The following exercise demonstrates the use of linear regression with differential equations and integral curves in making predictions of the cooling rate of water at constant room temperature.


A glass of water is heated to 180º F and then left to cool in a room with a constant temperature of 80º F. Temperature measurements are made each minute resulting in the following data:

Time in minutes 0 1 2 3 4 5 7 10

Temp. in º F 180 177 173 170 166 163 159 149

From this data, we will find an equation for an integral curve, with constants k and C. We will then employ methods of linear regression to find the value of these constants and assess the equation’s accuracy in estimating the water’s temperature after a time t.


 

 

 

The differential equation

 

for the water’s cooling represents the change in temperature over time. This change is proportional to some constant k and is due to the constant temperature of the surrounding environment. If the room temperature is a constant 80º F, as above, then it is sensible to say that the water temperature y can never drop below 80º. Applying this fact to equation (1) above shows that the change in temperature is always negative. Thus, the temperature is always decreasing.

This idea is characteristic of a situation that involves Newton’s Law of Cooling. In this law, Newton states that the "rate at which an object’s temperature is changing is proportional to the difference between the temperature of the surrounding environment and the temperature of the object at time t."

Proceeding toward our goal of estimating the water’s cooling, we reorganize equation (1) by combining terms containing the variable y. This yields

 

Integrating each side of the equation (2):

 

we have

 

 

where k and C are constants.

Next we transform our data, by substituting the values of y into the left-hand side of equation (4), leaving

 

 

T 0 1 2 3 4 5 7 10

Y -4.605 -4.575 -4.533 -4.500 -4.454 -4.419 -4.369 -4.234

 

 

where T = t and

We then use linear regression to estimate the values of k and C, finding that k = .0366 and C = -4.607. Our equation then becomes

which is the equation of the least squares regression line of Y on T. From Minitab, we obtain the regression plot that shows that equation (6) provides a good estimation of our transformed data. As can be seen from the figure below, the least squares regression line fits very closely to each of the observations (T, Y).


 

 

 

 

Knowing that our equation will be a good estimator of the temperature Y, we substitute equation (5) into equation (6), yielding the integral curve:

 

Continuing, we solve equation (7) for y (as a function of time t), to find the function

This function will predict the temperature of the water at any time as it cools from 180º F. Below is a graphical representation of equation (8) called an approximation model. Included on the graph are plots of the original data points and the estimate for time t=15 minutes. Note that as the time increases to the right, the temperature of the water approaches 80º F.

 

 

As seen above, differential equations can be used with techniques of linear regression and integral curves to provide a respectable estimation of water cooling over time.

 

 

This exercise is presented by Jeremy Touchstone, Barry Underwood, and Dragan Vukosavljevic for MATH 3200, Differential Equations under

Dr.Lyndell Kerley.