ThreeDiffEquations.html

Three Differential Equations

MAT - NYB -- Spring 2001

> restart:with(DEtools):

Exponential Decay

A radioactive sample decays at a rate that is proportional to the amount present at time t . That is, if A is the amount of a radioactive element present at time t then .

e.g. (#8, page 610)

Polonium-210 has a half-life of 140 days.

a) If a sample has a mass of 200 mg, find a formula for the mass that remains after t days.

b) Find the mass after 100 days.

c) When will the mass be reduced to 10 mg?

d) Sketch the graph of the mass function.

> DE:=diff(A(t),t)=k*A(t);

> dsolve(DE);

Next line puts the solution A(t) into an easier to handle functional form:

> M:=t->K*exp(1)^(k*t);

Find K :

> K:=solve(M(0)=200,K);

> print(M(t));

Find k :

> k:=solve(M(140)=100);

> print(M(t));

> M(100);evalf(%);

=> About 122 gm after 100 days

> solve(M(t)=10,t);evalf(%);

=> It will take about 605 days (a little over 1.5 years) for the sample to decay to a mass of 10 gm.

> plot(M(t),t=0..1000,thickness=2,title=`Decay of Polonium-210` );

> restart:with(DEtools):

Newton's Law of Cooling

The rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not "too large."

e.g. A (dead) body (a victime of foul circumstances) is found at 12 noon in a temperature controlled room. The body's core temperature is measured to be 32 degrees C. Thirty minutes later it is 31 degrees. When was the dirty deed performed if the room is at the constant temperature of 25 degrees C. You may assume that the person was a healthy individual with a normal body temperatuire of 36.6 degrees C.

Let T be the temperature of the body at time t

> DE:=diff(T(t),t)=k*(T(t)-25);

> dsolve(DE);

> Temp:=t->25+K*exp(k*t);

> K:=solve(Temp(0)=32);

> Temp(t);

> k:=solve(Temp(30)=31);evalf(%);

> Temp(t);

> solve(Temp(t)=36.6);

=> The deed was done around 10:30 A.M. (98 minutes before noon)

> plot({Temp(t),37},t=-120..600,y=25..38,thickness=2,colour=[red,blue]);

> restart:with(DEtools):

Logistic Growth

This model assumes that a population is growing at a rate that is jointly proportional to the population at time t and the difference between the population at time t and the maximum possible population (the "carrying capacity" of the environment in question). The differential equation modelling this behaviour is therefore where K is the carrying capacity and k is a proportionality constant.

e.g. The world population was about 1.608 billions in 1900 and 2.517 billions in 1950. Use a logistic model to predict the 1992 population assuming a carrying capacity of 100 billion. Assume t = 0 refers to 1900.

> DE:=diff(P(t),t)=k*P(t)*(K-P(t));