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Homework set 3 - Solutions
Math 407 – Renato Feres
1. (Exercise 2.2, page 24 of notes.) The locus of the points on the coordinate plane satisfying the algebraic equation
¢
¡
(x 2 + y 2 )2 − a 2 x 2 − y 2 = 0
is a curve called a lemniscate. Give a parametrization of the curve in polar coordinates. That is, express the curve
as a function of the angle parameter θ by writing
x(θ) = ρ(θ) cos θ, y(θ) = ρ(θ) sin θ.
Draw a careful sketch of the curve.
Solution. Let us first note that the geometric curve C represented by this equation has the following symmetries:
(x, y) ∈ C ⇒ (−x, y), (x, −y) ∈ C .
So it is enough to know what the curve is like on the positive quadrant of the coordinate plane. Also observe that
y 2 ≤ x 2 , so it is enough to consider the range θ ∈ [0, π/4]. Now, simple substitution gives ρ 4 − a 2 ρ 2 cos(2θ) = 0.
Disregarding the trivial case ρ(θ) = 0 we obtain
ρ(θ)2 = a 2 cos(2θ), 0 ≤ θ ≤ π/4.
The above graph was plotted using R. Here is the script:
a=1
theta=seq(-pi/4,pi/4,length.out=200)
rho=a*sqrt(cos(2*theta))
x1=rho*cos(theta)
y1=rho*sin(theta)
x2=-x1
y2=y1
x=c(x1,x2)
y=c(y1,y2)
plot(x,y,type=’l’,asp=TRUE)
♦
2. (Exercise 2.4, page 26 of notes.) Show that a regular curve f (s) between two points p, q ∈ Rn with minimal length
is necessarily the line segment from p to q. Suggestion: Let u = q − p and consider the Schwarz inequality
〈u, f 0 (s)〉 ≤ kukk f 0 (s)k.
Solution. Let [a, b] be the domain of f , so f (a) = p and f (b) = q. Integrating both sides of the inequality
〈u, f 0 (s)〉 ≤ kukk f 0 (s)k from a to b gives
¿ Z
kq − pk = 〈u, q − p〉 = u,
2
b
a
0
À
Z
f (s) d s =
b
a
Z
0
〈u, f (s)〉 d s ≤ kuk
b
a
k f 0 (s)k d s = kq − pk × length( f ).
This shows that the length of f is greater than or equal to the length kq − pk of the straight line between p and
q. By the assumption of minimality, the two curves must have the same length. This implies that the Schwarz
inequality must be an equality for all s: 〈u, f 0 (s)〉 = kukk f 0 (s)k. It follows that u and f 0 (s) must be collinear:
f 0 (s) = k f 0 (s)ku/kuk. Therefore, f (s) must be a straight line. (Note that the parametrization does not have to be
uniform.)
♦
3. (Exercise 2.5, page 32 of notes.) Show that a plane curve has constant curvature κ if and only if it is part of a
circle of radius 1/|κ| (if κ 6= 0) or a line segment (if κ = 0.) Note: if κ(s 0 ) 6= 0, then c(s) + κ(s10 ) u 2 (s) is constant if
and only if c(s) is part of a circle of radius 1/|κ(s 0 )|.
Solution. It is not difficult to show that if a curve c(s) is part of a circle of radius R, then it has constant curvature
equal to 1/R. The interesting direction to prove is that is κ(s) = κ is constant then the curve is part of a circle.
Consider the curve f (s) := c(s) + κ1 u 2 (s). Differentiating both sides gives
f 0 (s) = c 0 (s) +
1 0
1
u (s) = u 1 (s) + (−κu 1 (s)) = 0.
κ 2
κ
Therefore,
c(s) = a −
1
u 2 (s)
κ
where a is a constant vector. But u s (s) is a unit vector for all s so that
kc(s) − ak =
1
.
κ
This shows that c(s) is part of a circle of center a and radius 1/κ.
