Math 2250 Lab 1
Name/Unid:
1. (20 points) Verify that the exponential integral
Z
y(x) = Ei(x) =
x
∞
e−u
du
u
with x > 0, satisfies the ODE:
d2 y 1 dy
= 0.
+
1
+
dx2
x dx
Ra
Rx
Rb
d
f (u)du = f (x) are
Note that the integral identities a f (u)du = − b f (u)du and dx
a
useful. The exponential integral and its ODE are important in modeling water flow to
a pumping well through a porous ground medium.
Solution: The first derivative of the exponential integral
Z
Z x −u i
dh
e
d h ∞ e−u i
du =
−
du
dx x
u
dx
∞ u
e−x
.
=−
x
And the second derivative
d h e−x i
−
dx
x
(1)
(2)
(3)
=
e−x e−x 1 e−x
+ 2 = 1+
x
x
x x
(4)
Putting the pieces together:
d2 y 1 dy
+
1
+
dx2
x dx
1 e−x 1 e−x
= 1+
− 1+
x x
x x
=0
So the exponential integral does satisfy the ODE.
(5)
(6)
(7)
2. (20 points) Suppose a car skids 20 meters if it is moving at 50 km/h when the brakes
are suddenly applied. Assuming the car has a constant deceleration, how far will the car
skid if it is originally moving at 100km/h when the brakes are applied?
Solution: The acceleration a is constant, but unknown, so we integrate to solve for
position, which we do know, to then determine a. Before we start, it is useful to
convert km/h to m/sec: 50km/h = 50 × 1000/(602 ) = 13.89m/sec
dv
=a
dt
(8)
ads = at
(9)
=⇒ v(t) = at + v(0) = at + 13.89
(10)
(11)
Z
v(t) − v(0) =
t
0
At time t∗ the car will come to a stop: v(t∗ ) = 0. We solve for t∗ :
0 = at∗ + 13.89
=⇒ t∗ = −13.89/a
(12)
(13)
In order to solve for t∗ and a, we need to incorporate the additional piece of information that the car’s stopping distance was 20 meters:
Z t
x(t) − x(0) =
v(s)ds
(14)
0
Z t
a
x(t) =
as + 13.89ds = t2 + 13.89t
(15)
2
0
a
20 = (t∗ )2 + 13.89t∗
(16)
2
a
20 = (13.89/a)2 − 13.892 /a
(17)
2
−13.892
20 =
(18)
2a
−13.892
a=
≈ −4.82 [m/sec2 ]
(19)
40
Which is very close to half-g—a very sudden stop. To solve for the stopping distance
from 100kph, we integrate to find the stopping time, then integrate to find the
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accumulated distance. First, in m/sec: 100km/h = 100 × 1000/(602 ) = 27.78m/sec
dv
= −4.82
dt
(20)
ads = −4.82t
(21)
=⇒ v(t) = at + v(0) = at + 27.78
(22)
(23)
Z
v(t) − v(0) =
t
0
At time t∗ the car will come to a stop: v(t∗ ) = 0. We solve for t∗ :
0 = −4.82t∗ + 27.78
=⇒ t∗ = 27.78/4.82 = 5.76seconds
(24)
(25)
Integrating to find position:
Z
x(t) =
−4.82s + 27.78ds
t2
+ 27.78t
2
(5.76)2
stopping distance = −4.82
+ 27.78 × 5.76 = 79.8 meters
2
= −4.82
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(26)
(27)
(28)
3. (20 points) (a) A 1000 kg spacecraft can use the force imparted from solar radiation to
“sail” through space using a reflective sail made of aluminum foil. The suppose the
solar force is constant, equal to 0.0001 Newton per square meter of sail. Suppose
the craft’s sail is 200 m2 . If the spacecraft starts with zero initial velocity at postion
x = 0, how long will it take for the spacecraft to reach one tenth the speed of light
(c ∼ 3 × 108 m/sec)?
(b) Suppose a 100kg rocket-powered spacecraft fires its rockets at the same time and
position as the solar sail craft in the problem above. The rocked produces a trust
force Fr = 1/(1 + t) over time, in which the force dissipates as the rocket burns
down all its fuel. At what position x will the sail craft overtake the rocket?
Solution:
(a) F = ma is the principal equation dictating how a force produces motion. In
this case F = 0.0001 × 200 = 0.02 Newtons. The acceleration then is a(t) = dv
:
dt
F
dv
=
= 2 × 10−5 .
dt
m
We get velocity as a function of time through integration
Z t
dv
ds = v(t) − v(0) = v(t) = 2 × 10−5 t
dt
0
The time t∗ when it reaches 0.1c is
2 × 10−5 t∗ = 0.1c
3
t∗ = 3 × 107 /2 × 10−5 = × 1012
2
which is in seconds.
(b) The position of the sail craft is obtained by integrating again.
xsail (t) = 10−5 t2
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(30)
To get the position of the rocket, again, we use F = ma:
vrocket (t) − vrocket (0) = vrocket (t)
Z t
arocket (s)ds
=
0
Z t
1
1
=
ds
100 0 1 + s
1
ln(1 + t)
=
Z100
(31)
xrocket (t) − xrocket (0) =
(35)
(32)
(33)
(34)
t
vrocket (s)ds
Z t
1
=
ln(1 + s)ds
100 0
Z t+1
1
ln(s)ds
=
100 1
= [s ln(s) − s]t+1
1
= (t + 1) ln(t + 1) − (t + 1) − (−1)
= (t + 1) ln(t + 1) − t
0
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(36)
(37)
(38)
(39)
(40)
4. (20 points) Let y(t) be a population of bacteria in a petri dish over time. Using computer
graphical aid or drawing by hand, draw a series of 4 of slope fields for the following ODE
dy
= y(1 − y) − h
dt
where h = 0, 0.1, 0.2, 0.3 is a rate of harvesting from the petri dish. On each slope
field defined by h, draw in approximate solution curves for the initial conditions y(0) =
0, 0.3, 0.7, 1, 1.5. Indicate what happens to the population y of bacteria for each initial
condition. Is their a critical value of h where all bacteria populations go extinct? Plot
the slope fields and carefully describe your observations. Be sure to distinguish what
happens with the different values of h and what it means for the bacteria populations.
What harvesting rate appears sustainable over time? The Maple code
with(DEtools):
DEplot(diff(y(t), t) = y(t)*(1-y(t))-0, [y(t)], t = 0..10, y = -1..2, [[y(0)
= 0], [y(0) = .3], [y(0) = .7], [y(0) = 1], [y(0) = 1.5]]);
can be a timesaver for creating all the plots efficiently. Print out graphs and attach them
to your worksheet with your observations and conclusions.
Solution:
For h = 0, all populations eventually settle to y = 1 except y(0) = 0 which stays
zero.
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For h = 0.1, all populations eventually settle to y = 1 except y(0) = 0, which
diverges negative, which is absurd.
For h = 0.2, all populations eventually settle to a positive value that is less than
one y ≈ 0.75 but y(0) = 0 still diverges negative.
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For h = 0.3, all populations eventually diverge negative. All bacteria populations
go extinct. With harvesting as high as h = 0.3, no bacteria will survive.
Conclusion: The harvesting rate can’t be too high. It appears that h = 0.2 will
sustain a healthy population of bacteria to be harvested from continually. Going to
h = 0.3 will cause extinction and is unsustainable for the particular choices of
initial bacteria populations.
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