Accelerated Calculus
Differential Modeling
1. Translating into differential equation language:
Let B = the population of bacteria
dB
dt
dB
B

 kB , B (0)  200 and B(5)  500
 kdt
dB
B
  kdt
ln B  kt  C
B  e kt C  Ae kt , so the general solution is B(t )  Aekt .
Utilizing the initial conditions to solve for A and k:
B(t )  Aekt
B(0)  Ae0  200  A  200
B(t )  200ekt
1 5
B(5)  200e5 k  500  k  ln  
5 2
B(t )  200e
1 5
ln  t
5  2
Is this plausible? Take a look at the solution drawn over the slope field.
Accelerated Calculus
Differential Modeling
2. We are given
 0.003N(500  N) to be the differential equation.
dN
dt
a) We have been tasked with finding a solution to this equation,
dN
dt
 0.003N(500  N)
Let’s take a look at the slope field generated by this differential equation (jus to observe some
 interesting features about the curve we will find):

y






x



We are looking for a particular solution with an initial condition, N (0)  50 . Once we have
completed this task, we should be able to overlay our solution on this slope field and see if we it
behaves as expected.
So, to begin, using our separation of variables approach:

1
N (500  N )
dN   0.003 dt

1
N (500  N )
dN  0.003t  C
Here we want to recall our “partial fraction decomposition” integration method to be able to
evaluate the left-side of this equation.
Let’s take a moment to do some necessary algebra to re-write
1
N (500  N )
as
A
N
 500B N so we can
integrate a much simpler expression!
1
N (500  N )

A
N
 500B N
1  A(500  N )  B( N )  note: we only need to look at the numerator
Then, remembering an efficient approach:
let N  0 ; 1  A(500  0)  B(0)  1  A(500)  A 
1
500
let N  500 ; 1  A(500  500)  B(500)  1  B(500)  B 
So,
1
N (500  N )

1
500
N
1
500
1
 500500 N
Now, back to the calculus:

1
N (500 N )
500
1
1
1
dN  0.003t  C   500
N  500 N dN  0.003t  C  500  N  500  N dN  0.003t  C
1
1
Accelerated Calculus
Differential Modeling
1
500
ln N  ln(500  N )  0.003t  C
ln N  ln(500  N )  0.003(500)t  C
ln(500  N )  ln N   0.003(500)t  C  note: multiply both sides by -1, to make upcoming work efficient
 500  N 
ln 
  0.003(500)t  C  note: remember your log properties!
 N 
500  N
 e0.003(500)t C
N
500  N
 Ae0.003(500) t  note: remember what we can do with our exponent rules!
N
500
 1  Ae0.003(500) t
N
500
 1  Ae 0.003(500) t
N
N t  
500
1  Ae 0.003(500) t
Now, given the initial condition: N (0)  50
So, N  0  
500
500
 50 
 50  A  9
0.003(500)(0)
1  Ae
1 A
Our final solution to the differential equation
N (0)  50 :
N t  
500
1  9e 0.003(500) t
dN
dt
 0.003N(500  N) with initial condition,

Now, we see that our solution exhibits the expected behavior when placed on top of the slope
field for our differential equation!
y






x



Accelerated Calculus
Differential Modeling
b) The maximum number of fish in the pond is 500 (check the end-behavior: t   )
c) The rate of growth in the fish population is greatest at the inflection point of the curve N  t  . We
can find this by determining when and at what population the second derivative changes sign.
We were given
 0.003N(500  N) , the derivative of N  t  .
 0.003N (1)  (500  N )0.003
d2N
dt 2
So,
dN
dt
ThenddtN2  0.003(500  2 N )
2
Notice, at N  250 , this second derivative changes sign (from + to – as N increases)
Therefore, the rate of growth of the fish population grows fastest when the population is 250
(check your graphical solution to verify this).
3. Translating into differential equation language:
Let y = the fraction of population that has heard the rumor
a)
dy
dt
80
 ky 1  y  , y(0)  600
and y(230)  12 , where t is measured in minutes
Here is the slope field provided by the above differential equation.
y

x








This differential equation matches the structure of the following equation:
We therefore know the general solution is a logistic growth model:
P t  

dP
dt

 kP  M  P 
M
1  Ae  ( k )( M )t
So, we can safely state that the general solution to the differential equation boxed above is:
Accelerated Calculus
Differential Modeling
y t  
1
1  Ae  kt
We now need to find the solution given conditions laid out in the exercise.
So, using y(0) 
80
600
80
600
:

