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Math 122 - Calculus for Biology II
Fall Semester, 2001
Linear Differential Equations - Examples
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© 2000, All Rights Reserved, SDSU
& Joseph M. Mahaffy
San Diego State University -- This page last updated 30-Oct-01
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Linear Differential Equations- Examples
- Malthusian Growth
- Radioactive Decay and Hyperthyroidism
- Newton's Law of Cooling
- Pollution in a Lake
Example
2: Malthusian Growth.
- Suppose that a culture of Escherichia coli is growing
according to the Malthusian growth model
P '(t) = rP(t)
- If the population doubles in 30 minutes, then find the value
of r
- If the initial population is 50,000, then find the population
after 45 min and 2 hr.
Solution:
- If the initial population satisfies P(0) = P0, then the solution
of the Malthusian growth model is
P(t) = P0 ert.
- This population doubles when
2P0 =
P0 e30r
e30r= 2
30r = ln(2)
r
= ln(2)/30 = 0.0231 min-1
- With the initial condition P0 =
50,000, the solution can be written:
P(t) = 50000e0.0231t
P(45)
= 141,421 and P(120)
= 800,000
- Note that the second case follows easily from the fact that
2 hours is 4 doubling times or 24 = 16 times the initial
population
Example
3: Radioactive Decay and Hyperthyroidism.
- Hyperthyroidism
is a serious health problem caused by an overactive thyroid
- The primary hormone released is thyroxine, which has a major stimulating effect
on the release of other hormones in the endocrine system
- In hyperthyroidism, the excess thyroxine causes the patient
to release too many other hormones in the body, such as insulin and the sex
hormones
- The result can be low blood sugar causing lethargy or mood
disorders and sexual dysfunction from too many sex hormones
- One of the most common treatments for an over-active thyroid
is to ablate the thyroid by giving
a large dose of radioactive iodine,
131I
- The 131I undergoes both
b and g radioactive decays, which destroy tissue
- The thyroid concentrates iodine brought into the body
- The 131I, which is taken up by the thyroid, decays, and the energetic
decays cause this gland to be destroyed
- Afterwards the patient is given medicine to supplement the
loss of thyroxine caused by this non-surgical procedure
- Based upon the thyroid condition and body mass, a standard
dose for this procedure ranges from 110-150 mCi (milliCuries), which is given
in a special "cocktail"
- It is assumed that almost 100% of the 131I is absorbed by the blood from the gut, then the thyroid
uptakes 30% of this isotope of iodine, peaking around 3 days
- The remainder is excreted in the urine
- The half-life of 131I is 8 days, so this isotope fairly rapidly decays
- Still the patient must remain in a designated room for 3-4
days for this procedure, so that he or she does not irradiate the public from
his or her treatment
- Assume that a patient is given a 120
mCi cocktail of 131I and that 30%
is absorbed by the thyroid
- Find the amount of 131I in the thyroid (in mCi), if the patient is released four days after swallowing
the radioactive cocktail
- Calculate how many mCis the patient's thyroid retains after 30 days, assuming
that it was taken up by the thyroid and not excreted in the urine
Solution:
- It is assumed for simplicity of the model that the 131I is immediately
absorbed into the thyroid, then stays there until it undergoes radioactive
decay
- Since the thyroid uptakes 30% of the 120 mCi, we will
assume that the thyroid has 36 mCi immediately after the procedure
- Obviously, this is an oversimplification as it takes time
for the 131I to accumulate in the thyroid
- However, it allows us to produce the simple model
R'(t) = -kR(t), with initial
condition R(0)
= 36 mCi
- The solution of this differential equation is
R(t) = 36e-kt
- Since the half-life of 131I is 8
days, then we know that after 8 days there will be 18 mCi
of 131I
- Thus, we have
R(8)
= 18 = 36e-8k, so e8k =
2 or 8k = ln(2).
