Math 227 Problem Set I Solutions
1. Find the velocity, speed and acceleration at time t of the particle whose position is ~r(t). Describe the
path of the particle.
a) ~r(t) = a cos t ı̂ + a sin t ̂ + ct k̂
b) ~r(t) = a cos t sin t ı̂ + a sin2 t ̂ + a cos t k̂
Solution. a)
~r(t) = a cos t ı̂ + a sin t ̂ + ct k̂
~v (t) = −a sin t ı̂ + a cos t ̂ + c k̂
~a(t) = −a cos t ı̂ − a sin t ̂
The path is a helix with radius a and with each turn having height 2πc.
b)
~r(t) = a cos t sin t ı̂ + a sin2 t ̂ + a cos t k̂
=
a
2
2t
sin 2t ı̂ + a 1−cos
̂ + a cos t k̂
2
~v (t) = a cos 2t ı̂ + a sin 2t ̂ − a sin t k̂
~a(t) = −2a sin 2t ı̂ + 2a cos 2t ̂ − a cos t k̂
The (x, y) coordinates go around a circle of radius a2 and centre 0, a2 counterclockwise. At the same
time the z coordinate oscillates half as fast. As t runs from 0 to π,
◦ the (x, y) coordinates go around the circle once, starting at (0, 0), while
◦ z starts at a when (x, y) is at (0, 0), then decreases to 0 when (x, y) is at (0, a), and continues
decreasing to −a when (x, y) is back at (0, 0).
Then as t runs from π to 2π,
◦ the (x, y) coordinates go around the circle a second time, starting at (0, 0), while
◦ z starts at −a when (x, y) is at (0, 0), then increases to 0 when (x, y) is at (0, a), and continues
increasing to a when (x, y) is back at (0, 0).
Then the pattern repeats.
2. A projectile falling under the influence of gravity and slowed by air resistance proportional to its speed
has position satisfying
d2 ~
r
d~
r
dt2 = −g k̂ − α dt
d~
r
dt
where α is a positive constant. If ~r = ~r0 and
r
~u(t) = eαt d~
dt (t).)
= ~v0 at time t = 0, find ~r(t). (Hint: consider
r
Solution. Define ~u(t) = eαt d~
dt (t). Then
d~
u
dt (t)
2
αt d ~
r
r
= αeαt d~
dt (t) + e dt2 (t)
αt
αt d~
r
r
= αeαt d~
dt (t) − ge k̂ − αe dt (t)
= −geαt k̂
Integrating both sides of this equation from t = 0 to t = T gives
αT
~u(T ) − ~u(0) = −g e
−1
k̂
α
αT
⇒ ~u(T ) = ~u(0) − g e
−1
k̂
α
=
d~
r
dt (0)
r
−αT
Subbing in ~u(T ) = eαt d~
dt (T ) and multiplying through by e
d~
r
dt (T )
= e−αT ~v0 − g 1−eα
−αT
1
k̂
αT
− ge
−1
k̂
α
αT
= ~v0 − g e
−1
k̂
α
Integrating both sides of this equation from T = 0 to T = t gives
~r(t) − ~r(0) =
=⇒
e−αt −1
v0
−α ~
~r(t) = ~r0 −
− g αt k̂ + g e −α−1
k̂
2
−αt
e−αt −1
~v0
α
+ g 1−αt−e
α2
−αt
k̂
3. Find the specified parametrization of the first quadrant part of the circle x2 + y 2 = a2 .
a) In terms of the y coordinate.
b) In terms of the angle between the tangent line and the positive x–axis.
c) In terms of the arc length from (0, a).
√
Solution. a) x(t), y(t) =
a 2 − t2 , t , 0 ≤ t ≤ a .
b) Let θ be the angle between the radius vector (a cos θ, a sin θ) and the positive x–axis. The tangent
line to the circle at (a cos θ, a sin θ) is perpendicular to the radius vector and so makes angle φ = π2 + θ
with the positive x axis. The desired parametrization is
x(φ), y(φ) = a cos(φ − π2 ), a sin(φ − π2 ) = a sin φ, −a cos φ ,
π
2
≤φ≤π
c) Let θ be the angle between the radius vector (a cos θ, a sin θ) and the positive x–axis. The arc from
(0, a) to (a cos θ, a sin θ) subtends an angle π2 − θ and so has length s = a π2 − θ . Thus θ = π2 − as and
the desired parametrization is
x(φ), y(φ) = a cos( π2 − as ), a sin( π2 − as ) , 0 ≤ s ≤
π
2a
4. Find the length of the parametric curve
y = a sin2 t
x = a cos t sin t
z = bt
between t = 0 and t = T > 0.
