Module MA1132 (Frolov), Advanced Calculus
Homework Sheet 2
Each set of homework questions is worth 100 marks
Due: at the beginning of the tutorial session Thursday/Friday, 4/5 February 2016
Name:
1. Consider the vector function (with values in R3 )
√
(3 − t)2
k
(1)
r(t) = ln(3 − t) i + (1 + t) j +
4
(a) Find the arc length of the graph of r(t) if 1 ≤ t ≤ 4.
Solution: The arc length of the graph of r(t) is given by the definite integral
√ Z 4
Z 4
Z 2
dr
1
3− t
1
3−v
1
√
√ +
| | dt =
L=
dt =
+
dv
dt
2
3−v
2
3− t
1
1 2 t
1
(3 − v)2 2 3
=
− ln(3 − v) −
= + ln 2 ,
4
4
1
√
where we have made the substitution v = t.
(b) Find a negative change of parameter from t to s where s is an arc length parameter
of the curve having r(4) as its reference point.
Solution: The arc length parameter s can be found as follows
√ Z t
Z 4
Z 2 dr
1
3−v
1
3− u
1
√
√ +
du = √
+
dv
| | du = =
s=−
du
2
3−v
2
2 u 3− u
t
t
4
√
√
(3 − v)2 2
1 (3 − t)2
=
− ln(3 − v) −
+ ln(3 − t) ,
√ = − +
4
4
4
t
√
where we have made the substitution v = u.
√
√
2. Consider parabolic coordinates (µ, ν)
1
y = (µ2 − ν 2 ) .
(2)
2
(a) What are the dimensions of µ and ν?
√
√
Solution: It is the same as the dimension of x and y.
(b) Recall that if a parabola is described by the equation y = a(x − h)2 + k, then its
vertex is at (h, k) and its focus is at (h, k + 1/(4a)).
Show that curves of constant ν form confocal parabolae that open upwards (i.e.,
towards +y), while curves of constant µ are confocal parabolae that open downwards
(i.e., towards −y). Where are the foci of all these parabolae located?
Solution: It follows from (Derive!)
x = µν ,
1 x2
y = ( 2 − ν 2) ,
2 ν
The foci are located at (0, 0).
1
1
x2
y = (µ2 − 2 ) .
2
µ
(3)
(c) Show that in parabolic coordinates a curve given by the parametric equations µ =
µ(t), ν = ν(t) for a ≤ t ≤ b has arc length
v
2 2 !
Z bu
u
dµ
dν
t(µ2 + ν 2 )
+
dt .
(4)
L=
dt
dt
a
Solution: Straightforward computation
3. Consider elliptic coordinates (u, v)
x = luv,
y 2 = l2 (u2 − 1)(1 − v 2 ) ,
u ≥ 1 , −1 ≤ v ≤ 1 , l is a dimensionfull constant
(a) What is the dimension of l?
Solution: It is the same as the dimension of x and y.
(b) Show that curves of constant u are ellipses, while curves of constant v are hyperbolae.
Solution: It follows from (Derive!)
y2
x2
+
= 1,
l2 u2 l2 (u2 − 1)
x2
y2
−
= 1.
l2 v 2 l2 (1 − v 2 )
(5)
(c) Show that in the elliptic coordinates a curve given by the parametric equations
u = u(t), v = v(t) for a ≤ t ≤ b has arc length
s
2
2
Z b
u2 − v 2 dv
u2 − v 2 du
L=l
+
dt .
(6)
u2 − 1 dt
1 − v 2 dt
a
Solution: Straightforward computation
4. Consider the vector function
r(t) = e−t i + e−t cos t j − e−t sin t k .
(7)
(a) Find T(t), N(t), and B(t), at t = 0.
Solution: The unit tangent vector is
dr
dt
dr
| dt |
−e−t i − e−t (cos t + sin t) j + e−t (sin t − cos t) k
i + (cos t + sin t) j + (cos t − sin t) k
p
√
=−
T(t) =
=
3
e−t 1 + (cos t + sin t)2 + (sin t − cos t)2
(8)
The unit tangent vector at t = 0 is
T ≡ T(0) = −
i+j+k
√
.
3
(9)
The unit normal vector is
N(t) =
dT
dt
dT
| dt |
(cos t − sin t) j + (− sin t − cos t) k
(− cos t + sin t) j + (sin t + cos t) k
√
=−p
=
2
(cos t − sin t)2 + (sin t + cos t)2
(10)
2
The unit normal vector at t = 0 is
N ≡ N(0) =
−j + k
√
.
2
(11)
The unit binormal vector at t = 0 is
B=T×N=
−2i + j + k
√
.
6
(12)
(b) Find equations for the TN-plane at t = 0.
Solution: The TN-plane is normal to B. Thus, its equation at t = 0 is
(r − r(0)) · B = 0 ,
r = xi + yj + zk,
r(0) = i + j ,
(13)
or, explicitly
2(x − 1) − (y − 1) − z = 0 .
(14)
(c) Find equations for the NB-plane at t = 0.
Solution: The NB-plane is normal to T. Thus, its equation at t = 0 is
(r − r(0)) · T = 0 ,
r = xi + yj + zk,
r(0) = i + j ,
(15)
or, explicitly
(x − 1) + (y − 1) + z = 0 .
(16)
(d) Find equations for the TB-plane at t = 0.
Solution: The TB-plane is normal to N. Thus, its equation at t = 0 is
(r − r(0)) · N = 0 ,
r = xi + yj + zk,
r(0) = i + j ,
(17)
or, explicitly
−(y − 1) + z = 0 .
(18)
5. Prove the Serret-Frenet formulas
= κ(s)N(s)
(a) dT
ds
Solution: By definition
N(s) =
dT
ds
| dT
|
ds
=
dT
ds
κ(s)
.
(19)
(b) dB
= −τ (s)N(s), where τ is a scalar called the torsion of r(s).
ds
Solution: By definition
dB
dT
dN
dN
=
×N+T×
=T×
.
(20)
ds
ds
ds
ds
Thus dB
is normal to T. Since B is normal, dB
is also normal to B. Thus, it is
ds
ds
proportional to N.
(c) dN
= −κ(s)T(s) + τ (s)B
ds
Solution: We know that
N = B × T,
B = T × N,
T = N × B.
(21)
Thus
dN
dB
dT
=
×T+B×
= −τ (s)N × T + κ(s)B × N = −κ(s)T(s) + τ (s)B . (22)
ds
ds
ds
3