Here's a version of Newton's Law of Cooling that could apply to an object that is in a room with a temperature of 20.

Now, in this equation capital T is temperature and lower case t is time.

And, we 're interested in how capital T changes over time. So, given that the initial temperature of the beverage is 5 degrees,

I would like to know what is the temperature at all other times.

So, given that T(0) is five. That means the temperature at time little t=0, is 5.

That's the initial temperature. We would like to know what is T(t). How does the temperature vary; how does it change as a function of time?

So, if we can find this, we'll say we've solved the equation or found our solution.

Finding this would be like finding the orbit or itinerary of an iterated function.

So, in this sub-unit, I'll describe some qualitative methods for figuring out the general behavior of these types of differential equations.

And the key is going to be to graph the right hand side of this equation.

So, let me do that and we'll see what it looks like and what it tells us.

So, what I've done is I've drawn a plot/a graph of the right hand side of this equation.

So this purple line is this function.

If we don't see how to make this graph right away, don't worry. You can graph it on a computer, but for this course I'll provide you graphs like this.

Okay! So, let's think about what this tells us. On the horizontal axis is temperature in degrees Celsius.

So that's T as we vary t.

And then we interpret this as the derivative, the rate of change of the temperature. So this would have units of degree Celcius per minute.

How fast is the temperature warming up or cooling off.

So, if we have water that's right around 0, then it would be warming up at 4 degrees per minute.

So, we use this graph to read not the direct value

of the temperature. But if we know the temperature, we can

figure out how fast the temperature is changing.

If we are at 10 degrees C, then this says, and sorry the scale is off a little bit,

but, if we are at 10 degrees C, then we are warming up at 2 degrees per minute.

If we are at 20 degrees C, then we are not warming up at all.

The rate of change of the temperature is 0, because the purple line

goes through 0.

Or, if you plug 20 here, you get 20 minus 20; that's 0, so the rate of change

, i.e. the left hand side of this equation, is 0.

If we had something at 30, maybe a warm cup of tea,

then its derivative, its rate of change according to this function is -2.

So, it's warming up at -2 degrees per minute.

Or you would say it's cooling off at 2 degrees per minute.

Its temperature is decreasing at that instant at 2 degrees per minute.

So, from this type of graph we can go immediately to a phase line for the solutions for this differential equation.

So, let me draw that.

There is one fixed point or equilibrium value and that's at 20.

20 is fixed because the rate of change when the temperature is 20, is 0.

If you put a glass of water that's at room temperature in a room of 20 degrees, it will stay at 20 degrees.

If we are below 20, cooler than 20, we know that the water, the object will warm up. And we know that from everyday experience.

We can also see that from this graph. The graph, which we interpret as the derivative, is positive,

that means that capital T, the temperature is increasing so we move this way. And it increases until we get to 20.

And if capital T is over here, so at 30 or 40 degrees, then we know that the temperature will decrease, it will cool off to room temperature.

And we know that from everyday experience, but we also can see that from the function.

The purple line is negative here, that means that dT/dt, the rate of change of temperature, is negative.

The temperature is getting smaller. If I plug in 30 here, I'll get a negative number,meaning the temperature was decreasing.

So, immediately from this graph, we can get a lot of information.

We see that there is a fixed point, or an equilibrium at 20.

And we see that it's stable or attracting.