Download Lecture 3 - Some (relatively) simple applications

Lecture 3 - Some (relatively) simple applications
Mixtures:
Suppose a tank contains Q0 lbs of salt dissolved in 100 gal of water at time t = 0. Water
containing 1/4 lb of salt/gal enters the tank at a rate of r gal/minute, and r gal/minute of
the well-stirred mixture are drawn off per minute.
(a) Write down and solve an ODE for Q(t), the amount (in lbs) of salt in the tank at time t.
Solution: Time is being measured in minutes, so the rate of change of Q per unit time is
dQ
= # lbs of salt in/minute − # lbs of salt out/minute
dt
= rate in − rate out
The rate in = 1/4 lb/gal × r gal/min = r/4 lbs/min.
The rate out
= concentration in lbs/gal at time t × # gal/min out
Q =
×r
100
r
=
Q.
100
So the ODE is
dQ
r
r
= −
Q
dt
4 100
which is a linear equation with the solution (work it out):
rt
Q(t) = 25 + Ce− 100 .
Since r > 0, no matter what the value of C we still have
lim Q(t) = Q∞ = 25,
t→∞
which is reasonable (why?)
(b) If Q0 = 5, r = 3, find the time T at which Q(t) gets within 1% of Q∞ .
Solution: Given the values of Q0 and r, substitution yields C = −20. Since Q(t) < 25,
we’re looking for the time T at which Q(T ) = .99(25) = 25 − 20e−3T /100 . Solving this gives
T ≈ 146 minutes.
(c) Now let’s change the problem slightly: suppose all is as above except that 4 gal/minute
of the salt solution flows in, while 3 gal/minute flows out. Suppose the tank has a capacity
of 200 gallons. How much salt is in the tank at the moment it begins to overflow?
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Solution: What’s changed here is that the volume liquid in the tank is no longer constant.
Initially, there are 100 gals in the tank, and each minute 4 gallons flow in and 3 flow out. So
the volume of liquid in the tank at time t is V (t) = 100 + t. Therefore, the concentration of
salt in the tank at time t is going to be Q/(100 + t) = # lbs salt in the tank/ # gallons in
tank.
So the rate in will be (4 gal/min)(1/4 lb/gallon) = 1 (lb/min), while the rate out will be the
concentration of salt in the tank times the number of gals/minute out or 3Q/(100 + t). And
the new DE is
dQ
= 1 − 3Q/(100 + t).
dt
This is still linear:
dQ
3
+
Q = 1.
dt
100 + t
We have
R
p = ln (100 + t)3 , so the integrating factor is µ(t) = (100 + t)3 . So
d
Q(100 + t)3 = (100 + t)3
dt
1
Q(100 + t)3 = (100 + t)4 + C
4
1
Q = (100 + t) + C(100 + t)−3 .
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If, as before, Q0 = 5, we find C = −20 × 106 . Now the tank overflows when V = 200 or
when t = 100. At that time, we have
Q(100) = 50 − 20 · 106 (200)−3 = 50 − 2.5 = 47.5 (lbs)
Question: What happens at this point? That is, write down and solve a DE that describes
Q(t) for t ≥ 100.
Newton’s 2nd Law:
For one-dimensional mechanics problems, the relevant DE is
ma = m
d2 x
= f (t, x, ẋ),
dt2
(1)
where x(t) denotes the position of the “particle” with mass m at time t. The forces, which
may depend on position, velocity and time, are on the other side of the equation, and the
general rule is that the forces add vectorially. So in one dimension, if there’s a force f1
due to gravity, and another force f2 due to friction, then the total force on the mass is just
f1 + f2 . Any force has a sign, which is positive if the force acts in the positive x-direction
and negative otherwise.
In the following two examples, we consider objects moving in the earth’s gravitational field.
The x-axis is perpendicular to the surface of the earth, and the positive direction is upward.
