MATH 2280-002
Lecture Notes: 01/17/2013
Math 2280 Section 002 [SPRING 2013]
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Logistic Models Summary
A logistic DE
dP
= k(M − P )P
dt
has two critical points, P = 0, M . The solution curves exhibit very different behavior depending on
whether k > 0 or k < 0.
Let’s first consider what happens when k > 0. Actually, the mice colony we examined last time is a
good example of this scenario.
Exercise: Use a phase diagram to show that, in this case, P = 0 is unstable and P = M is stable.
If P (0) > 0, then P (t) approaches the carrying capacity M as t → ∞. Slight increases or decreases
in the population do not have dramatic effects in the long run. The slope field and typical solutions
curves will look something like the plot below.
In fact, this is the slope field for P 0 = 10(40 − P )P . We see that the solution curves approach the
line P (t) = 40 as time progresses.
Now let’s first consider what happens when k < 0.
Exercise: Use a phase diagram to show that, in this case, P = 0 is stable and P = M is unstable.
M is called the threshold. If the population starts above the threshhold (P (0) > M ), then the graph
of P (t) has a vertical asymptote at some positive value T and P (t) → ∞ as t → T . For this starting
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MATH 2280-002
Lecture Notes: 01/17/2013
value, the population will experience doomsday, a population boom. If the population starts below
the threshhold (0 < P (0) < M ), then P (t) → 0 as t → ∞. For this starting value, the population
will experience extinction, a population bust. (If P (0) = M , then P (t) will just be identically equal
to M everywhere). Increases or decreases in the population that cause P (t) to cross the threshhold
have very dramatic effects in the long run.
The slope field and typical solutions curves will look something like the plot below.
This is the slope field for P 0 = −10(40 − P )P . The solution curves diverge away from the line
P (t) = 40 as time progresses. Before moving on, I’ll leave you with one last exercise.
Exercise: Suppose a population P (t) of alligators is modeled by
dP
= 0.0012P 2 − 0.24P.
dt
• Scenario 1. Find P (t) and describe the behavior of this model over time when P (0) = 100.
• Scenario 2. Find P (t) and describe the behavior of this model over time when P (0) = 300.
In the above exercise, the threshhold is 200 alligators, so we already know what happens to P (t)
over time, depending on P (0). What I want you to do is to actually calculate P (t) for the two initial
conditions and then verify that the behavior of the function coincides with our predictions.
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Improved Velocity-Acceleration Models
Recall the following variables used in our velocity-acceleration models:
• x(t) = position of an object at time t;
• v(t) = x0 (t) = velocity ; and
• a(t) = v 0 (t) = x00 (t) = acceleration.
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MATH 2280-002
Lecture Notes: 01/17/2013
Newton’s second law relates the total force on an object to its mass and acceleration.
F = ma = mv 0
Until now, we’ve assumed no air resistance, so we got that the force on a falling body is
F = −mg,
where m is the mass of the body and g is the acceleration due to gravity (≈ 9.8 m/s2 near the
Earth’s surface). This is usually not realistic, so we need to refine our models for falling bodies.
We’ll assume the easiest case – resistance proportional to velocity – or the next easiest case –
resistance proportional to the square of velocity. Even though this approximation of drag seems
simplistic; it’s not that unreasonable, especially for small, spherical objects moving relatively slowly
(see Stokes’ Law ).
