Math 1320 - Lab 3
uNiD:
Name:
Due: 09/19
Today you are going to consider problems in aeronautics and use differential equation to describe them.
Answer the questions below and be sure to explain your reasoning behind your approach to these questions
with appropriate diagrams and complete sentences. In your writing, try to be complete enough so that
someone who has completed Math 1310 would understand what you are doing and why you are doing it.
1. (10 points) Hydrogen is a light and extremely powerful rocket propellant used in the space shuttles. On
a routine inspection of a liquid hydrogen tank, it is noticed that the hydrogen level is well above the
expected amount in the tank. Upon further inspection, a leak from a neighboring tank was pinpointed
as the source of the extra hydrogen in the tank. You are brought in to help prevent the tank from
overflowing. You are provided 6 volume measurements of the liquid hydrogen in the tank right after the
leak was discovered.
time (mins)
volume (m3 )
30
56094
60
62930
90
70599
120
79204
150
88857
180
99686
You are also told that an initial empirical test on these measurements suggests the hydrogen level
increase can be modeled with an exponential function with an unknown growth rate, k. Find the value
of k using this empirical assumption. What is the initial volume of the liquid hydrogen if we assume
the leak started at the same time the measurements commenced, (i.e., volume at t = 0 min)? Find the
differential equation that models the rate of change of the liquid hydrogen in the tank?
Solution: You have been told that empirical tests suggest the process is an exponential process.
We can therefore model the volume after the leak starts with a generic exponential process as:
V (t) = V0 ekt
where V0 is the initial volume just right before the leak starts (t = 0 mins), k is the exponential
growth constant, and t is time. We can use the data we’ve been given to solve for V0 and k. If
pairs of the data can be modeled with the same k we can conclude that the process is indeed an
exponential process. Lets pick t = 30mins and t = 60mins as the two time values, then we have the
following pair of equation:
56094 = V0 e30k
(1)
62930 = V0 e60k
(2)
Divide (2) by (1) gives
62930
V0 e60k
=
56094
V0 e30k
1.1219 = e60k−30k
ln(1.1219) = 30k
k = 0.0038 min−1
You can confirm that choosing any other pair of time instances from the table gives you the same k
value. Now to get V0 we just plug in the value of k in (1) or (2).
56094 = V0 e30(0.0038)
56094
V0 = 0.114 = 50, 050 m3
e
To get the differential equation we differentiate V (t)
d
V (t) = kV0 ekt = kV (t)
dt
2. (10 points) A parachute is one of the basic safety devices on any aeronautic vehicle. When a person
wearing a parachute jumps off a plane, there are two forces acting on the person, the gravitational force,
FG = −mg, and the resistance force, FR = −kv. where m is the mass, g is the acceleration due to
gravity, k is a positive constant and v is the velocity. We have assumed a coordinate system such that
the ground is at y = 0. Note that FR acts against motion.
(a) Use Newton’s Second Law to write down a differential equation that relates the velocity to the net
force on the person.
(b) Solve this differential equation for the velocity as a function of time during the fall. Assume the
initial velocity, v(0) = v0
(c) The speed of the person falling under the influence of air resistance does not increase indefinitely,
instead, it approaches a finite limiting speed or terminal speed. What is this terminal speed, vτ of
the person? Draw a slope field for the differential equation you obtained in part (a).
(d) Derive the equation that describes the vertical height, y, of the person during the fall. Assume the
initial vertical position y(0) = y0
Solution:
(a) The net force acting on a body is the sum of all the forces acting on the body. In this case there
are two forces namely the gravitation force and the resistance force.
Fnet = FG + FR = −mg − kv
Newtons Second Law gives the relationship between force and acceleration so we can write the
net force in term of acceleration.
