Math 227 Problem Set II Solutions
1. Find the curvature of the plane curve y = ex .
Solution. This curve has equation y = f (x) with f (x) = ex . So
κ(x) =
f ′′ (x)
ex
=
′
2
3/2
[1 + f (x) ]
[1 + e2x ]3/2
2. Find the minimum and maximum values for the curvature of the ellipse x(t) = a cos t, y(t) = b sin t.
Here a > b > 0.
Solution. For the given ellipse
r(t) = a cos t ı̂ı + b sin t ̂
v(t) = −a sin t ı̂ı + b cos t ̂
|v(t)| =
a(t) = −a cos t ı̂ı − b sin t ̂
p
a2 sin2 t + b2 cos2 t
v(t) × a(t) = ab k̂
ab
|v(t) × a(t)|
= 2 2
κ(t) =
3
|v(t)|
[a sin t + b2 cos2 t]3/2
Hence the maximum (minimum) curvature is achieved when the denominator is a minimum (maximum)
which is the case when sin t = 0 (cos t = 0). So κmax = ba2 and κmin = ab2 .
3. Let r = f (θ) be the equation of a plane curve in polar coordinates. Find the curvature of this curve at
a general point θ.
Solution. Think of θ as a time parameter. Then the given curve has
r(θ) = f (θ) cos θ ı̂ı + sin θ ̂
v(θ) = f ′ (θ) cos θ ı̂ı + sin θ ̂ + f (θ) − sin θ ı̂ı + cos θ ̂
p
|v(θ)| = f ′ (θ)2 + f (θ)2
a(θ) = f ′′ (θ) − f (θ) cos θ ı̂ı + sin θ ̂ + 2f ′ (θ) − sin θ ı̂ı + cos θ ̂
v(θ) × a(θ) = 2f ′ (θ)2 − f (θ)[f ′′ (θ) − f (θ)] k̂
f (θ)2 + 2f ′ (θ)2 − f (θ)f ′′ (θ)
|v(θ) × a(θ)|
=
κ(θ) =
|v(θ)|3
[f (θ)2 + f ′ (θ)2 ]3/2
The computation of the cross product v(θ) × a(θ) (and also of |v(θ)|) has been facilitated by the
observation that êr (θ) = cos θ ı̂ı + sin θ ̂ and êθ (θ) = − sin θ ı̂ı + cos θ ̂ are mutually perpendicular unit
vectors obeying êr (θ) × êθ (θ) = k̂ and êr (θ) × êr (θ) = êθ (θ) × êθ (θ) = 0.
4. Find the curvature of the cardiod r = a(1 − cos θ).
Solution. By the previous problem with
f (θ) = a(1 − cos θ)
we have
κ(θ) =
f ′ (θ) = a sin θ
f (θ)2 + 2f ′ (θ)2 − f (θ)f ′′ (θ)
f ′′ (θ) = a cos θ
[f (θ)2 + f ′ (θ)2 ]3/2
2
a − 2a2 cos θ + a2 cos2 θ + 2a2 sin2 θ − a2 cos θ + a2 cos2 θ
=
[a2 − 2a2 cos θ + a2 cos2 θ + a2 sin2 θ]3/2
3
3
3a2 − 3a2 cos θ
= p
= 3/2 √
=
2
2
3/2
[2a − 2a cos θ]
2 a 1 − cos θ
2 2ar(θ)
1
5. Find the unit tangent, unit normal and binormal vectors and the curvature and torsion of the curve
r(t) = tı̂ı +
t2
2
̂ +
t3
3
k̂
Solution. For the specified curve
r(t) = tı̂ı +
t2
2
̂ +
t3
3
k̂
2
v(t) = ı̂ı + t ̂ + t k̂
a(t) = ̂ + 2t k̂
v(θ) × a(θ) = t2 ı̂ı − 2t ̂ + k̂
a′ (t) = 2 k̂
From this, we read off
ı̂ı + t ̂ + t2 k̂
v(t)
= √
|v(t)|
1 + t2 + t4
√
|v(t) × a(t)|
1 + 4t2 + t4
κ(t) =
=
|v(t)|3
[1 + t2 + t4 ]3/2
T̂(t) =
t2 ı̂ı − 2t ̂ + k̂
v(t) × a(t)
=√
|v(t) × a(t)|
1 + 4t2 + t4
−(t + 2t3 )ı̂ı + (1 − t4 ) ̂ + (2t + t3 )k̂
√
√
N̂(t) = B̂(t) × T̂(t) =
1 + t2 + t4 1 + 4t2 + t4
′
v(t) × a(t) · a (t)
2
=
τ (t) =
2
|v(t) × a(t)|
1 + 4t2 + t4
B̂(t) =
6. (a) Show that if a curve has curvature κ(s) = 0 for all s, then the curve is a straight line.
(b) Show that if a curve has torsion τ (s) = 0 for all s, then the curve lies in a plane.
(c) Show that if a curve has curvature κ(s) = κ0 , a strictly positive constant, and torsion τ (s) = 0 for
all s, then the curve is a circle.
dr
Solution. (a) If κ(s) ≡ 0, then ddsT̂ = κ(s)N̂(s) ≡ 0 so that T̂ is a constant. As a result ds
(s) = T̂ and
r(s) = sT̂ + r(0) so that the curve is the straight line with direction vector T̂ that passes through r(0).