- Thus, k
= ln(2)/8 = 0.0866 day-1
- At the time of the patient's release t
= 4 days, so
R(4)
= 36e-4k =
36/21/2 = 25.46 mCi
in the thyroid
R(30)
= 36e-30k =
2.68 mCi in the thyroid
Example
4: Newton's Law of Cooling.
- We would like to determine whether a cup of tea cools more
rapidly by adding cold milk right after brewing the tea or if you wait 5 minutes
to add the milk
- Assume that you begin with 4/5 cup of boiling hot tea, T(0)
=100oC, and that the tea cools according to Newton's law of
cooling
T '(t) = -k(T(t) - Te),
- where Te = 20oC is the temperature of the room and k is a constant based on the properties of the cup
and will be determined below
a. In the first scenario, you let the tea cool for 5
minutes, then add 1/5 cup of cold milk, 5oC. Assume that after 2
minutes the tea has cooled to a temperature of 95oC, and determine the
value of k (which we will assume holds throughout
the solution of this problem). When you mix in the milk, assume that the temperature
mixes perfectly in proportion to the volume of the two liquids. Find how long
until the tea reaches a temperature of 70oC.
b. In the second case, you add 1/5 cup of cold milk,
5oC, immediately and mix it thoroughly, then allow the tea/milk mixture
to cool to 70oC. Once again, find
how long this takes.
Which situation gets your tea cooled the fastest?
Alice sighed wearily. `I think you might do something better
with the time,' she said, `than waste it in asking riddles that have no answers.'
Solution:
- Begin by solving the initial value problem in a.
in order to find the rate constant k
- The information in the problem gives
T '(t) = -k(T(t) - 20),
T(0) =100 and T(2)
=95.
- As noted in the lecture notes, the easiest way to solve
this differential equation is to make the substitution,
z(t) = T(t) - 20,
so z(0)
=T(0) -20 =80
- Since z'(t) = T '(t), the initial value
problem above reduces to the following:
z'(t) = -kz(t),
z(0) =80
z(t) = 80e-kt =
T(t) - 20,
so
T(t) = 80e-kt + 20.
- From the condition, T(2)
=95, we have
95 = 80e-2k +
20 or e2k =
80/75
- Thus, k
= (ln(80/75))/2 = 0.03227. We substitute this value of k
into the equation for T(t), then evaluate
T(5)
= 80e-5k +
20 = 88.1oC
- We now mix the 4/5 cup of tea at 88.1oC with the 1/5 cup
of milk at 5oC, which gives one
cup of the tea/milk mixture at a temperature just following 5 minutes of
T+(5) =
88.1(4/5)
+ 5(1/5) = 71.5oC
- This results in the new initial value problem:
T '(t) = -k(T(t) - 20), T+(5) =
71.5oC
- Following the procedure above with the same substitution,
we have
z'(t) = -kz(t), z+(5) = 51.5
z(t) = 51.5e-k(t-5)= T(t) - 20, so
T(t) = 51.5e-k(t-5)+ 20
- It is easy to check that this is the solution of the new
initial value problem.
- To find when the tea is 70oC, we must solve
70 = 51.5e-k(t-5)+ 20
or ek(t-5)= 51.5/50.
- Thus, k(t - 5) = ln(51.5/50), so
t
= 5 + (ln(51.5/50))/k = 5.92 min.
- b. In this version of the problem, we begin with
the mixing of the tea and milk to find the initial temperature for our computations
- The mixing of 4/5 cup of tea at 100oC with 1/5 cup of milk
at 5oC gives one cup of
the tea/milk mixture at a temperature of
T(0)
= 100(4/5)
+ 5(1/5) = 81oC
- This results in the new initial value problem:
T '(t) = -k(T(t) - 20),
T(0) = 81oC
- Using the same z(t) substitution, we have
z'(t) = -kz(t),
z(0) = 61.
z(t) = 61e-kt =
T(t) - 20,
so
T(t) = 61e-kt+ 20
- To find when the tea is 70oC, we must solve
70 = 61e-kt +
20 or ekt
= 61/50.
t
= (ln(61/50))/k = 6.16 min
- It follows that by waiting to pour in the milk for 5 minutes,
you save about 15 seconds in cooling time.