Solution.
x′ (t) = a cos2 t − sin2 t = a cos 2t
y ′ (t) = 2a sin t cos t = a sin 2t
z ′ (t) = b
So
p
p
x′ (t)2 + y ′ (t)2 + z ′ (t)2 = a2 + b2
=⇒
length=
√
a2 + b 2 T
5. Reparametrize the curve
~r(t) = a cos3 t ı̂ + a sin3 t ̂ + b cos 2t k̂, 0 ≤ t ≤
π
2
with the same orientation, in terms of arc length measured from the point where t = 0.
Solution.
=⇒
ds
dt
~v (t) = − 3a cos2 t sin t, 3a sin2 t cos t, −2b sin 2t
1/2
= k~v(t)k = 9a2 cos4 t sin2 t + 9a2 sin4 t cos2 t + 4b2 sin2 2t
1/2
= 9a2 sin2 t cos2 t + 4b2 sin2 2t
1/2
1/2
| sin 2t| = 94 a2 + 4b2
sin 2t for 0 ≤ t ≤
= 49 a2 + 4b2
2
π
2
Integrating (and recalling that s = 0 corresponds to t = 0)
s(t) =
Setting K =
hence
9
4a
2
+ 4b2
~r(s) = a 1 −
1/2
s 3/2
K
9
4a
2
+ 4b2
1/2 1−cos 2t
2
, we have sin t =
ı̂ + a
s 3/2
K
ps
9
4a
cos t =
K,
̂ + b 1 −
=
2s
K
2
+ 4b2
p
1−
1/2
s
K,
sin2 t
cos 2t = 1 − 2 sin2 t = 1 −
k̂, 0 ≤ s ≤ K, where K =
9
4a
2
+ 4b2
2s
K
and
1/2
6. The plane z = 2x + 3y intersects the cylinder x2 + y 2 = 9 in an ellipse. Find a parametrization of the
ellipse. Express the circumference of this ellipse as an integral. You need not evaluate the integral.
Solution.
ds
dθ
x = 3 cos θ, y = 3 sin θ, z = 6 cos θ + 9 sin θ, 0 ≤ θ ≤ 2π
p
= x′ (θ)2 + y ′ (θ)2 + z ′ (θ)2
p
= 9 sin2 θ + 9 cos2 θ + 36 sin2 θ + 81 cos2 θ − 108 sin θ cos θ
p
= 45 + 45 cos2 θ − 108 sin θ cos θ
R 2π √
s= 0
45 + 45 cos2 θ − 108 sin θ cos θ dθ
=⇒
7. A wire of total length 1000 cm is formed into a flexible coil that is a circular helix. If there are 10 turns
to each centimeter of height and the radius of the helix is 3 cm, how tall is the coil?
Solution. The parametrized equation of a helix is
~r(θ) = a cos θ ı̂ + a sin θ ̂ + bθk̂
The radius of the helix is 3 cm, so a = 3 cm. After 10 turns (i.e. θ = 20π) the height is 1 cm, so
1
1
1
cm/rad. Thus ~r(θ) = 3 cos θ ı̂ + 3 sin θ ̂ + 20π
θk̂ and ~r′ (θ) = −3 sin θ ı̂ + 3 cos θ ̂ + 20π
k̂ so that
b = 20π
q
q
ds
r′ (θ) = 9 + 1 2 . If θ varies from θ = 0 to θ = θF , then the wire has length 9 + 1 2 θF .
dθ = ~
400π
400π
−1/2
1
. This corresponds to a height
This must be 1000 cm. So θF = 1000 9 + 400π2
bθF =
1
20π 1000
9+
−1/2
1
400π 2
≈ 5.3 cm
8. Show that, if the position and velocity vectors of a moving particle are always perpendicular, then the
path of the particle lies on a sphere.
Solution.
d
r (t)k2
dt k~
=
d
dt
~r(t) · ~r(t) = 2~r(t) · ~r′ (t) = 0
since we are told that ~r(t) ⊥ ~r′ (t) for all t. Consequently, k~r(t)k2 is a constant, say A, independent of
√
time and ~r(t) always lies on the sphere of radius A centred on the origin.
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