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Unformatted text preview: 18.03 Problem Set 2 Fall 2009 Solutions 1. (15 points) This problem is based on the example on pages 57–58 of EP. Suppose that the outdoor temperature in Cambridge during September is given by A ( t ) = 70 a cos( πt/ 12) . ( OUTSIDE ) You can think of A as standing for “ambient.” Here a is a positive constant, and t is the time in hours after midnight (on September 1, say). This means that the temperature oscillates daily with a minimum value of 70 a at midnight and a maximum value of 70 + a at noon. Your dormitory room has no source of heat or cooling, so according to Newton’s law of cooling, the temperature in your room satisfies the differential equation dT ( t ) dt = k ( T ( t ) A ( t )) . ( ODE ) Here k is a positive constant that depends on how wellinsulated your dorm room is. a) If you insulate your room better, will k increase or decrease? Explain why. If you insulate your room better, k decreases. When k is a small positive constant, the indoor temperature changes at a rate which is only a small amount times the difference in indoor/outdoor temperatures; this corresponds to good insulation. (In the limiting case of k = 0, we have dT/dt = 0, the case of perfect insulation, where the indoor temperature doesn’t depend on the outside temperature.) When k is large, the indoor temperature changes quickly in response to the temperature difference; this corresponds to poor insulation. (In the limiting case of infinite positive k , the indoor temperature would instantaneously equalize with the outdoors.) b) The text says that “typical” values of k might range from . 2 to . 5 . How would these values change if we used Celsius instead of Fahrenheit? What about Kelvin? What if we measured t in days instead of hours? Celsius is Fahrenheit translated and rescaled. Thus T ( t ) A ( t ) is rescaled. However, dT ( t ) /dt is also rescaled (and the derivative ignores the translation), and so k is unaffected....
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 Spring '09
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 Exponential Function, dt, Kelvin

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