ch1 - CHAPTER 1 Section 1 Differential Equation Models 1.1...

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CHAPTER 1 Section 1. Differential Equation Models 1.1. Let y(t) be the number of bacteria at time t . The rate of change of the number of bacteria is y (t) . Since this rate of change is given to be proportional to y(t) , the resulting differential equation is y (t) = ky(t) . Note that k is a positive constant since y (t) must be positive (i.e. the number of bacteria is growing). 1.2. Let y(t) be the number of field mice at time t . The rate of change of the number of mice is y (t) . Since this rate of change is given to be inversely proportional to the square root of y(t) , the resulting differential equation is y (t) = k/ y(t) . Note that k is a positive constant since y (t) must be positive (i.e. the number of mice is growing). 1.3. Let y(t) be the number of ferrets at time t . The rate of change of the number of ferrets is y (t) . Since this rate of change is given to be proportional to the product of y(t) and the difference between the maximum population and y(t) (i.e. 100 y(t) ), the resulting differential equation is y (t) = ky(t)( 100 y(t)) . Note that k is a positive constant since y (t) must be positive (i.e. the number of ferrets is growing provided y(t) < 100). 1.4. Let y(t) be the quantity of radioactive substance at time t . The rate of change of the material is y (t) . Since this rate of change (decay) is given to be proportional to y(t) , the resulting differential equation is y (t) = − ky(t) . Note that k is a positive constant since y (t) must be negative (i.e. the quantity of radioactive material is decreasing). 1.5. Let y(t) be the quantity of material at time t . The rate of change of the material is y (t) . Since this rate of change (decay) is given to be inversely proportional to y(t) , the resulting differential equation is y (t) = − k/y(t) . Note that k is a positive constant since y (t) must be negative (i.e. the quantity of material is decreasing). 1.6. Let y(t) be the temperature of the potato at time t . The rate of change of the temperature is y (t) . Since this rate of change is given to be proportional to the difference between the potato’s temperature and that of the surrounding room (i.e. y(t) 65), the resulting differential equation is y (t) = − k(y(t) 65 ) . Note that k is a positive constant since y (t) must be negative (i.e. the potato is cooling) and since y(t) 65 > 0 (i.e. the potato is hotter than the surrounding room). 1.7. Let y(t) be the temperature of the thermometer at time t . The rate of change of the temperature is y (t) . Since this rate of change is given to be proportional to the difference between the thermometer’s temperature and that of the surrounding room (i.e. 77 y(t) ), the resulting differential equation is y (t) = k( 77 y(t)) . Note that k is a positive constant since y (t) must be positive (i.e. the thermometer is warming) and since 77 y(t) > 0 (i.e. the thermometer is cooler than the surrounding room).
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