Homework 2
Math2023
Exercise 11.1
Homework 2
Math2023
Exercise 11.3
Qu. 2 On the first quadrant part of the circle x2 + y 2 = a2 , we have y =
Qu. 12 Position: r(t) = at cos ωt i + at sin ωt j + b ln t k
√
required parametrization is
′
Velocity: v = r (t) = a(cos ωt − ωt sin ωt) i + a(sin ωt + ωt cos ωt) j + (b/t) k
Acceleration: a = r′′ (t) = −aω(2 sin ωt + ωt cos ωt) i + aω(2 cos ωt − ωt sin ωt) j − (b/t2 ) k
1/2
Speed: kvk = a2 (1 + ω 2 t2 ) + b2 /t2
r = r(x) = x i +
p
a2 − x2 j,
a2 − x2 , 0 6 x 6 a. The
(0 6 x 6 a)
Let
x = at cos ωt
(1)
y = at sin ωt
(2)
z = b ln t
⇒
t = ez/b
(3)
From (1) and (2)
y2
x2
+ 2 2 =1
a2 t2
a t
Qu. 4 In terms of the parameter t,
x2 + y 2 = a2 t2 = a2 e2z/b
∴ Path: a spiral on the surface (a“cone” - see page 6)
x2 + y 2 = a2 e2z/b .
∴
π
x = a sin t
06t6
y = a cos t
2
Z t
s
a dτ = at ⇒ t =
s=
a
0
s
s
s
π
r(s) = a sin i + a cos j, where 0 6 6 .
a
a
a
2
Qu. 24 Note that if
r · r′ = 0
d
(r · r) = 0
dt
Qu. 16
x = a cos t sin t =
2
krk = const.
⇒
y = a sin2 t =
krk = const. = a
a
sin 2t
2
a
(1 − cos 2t)
2
z = bt
i.e. r(t) lies on a sphere centred at the origin with radius a.
∴
Qu. 29
∴
d
[u · (v × w)] = u′ · (v × w) + u · (v′ × w) + u · (v × w′ ).
dt
a2
a 2
= , note also that as t increases, z increases too.
x2 + y −
2
4
The curve is a circular helix lying on the cylinder
a 2
a2
x2 + y −
=
2
4
(see page 6)
It’s length, from t = 0 to t = T , is
Qu. 32
d
d
d
[(u × u′ ) · (u′ × u′′ )] = (u × u′ ) · (u′ × u′′ ) + [ (u × u′ )] · (u′ × u′′ )
dt
dt
dt
′
′′
′′
′
′′′
′
′
L=
′′
′
′′
= (u × u ) · [u × u + u × u ] + [u × u + u × u ] · (u × u )
= (u × u′′ ) · (u′ × u′′ ) + (u × u′ ) · (u′ × u′′′ ).
–1–
Z
T
0
=T
p
p
a2 cos2 2t + a2 sin2 2t + b2 dt
a2 + b2 (units)
–2–
Homework 2
Math2023
Qu. 18 (see page 6 also)
Homework 2
Math2023
Qu. 27 Since r1 (t) = r2 (u(t)), where u is a function from [a, b] to [c, d], having u(a) = c and u(b) = d.
We assume u is differentiable. Since u is one-to-one and orientation-preserving.
du
> 0 on
dt
[a, b].
By the Chain rule:
d
du
d
r1 (t) =
r2 (u)
dt
du
dt
and so
One-eighth of the curve C lies in the first octant.
x2 + y 2 + z 2 = 1
2
2
x + 2z = 1
(sphere)
(1)
(elliptic cylinder)
(2)
Z b
Z b
du
d
d
dt
r2 (u)
r1 (t) dt =
dt
a du
a dt
Z d
d
=
r2 (u) du.
c du
From (2), Let

 x = cos t
1
 z = √ sin t
2
06t6
π
2
From (1),
Extra questions
y=
r
1 − cos2 t −
1
= √ sin t
2
1
sin2 t
2
Qu. 1 In cylindrical coord.

