s, we can compute the curvature κ via
κ=
kT0 (t)k
kr0 (t)k
0
dT dt dT 1 T (t) dT kT0 (t)k
(Proof) κ := ds = dt ds = dt ds/dt = r0 (t) = kr0 (t)k
(b) Without finding T, we can compute the curvature κ via
κ=
kr0 (t) × r00 (t)k
kr0 (t)k3
(Proof) We start from the right side of the equation and yield the left side. Notice first that from T =


 r0
=
r0
,
kr0 k
ds
T
dt
d s
ds 0
T+
T
dt2
dt
kr0 kT =
2

 r00
Hence,
kr0 × r00 k
=
ds T ×
dt
=
2
ds d2 s
ds
0
T
×
T
+
T
×
T
dt dt2
dt
=
2
kT0 k
!
because T/ /T, T ⊥ T0 and kTk = 1
kr0 k2 kT0 k
=
Now,
ds
dt
d2 s
ds 0
T+
T
dt2
dt
=
kr0 × r00 k
kr0 k2 kT0 k
kT0 k
=
=
= κ.
kr0 k3
kr0 k3
kr0 k
Example Find the curvature of the curve :
(a) r(t) =
t,
t2 2
,t
2
(b) r(t) = 3ti + 4 sin tj + 4 cos tk
(Answers) (a)
√
5
,
1 + 5t2
(b)
4
25
Formulas For a curve y = f (x) in xy−plane, we can compute the curvature κ via
κ(x) =
|f 00 (x)|
[1 + (f 0 (x))2 ]3/2
(Proof) Let r(x) =< x, f (x) > and apply the formula in above (b). Since r0 (x) =< 1, f 0 (x) > and r00 (x) =< 0, f 00 (x) >,
i
r0 × r00 = 1
0
j
f0
00
f
k
0
0
= f 00 (x)k.
⇒
κ=
kr0 (t) × r00 (t)k
|f 00 (x)|
= p
and this proves the formula.
kr0 (t)k3
( 1 + f 0 (x))3
Example Find the curvature of the curve in xy−plane y = cos x at x = 0,π/4, and π/2.
√
(Answers) κ(0) = 1,
κ(π/4) =
2+
2
√ ,
2
κ(π/2) = 0.
3
The Normal and Binormal Vectors (T, N, B system)
Definition (The unit normal and binomial vectors)
(a) The unit vector in the direction of T0 is called the unit normal vector and we denote it by N. That is,
N=
T0
kT0 k
(Notice that T0 ⊥ T because T · T = 1 and by differentiating both sides with respect to t, we get
2T0 · T = 0.)
(b) We define the binormal vector of r as the cross product of T and N and denote it by B. That is,
B=T×N
Because T and N are unit vectors and perpendicular to each other, B is also a unit vector.
Remark Given a smooth curve C and its parametrization r, the three vectors T, N, and B provide a local axis
system like x,y,z axes do.
Definition (The normal plane and the osculating plane) Assume r(t) as before and let P = r(to ) be a point on the curve
drawn by r. Then
(a) the plane containing B and N is called the normal plane of r at r(to ),
(b) and the plane containing T and N is called the osculating plane of r at r(to ).
Example Consider the curve and the point P on the curve
r(t) =< cos t, sin t, ln cos t > at P (1, 0, 0)
(a) Find the vectors T, N, and B at P .
(b) Find equations of the normal plane and osculating plane of the curve at P .
1
1
N = √ < −1, 0, −1 >,
B = √ < −1, 0, 1 >
2
2
(b) Normal plane y = 0 and osculating plane x − z = 1.
(Answers) (a) T =< 0, 1, 0 >,
4