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Unformatted text preview: 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)) ....
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This note was uploaded on 03/05/2008 for the course MATH 33a taught by Professor Lee during the Spring '08 term at UCLA.
 Spring '08
 lee
 Rate Of Change

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