1
1  Ae k 0

2
15

1

1 A
15
2
 1 A 
A  132
Now, using y(230)  12 :
1
2

1
1  e  k  230
k 
 2  1  132 e
13
2
 k  230 

2
13
e
 k  230 
 ln 132   k  230 
ln 132
 0.008138
230
y t  
Therefore, our model is:
1
1  e 0.008138t
13
2
Let’s check for reasonability of our model by graphing our model on our slope field provided
by the differential equation.
y

x









b) We want to find out where 90% of the population of 600 students has heard the rumor.
So, we want to solve the following for t:
0.9 
1
1  e 0.008138t
This happens at t  500 minutes (around 4:30pm).
13
2

Accelerated Calculus
Differential Modeling
4. Given the Differential equation:
dP
 0.0004 P(250  P) . With P=28 when t=0 ( 1970)
dt
dP
 0.0004 P(250  P) is separable
dt
(a)
1
dP  0.0004dt
P (250  P )
1
 P(250  P) dP   0.0004dt
1 1
1

dP  0.0004t  C

250 P 250  P
1
P
ln
 0.0004t  C
250 250  P
ln
P
 250  0.0004t  C 
250  P
250  P
 0.1t  C
P
250  P
 e 0.1t C  Ae 0.1t
P
250  P  PAe 0.1t
ln
(different C )
250
1  Ae 0.1t
250
222
28 
 A
 7.929
1 A
28
250
P
1  7.929e 0.1t
P
(b) Rate of change greatest when
d 2P
0
dt 2
d 2 P d  dP  d
dP
dP
 
 0.0004 P(250  P)   (.1  0.0008P)

2
dt
dt  dt  dP
dt
dt
2
d P
.1
 0 when P 
 125 P  0 or P  250
2
dt
.0008
From the slope field we see greatest rate of change when P=125
125 
250
1
 e.1t 
 t  20.7
0.1t
1  7.929e
7.929
ie in 1990.
The rate at this point was .0004*1252= 6.25 gorillas per year
Accelerated Calculus
Differential Modeling
5. $1000 invested at 2%
dA
1
1
 .02 A  dA  .02dt   dA   .02dt
dT
A
A
.02 t  C
.02 t
ln A  .02t  C  A  e
 A0e

A0 =1000  A  1000e.02t
When t=10 A  1000e.2  1000*1.2214 therefore the amount after 10 years is $1221.40
6. Translating into differential equation language:
Let F = temperature (in degrees F) of the roast
dF
dt
 k  350  F  , F (0)  40 and F (60)  90 , where t is measured in minutes
Here is the slope field provided by the above differential equation.

y








































x
         

Note: this differential equation does NOT fit the structure of that leading to an exponential growth or
logistic growth model. We must do Calculus to solve for F (t ) .
dF
dt
 k  350  F  

1
350 F
ln 350  F  kt  C 
dF   kdt   ln 350  F  kt  C
350  F  Ae kt
Given the initial condition, F (0)  40 :
So, 350  F  310e kt
350  40  Ae
 k  0
 310  A
 350  F  310e kt  F t   350  310e kt
F  t   350  310e kt
Now, to find k, use our other condition, F (60)  90 :
 k  60
90  350  310e
Accelerated Calculus
Differential Modeling
90  350  310e
 k  60 

260
310
e
 k  60 
260
 ln  310
  60k  k 
260
ln  310

60
k  0.002932
So, our model is F  t   350  310e0.002932t
Let’s check for reasonability of our model by graphing our model on our slope field provided
by the differential equation.

y








































x
         

We wish to find the time at which the roast will be done (reaching a temp of 140 degrees).
140  350  310e0.002932t
So,
t
210
ln  310

