Module MA1132 (Frolov), Advanced Calculus
Tutorial Sheet 2
To be solved during the tutorial session Thursday/Friday, 28/29 January 2016
1. Consider the vector function (with values in R3 )
r(t) = ln t i − t j +
t2
k
4
(a) Find the arc length of the graph of r(t) if 1 ≤ t ≤ 2.
Solution: The arc length of the graph of r(t) is given by the definite integral
Z 2
Z 2
1 t
dr
t2 2 3
+
L=
|| || dt =
dt = ln t +
= + ln 2 .
1
dt
t 2
4
4
1
1
(1)
(b) Find a positive change of parameter from t to s where s is an arc length parameter
of the curve having r(1) as its reference point.
Solution: The arc length parameter s can be found as follows
Z t
Z t
dr
1 u
u2 t
t2 − 1
s=
|| || du =
+
+ ln t .
(2)
du = ln u +
=
1
du
u 2
4
4
1
1
2. Show that in cylindrical coordinates a curve given by the parametric equations r = r(t),
θ = θ(t), z = z(t) for a ≤ t ≤ b has arc length
s
2 2
Z b 2
dθ
dz
dr
L=
+ r2
+
dt .
(3)
dt
dt
dt
a
Hint: x = r cos θ, y = r sin θ.
Solution: It follows from
2 2 2 2 2
2
dx
dy
dr
dθ
dr
dθ
dr
dθ
2
+
=
cos θ − r sin θ
+
sin θ + r cos θ
=
+r
. (4)
dt
dt
dt
dt
dt
dt
dt
dt
3. Show that in spherical coordinates a curve given by the parametric equations ρ = ρ(t),
θ = θ(t), φ = φ(t) for a ≤ t ≤ b has arc length
s
2
2
Z b 2
dρ
dθ
dφ
2
2
2
L=
+ ρ sin φ
+ρ
dt .
(5)
dt
dt
dt
a
Hint: x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ.
1
Solution: It follows from
2 2 2 2
dx
dy
dz
dρ
dφ
dθ
+
+
=
sin φ cos θ + ρ cos φ cos θ
− ρ sin φ sin θ
dt
dt
dt
dt
dt
dt
2
dφ
dθ
dρ
sin φ sin θ + ρ cos φ sin θ
+ ρ sin φ cos θ
+
dt
dt
dt
2
dφ
dρ
cos φ − ρ sin φ
+
dt
dt
2
2
2
dφ
dθ
dρ
2
2
2
+ ρ sin φ
+ρ
.
=
dt
dt
dt
(6)
4. Consider the vector function
r(t) = e−t cos t i − e−t sin t j + e−t k .
(7)
(a) Find T(t), N(t), and B(t), at t = 0.
Solution: The unit tangent vector is
T(t) =
dr
dt
dr
| dt |
=
(− cos t − sin t) i + (sin t − cos t) j − k
√
3
(8)
The unit tangent vector at t = 0 is
T ≡ T(0) =
−i − j − k
√
.
3
(9)
The unit normal vector is
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
2
2
(cos t − sin t) + (sin t + cos t)
(10)
The unit normal vector at t = 0 is
N(t) =
N ≡ N(0) =
−j + k
√
.
2
(11)
i + j − 2k
√
.
6
(12)
The unit binormal vector at t = 0 is
B=T×N=
(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 + k ,
(13)
or, explicitly
(x − 1) + y − 2(z − 1) = 0 .
2
(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 + k ,
(15)
or, explicitly
(x − 1) + y + z − 1 = 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 + k ,
(17)
or, explicitly
−y + z − 1 = 0 .
3
(18)