1
(
)=(
+
)
(
)=(
+
√
)
h
y
Spring 2019
Assignment Section 2.3 Modeling with First Order
1. (1 point)
Newton’s law of cooling says that the rate of cooling of an
object is proportional to the difference between the
temperature of the object and that of its surroundings
(provided the difference is not too large).
If T = T(t) represents the temperature of a (warm) object at
time t, A represents the ambient (cool) temperature, and k is a
negative constant of proportionality, which equation(s)
accurately characterize Newton’s law?
•A. dTdt = kT(T −A)
•B. dTdt = k(T −A)
•C. dTdt = kT(1−T/A)
•D. dTdt = k(A−T)
•E. All of the above
•F. None of the above
Answer(s) submitted:
•B
(correct)
Correct Answers:
•B
2. (1 point)
Water leaks from a vertical cylindrical tank through a small
hole in its base at a rate proportional to the square root of the
volume of water remaining. The tank initially contains 275 liters
and 19 liters leak out during the first day.
A. When will the tank be half empty? t =days
B. How much water will remain in the tank after 3 days?
volume = Liters
Solution:
SOLUTION
Let V(t) be the volume of water in the tank at time t, then dV
√
= k V. dt
This is a separable equation which has the solution
kt
V .
Since V(0)= 275 this gives 275 =C2 so
V 2.
However, V(1)= 256, and so
k √ 2
256 =( + 275) ,
2
√ √
so that k = 2( 256− 275)= −1.1662. Therefore,
√
V(t)=(−0.5831t + 275)2.
The tank will be half-empty when V(t)= 137.5, so we solve
√
137.5 =(−0.5831t + 275)2
to obtain t = 8.329 days. The tank will be half empty in 8.329
days.
The volume after 3 days is V(3) which is approximately
220.042 liters.
Answer(s) submitted:
•8.326
•219.96
(correct)
Correct Answers:
•(sqrt(275)-sqrt(275/2))/(sqrt(275) - sqrt(275-19))
•(sqrt(275) - (sqrt(275) - sqrt(275-19))*3)ˆ2
3. (1 point) A tank contains 1040 L of pure water. Solution
that contains 0.07 kg of sugar per liter enters the tank at the
rate 9 L/min, and is thoroughly mixed into it. The new solution
drains out of the tank at the same rate. (a) How much sugar
is in the tank at the begining?
y(0)= (kg)
(b) Find the amount of sugar after t minutes.
y(t)= (kg)
(c) As t becomes large, what value is y(t) approaching ? In other
words, calculate the following limit. lim y(t)= (kg)
t→∞
Solution: (a) Since the tank originally contains pure water,
the initial amount of sugar is y(0)= 0.
(b) First note that, since the incoming and outgoing flows of
water are the same, the amount of water in the pool remains
constant at 1040 L. We have
dy
2
= rate in−rate out
dt
where ”rate in” and ”rate out” refer to the rates at which the
sugar enters and exits the tank, respectively. The rate at
which the sugar enters the tank is given by rate in =(9
L/min)(0.07 kg/L)= 0.63 kg/min
The concentation of sugar in the tank is 1040 y kg/L, so
the rate of flow out is
3
i 9 y kg/min rate out
=
.
−
y
=
−
(
y
−
.
)
Z
y
−
.
=
−
Z
|
y
−
.
|
=
−
+
y
=
.
+
−
y
(
)=
=
−
.
y
=
.
−
.
−
→
.
−
.
−
=
.
•
0
•
•
•
0
•
•
.
kg of salt per liter
y
=
y
(
)=
c) Solve the initial value problem in part (b
y
(
)=
4
=(9 L/min) kg/L = 1040 Solution: (a) Since the tank
originally contains 50 L, the initial concentration is
05kg/L.
(b) First note that, since the incoming and outgoing flows
ofwater are the same, the amount of water in the pool
remains constant at 1000 L. We have
dy
= rate in−rate out dt where ”rate in”
and ”rate out” refer to the rates at which the salt enters and
exits the tank, respectively. The rate at which the salt enters
the tank is given by rate in =(10 L/min)(0.025 kg/L)= 0.25
kg/min
The concentation of salt in the tank is 1000 y kg/L, so the rate
of flow out is
i rate out =(10 L/min)kg/L = 0.01y kg/min
Thus we obtain the differential equation
dy
= 0.25−0.01y dt
where each term has the units kg/min. The initial condition is
y(0)= 50.
(c) We rewrite the differential equation as
dy
= −0.01(y−25) dt
Separating the variables yields
dt
so
ln|y−25| = −0.01t +K
and y = 25+Ce−0.01t
The initial condition y(0)= 50 gives C = 25, so
y(t)= 25+25e−0.01t
(d) The amount of salt after 4.5 hours is the given by
y(270)= 25+25e−0.01(270) = 26.6801
(e) The concentration in the tank at any time t is given by
tso
Note that this limiting concentration equals the incoming
concentration.
