Radioactivity
It is well know (Due to the work of Henri Becwuerel, Marie Sklodowska Curie and Pierre
Curie) that rate of decay of a radioactive material is directly proportional to the amount of
radioactive material at any given time. Mathematically this can be written as,
dy(y)
= ky(t)
dt
where k is constant depending on the material, which can be found from the half life of the
material. Note that k is always negative as the amount of radioactive material always decays.
If we assume that initial amount of radioactive material is y0 then we can find the amount of
material at any given time if k is know. Consider the following example:
If half-life of plutonium P o209 (a radioactive substance) is 100 years, determine the percentage
of plutonium left after 50 years.
If y0 is the initial amount then plutonium at time t is y(t) = y0 ekt . From the given conditions, y(100) = 21 y0 , which implies ek = (1/2)100 (clearly k is negative). Nowy(50) =
y0 (1/2)(50/100) = 0.7071y0 . So 70.71% plutonium is left after 50 years.
Population dynamics
Let t : time, x(t) : number individual population at time t, b : birth rate of population, d :
death rate. Then we have the following discrete model which is due to Maltus.
Maltus Model: The increment of the population x(n + 1) − x(n) = bx(n) − dx(n) between
t = n and t = n + 1 is equal to the number of new born individuals bx(n) minus the number
of deaths dx(n).
i.e., x(n + 1) − x(n) = bx(n) − dx(n) = (b − d)x(n).
When we pass to continuous time variables, we take an infinitesimal change of time ∆t. Then
the change of population x(t + ∆t) − x(t) between t and ∆t given by the unit growth rate k
times the population size x(t), times the interval of time ∆t. Thus
x(t+ M t) − x(t) = (b − d)x(t)∆t.
Dividing by ∆t and taking limit ∆t → 0, we get x0 (t) = (b − d)x(t). Now define k = b − d as
the growth rate. The solution is x(t) = x0 ekt . So if b = d i.e. k = 0, x(t) = x0 for all t ≥ 0 as
we expect. If b > d, then x(t) grows exponentially to ∞. On the other hand, if b < d, then
x(t) → 0 exponentially. This model is rather rough in the sense that it does not take into
account the fact that b, d might depend on the population size. Later we will see the logistic
model which is more realistic model.
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Newton’s Law of heating
A solid body loses heat to its surrounding medium. The rate of heat loss is observed to be
proportional to the temperature difference between the solid body and surrounding medium.
Typically
dT
mcp
= −hA(T − Ta )
dt
models the temperature T (t), where m, cp , h, A and Ta denote respectively the mass of the
body, specific heat of the material, heat transfer coefficient, surface area of the body and
temperature of the surrounding medium. The above equation may be written as
dT
= −k(T − Ta )
dt
and k is constant of proportionality which can be fitted from observation.
Note that all the above models cover wide class of scientific problems, however underlined
mathematical models are similar and depends on the fact that the rate of change of quantity
depends the quantity linearly. We will now consider the cases when other factors are effecting
the rate of change.
Mixture Problem
Let us consider a tank containing V0 liters of brine solutions (water and salt mixture). Assume
that a brine solution with the concentration of S1 Kg/Ltrs is allowed to flow inside the tank
at the rate of R1 liters per minutes while a well-stirred (i.e. we assume the concentration is
constant in the tank) mixture flows out of tank with the rate R2 liters per minutes. If y(t)
represent the amount of salt in the tank at any time t, then the equation describing the rate
of change of y is,
dy
= (Rate at which salt enter the tank)- (Rate at which tank leaves the salt)
dt
= (S1 R1 )kg/min −
y(t)
R2 kg/min
V (t)
where V (t) is the total volume of solution in the tank. Note that y(t)/V (t) represent
the concentration of the brine solution in the tank at any time t (assuming continuous well
stirring of the solution in the tank).
Also V (t) changes as follows:
dV (t)
= (Rate at which solution enters the tank)-(Rate at which solution leaves the tank)
dt
= (R1 − R2 )Ltr/Min.
2
Note that second equation can be easily solved, which allows us to solve first (linear)
equation. Also we need to initial condition. To solve second equation we need initial volume
of solution in the tank V0 and to solve first equation we need initial concentration y0 of the
brine solution in tank.
Population dynamics
The following equation due to P.F. Verhulst where one assumes that the variation x0 of the
number of individuals in a population is proportional to x, but also considers the resources
available. This leads to the equation,
1
x = kx 1 − x ,
K
0
where K is carrying capacity of the population. This is defined as the constant population
which can be maintained indefinitely given the resources. Rewriting the equations we get,
x0 = x(α − βx), α, β > 0,
αc1 eαt
with α = k and β = k/K. By separation of variables x(t) =
.
1 + βc1 eαt
α
Now it is easy to see that as t → ∞, x(t) → = K, the carrying capacity of the population.
β
Here again, one can check using the existence and uniqueness theorem, the solution exists
and unique. Hence the solution is
x(t) =
αc1 eαt
.
1 + βc1 eαt
Applications
1. Falling Body: Consider a stone falling freely through the air. Assuming that the
air resistance is negligible and that the acceleration due to gravity is g = 9.8 m/s2 ,
construct the resulting ODE and find the solution of the ODE, if the initial position is
h0 and the initial velocity is v0 .
2. Suppose that an object with mass 1 Kg is thrown downward with an initial velocity of
2m/s with an attached parachute which provides resistance of v 2 .
3. Subsonic Flight: The efficiency of engines of subsonic airplanes depends on the air
pressure and (usually) is maximum at 36000 ft. The rate of change of air pressure y 0 (x)
is proportional to air pressure y(x) at height x. If y0 is the pressure at sea level and the
pressure decreases to half at 18000 ft, then find the air pressure at 36000 ft.
4. Dryer: In a laundry dryer loss of moisture is directly proportional to moisture content
of the laundry. If wet laundry loses one fourth of its moisture in the first 10 minutes,
when will the laundry be 95% dry.
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5. Find all the curves in the xy-plane whose tangents pass through the point (a, b).
6. Suppose a bacteria culture has an initial population of 100. If the population doubles
every 3 days, determine the population after 30 days using Malthus’s model. How much
time is required for population to reach 5000.
7. Suppose only two third of the cells are remaining after 1 days in a cell culture. Find
the number of days until only one third of initial population is left.
8. Suppose half life of an element is 1000 Hrs. If initially we have 100g of it, how much is
left after 1 Hr. How much is left after 500 Hrs.
9. If an artifact contains 40% of the amount of Th230 as a present day smple, what is the
age of the artifact. Note that the half age of h230 is 75000 Years.
10. One student in a IIT Delhi hostel of 200 students proceeds to spread a rumor. If the
rate at which rumor spreads is proportional to the number of student knows as well
as number of student do not know, then solve the IVP to find the number of students
informed of the rumor after t days if 50 students knows the rumor after 1 days. Will
all the students eventually be informed of the rumor.
11. A hot cup of tea is initially 100 degree Celsius when poured. How long does it take for
the tea to reach temperature of 50 degree Celsius if it at 80 degrees after 15 Minutes
and room temperature is 30 degree Celsius.
12. Suppose that during month of July in Delhi the outside temperature for a day is given
by C(t) = 37 − 5 cos(πt/12), 0 ≤ t ≤ 24. Determine the temperature in a building that
has an initial temperature of 30 degree Celsius if k = 1/4 (constant of proportionality).
Assume no heating or cooling in the building.
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