Math 227 Problem Set III Solutions
1. Let Y (x) be an infinitely differentiable function on −1 < x < 1. Parametrize the curve y = Y (x),
−1 < x < 1, by x. That is, set r(t) = tı̂ı + Y (t)̂.
(a) Develop formulae for κ(t), T̂(t), N̂(t), B̂(t) and the centre of curvature c(t). (View the xy–plane as
the plane z = 0 in IR3 .)
(b) Give an example of an infinitely differentiable function Y (x) on −1 < x < 1 for which N̂(t), B̂(t)
and c(t) are infinitely differentiable except at t = 0, where they are not even continuous. Sketch the
curve, a circle of curvature for one x < 0 and a circle of curvature for one x > 0.
Solution. Since
v(t) = ı̂ı + Y ′ (t) ̂
r(t) = tı̂ı + Y (t) ̂
a(t) = Y ′′ (t) ̂
v(t) × a(t) = Y ′′ (t) k̂
and
ds
dt (t) T̂(t)
v(t) =
v(t) × a(t) = κ(t)
3
ds
dt (t)
B̂(t)
N̂(t) = B̂(t) × T̂(t)
we have
ds
dt (t)
=
Y ′′ (t) k̂
ı̂ı + Y ′ (t) ̂
B̂(t) =
T̂(t) = p
= sgn Y ′′ (t) k̂
′′
′
2
|Y (t)|
1 + Y (t)
′
−Y (t)ı̂ı + ̂
N̂(t) = sgn Y ′′ (t) p
1 + Y ′ (t)2
p
1 + Y ′ (t)2
|Y ′′ (t)|
κ(t) = 3/2
1 + Y ′ (t)2
and
c(t) = r(t) +
1
κ(t) N̂(t)
= tı̂ı + Y (t) ̂ +
1 + Y ′ (t)2 − Y ′ (t)ı̂ı + ̂
′′
Y (t)
Here, sgn Y ′′ (t) is the sign of Y ′′ (t) (that is, +1 when Y ′′ (t) > 0 and −1 when Y ′′ (t) < 0) and B̂(t),
N̂(t) and c(t) are all not defined when Y ′′ (t) = 0.
(b) Let Y (x) = sin(πx) for −1 < x < 1. Then Y ′ (x) = π cos(πx) , Y ′′ (x) = −π 2 sin(πx) and
sgn(Y ′′ (x)) = −sgn(x). So
B̂(t) = k̂
n
1
if −1 < t < 0
−1 if 0 < t < 1
o
and
c(t) = tı̂ı + sin(πt) ̂ −
Here is the sketch for Y (x) =
nicer.
1
2
o
−π cos(πt)ı̂ı + ̂ n 1
if −1 < t < 0
N̂(t) = p
1 + π 2 cos2 (πt) −1 if 0 < t < 1
1 + π 2 cos2 (πt) − ı̂ı π cos(πt) + ̂
2
π sin(πt)
sin(πx). I have just put in the extra factor of
1
1
2
to make the sketch look
2. In this problem, we define what we mean by “the circle that fits the parametrized curve C best near r0 ”.
◦ For convenience, let r(s) be a parametrization of C by arc length with r(0) = r0 . Define T̂, N̂ and
d2 r d2 r
κ by T̂ = dr
ds (0), κ = ds2 (0) and κN̂ = ds2 (0). In this problem, we only consider the case that
κ > 0, so that N̂ is a well defined unit vector that is perpendicular to T̂.
◦ Pick any c ∈ IR3 , any ρ′ > 0 and any two mutually perpendicular unit vectors T̂′ and N̂′ . Then
R(s) = c − ρ′ cos ρs′ N̂′ + ρ′ sin ρs′ T̂′
is a circle, parametrized by arc length. We may parametrize any circle by choosing c, ρ′ , T̂′ and
N̂′ appropriately. (See the notes “Parametrizing Circles”.)
◦ Set D(s) = |R(s) − r(s)|2 . It is, of course, the square of the distance from the point R(s) on the
circle to the point r(s) on C.
◦ We’ll say that the circle above fits C best near r0 if D(0) = D′ (0) = D′′ (0) = D(3) (0) = D(4) (0) = 0.
p
(These conditions imply that the distance from r(s) to R(s), i.e. D(s), goes to zero faster than
order s2 as s tends to zero.)
Prove that D(0) = D′ (0) = D′′ (0) = D(3) (0) = D(4) (0) = 0 if and only if T̂′ = T̂, N̂′ = N̂, ρ′ = κ1 and
c = r0 + κ1 N̂.
Solution. Observe that
R(0) = c − ρ′ N̂′
R′ (0) = sin ρs′ N̂′ + cos ρs′ T̂′ s=0 = T̂′
R′′ (0) = ρ1′ cos ρs′ N̂′ − sin ρs′ T̂′ s=0 = ρ1′ N̂′
First, D(0) = 0 if and only if R(0) = r(0), which is the case if and only if
c − ρ′ N̂′ = r0
(1)
Next,
D′ (0) = 2 R′ (s) − r′ (s) · R(s) − r(s) s=0 = R′ (0) − r′ (0) · R(0) − r(0)
is automatically zero if R(0) = r(0). Next, assuming that R(0) = r(0),
h
i
D′′ (0) = 2 R′′ (s)−r′′ (s) · R(s)−r(s) + 2 R′ (s)−r′ (s) · R′ (s)−r′ (s)
s=0
′
2
′
= 2 R (0) − r (0)
is zero if and only if R′ (0) = r′ (0), which is the case if and only if
T̂′ = T̂
(2)
Next,
h
i
D(3) (0) = 2 R(3) (s)−r(3) (s) · R(s)−r(s) + 6 R′′ (s)−r′′ (s) · R′ (s)−r′ (s)
s=0
is automatically zero if R(0) = r(0) and R′ (0) = r′ (0). Finally, assuming that R(0) = r(0) and
R′ (0) = r′ (0),
h
D(4) (0) = 2 R(4) (s)−r(4) (s) · R(s)−r(s) + 8 R(3) (s)−r(3) (s) · R′ (s)−r′ (s)
i
+ 6 R′′ (s)−r′′ (s) · R′′ (s)−r′′ (s)
s=0
′′
2
′′
= 6 R (0) − r (0)
is zero if and only if R′′ (0) = r′′ (0), which is the case if and only if
′
1
ρ′ N̂
= κN̂ ⇐⇒ ρ′ = κ1 , N̂′ = N̂
Combining (1,2,3) gives all of the desired conditions.
2
(3)
3. The Frenet–Serret formulae may be written
 