Answer(s) submitted:
• 0.05
• 0.25-(y/100)
• 50
• 25+(25/(exp((1/100)t)))
• 26.68
• 0.025
(correct)
Correct Answers:
• 0.05
• 10*0.025-10*y/1000 • 50
• 25+25*exp(-0.01*t)
• 26.6801378184937
• 0.025
5. (1 point) A tank contains 60 kg of salt and 2000 L of water.
Pure water enters a tank at the rate 8 L/min. The solution
is mixed and drains from the tank at the rate 4 L/min.
(a) Write an initial value problem for the amount of salt, y, in
kilograms, at time t in minutes: dy
= (kg/min) y(0)= kg. dt
(b) Solve the initial value problem in part (a) y(t)=
kg.
(c) Find the amount of salt in the tank after 4.5 hours.
amount = (kg)
(d) Find the concentration of salt in the solution in the tankas
time approaches infinity. (Assume your tank is large enough
to hold all the solution.)
concentration = (kg/L)
Solution: (a) First note that, since the incoming and
outgoing flows of water are not the same, the amount of water
in the tank is not constant. In fact the amount of water in the
tank increases at a rate of 4 L/min and therefore the amount
of water at any given time time t is given by 2000+4t.
Let y(t) be the amount of salt (in kg) at time t minutes.
We have
dy
= rate in−rate out dt where ”rate in”
and ”rate out” refer to the rates at which the salt enters and
exits the tank, respectively. Since pure water enters the tank,
the rate at which the salt enters the tank is given by rate in =(8
L/min)(0 kg/L)= 0 kg/min
The concentation of salt in the tank is 2000 y+4t kg/L, so the
rate of flow out is
y y
5
rate out =(4 L/min) kg/L = 4 kg/min
2000+4t 2000+4t
Thus we obtain the differential equation
dy 4y
= − dt
2000+4t
where each term has the units kg/min. The initial condition is
y(0)= 60.
(b) Separating the variables gives
dy 4
= − dt y
2000+4t
Integrating
ln|y| = −ln|2000+4t|+K
and
C
y =
2000+4t
The initial condition y(0)= 60 gives C = 120000, so
120000
y(t)=
2000+4t
(d) The amount of salt after 4.5 hours, i.e. the amount of salt
after 270 minutes, is the given by y
(d) The concentration in the tank at any time t is given by
.
t
The limiting concentration is zero, as expected, since pure
water is entering the tank.
Answer(s) submitted:
• -y/(500+t)
• 60
• 30000exp(-ln(500+t))
• 38.96
• 0
(correct)
Correct Answers:
• -4*y/(2000+4*t)
• 60
• 120000/(2000+4*t)
• 38.961038961039
• 0
6. (1 point) Newton’s law of cooling states that the
temperature of an object changes at a rate proportional to the
difference between its temperature and that of its
surroundings.
Suppose that the temperature of a cup of coffee obeys
Newton’s law of cooling. Let k > 0 be the constant of
proportionality. Assume the coffee has a temperature of 205
degrees Fahrenheit
when freshly poured, and 3 minutes later has cooled to 189
degrees in a room at 70 degrees.
(a) Write an initial value problem for the temperature T
of the coffee, in Fahrenheit, at time t in minutes. Your answer
will contain the uknown constant k: dT
dt = T(0)=
7.
(1
point)
A
curve
passes
through
the
point
(0,6)
and
has
the
prop-
erty that
the
slope
of
the
curve
at
every
point
P
is
twice
the
y-coordinate
of
P.
What
is
the
equation
of
the
curve?
yx)
=
Solution:
y(x)
satisfies
the
differential
equation
dy
dx
with
solution
v(x)
=Ce
Since
the
curve
passes
through
the
point
(0,6),
we
have
6
=C
so
y(x)
=
6c"
Answer(s)
submitted:
e@
6exp
(2x)
(correct)
Correct
Answers:
e
6 *
e7(2
*
x)
8.
(1
point)
A
thermometer
is
taken
from
a
room
where
the
temperature
is
22°C
to
the
outdoors,
where
the
temperature
is
—4°C.
After
one
minute
the
thermometer
reads
11°C.
(a)
What
will
the
reading
on
the
thermometer
be
after
3
more
minutes?
(b)
When
will
the
thermometer
read
—3°C?
minutes
after
it
was
taken
to
the
outdoors.
Solution: Let
T(t)
be
the
temperature
at
time
¢,
measured
in
minutes.
T(t)
satisfies
the
differential
equation
aT
—
=-k(T+4
dt
(T+4)
with
solution
T(t)=—4+Ce™
Since
T(0)
=
22,
we
have
T(t)
=—4426e
*
After
one
minute
the
temperature
is
11,
thus
11=—4+426e*,
which
gives
k
=
—In
(52)
+
0.550046.