We have two forces acting on a falling body – the force of gravity FG and the force of air resistance
FR . Just as before, FG = −mg. The negative sign just means that we’re treating the direction
“downward” as negative and the direction “upward” as positive. Under this convention, velocity is
negative for a falling body. Since the force of air resistance opposes the force of gravity, FR should
be positive for a falling body. Assuming linear drag,
FR = |kv| = −kv
(the negative sign is needed because v < 0), where k is a positive constant. Assuming quadratic
drag,
FR = |kv 2 | = kv 2 ,
where k is again a positive constant. In summary, we’ve considered the following three scenarios:
• No air resistance. F = −mg
mv 0 = −mg =⇒ v0 = −g
• Linear drag. FG + FR = −mg − kv
mv 0 = −mg − kv =⇒ v0 = −g −
k
v
m
• Quadratic drag. F = FG + FR = −mg + kv 2
mv 0 = −mg + kv 2 =⇒ v0 = −g +
3
k 2
v
m
Terminal Velocity
We know from physics that the speed of an object falling through our atmosphere doesn’t increase
indefinitely; instead, it will approach a finite limiting speed – called the terminal velocity. We expect
that the terminal velocity should correspond to a stable critical point in our differential equation.
Question. Assuming linear drag, what is the terminal velocity of a falling body?
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MATH 2280-002
The DE v 0 = −g −
Lecture Notes: 01/17/2013
k
mv
has a single critical point at
vT = −
gm
.
k
We have the following phase diagram:
As expected, the critical point is stable, and the behavior predicted by our differential equation
matches the physics we’re trying to model.
Question. Assuming quadratic drag, what is the terminal speed of a falling body?
The DE v 0 = −g +
k 2
mv
has critical points at
r
r
gm
gm
vT = −
and − vT =
.
k
k
We have the following phase diagram:
The stable critical point vT is the terminal velocity. Of course, this makes sense since −vT couldn’t
possibly be the terminal velocity; because −vT is positive, that would mean falling objects eventually
fly back upward!
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Gravitational Attraction
Treating gravity as constant is reasonable near the surface of the Earth, but if we want to improve our
velocity-acceleration models further we need to account for the fact that gravitational acceleration
is actually variable.
According to Newton’s law of gravitation,
FG =
GM m
.
r2
Here we have that
• G = 6.6726 × 10−11 N·(m/kg)2 (gravitational constant)
• M = the first mass (for example, a planet)
• m = the second mass (a projectile)
• r = the distance between the two masses
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MATH 2280-002
Lecture Notes: 01/17/2013
In the next example, we’ll use this law to compute the escape velocity from the moon.
Example: Given that the mass of the moon is 7.35 × 1022 kg and its radius is 1.74 × 106 m, what
is the escape velocity for the moon?
On the moon, we can disregard air resistance, so the force on a projectile is
GM m
,
r2
where M is the mass of the moon, m is the mass of the projectile, and r is the distance between
the projectile and the center of the moon. According to Newton’s second law (F = ma)
−
m
dv
GM m
.
=−
dt
r2
Notice that both v and r are functions of t. Use the chain rule to rewrite
dv
dt .
dv
dv dr
dv
=
=
v
dt
dr dt
dr
Our original DE becomes
dv
GM
=− 2 .
dr
r
Now we don’t have to worry about the variable t, and we have a nice separable DE that we can
solve by applying separation of variables.
Z
Z
GM
dr
vdv =
r2
v2
GM
=
+C
2
r
Solve for C given an initial velocity v0 and an initial distance R = radius of the moon.
v
C=
v02 GM
−
.
2
R
In other words,
v2
GM
v 2 GM
=
+ 0 −
.
2
r
2
R
We have a maximum height when v = 0. This occurs when
v 2 GM
GM
+ 0 −
=0
r
2
R
We don’t want a max height; we want our projectile to escape the gravity of the moon. We have
no maximum height if
GM
v 2 GM
+ 0 −
>0
r
2
R
for all r > 0. Since GM
r is always positive, this means
v02 GM
−
≥0
2
R
r
v0 ≥
5
2GM
.
R
MATH 2280-002
Lecture Notes: 01/17/2013
So the escape velocity of the moon is
r
2GM
≈ 2375 m/s = 5000 mph.
R
Notice that the escape velocity doesn’t depend on the mass of the projectile! Also, there was
nothing unique about this particular example. Assuming no air resistance, the escape velocity for
a moon or planet (or death star) is
r
2GM
R
where M is the mass and R is the radius of the planetary body.
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