Chapter 3
Initial Value Problems





3.2 An Initial Value Problem: A Cooling Body

3.2.2 Formulating the Initial Value Problem

Since the body of the murder victim has been cooling off since the time of death, our detective can determine how long the industrialist has been dead by examining the temperature of the body. In the remainder of this section, we study an initial value problem that describes heat loss from a cooling body. Our analysis involves the following steps:

Physics provides us a theoretical model (verifiable by simple experiments) called Newton's Law of Cooling:

Our detective can determine the surrounding (room) temperature by checking the thermostat: 21°C.

Our next task is to write the physical law in mathematical notation. We'll let T = T ( t ) represent the temperature of the cooling body in degrees Celsius at time `t` in hours. Then Newton's Law of Cooling may be written

d T d t = - k ( T - 21 ) .

We have chosen to write a negative sign in front of the "`k`" because we know the proportionality constant must be negative.

That's our differential equation - except we don't know `k` yet. Now we need to find an initial value condition. We'll set t = 0 at the time the detective takes the first measurement. Let's suppose this first measurement is taken at 8:50 A.M. and is 30°C. Then our initial value problem is

d T d t = - k ( T - 21 )   with   T = 30  at  t = 0 .

Now we have to

  1. solve the initial value problem,

  2. determine `k` - to make the solution an explicit function of time - and

  3. use the solution to figure out the time of death.


Activity 2

  1. For `k= 1, 0.5`, and `0.25`, use a graphing tool to plot the unique solution T ( t ) of the corresponding differential equation with T ( 0 ) = 30 .

  2. What features do these three solution functions have in common?

  3. How does reducing the size of `k` affect the solution?

Comment 2Comment on Activity 2


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