Hence
T
(t)
=
—4
+. 26e~
09900461
(a)
The
reading
on
the
thermometer
after
3
more
minutes
will
be
T(4)
=
—4
+26¢~9-550046(4)
~
1.11965
(a)
To
find
when
the
termomether
will
read
—3,
we
solve
the
equation
—3
=
—4
4.26
e-0-5500461
6
(b) Solve the initial value problem in part (a). Your
7
answer will contain the unknown constant k. T(t)=
(c) Determine the value of the constant k k =
minutes.
(d) Determine when the coffee reaches a temperature of
159 degrees.
minutes.
Solution: (a) T(t) satisfies the initial value problem
dT
= −k(T −70) T(0)= 205 dt
(b) Solving the differential equation gives
T(t)= 70+Ce−kt
Since T(0)= 205, we have
T(t)= 70+135e−kt
(c) After 3 minutes the temperature is 189, thus
189 = 70+135e−3k,
which gives
k .
(d) The temperature is given by
T(t)= 70+135e−0.0420504t
To find when the coffee will reach a temperature of 159, we
solve the equation
159 = 70+135e−0.0420504t
This gives t = 9.90807 hours.
Answer(s) submitted:
• -k(T-70)
• 205
• 135exp(-kt)+70
• 0.04205
• 9.908
(correct)
Correct Answers:
• -k(T-70)
• 205
• 70+135 eˆ(-k*t)
• 0.0420504284423
• 9.90806572346784
This
gives
t
=
5.92332
minutes.
Answer(s)
submitted:
e
-1.119
e@
5.923
(correct)
Correct
Answers:
e@
-1.11965179790624
@
5.92331285445804
9.
(1
point)
Dead
leaves
accumulate
on
the
ground
in
a
forest
at
a
rate
of
3 grams
per
square
centimeter
per
year.
At
the
same
time,
these
leaves
decompose
at
a
continuous
rate
of
50
percent
per
year.
A.
Write
a
differential
equation
for the
total
quantity
Q
of
dead
leaves
(per
square
centimeter)
at
time
¢:
d
“
B. Sketch
a
solution
to
your
differential
equation
showing
that
the
quantity
of
dead
leaves
tends
toward
an
equilibrium
level.
Assume
that
initially
(¢
=
0)
there
are
no
leaves
on
the
ground.
What
is
the
initial
quantity
of
leaves?
Q(0)
=
What
is
the
equilibrium
level?
Qe,
=
Does
the
equilibrium
value
attained
depend
on
the
initial
condi-
tion?
e
A.
yes
e
B.no
Solution:
SOLUTION
Let
Q(t)
be
the
quantity
of
dead
leaves,
in
grams
per square
centimeter.
Then
ae
=
3-—0.5Q,
where
is
in
years.
If
we
suppose
that
the
initial
quantity
is
Q(0)
=
0,
then
we
can
guess
what
the
solution
will
look
like
without
solving
the
differential
equation:
initially
40
=
3,
so
the
quantity
is
increas-
ing.
However,
as
it
increases
the
rate
of
decay
—0.5Q
increases,
so
Q
will
increase
until
the
decay
rate
is
equal
to
the
accumula-
tion
rate.
This
occurs
when
3
=
0.5Q,
or
Q
=
6.
Note
that
Q
=
is
the
equilibrium
solution,
which
is
where
Qis
constant:
when
is
constant,
a0
=0,
which
gives
the
same
condition
as
we
solved above.
Therefore
we
know
that
the
solution
will
look
like
the
figure
below,
which
shows
the
solution
in
blue
and
the
equilibrium
in
red.
(Click
on
the
graph
for
a
larger
version.)
We
can
also
solve
the
differential
equation
using
separation
of
variables.
dQ
=3-05
dt
2.
so
3-050
as
=
[a
and
—
ps
ini3
-
0.59|
=t+C.
Solving
for
Q,
we
get
Q=6-Ae°™,
.
With
Q(0)
=
ont
Note
that
no
matter
what
the
initial
condition
is,
the
same
equilibrium
solution
will
be
attained.
Answer(s)
submitted:
©
3-0.50
e
0
e6
eB
where
A
=
+4>—
5
(correct)
Correct
Answers:
e
3
-
0.5*0
e0
©
3/0.5
eB
10.
(1
point)
According
to
a
simple
physiological
model,
an
athletic
adult
male
needs
20
calories
per
day
per
pound
of
body
weight
to
maintain
his
weight.
If
he
consumes
more
or
fewer
calories
than
those
required
to
maintain
his
weight,
his
weight
changes
at
a
rate
proportional
to
the difference
between
the
number
of
calo-
ries
consumed
and
the
number
needed
to
maintain
his
current
weight;
the
constant
of
proportionality
is
1/3500 pounds
per
calorie.