T̂(s)
0
κ(s)
d 
0
N̂(s)  =  −κ(s)
ds
0
−τ (s)
B̂(s)


T̂(s)
0
τ (s)   N̂(s) 
0
B̂(s)
This is a system of first order linear ordinary differential equations. In general, a system of first order
linear ordinary differential equations is one of the form dx
ds (s) = M (s) x(s), where for each s ≥ 0,
n
x(s) ∈ IR is an n–component vector and M (s) is an n × n matrix. Let
◦ x0 ∈ IRn
◦ for each s ≥ 0, M (s) be an n × n matrix
Assume that each matrix element of M (s) is continuous in s. In the case of the Frenet–Serret formulae,
we also have that
◦ M (s) is antisymmetric, meaning that M (s)ij = −M (s)ji for all 1 ≤ i, j ≤ n and s ≥ 0
(a) Assume that M (s) is antisymmetric. Prove that if x(s) obeys dx
ds (s) = M (s) x(s) for all s > 0, then
|x(s)| is a constant, independent of s.
(b) Again assume that M (s) is antisymmetric. Prove that if both xa (s) and xb (s) solve the initial value
problem dx
ds (s) = M (s) x(s), x(0) = x0 , then xa (s) = xb (s) for all s ≥ 0.
(c) Let κ(s) and τ (s) be continuous functions. Assume that T(s), N(s), B(s) obeys the Frenet–Serret
formulae and that T(0), N(0) and B(0) are mutually perpendicular unit vectors. Prove that, for each
s > 0, T(s), N(s) and B(s) are also mutually perpendicular unit vectors. Hint: It is true, but beyond
the scope of this course to prove, that there is a unique solution to the initial value problem x(0) = x0 ,
dx
ds (s) = M (s) x(s). You may use this result.
Solution. (a)
2
d
ds |x(s)|
=
d
ds [x(s)
· x(s)] = M (s)x(s) · x(s) + x(s) · M (s)x(s)
X X
M (s)ij x(s)j x(s)i +
x(s)j M (s)ji x(s)i
=
1≤i,j≤n
=
X
1≤i,j≤n
=0
1≤i,j≤n
x(s)j M (s)ij + M (s)ji x(s)i
(since M (s) is antisymmetric)
(b) Let x(s) = xa (s) − xb (s). Then x(0) = 0 and
dx
ds (s)
=
dxa
ds (s)
−
dxb
ds (s)
= M (s) xa (s) − M (s) xb (s) = M (s)x(s)
So |x(s)| is a constant, independent of s, and that constant is 0, since x(0) = 0. So |x(s)| = 0 for all s
and xa (s) = xb (s) for all s ≥ 0.
(c) Set


 
T(s) · T(s)
1
 N(s) · N(s) 
1


 
 B(s) · B(s) 
1
x(s) = 
x0 =  

 T(s) · N(s) 
0


 
T(s) · B(s)
0
N(s) · B(s)
0
3
and
0
0
0
0
0
 0

0
0
 0
M (s) = 
0
 −κ(s) κ(s)