Example
5: Pollution in a Lake.
- Suppose that a new industry starts up river from a lake
at t = 0 days, and this industry starts dumping a toxic pollutant,
P(t), into the river at a rate of 7 g/day, which flows directly into the lake
- The flow of the river is 1000 m3/day, which goes into the lake that maintains a constant
volume of 400,000 m3
- The lake is situated in a hot area and loses 50
m3/day of water to evaporation
(pure water with no pollutant), while the remainder of the water exits at
a rate of 950 m3/day through a river
- We assume that all quantities are well-mixed and that there
are no time delays for the pollutant reaching the lake from the river
- a. Write a differential equation that describes the
concentration, c(t), of the pollutant in the lake, using units of mg/m3 (Note that 1000 mg = 1 g.),
then solve the differential equation
- If a concentration of only 2 mg/m3 is toxic to the fish population, then find how
long until this level is reached
- If unchecked by regulations, then find what the eventual
concentration of the pollutant is in the lake, assuming constant output by
the new industry
- b. Suppose that the lake is at this limiting level
of pollutant and a new environmental law is passed that shuts down the industry
at a new time t = 0 days
- Write a new differential equation describing the situation
following the shutdown of the industry and solve this equation
- Calculate how long it takes for the lake to return to a
level that allows fish to survive.
Solution:
- a. In this example we need to return to the original
derivation of the lake pollution problem, where we used conservation of mass
to balance the amounts of pollutant entering and leaving the lake
- Recall that
The change in
amount of pollutant = Amount entering - Amount leaving
- If we let P(t) be the amount of pollutant in the lake at any time t,
then the change in amount is given by P '(t)
- The concentration is given by c(t) = P(t)/V and c'(t) = P '(t)/V
- The amount entering
is the constant rate of pollutant dumped into the river, which is given by
k = 7000 mg/day
- The amount leaving is given by the concentration of
the pollutant in the lake, c(t) (in mg/m3), times the flow of water out of the lake, f
= 950 m3/day
- Thus, we have the conservation of amount of pollutant given
by the equation:
P '(t) = k - fc(t) = 7000 - 950 c(t)
- Note that the evaporation from the lake indirectly affects
this amount equation by keeping the volume constant, but does not enter directly
as no pollutant is assumed to leave from evaporation in the amount equation
above
- The concentration equation follows by dividing the equation
above by the constant volume, V = 400,000
m3
- Thus,
P '(t)/V = k/V - fc(t)/V = 7/400
- 950 c(t)/400,000
or
- We make the change of variables,
z(t) = c(t) - k/f, with
z(0)
= c(0) - k/f = -k/f,
- which gives the differential equation
z(t) =-(k/f)e-f t/V
= c(t) - k/f or
- If a concentration of 2 mg/m3
is toxic to the fish population, then we want to find when c(t) = 2
mg/m3
- Thus, we need to solve
2 = 7.368(1-e-0.002375t )
or e0.002375t = 7.368/5.368 ~ 1.3726
t
= ln(1.3726)/0.002375 ~ 133.3 days
- The limiting concentration has the exponential term go to
zero. Thus, we have
- b. The new environmental law shuts down the industry
at a new time t
= 0 days, so the equation above follows with k = 0 mg/day with c(0)
= 7000/950 ~ 7.368 mg/m3
- Thus, we need to solve
- To find when the concentration is reduced to 2 mg/m3, we evaluate
2 = 7.368 e-0.002375t or
e0.002375t = 7.368/2 ~ 3.684
t
= ln(3.684)/0.002375 ~ 549 days