Question

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. If we measure temperature in degrees Celsius and time in minutes, the constant of proportionality k equals 0.3. Suppose the ambient temperature $T_A(t)$ is equal to a constant 47 degrees Celsius. Write the differential equation that describes the time evolution of the temperature $T$ of the object. (a) $frac{dT}{dt} = -0.3(T - 47)$ \ Suppose the ambient temp $T_A(t) = 47cos(frac{pi t}{30})$ degrees Celsius (time measured in minutes). Write the DE that describes the time evolution of temperature $T$ of the object. (b) $frac{dT}{dt} = -0.3 left( T - 47 cos left( frac{pi t}{30} ight) ight)$ \ If we measure time in hours the differential equation in part (b) changes. What is the new differential equation? Use the letter s to denote time in hours. (c) $frac{dT}{ds} = -frac{1}{200} left( T - 47 cos left( frac{pi t}{30} ight) ight)$ \ If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more. What is the new differential equation? Use the letter F to denote temperature in degrees Fahrenheit. (d) $frac{dF}{ds} = -frac{9}{1000} left( T - 47 cos left( frac{pi t}{30} ight) ight) + 32$

          Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. If we measure temperature in degrees Celsius and time in minutes, the constant of proportionality k equals 0.3. Suppose the ambient temperature $T_A(t)$ is equal to a constant 47 degrees Celsius. Write the differential equation that describes the time evolution of the temperature $T$ of the object.
(a) $frac{dT}{dt} = -0.3(T - 47)$ \
Suppose the ambient temp $T_A(t) = 47cos(frac{pi t}{30})$ degrees Celsius (time measured in minutes). Write the DE that describes the time evolution of temperature $T$ of the object.
(b) $frac{dT}{dt} = -0.3 left( T - 47 cos left( frac{pi t}{30} 
ight) 
ight)$ \
If we measure time in hours the differential equation in part (b) changes. What is the new differential equation? Use the letter s to denote time in hours.
(c) $frac{dT}{ds} = -frac{1}{200} left( T - 47 cos left( frac{pi t}{30} 
ight) 
ight)$ \
If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more. What is the new differential equation? Use the letter F to denote temperature in degrees Fahrenheit.
(d) $frac{dF}{ds} = -frac{9}{1000} left( T - 47 cos left( frac{pi t}{30} 
ight) 
ight) + 32$

$Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. If we measure temperature in degrees Celsius and time in minutes, the constant of proportionality k equals 0.3. Suppose the ambient temperature TA(t) is equal to a constant 47 degrees Celsius. Write the differential equation that describes the time evolution of the temperature T of the object. (a) fracdTdt = -0.3(T - 47) Suppose the ambient temp TA(t) = 47cos(fracpi t30) degrees Celsius (time measured in minutes). Write the DE that describes the time evolution of temperature T of the object. (b) fracdTdt = -0.3 left( T - 47 cos left( fracpi t30 ight) ight) If we measure time in hours the differential equation in part (b) changes. What is the new differential equation? Use the letter s to denote time in hours. (c) fracdTds = -frac1200 left( T - 47 cos left( fracpi t30 ight) ight) If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more. What is the new differential equation? Use the letter F to denote temperature in degrees Fahrenheit. (d) fracdFds = -frac91000 left( T - 47 cos left( fracpi t30 ight) ight) + 32$

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