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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.
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Now, in this equation capital T is temperature and lower case t is time.
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And, we 're interested in how capital T changes over time. So, given that the initial temperature of the beverage is 5 degrees,
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I would like to know what is the temperature at all other times.
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So, given that T(0) is five. That means the temperature at time little t=0, is 5.
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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?
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So, if we can find this, we'll say we've solved the equation or found our solution.
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Finding this would be like finding the orbit or itinerary of an iterated function.
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So, in this sub-unit, I'll describe some qualitative methods for figuring out the general behavior of these types of differential equations.
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And the key is going to be to graph the right hand side of this equation.
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So, let me do that and we'll see what it looks like and what it tells us.
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So, what I've done is I've drawn a plot/a graph of the right hand side of this equation.
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So this purple line is this function.
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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.
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Okay! So, let's think about what this tells us. On the horizontal axis is temperature in degrees Celsius.
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So that's T as we vary t.
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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.
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How fast is the temperature warming up or cooling off.
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So, if we have water that's right around 0, then it would be warming up at 4 degrees per minute.
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So, we use this graph to read not the direct value
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of the temperature. But if we know the temperature, we can
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figure out how fast the temperature is changing.
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If we are at 10 degrees C, then this says, and sorry the scale is off a little bit,
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but, if we are at 10 degrees C, then we are warming up at 2 degrees per minute.
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If we are at 20 degrees C, then we are not warming up at all.
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The rate of change of the temperature is 0, because the purple line
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goes through 0.
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Or, if you plug 20 here, you get 20 minus 20; that's 0, so the rate of change
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, i.e. the left hand side of this equation, is 0.
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If we had something at 30, maybe a warm cup of tea,
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then its derivative, its rate of change according to this function is -2.
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So, it's warming up at -2 degrees per minute.
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Or you would say it's cooling off at 2 degrees per minute.
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Its temperature is decreasing at that instant at 2 degrees per minute.
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So, from this type of graph we can go immediately to a phase line for the solutions for this differential equation.
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So, let me draw that.
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There is one fixed point or equilibrium value and that's at 20.
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20 is fixed because the rate of change when the temperature is 20, is 0.
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If you put a glass of water that's at room temperature in a room of 20 degrees, it will stay at 20 degrees.
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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.
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We can also see that from this graph. The graph, which we interpret as the derivative, is positive,
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that means that capital T, the temperature is increasing so we move this way. And it increases until we get to 20.
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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.
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And we know that from everyday experience, but we also can see that from the function.
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The purple line is negative here, that means that dT/dt, the rate of change of temperature, is negative.
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The temperature is getting smaller. If I plug in 30 here, I'll get a negative number,meaning the temperature was decreasing.
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So, immediately from this graph, we can get a lot of information.
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We see that there is a fixed point, or an equilibrium at 20.
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And we see that it's stable or attracting.