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Example 1: An object is released at a height of h0 meters above the surface of the earth
with an initial velocity v0 , and it falls toward the surface. The forces acting on it are (a)
the gravitational force (which points downward) and (b) a force due to air resistance which
opposes the motion and therefore is directed upwards. Suppose that this force due to air
resistance is proportional to the square of the particle’s speed. Write down a differential
equation describing the motion of the object, assuming the gravitational force is constant
over the small distance considered. What can you conclude about the long-time behavior?
Solution: We start with ma = f . Notice that in this problem, the only forces are the
(constant) gravitational force, and (b) a force depending on v. So there’s no x in the
= f . The magnitude
problem - that is, we can just write down a first-order equation: m dv
dt
of the gravitational force is g = 9.8m/s2 . The force due to air resistance will have the form
kv 2 assuming that the initial velocity is negative 1 . So the DE is simply
m
dv
= kv 2 − mg,
dt
which is a separable equation. Plodding along, we get
dv
k
= v2 − g
dt
m
m/k
dv = dt, or
v 2 − gm/k
Z
Z
α2 /g
dv = dt, (α2 = gm/k)
2
2
v −α
As you’ll no doubt recall, the left hand side can be integrated using partial fractions:
1
1
1
1
=
−
.
v 2 − α2
2α v − α v + α
Using the fact that v < 0 initially, and is smaller in magnitude than α, the integration yields
α − v ln
= 2gt/α + c.
α+v
Exponentiation and a bit of algebra leads to
v=
α(1 − Ce2gt/k
.
1 + Ce2gt/k
The constant C can of course be evaluated by giving an initial condition. A more interesting
question is what happens to v as t → ∞? Method 1: divide numerator and p
denominator
2gt/k
by e
and then let t → ∞; the answer you should get is vterm = −α = − gm/k. For
obvious reasons, this is called the terminal velocity. Method 2: How can you deduce the
same result by “inspection” of the original DE?
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More accurately, the force due to resistance should have the form kv|v|. Why?
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Example 2: Escape velocity
Now we want to consider what happens if we launch an object upward from the surface of the
earth with sufficient initial velocity for it to potentially escape from the earth’s gravitational
field. Here our coordinate line is the x-axis, with x = 0 at the center of the earth, and
x = x0 denoting the earth’s surface (x0 ≈ 4000 miles). We ignore all forces except that of
gravitation, and use the fact that for a roughly spherical body like the earth, the gravitational
field is the same as if the earth were a point mass located at the earth’s center. Newton’s
2nd law then reads
GmM
dv
m =−
.
dt
x2
Here m is the mass of our object, M is the mass of the earth, and G (not g!) is Newton’s
gravitational constant. The first thing we can do is to get rid of the GM : we know that the
acceleration due to gravity at the surface of the earth x = x0 is g, so this gives us
GM m
, and so
x20
GM = gx20
ma = −g = −
The DE looks impossible, since there seem to be 3 variables (t, x, v). But we can get rid of
t using the chain rule:
dv dx
dv
dv
=
=v .
dt
dx dt
dx
Making this substitution, our DE now reads
dv
GM m
=−
.
dx
x2
Cancelling m from both sides, we easily integrate this separable equation:
Z
Z
GM
v dv = −
dx
x2
GM
v2
=
+ C, and from the above,
2
x
v2
gx2
= 0 +C
2
x
The situation we have in mind is that the mass m is launched upward from the surface x−x0
with some initial velocity v0 . Plugging this in, we get
mv
C=
v02
− gx0 .
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Rewriting, we now have
1 2
x0
v − v02 = gx0
−1 .
2
x
Suppose we want to know how high our object will go? At the top of its trajectory, we’ll
have v = 0. So we put v = 0 and solve for x to obtain
x=
2gx20
.
2gx0 − v02
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The object will escape the earth’s gravitational field if, in the above, x = ∞, which happens
if the denominator vanishes, i.e., when
p
v0 = 2gx0 .
If you plug in some numbers, and get the units right, this is a bit more than 7 miles/second.
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