Math 1321
Week 6 Lab Worksheet
Due Thursday 02/21
1. Arc Length & Curvature: Consider a helix r(t) = hR cos(ωt), R sin(ωt), hti, where
ω and h are numerical parameters. The arc length of one turn of the helix is a function
of the parameter ω, (i.e. L = L(ω)), and the curvature at any fixed point of the helix is
also a function of ω, (i.e. κ = κ(ω)).
(a) Find the limits of L(ω) and κ(ω) as ω → ∞.
(b) What does each turn of the helix start to approximate as ω → ∞?
Solution: (a) The vector function r(t) traces out one turn of the helix when t ranges
over the period of cos(ωt) or sin(ωt) (i.e., over the interval of length 2π/ω). Thus,
the helix rises by 2πh/ω along the z axis per each turn.
Z
2π/ω
L(ω) =
k r0 (t) k dt
0
2π p
(Rω)2 + h2
=
ωp
= 2π R2 + (h/ω)2 → 2πR
k r0 (t) × r00 (t) k
k r0 (t) k3
Rω 2 [(Rω)2 + h2 ]1/2
=
[(Rω)2 + h2 ]3/2
1
R
→
as
= 2
2
R + (h/w)
R
as
ω→∞
κ(ω) =
ω→∞
(b) When ω → ∞, the height H(ω) → 0 so that each turn of the helix becomes
closer and closer to a circle of radius R.
2. Kepler’s Laws: (Reference Section 10.4 of Text).
Johannes Kepler stated the following three laws of planetary motion on the basis of
masses of data on the positions of the planets at various times. These laws are as
follows:
1. A planet revolves around the sun in an elliptical orbit with the sun at one focus.
2. The line joining the sun to a planet sweeps out equal areas in equal times.
3. The square of the period of revolution of a planet is proportional to the cube of the
length of the major axis of its orbit.
Kepler formulated these laws because they fitted the astronomical data. He wasn’t able
to see why they were true or how they related to each other. But Sir Isaac Newton, in
his Principia Mathematica of 1687, showed how to deduce Kepler’s three laws from two
of Newton’s own laws, the Second Law of Motion and the Law of Universal Gravitation.
In Section 10.4 of your text, Calculus: Concepts and Contexts, 4Ed, Kepler’s First Law
is proved using the calculus of vector functions. You will be proving Kepler’s Second
Law in the following questions.
Use the following steps to prove Kepler’s Second Law. The notation is the same as in
the proof of the First Law in Section 10.4. In particular, use polar coordinates so that
r = (r cos θ)i + (r sin θ)j.
k. (Hint: h = r × r0 )
(a) Show that h = r2 dθ
dt
Solution:
h = r × r0
h
dθ dθ i
= [(r cos θ)i + (r sin θ)j] × r0 cos θ − r sin θ
i + r0 sin θ + r cos θ
j
dt
dt
h
i
0
2
2 dθ
0
2
2 dθ
= rr cos θ sin θ + r cos θ − rr cos θ sin θ + r sin θ
k
dt
dt
dθ
= r2 k
dt
(b) Deduce that r2 dθ
=h
dt
Solution:
h =k h k= r2
dθ
dt
(c) If A = A(t) is the area swept out by the radius vector r = r(t) in the time interval
[t0 , t] as shown in the figure below,
show that,
dA
1 dθ
= r2
dt
2 dt
Solution:
1
A(t) =
2
Z
θ
k r k2 dθ
θ0
In polar coordinates, we have
1
A(t) =
2
Z
t
r2 (dθ/dt)
t0
By the Fundamental Theorem of Calculus,
dA
r2 dθ
=
dt
2 dt
= 12 h = constant. This says that the rate at which A is swept out
(d) Deduce that dA
dt
is constant and proves Kepler’s Second Law.
Solution:
dA
r2 dθ
h
=
= = constant since h is a constant vector and h =k h k
dt
2 dt
2