0
0
0
0
−τ (s) τ (s)

Then, x(0) = x0 and, by the Frenet–Serret formulae,
d
ds N(s)

2κ(s)
0
0
−2κ(s)
0
2τ (s) 

0
0
−2τ (s) 

0
τ (s)
0 

−τ (s)
0
κ(s)
0
−κ(s)
0
dx
ds (s)
= M (s) x(s). For example
· B(s) = N′ (s) · B(s) + N(s) · B′ (s)
= [−κ(s)T(s) + τ (s)B(s)] · B(s) + N(s) · [−τ (s)N(s)]
= −τ (s)N(s) · N(s) + τ (s)B(s) · B(s) − κ(s)T(s) · B(s)
Since M (s)x0 = 0 for all s, the constant function x̃(s) = x0 also solves the initial value problem
x(0) = x0 , dx
ds (s) = M (s) x(s). By uniqueness of solutions to initial value problems x(s) = x̃(s) = x0
for all s ≥ 0. This is precisely the statement that T(s), N(s) and B(s) are mutually perpendicular unit
vectors.
4. Do either part (a) or part (b) of the following question.
Let the curve r(s) be parametrized by arc length and have κ(s) > 0 and τ (s) 6= 0.
(a) Suppose that the curve lies on the sphere with centre c and radius R. Prove that
r(s) − c = −ρ(s)N̂(s) − ρ′ (s)σ(s) B̂(s)
1
1
where ρ(s) = κ(s)
and σ(s) = τ (s)
. In particular R2 = ρ(s)2 + ρ′ (s)2 σ 2 .
2
(b) Prove that, conversely, if ρ(s) + ρ′ (s)2 σ(s)2 is a constant and ρ′ (s) 6= 0, then r(s) lies on a sphere.
Hint: Prove that r(s) + ρ(s) N̂(s) + ρ′ (s)σ(s) B̂(s) is a constant.
Solution. (a) Since |r(s) − c|2 = R2 is a constant, the Frenet-Serret formulae give
0=
0=
d
ds r(s) − c ·
d2
ds2 r(s) − c ·
r(s) − c = 2 r(s) − c · T̂(s)
r(s) − c = 2T̂(s) · T̂(s) + 2 r(s) − c · κ(s)N̂(s)
⇒
⇒
r(s) − c · T̂(s) = 0
1
r(s) − c · N̂(s) = − κ(s)
Differentiating yet again gives
0 = 4κ(s)N̂(s) · T̂(s) + 2T̂(s) · κ(s)N̂(s) + 2 r(s) − c · κ′ (s)N̂(s)
+ 2 r(s) − c · κ(s) − κ(s) T̂(s) + τ (s) B̂(s)
′
(s)
= −2 κκ(s)
+ 2κ(s)τ (s) r(s) − c · B̂(s)
which implies that
r(s) − c · B̂(s) =
κ′ (s)
κ(s)2 τ (s)
= −ρ′ (s)σ(s)
So the component of r(s) − c in the T̂(s) direction is 0, in the N̂(s) direction is −ρ(s) and in the B̂(s)
direction in −ρ′ (s)σ(s). That’s what we were to show.
(b) By hypothesis, ρ(s)2 + ρ′ (s)2 σ(s)2 is a constant (call it R2 ) so that
0 = 2ρ(s)ρ′ (s) + 2ρ′ (s)ρ′′ (s)σ(s)2 + 2ρ′ (s)2 σ(s)σ ′ (s)
Since ρ′ (s) and σ(s) are nonzero, we may divide by 2ρ′ (s)σ(s) to give
ρ(s)τ (s) + ρ′′ (s)σ(s) + ρ′ (s)σ ′ (s) = 0
4
Consequently, recalling that ρ(s)κ(s) = 1 and σ(s)τ (s) = 1,
d
ds
r(s) + ρ(s) N̂(s) + ρ′ (s)σ(s) B̂(s)
= T̂(s) + ρ′ (s) N̂(s) + ρ(s) − κ(s) T̂(s) + τ (s) B̂(s) + ρ′′ (s)σ(s) B̂(s) + ρ′ (s)σ ′ (s) B̂(s)
+ ρ′ (s)σ(s) − τ (s) N̂(s)
= ρ(s)τ (s) B̂(s) + ρ′′ (s)σ(s) B̂(s) + ρ′ (s)σ ′ (s) B̂(s)
=0
So c = r(s) + ρ(s) N̂(s) + ρ′ (s)σ(s) B̂(s) is a constant vector and
2
|r(s) − c|2 = ρ(s) N̂(s) + ρ′ (s)σ(s) B̂(s) = ρ(s)2 + ρ′ (s)2 σ(s)2 = R2
5