In this lab we study situations that may be modeled by (differential) equations of the form

In particular, the quantity y has an equilibrium value at y = a, and this equilibrium value is stable because dy/dt is negative when y > a and positive when y < a, and zero when y=a.  In words, the differential equation says that y approaches a at a rate proportional to how far it is from a. Given data that appear to possibly  fit this model, our objectives will be

• to determine if the data really do fit the model, and, if so,
• to determine values of the parameters k and a, and
• given k and a, find a formula which describes the relation between y and t

Enter the data in lists L1 and L2 on your calculator.  Plot and describe the graph.  Can you make a rough guess regarding the room ("ambient") temperature?

This experiment should satisfy Newton's Law of Cooling: The rate at which an object cools is proportional to the difference between its temperature and the ambient temperature. In mathematical symbols, that's the differential equation

where a is the ambient temperature.  Note that T and t are variables, a is a constant given by the physical situation, and k is the "proportionality constant."  We put a minus in front of k since we know that dT/dt is actually negative because of the decreasing temperature, so the constant is actually "-k" for some positive number k.   We will try to determine k and a from the data.

There are at least two ways to examine whether our data are actually modeled by Newton's Law of Cooling.

Option 1: we could make a change of variable, y = T - a -- which is the same thing as resetting 0 on the vertical scale at the ambient temperature. Then dy/dt = dT/dt, and the differential equation becomes

That's the defining equation for decaying exponential functions (we did something similar in the Radioactive Decay lab), and we can determine whether the data fit such a model by looking at a "semilog" graph (plotting log(y) vs. x as in the Decay lab). If we see a straight line, we know the model fits -- and the slope of the line determines the decay constant.   However, in order to do this, we would need to know the room temparature a, and we don't yet know this number, so we will try...

Option 2: we could estimate values of dT/dt from the data, by using "symmetric difference" quotients:

where we are using t_0, t_1, t₂ , t₃...   for time = 0, 5, 10, 15...    For example, using i=1, will give us and estimate for the slope at 5 minutes, using t₀=0 and t₂=10 minutes.  Note that this works for 5,10,15,...,55 minutes, but we will not have an estimate at 0 and 60 minutes (that is OK).
We can plot (see below) these slope estimates versus T_i to see if we get a straight line -- after all, Newton's Law of Cooling says the slope dT/dt should be a linear function of T. Furthermore, as we will see, the slope and y-intercept of the line will give us both parameters in the model.  For this second method, we don't need to know the ambient temperature.

• Explain why (T_i+1 - T_i-1) / (t_i+1 - t_i-1) is a good estimate of dT/dt at t = t_i.  Draw a graph to show why this might be a better estimate than (T_i+1 - T_i) / (t_i+1 - t_i).  

 

Enter the estimates for dT/dt into list L4, and copy the values for list L2 into list L3 (L3=L2).  Delete the first and last entries in L3, since you will not have corresponding estimates for dT/dt when t = 0 or t = 60.  Plot  dT/dt vs. T.  Do these points appear to approximate straight line?  Roughly estimate the slope, and then use the linear regression option to get the calculator's estimate for the best fit line (remember that the data are in lists L3 and L4.)

Since you are looking at data that could represent the equation dT/dt = - k(T - a) with T on the horizontal  axis and dT/dt on the vertical axis, you should be able to use the slope and intercept on the regression line to provide values for k and a.  Does your value for a  make sense, keeping in mind that this represents the room temperature (deg. C.)?
 
 

• Explain why the solution of the initial value problem dT/dt = - k(T - a), T(0) = T₀ is T = a + (T₀ - a) e^-kt  or equivalently as T = a + (T₀ - a) b^t  for some constant value of b.  You don't need to work with both of these equations, but note that the relation between the two is given by b = e^-k and that k = -ln(b).  (Hint: take the derivative in the 2nd  equation for T, and substitute T in the 2nd equation and the derivative of T you have just found into the first equation.  You should also check that the solution has the "correct" value T(0) = T₀  when t=0.)

Find a formula for T using the solution given above.  Plot this on the same set of axes with the original data points.

Does this support the Newton's Law of Cooling model?

As a further check, use the value of a you calculated above and enter data values for y=T-a in list L5  (L5=L2-a).  Since these should satisfy the equation dy/dt = -ky, what should be the shape of a curve approximating a plot of y vs. time?    How should a graph of log(y) vs. time appear?  Enter log(y) in list L6 and plot log(y) (in L6) vs. time (in L1).  Does this confirm your hypothesis?

This lab has been adapted from materials developed by David Smith, Duke University, for the  Connected Curriculum Project with funding from the National Science  Foundation.  Used by permission.