|UBC Calculus Online Course Notes|
Other differential equationsWe have examined the behaviour of two simple differential equations so far, one for population growth, and one for the radioactive decay of a substance. The methods we have developed are actually useful for many other interesting problems, and can help us to make predictions about other systems that, at first sight, do not seem at all related. We will find that the common thread in all these systems is the simple differential equation of the form
This equation is of interest for either positive or negative values of the constant . In fact, in the examples studied so far, we looked at one case in which , and another case in which .
Before continuing, let us recall that the behaviour of the solution(s) to this equation depend on whether the constant is positive or negative:
With this in mind, let us examine another realization of a differential equation, this time connected with the property of cooling (or loss of heat) of a warm object in a colder environment. The following "Law" is an approximate description of experimentally observed behaviour.
Newton's Law of Cooling
Newton's Law makes a statement about an instantaneous rate of change of the temperature. We will see that when we translate this verbal statement into a differential equation, we arrive at a differential equation. The solution to this equation will then be a function that tracks the complete record of the temperature over time. Newton's Law would enable us to solve the following problem.
Solution: Let us summarize the information
briefly and define notation for this problem.
Given: The rate of change of the temperature , is (by Newton's Law of Cooling) proportional to the difference between the temperature of the soup and the ambient temperature This means that:
Here a bit of care is needed: Clearly if the soup is hotter than the water in the sink , then the soup is cooling down which means that the derivative should be negative. (Remember the connection between a decreasing function and the sign of the derivative ?). This means that the equation we need has to have the following sign pattern:
is a positive constant.
For your consideration:
Note that if we take a derivative of , and use the Newton's law of cooling, we arrive at
(We have used the fact that is constant to eliminate its derivative, and we plugged in for in the last step.) What a nice surprize ! By defining this new variable, we have arrived once more at the familiar equation
whose solution is well known to us, namely:
We can use this result to conclude (by plugging in and ) that
It follows that
We found the solution in general form, but it looks quite complicated. Let's try to understand this expression and its predictions in the case of the problem described above.
We also know that after 10 minutes, the soup cools to 60 degrees, so that . Plugging into the last equation, we find that
(The steps are much the same as in our previous work in the example on radioactive decay. In the last step we took a reciprocal of both sides of the equation. This just makes all the quantities come out to be positive in the next step, so it is done for convenience, though it is not an essential step). We have found that
Taking the natural logarithm of both sides, and solving for , we find that
So we see that the constant which governs the rate of cooling is per minute. Now we can specify the solution fully, since all constants have been determined from the information in the problem. The prediction is that the temperature of the pot of soup at time t will be
For your consideration:
This equation can be solved for in much the same way as before. Subtracting 5 from both sides and dividing by 95 we get:
Taking logarithms of both sides, we find that
Thus, using the fact that we have
Thus, it will take a little over half an hour for Jim's soup to cool
off enough to be put into the refrigerator.