♦
4. (Exercise 2.6, page 33 of notes.) The cardioid is the parametric curve with coordinates
x(t ) =
a
a
(2 cos t − cos 2t ), y(t ) = (2 sin t − sin 2t ).
2
2
2
(a) Draw a careful sketch of this curve.
(b) Compute its curvature function, as a function of the parameter t .
♦
Solution. (a) The following graph was obtaining using the R-script:
a=2
t=seq(0,2*pi,length.out=400)
x=2*cos(t)-cos(2*t)
y=2*sin(t)-sin(2*t)
plot(x,y,type=’l’,asp=TRUE)
Note that I used a = 2.
(b) Recall that if c(s) is a curve parametrized by arclength, then u 1 (s) = c 10 (s) and u 10 (s) = κ(s)u 2 (s). Now let
c(t ) = (x(t ), y(t )), where t is not arclength. The unit tangent vector field to c(t ) is then u 1 (s(t )) = c 0 (t )/ṡ(t ), where
ṡ = kc 0 (t )k, and u 2 is such that u 1 , u 2 form a positive orthonormal basis of R2 . Differentiating c 0 (t ) = ṡu 1 (s(t ))
gives
c 00 (t ) = s̈u 1 + ṡ 2 κu 2 , κ = 〈c 00 , u 2 〉/ṡ 2 .
The first derivative of c(t ) is
c 0 (t ) = a (sin(2t ) − sin t , cos t − cos(2t )) , ṡ 2 = kc 0 (t )k2 = 2a 2 (1 − cos t ) = 4a 2 sin2 (t /2).
So u 1 = c 0 (t )/ṡ and
u1 =
1
1
(sin(2t ) − sin t , cos t − cos(2t )) , u 2 =
(− cos t + cos(2t ), sin(2t ) − sin t ) .
2 sin(t /2)
2 sin(t /2)
The second derivative of c(t ) is
c 00 (t ) = a (2 cos(2t ) − cos t , 2 sin(2t ) − sin t ) .
3
The dot-product of c 00 and u 2 is easily computed: 〈c 00 , u 2 〉 = 3a sin(t /2). Given the already computed value of ṡ 2
we arrive at
κ=
3
.
4a sin(t /2)
♦
5. (Exercise 2.7, page 34 of notes.) A helix is a curve in R3 for which we can find a unit vector u such that the angle
between u and tangent vector to the curve is constant. Thus T (s) · u = C for some constant C . If the Frenet
curve c(s) parametrized by arclength is a helix, show that the ratio κ/τ is constant. (Note: the curve being Frenet
implies κ > 0.)
Solution. We have T · u constant if and only if κN · u = T 0 · u = 0 if and only if N · u = 0 (since κ > 0) if and only if
u = c 1 T + c 2 B where c 1 , c 2 are constants. Taking derivatives:
0 = c 1 T 0 + c 2 B 0 = c 1 κN + c 2 (−τ)N .
Therefore, c 1 κ = c 2 τ, which shows that the ratio κ/τ is constant.
♦
6. (This problem is optional. It is Exercise 2.9, page 34 of notes.) A material particle of mass m is subject to a force
field F and its motion, given by a time-parametrized curve γ(t ), is governed by Newton’s equation
mγ00 (t ) = F (t , γ(t ), γ0 (t )).
(a) Show that, with respect to the Frenet frame (T, N , B ), the tangent and normal components of the acceleration γ̈ are given by
1
1
F · T = s̈,
F · N = (ṡ)2 κ
m
m
where s(t ) is the arclength parametrized by time and the the dots on top of letters represent time derivative.
(b) A particle of mass m and electric charge q in an electric field E and magnetic field B is subject to the
Lorentz force, given by
£
¤
F (t , γ(t ), γ0 (t )) = q E (t , γ(t )) + γ0 (t ) × B(t , γ(t ))
where the curve γ(t ) ∈ R3 gives the position of the particle at time t . Show that
m s̈ = qE · T
m ṡ 2 κ = q (E · N − ṡB · B )
0 = E · B + ṡB · N .
