Calculus II Class 04/18

The figure below is an attempt to depict the temperature T of an object vs. clock time t as the temperature approaches room temperature.  We let Troom stand for room temperature.

• Tdiff = T - Troom is the difference between temperature T and room temperature Troom.  Tdiff is depicted in the equation below as an arrow from Troom to T.
• The rate at which temperature changes with respect to clock time is denoted dT/dt and is represented by the slope of the graph.

Tdiff and the slope are depicted on the graph at two different clock times.

Newton's Law of Cooling tells us that the rate of change of the temperature of an object is proportional to Tdiff.  To understand this in terms of the graph we compare Tdiff and the slope at our two points.

• If slope and Tdiff remain in constant proportion then the ratio of two slopes will be the same as the ratio of the corresponding temperature differences.
• In the figure below we see that our first Tdiff is about 3 times the second, judging by the lengths of the two Tdiff arrows.
• It slope and Tdiff are proportional then the slopes should be in the same ratio.  However the in the figure below it does not appear that the first slope is in fact 3 times the second, as should be the case (it looks like the first slope is perhaps double the second).  This means that the temperature curve was not drawn quite right.   If the temperature curve is drawn correctly the slopes and the Tdiff values will remain proportional.

How would we write the statement 'the slope is proportional to Tdiff'?

If Tdiff is temperature difference then  if slope is proportional to temperature difference we can say that

• slope = k * Tdiff 

for some constant k.

If T is the temperature of the object and t stands for clock time then what expression stands for the instantaneous rate at which T is changing?

• The rate is T ' (t) = dT / dt.

If room temperature is Troom and temperature is T, how then do we write slope = k * Tdiff as a differential equation?

Tdiff is T - Troom and slope is T ' = T ' (t) = dT/dt.  So the equation

• slope = k * Tdiff

means

• dT/dt = k * (T - Troom)

This is a differential equation.

How do we separate the variables of the differential equation dT/dt = k * (T - Troom)?

The variables are temperature T and clock time t.

Note that Troom and k are constants.

To separate variables we have to get all T expressions on one side of the equation, all t expressions on the other.

Starting with

• dT/dt = k * (T - Troom) w

e can rewrite this as

• dT = k * (T - Troom) * dt. 

At this point dT and T are on different sides of the equation.  We fix this if we divide both sides by T - Troom to get

• dT / (T - Troom) = k dt.

We can integrate both sides of the resulting equation dT / (T - Troom) = k dt.  What is the integral of dT / (T - Troom)?

What is the integral of k dt?

What therefore is our equation relating temperature T to clock time t?

Solve the resulting equation ln | T - Troom | = k t + c for T as a function of t.

If we apply the exponential function to both sides we obtain

| T - Troom | = e^(k t + c).  Applying the laws of exponents we have

| T - Troom | = e^c * e^(k t).  Since e^c can be any number greater than zero we write is as A, obtaining

| T - Troom | = A e^(k t), A > 0.   Since | x | = b means x = + b or x = -b this can be written as

T - Troom = A e^(k t), A > 0 or A < 0, or in more abbreviated form

T - Troom = A e^(k t), A not zero.  We write this in its final form as

T = A e^(k t) + Troom , A not zero.

This equation has two constants, A and k, which can be manipulated to fit a given situation.