0.002932

210
310
 e0.002932t
210
 ln  310
  0.002932t
 t  132.8325
Therefore, the roast will be done after roughly 2 hours 13 minutes.
7. Newton’s Law of cooling states that given the temperature of the room is 68 and therefore the
body cannot get colder than that
dT
 k (68  T )
dt
 ln 68  T  kt  C
68  T  Ae
A few steps were skipped here. Refer to #6 above for the missing steps
 kt
Setting t=0 at 9:0am 68 – 90.3 = A = -22.3
T  68  22.3e  kt t  1 T  89 
21
 e  k  k  .06  T  68  22.3e .06t
22.3
At time of death T=98.6 therefore t=-5.27 Approximately 3:43 AM
Accelerated Calculus
Differential Modeling
Slope field confirmation of results in #7
The circle is the point (-5.27,98.6)
8. This problem is done in two parts firstly we calculate the solution of the problem of the rate by which
the body processes morphine. Then we add the new information of the continuous intravenous drip.
(a)
dP
 kP  P  P0e kt
dt
Half life is 2 hours therefore when
1
1
1  1  .693
P  P0 t  2  P0  P0e 2 k  k  ln   
 .3465
2
2
2 2
2
So now that morphine is being added we have
dP
 2.5  0.3465P  0.3465(7.21  P)
dt
(b) The horizontal line is P=7.21
As can be seen by the slope field the long term tendency of
all curves would be to maintain 7.21 mg of morphine in the
blood as at this level the rate of change is zero.
Accelerated Calculus
Differential Modeling
9.
dL
 3  .75L  .75(4  L)
dt
Solving this equation would lead to the expression: L  4  Ae .75t for some constant A, which
may be positive or negative depending on initial conditions.
In either case lim e .75t  0  L  4 over time
t 
From the initial equation as L  4
dL
0
dt
Therefore this is an equilibrium value. We can also see this from the slope field for the
differential equation:
Please note that the equilibrium line should be at 4 not 40
10. Assume that “water leaks out of a barrel” means that the height of the water in the barrel is
changing…. So let h = height of water in the barrel at time, t.
dh
 k h , with h(0)  36 and h(1)  35
dt

dh
 kdt
h 
2 h  kt  C
Use the first condition: 2 36  k  0  C  12
2 h  kt  12
And the second condition: 2 35  k 1  12  k  2 35  12  0.168
Accelerated Calculus
Differential Modeling
2 h  0.168t  12 or h(t )   0.084t  6 
2
To determine when the barrel will be empty, set h(t )   0.084t  6   0  t  71.496 hours
2
Graphically: This model looks like, which is reasonable given the initial conditions and
parameters of the problem.
11. Let s = the number of kilograms of salt at time, t.
ds
 salt into the tank  salt out of the tank
dt
ds
kg
L
s kg
L
ds
kg
 0.03  25

 25
(note that units of
will be
, as expected)
dt
L
min 5000 L
min
L
dt
ds
s
 0.75 
, with s (0)  20
dt
200

ds
0.75 
s
200
  dt
s 

200ln  0.75 
 t C
200 

s 
1

ln  0.75 
t C

200 
200

Accelerated Calculus
Differential Modeling
1

t
s
200
0.75 
 Ae
200
1
1

t

t

200
200
200  0.75  Ae
 s(t )
  150  Be


Use initial condition: s (0)  20 , s(0)  150  Be0  20  B  130
s(t )  150  130e

1
t
200
, so after 30 minutes s(30)  150  130e

1
(30)
200
 38.108 kg
12. Here is the most important observation about the way the banks work in “mixing” the
currency: every day 50M in mixed (new/old) bills go into the bank. Of that mixture, every
old bill gets exchanged for a new bill and thus the currency exchanges old bills for new bills.
This means that every day the rate of change in new bills is exactly equal to the number of
old bills that enter the bank on a given day!
So, let N = Amount of money (in millions) in new bills in circulation at time t.
Of the 10B = 10,000M in circulation,
circulation. And therefore 1 
N
is the proportion of new bills in the total
10,000
N
is the proportion of old bills in the total circulation.
10,000
dN
 number of old bills that enter the banks on a given day
dt
fraction that are old
dN

dt
million $/ day
50
N 

 1 
 the units would be “millions of old bills per day”
 10,000 
dN
N
 50 
, with N (0)  0
dt
200

dN
N
50 
200
  dt
N 

200ln  50 
 t C
200 

N 
1

ln  50 
t C

200 
200

1
t

t C
N
200
200
50 
e
 Ae
200
t
10,000  Be 200  N
Accelerated Calculus
Differential Modeling
With initial condition: N (0)  10,000  Be
 (0)
200
 0  B  10,000
t
t


N (t )  10,000  10,000e 200  10,000 1  e 200 


So, we want to know when the amount of new money will be 90% of the total, so
t
t


1
N (t )  10,000  10,000e 200  10,000 1  e 200   9000  t  200ln    460.517 days
 10 