Fnet = −mg − kv
ma = −mg − kv
The acceleration of a body is just the derivative of its velocity so we have a final expression as:
m
d
v = −mg − kv
dt
(b)
d
k
v = −g − v
dt
m
d
v = −g − ρv where k/m is ρ.
dt
We have a separable differential equation so we can solve it as follows.
Z
Z
dv
= − dt
g + ρv
Using a simple substitution u = g + ρv, du = ρdv, we have
Z
Z
1
du
= − dt
ρ
u
1
ln |u| = −t + C
ρ
ln |g + ρv| = −ρt + C
|g + ρv| = Ce−ρt
g + ρv = Ce−ρt
g
v = − + Ce−ρt
ρ
Use th eintitial velocity to find the constant C
g
v0 = − + C
ρ
g
C = v0 +
ρ
The final equation for v is:
g
g
v = − + v0 +
e−ρt
ρ
ρ
(c) The terminal velocity is reached when the acceleration is 0, i.e., when dv/dt = 0
d
v = −mg − kv = 0
dt
mg
g
vτ = −
=−
k
ρ
(d) Notice that the velocity is the derivative of the vertical height so we can obtain the vertical
height by integrating the velocity equation we derived in (b).
g
g
v = − + v0 +
e−ρt
ρ
ρ
g
g
d
y = − + v0 +
e−ρt
dt
ρ
ρ
Z
Z
g
g
dy = − + v0 +
e−ρt dt
ρ
ρ
g
1
g
y =− t−
v0 +
e−ρt + C
ρ
ρ
ρ
Using the initial condition, y(0) = y0 gives
g
v0 +
+C
ρ
1
g
C = y0 +
v0 +
ρ
ρ
1
y0 = −
ρ
This gives a final equation for the vertical height:
g
1
g
1
g
y =− t−
v0 +
e−ρt + y0 +
v0 +
ρ
ρ
ρ
ρ
ρ
g
1
g
y =− t−
v0 +
1 − e−ρt + y0
ρ
ρ
ρ
3. (10 points) According to Newton’s Law of Universal Gravitation, the gravitational force on an object of
mass m that has been projected vertically upward from the earth’s surface is
F =
mgR2
(x + R)2
where x = x(t) is the object’s distance above the surface at time t, R is the earth’s radius, and g is the
acceleration due to gravity. Also, by Newton’s Second Law,
F = ma = m(dv/dt)
and so
m
dv
mgR2
=−
dt
(x + R)2
(a) Suppose a rocket is fired vertically upward with an initial velocity v0 . Let h be the maximum height
above the surface reached by the object. Show that
r
2gRh
v0 =
R+h
[Hint: By the chain rule, m(dv/dt) = mv(dv/dx).]
(b) Calculate ve = lim v0 . This limit is called the escape velocity for the earth.
h→∞
(c) Use R = 3960 mi and g = 32 ft/s2 to calculate ve in feet per second and in miles per second.
Solution:
(a) We can use chain rule to represent the force equation in terms of the following
m
dx dv
dv
dv
=m
·
= mv
dt
dt dx
dx
We can now solve the differential equation and find the initial velocity v0 .
dv
mgR2
mv
=−
dx
(x + R)2
Z
Z
gR2
v dv = −
dx
(x + R)2
v 2 (x)
gR2
=
+C
2
x+R
r
2gR2
v(x) =
+C
x+R
At maximum height, h, v(h) = 0, so
r
2gR2
+C
h+R
2gR2
C=−
h+R
0=
Substitute C in v(x).
r
v(x) =
2gR2
2gR2
−
x+R h+R
We can now find the initial velocity at x = 0 as
r
2gR2
2gR2
v0 =
−
R
h+R
r
2ghR
v0 =
h+R
(b)
r
ve = lim v0 = lim
h→∞
h→∞
p
ve = 2gR
2ghR
h+R
(c)
1 mile = 5280 ft
r
ve =
p
2gR =
2 · 32
ft
5280 ft
ft
· 3960 mi ·
= 36581
2
s
1 mi
s