x = r cos θ


y = r sin θ


z=z
Note that the first octant part of C lies in the plane y = z, it must be a quarter of a circle of
radius 1 (why!). Thus the length of all of C is
8×
π
= 4π (units).
2
2
kr′ (t)k =
If you wish to use an integral, the length is
L=8
Z
=8
Z
π
2
0
π
2

dx
dr
dθ


 dt = cos θ dt − r sin θ dt




dr
dθ
dy
= sin θ
+ r cos θ

dt
dt
dt





 dz = dz
dt
dt
(take +ve root only).
r
sin2 t +
1
1
cos2 t + cos2 t dt
2
2
dx
dt
2
= cos2 θ
+
0
= 4π (units).
=
dr 2
dt
dy 2
dt
dr 2
dt
+ sin2 θ
dt
⇒
dz 2
dt
− 2r cos θ sin θ
dr 2
+ r2
+
dt
+ 2r cos θ sin θ
dθ 2
dt
+
dz 2
dt
Hence the answer.
–3–
dθ 2
dr dθ
+ r 2 sin2 θ
dt dt
dt
–4–
dθ 2 dz 2
dr dθ
+
+ r 2 cos2 θ
dt dt
dt
dt
Homework 2
Math2023
Qu. 2 In spherical coord.
Homework 2
Math2023
Ex 11.1 Qu. 12
x
x = ρ sin φ cos θ,
y = ρ sin φ sin θ,
z = ρ cos φ
10
0
-10
2
2
therefore,
0
0
0
dx
dρ
dφ
dθ
=
sin φ cos θ + ρ cos φ cos θ
− ρ sin φ sin θ
dt
dt
dt
dt
z
-2
-2
-4
-10
dρ
dφ
dθ
dy
=
sin φ sin θ + ρ cos φ sin θ
+ ρ sin φ cos θ
dt
dt
dt
dt
-4
0
-10
0
x
10
-10
-4
x
-10
-10
0
0
10
y
dρ
dφ
dz
=
cos φ − ρ sin φ
dt
dt
dt
dx 2 dy 2 dz 2
2
kr′ (t)k =
+
+
dt
dt
dt
2
dθ 2
dρ 2
2 dφ
+ρ
+ ρ2 sin2 φ
=
dt
dt
dt
z
z -2
10
Taking a = 2π and b = 2
0
10
y
2
10
y
2
2 2z/b
The surface x + y = a e
The surface and the curve
Ex 11.3 Qu. 16
y
0
0.5
1
y
0
1.5
0.5
2
1
y
0
1.5
0.5
2
1
1.5
2
Hence the answer.
15
15
15
10
Qu. 3
z 10
3
3
r = a cos t i + a sin t j + b cos 2t k,
π
06t6
2
z
5
5
5
-1
-0.5
0 x
2
2
v = −3a cos t sin t i + 3a sin t cos t j − 4b sin t cos t k
p
v = kvk = 9a2 + 16b2 sin t cos t
Z tp
s=
9a2 + 16b2 sin u cos u du
0
1p 2
9a + 16b2 sin2 t = K sin2 t
2
1p 2
where K =
9a + 16b2
2
-1
-0.5
0 x
0
Taking a = 2 and b = 2
-1
-0.5
0 x
0
0.5
0.5
1
0
=
z 10
0.5
1
1
The surface x2 + (y − a/2)2 = a2 /4
Ex 11.3 Qu. 18
x
Therefore sin t =
r
s
, cos t =
K
r
1−
s
2s
, cos 2t = 1 − 2 sin2 t = 1 − .
K
K
1
0.5
-1
0 -0.5
1
1
0.5
0.5
The required parameterization is
z
z
0
0.5
0
-1
s 3/2
s 3/2
2s
r=a 1−
k
j+a
j+b 1−
K
K
K
z
-0.5
-0.5
-0.5
-1
1p 2
9a + 16b2 .
2
0.5
-0.5
–5–
y
x
-0.5
0
0
0
for 0 6 s 6 K, where K =
-1
0
-0.5
0
-1
-1
-1
y
1
The surfaces x2 + y 2 + z 2 = 1
and x2 + 2z 2 = 1
–6–
-1
0.5
0
0.5
1
1
y
1
1
x