UNIT10 - Unit 10 Applications of rst order Dierential...

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Unit 10 Applications of first order Differential Equations We now look at three main types of question which involve the applica- tions of first order differential equations, namely (i)Heating and Cooling (ii) Mixing problems, and (iii) Falling objects We will discuss and look at an example of each type of problem. Heating and Cooling Problems For these types of problems we will be assuming that the question involves the temperature ( T )of a certain body placed in a medium of constant tem- perature ( M ) and as time ( t ) varies, so does T , ( so T has a rate of change with respect to t ). In this case Newtons law of cooling tells us the following: dT dt = k ( T M ) for some constant k Another way of saying this is that the time rate of change of temperature of the body is proportional to the difference in temperatures of the body and the medium in which it is placed. This law is the source of a popular miscon- ception, namely that when placed in a freezer, warm water will freeze into ice cubes faster than cold water. This is not true (of course) but what is true is that the ”rate of change” of temperature of the warm water is much faster than that of the cold water since the difference in temperature ( between the water and freezer) is much greater for the warm water. Thus the warm water will cool at a much faster rate, but the cold water will still freeze sooner. 1
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Example 1. A boiling (100 C ) solution is set on a table where room temper- ature is assumed to be constant at 20 C . The solution cooled to 60 C after five minutes. (a) Find a formula for the temperature ( T ) of the solution, t minutes after it is placed on the table. (b) Determine how long it will take for the solution to cool to 22 C . Solution: (a) We are asked to find an explicit formula for T in terms of t . We know this is a heating and cooling question so Newtons law of cooling tells us dT dt = k ( T M ) for some constant k So letting M=20, we have: dT dt = k ( T 20) for some constant k Recognizing this as a seperable differential equation: 1 T 20 dT = k ( dt ) 1 T 20 dT = k ( dt ) ln ( T 20) = kt + C e ln (
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