(c) Describe the motion when E and B are constant.
♦
Solution. (a) We write ṡ = kγ0 (t )k and γ0 (t ) = ṡT (t ). The second derivative is
γ00 (t ) = s̈T (t ) + ṡT 0 (t ) = s̈T (t ) + κṡ 2 N (t ).
4
(1)
This shows that the tangential component of the acceleration is s̈ and the normal component of the acceleration is κṡ 2 as claimed.
(b) Note that γ0 × B is perpendicular to B , so F · T = qE · T. Due to part (a) this implies that
m s̈ = mγ00 · T = qE · T
which proves the first equation. Taking now the inner product with N gives
mκṡ 2 = mγ00 · N = F · N = q[E · N + ṡ(T × B) · N ] = q[E · N − ṡB · (T × N )] = q(E · N − ṡB · B ).
This is the second identity. From equation 1 we see that γ00 does not have component in the direction of B ,
so 0 = mγ00 · B = mq[E + γ0 × B] · B. This implies that
0 = E · B + ṡ(T × B) · B = E · B − ṡ(T × B ) · B = E · B + ṡB · N .
Therefore,
0 = E · B + ṡB · N
as claimed.
(c) The case B = 0 is much easier than the general case and can be studied separately, so I will assume
B 6= 0. It is convenient to write v(t ) := γ0 and v̇ = γ00 . Then v̇ = z − Z v where
z :=
³q ´
q
E , Z := ω
B .
m
m
Recall that

0
−a 3

ω(a 1 , a 2 , a 3 ) =  a 3
0
a1
−a 2
a2


−a 1  .
0
This first order matrix-valued differential equation can be solved by the standard method of integrating
factor:
d ¡ tZ ¢
e v = e t Z v̇ + e t Z Z v = e t Z z.
dt
Integrating from 0 to t :
e
tZ
v(t ) − v(0) =
Therefore,
v(t ) = e −t Z v(0) +
t
Z
t
Z
0
0
e τZ z d τ.
e (τ−t )Z z d τ = e −t Z v(0) +
t
Z
0
e −τZ z d τ.
Let us define Ωc := qkBk/m. The absolute value of Ωc is known as the Larmor frequency or cyclotron
frequency. If w is any vector, we may write w = w ∥ + w ⊥ where w ∥ is the projection of w in the direction of
B and w ⊥ is the projection of w on the plane perpendicular to B. Note that
e
and that
t
Z
0
−t Ωc ω(B/kBk)
B
w = w ∥ + cos(−t Ωc )w ⊥ + sin(−t Ωc )
× w⊥
kBk
e −τΩc ω(B/kBk) w d τ = w ∥ t +
µ
¶
µ
¶
1
1
B
sin(t Ωc )w ⊥ +
(cos(t Ωc ) − 1)
× w⊥ .
Ωc
Ωc
kBk
5
Therefore,
v(t ) = v ∥ (0) +
¶
µ
¶
µ
B
E ⊥ + v ⊥ (0) × B
q
E⊥ × B
E ⊥ + v ⊥ (0) × B
+
cos(t
Ω
)
+
sin(t
Ω
)
.
E∥ t +
×
c
c
m
kBk2
kBk
kBk
kBk
Finally, we integrate again from 0 to t to obtain the curve γ(t ). Before doing it, let ζ :=
constant vector. Then, expressing γ(t ) in terms of its projections γ∥ (t ) and γ⊥ (t ),
1 q
E∥ t 2
2m
1
1
E⊥ × B
1
B
γ⊥ (t ) =
t−
× ζ.
ζ+
cos(Ωc t )ζ +
sin(Ωc t )
2
Ωc
kBk
Ωc
Ωc
kBk
γ∥ (t ) = γ(0) + v ∥ (0)t +
6
E ⊥ +v ⊥ (0)×B
.
kBk
This is a