(b) If τ (s) ≡ 0, then ddsB̂ = −τ (s)N̂(s) ≡ 0 so that B̂ is a constant. As T̂(s) ⊥ B̂,
d
ds (r(s)
− r(0)) · B̂ = T̂(s) · B̂ = 0
and (r(s) − r(0)) · B̂ must be a constant. The constant must be zero (set s = 0), so (r(s) − r(0)) · B̂ = 0
and r(s) always lies in the plane through r(0) with normal vector B̂.
(c) Parametrize the curve by arc length. Define the “centre of curvature” at s by
rc (s) = r(s) +
1
κ(s) N̂(s)
Since κ(s) = κ0 is a constant and τ (s) ≡ 0,
d
ds rc (s)
= T̂(s) +
1
κ0
τ (s)B̂ − κ(s)T̂ = T̂(s) +
1
κ0
0B̂ − κ0 T̂ = 0
Thus rc (s) = rc is a consant and r(s) − rc = κ10 lies on the sphere of radius
τ (s) ≡ 0, the curve also lies on a plane, so it is a circle.
2
1
κ0
centred on rc . Since
7. Solve the initial value problem
dr
dt
= k̂ × r
r(0) = ı̂ı + k̂
Describe the curve r = r(t).
Solution. The velocity vector dr
dt = k̂ × r is always perpendicular to k̂. Consequently, the curve lies in
a plane perpendicular to k̂. Furthermore, the velocity vector is always perpendicular to r. So the curve
lies on a sphere. (See problem #7 of Problem Set I.) The intersection of a sphere with a plane is a circle,
so the path is a circle. We are told that r(0) = ı̂ı + k̂. Guess r(t) = cos tı̂ı + sin t ̂ + k̂. As
cos tı̂ı + sin t ̂ + k̂
= ı̂ı + k̂
t=0
d
 + k̂ = − sin tı̂ı + cos t ̂
dt cos tı̂ı + sin t ̂
k̂ × cos tı̂ı + sin t ̂ + k̂ = cos t ̂ − sin tı̂ı
the guess satisfies both the required differential equation and the required initial condition.
8. Find the curvature κ as a function of arclength s (measured from (0, 0)) for the curve
x(θ) =
Z
0
θ
cos
2
1
2 πt
dt
y(θ) =
Z
θ
sin
0
2
1
2 πt
dt
Solution. The velocity vector is
v(θ) = x′ (θ)ı̂ı + y ′ (θ) ̂ = cos
2
1
2 πθ
ı̂ı + sin
2
1
2 πθ
̂
Consequently the speed
ds
dθ (θ)
= |v(θ)| = 1
=⇒
s(θ) = θ + s(0)
Since s(θ) is zero when θ = 0, we have s(θ) = θ and hence T̂(s) = v(s) = cos 21 πs2 ı̂ı + sin
that
T̂ (s) = − πs sin 21 πs2 ı̂ı + πs cos 21 πs2 ̂ = πs
κ(s) = dds
1
2
2 πs
̂ so
9. A frictionless roller–coaster track has the form of one turn of the circular helix with parametrization
(a cos θ, a sin θ, bθ). A car leaves the point where θ = 2π with zero velocity and moves under gravity to
the point where θ = 0. By Newton’s law of motion, the position r(t) of the car at time t obeys
mr′′ (t) = N r(t) − mg k̂
Here m is the mass of the car, g is a constant, −mg k̂ is the force due to gravity and N r(t) is the
force that the roller–coaster track applies to the car to keep the car on the track. Since the track is
frictionless, N r(t) is always perpendicular to v(t) = dr
dt (t).
1
2
(a) Prove that E(t) = 2 m|v(t)| +mgr(t)· k̂ is a constant, independent of t. (This is called “conservation
of energy”.)
(b) Prove that the speed |v| at the point θ obeys |v|2 = 2gb(2π − θ).
(c) Find the time it takes to reach θ = 0.
Solution. (a) By Newton’s law of motion
E ′ (t) =
2
d 1
dt 2 m|v(t)|
=0
+ mgr(t) · k̂ = mv(t) · v′ (t) + mgv(t) · k̂ = v(t) · N r(t) − mg k̂ + mgv(t) · k̂
since v(t) · N r(t) = 0. So E(t) is a constant, independent of t.
3
(b) By part (a),
E(t) = E(0)
2
1
2 m|v(t)|
=⇒
+ mgbθ(t) = mg(2πb)
=⇒
|v(t)|2 = 2gb 2π − θ(t)
(c) Now
v=
dr
dt
=
dr dθ
dθ dt
= − a sin θ, a cos θ, b
dθ
=⇒
2
|v|2 = a2 + b2 ] dθ
dt
2 1/2
1/2
= − a|v|
= − 2gb(2π−θ)
2 +b2
a2 +b2
=⇒
dt
dθ
dt
We have chosen the negative sign because θ must decrease from 2π to 0. The time required to do so is
Z
dt =
=
Z
0
2π
dt
dθ dθ
=−
a2 +b2 1/2 2gb
a2 +b2 1/2
2gb
− 2(2π −
Z
0
1
dθ
(2π−θ)1/2
2π
1/2 2π
θ)
0
4
=2
=
a2 +b2 1/2
2gb
a2 +b2 1/2
gb π
Z
2π
0
1
dθ
(2π−θ)1/2