Suppose
that
a
particular
person
has
a
constant
caloric
intake
of
H
calories
per
day.
Let
W(t)
be
the
person’s
weight
in
pounds
at
time
¢
(measured
in
days).
(a)
What
differential
equation
has
solution
W(t)?
dw
_
dt
(Your
answer
may
involve
W,
H
and
values
given
in
the
prob-
lem.)
(b)
Solve
this
differential
equation,
if
the
person
starts
out
weighing
150
pounds and
consumes
2900
calories
a
day.
We
(c)
What
happens
to
the
person’s
weight
as
t
—>
c0?
Ww
Solution:
SOLUTION
8
9
(a) Since the rate of change of the weight is equal to
(Intake − Amount to maintain weight) we
have
.
(b) Starting off with the equation
,
we separate variables and integrate:
.
Thus we have
C
so that
W
or, in other words
H −t/175
W Ae .
If the caloric intake is 2900 calories per day we know that H =
2900. Then, if the person initially weighs 150 pounds, we have
150 = 145+A,
so that A = 5. Thus
W = 145+5e−t/175.
(c) From our solution in (b), we know that as t → ∞, W →
145. This is seen in the graph of weight vs. time, shown
below.
Answer(s) submitted:
• (1/3500)(H-20W)
145+5exp(-t/175)
(correct)
Correct Answers:
• 1/3500*(H-20*W)
=
.
v
(
)
positive velocity). The gravitational constant is
=
.
/
s
v
=
a) The net force acting on the skydiver is given by
=
−
+
=
−
+
v
,
=
−
.
+
.
v
b) The terminal velocity is given by the equilibrium solution,
=
−
.
+
.
v
v
=
.
.
.
Since the velocity is downward, the limiting value is
v
=
−
r
.
.
≈−
.
•
•
•
•
10
• 2900/20+(150-2900/20)*eˆ(-t/175)
• 2900/20
12. (1 point)
A body of mass 4 kg is projected vertically upward with an
initial velocity 26 meters per second.
We assume that the forces acting on the body are the force of
gravity and a retarding force of air resistance with direction
opposite to the direction of motion and with magnitude
c|v(t)|
where c = 0.45kg s and v(t) is the velocity of the ball at time t.
The gravitational constant is g = 9.8m/s2.
a) Find a differential equation for the velocity v:
dv
= dt
b) Solve the differential equation in part a) and
find a formulafor the velocity at any time t: v(t)=
Find a formula for the position function at any time t, if the
initial position is s(0)= 0: s(t)=
How does this compare with the solution to the equation for
velocity when there is no air resistance?
If c = 0, then v(t) = 26 − 9.8t, and if s(0) = 0, then s(t) =
26t −4.9t2.
We then have that v(t) = 0 when t ≈ 2.653, and s(2.653) ≈
34.490, and that the positive t solution to s(t)= 0 is t ≈ 5.306,
which leads to v(5.306)= −26 meters per second.
Answer(s) submitted:
• -9.8-((0.45(v))/4)
• (1018/9)exp(-0.1125t)-(9.8/0.1125)
• (-81440/81)exp(-0.1125t)-(9.8/0.1125)t+(81440/81)
(correct)
Correct Answers:
• -9.8 - 0.45*v/4
• -4*9.8/0.45 + (26 + 4*9.8/0.45)*exp(-0.45*t/4)
Q0 =
(b) Solve this differential equation, assuming there is no drug in
the body initially. Your answer will contain r and k.
Q =
(c) What is the limiting long-run value of Q?
lim Q(t)=
t→∞
Solution:
SOLUTION
The differential equation for Q is
dQ
=
r−kQ, dt
so that
dt,
and, after solving,
r −kt
Q = +Ae
. k
When t = 0, Q = 0,
r −e−kt).
Q = (1 k
Thus,
r
Q∞= lim Q = .
t→∞ k
This is shown in the following graph.
(Click on the graph for a larger version.) Answer(s)
submitted:
• (4*26/0.45 + 4ˆ2*9.8/0.45ˆ2)-4*9.8/0.45*t - (4*26/0.45 + 4ˆ2*9.8/0.45ˆ2)*exp(-
0.45*t/4)
• r-kQ
11
13. (1 point) A drug is administered intravenously at a con- • (-r/k)*exp(-kt)+(r/k) stant rate of r
mg/hour and is excreted at a rate proportional to • r/k the quantity present, with constant of
proportionality k > 0. (correct)
Correct Answers:
(a) Set up a differential equation for the quantity, Q, in mil- • r-k*Q ligrams, of the drug in the
body at time t hours. Your answer • r/k*[1-eˆ(-k*t)] will contain the unknown constants r and
k. • r/k
