SalasSV_07_06

# SalasSV_07_06 - 7.6 EXPONENTIAL GROWTH AND DECAY 413 7.6...

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7.6 EXPONENTIAL GROWTH AND DECAY ± 413 ± 7.6 EXPONENTIAL GROWTH AND DECAY We begin by comparing exponential change to linear change. Let y = y ( t ) be a function of time t . If y is a linear function y ( t ) = kt + C , k , C constants, then y changes by the same additive amount during all periods of the same duration : y ( t + ± t ) = k ( t + ± t ) + C = ( kt + C ) + k ± t = y ( t ) + k ± t . During every period of length ± t , y changes by the amount k ± t . If y is an exponential function y ( t ) = Ce kt , k , C constants, then y changes by the same multiplicative factor during all periods of the same duration : y ( t + ± t ) = k ( t + ± t ) = kt + k ± t = kt · e k ± t = e k ± t y ( t ). During every period of length ± t , y changes by the factor e k ± t . An exponential function f ( t ) = kt has the property that its derivative f ± ( t ) is proportional to f ( t ): f ± ( t ) = Cke kt = kCe kt = kf ( t ). Moreover, it is the only such function: THEOREM 7.6.1 If f ± ( t ) = ( t ) for all t in some interval then f is an exponential function f ( t ) = kt where C is an arbitrary constant. PROOF We assume that f ± ( t ) = ( t ) and write f ± ( t ) ( t ) = 0.

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414 ± CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Multiplying this equation by e kt ,wehave ( ) e kt f ± ( t ) ke kt f ( t ) = 0. Observe now that the left side of this equation is the derivative d dt ± e kt f ( t ) ² . (verify this) Equation ( ) can therefore be written d dt ± e kt f ( t ) ² = 0. It follows that e kt f ( t ) = C for some constant C . Multiplication by e kt gives f ( t ) = Ce kt . ± Remark In the study of exponential growth or decay, time is usually measured from t = 0. The constant C is the value of f at time t = 0: f (0) = 0 = C . This is called the initial value of f . The exponential f ( t ) = kt is conveniently written f ( t ) = f (0) e kt . ± Example 1 Find f ( t ) given that f ± ( t ) = 2 f ( t ) for all t and f (0) = 5. SOLUTION The fact that f ± ( t ) = 2 f ( t ) tells us that f ( t ) = 2 t where C is some constant. Since f (0) = C = 5, we have f ( t ) = 5 e 2 t . ± Population Growth Under ideal conditions (unlimited space, adequate food supply, immunity to disease, and so on), the rate of increase of a population P at time t is proportional to the size of the population at time t . That is, P ± ( t ) = kP ( t ), where k > 0 is a constant, called the growth constant .† Thus, by our theorem, the size of the population at any time t is given by P ( t ) = P (0) e kt , and the population is said to grow exponentially . This is a model of uninhibited growth. In reality, the rate of increase of a population does not continue to be proportional to the † The growth constant of a population is often expressed as a percentage. In such instances, the percentage must be converted to its decimal equivalent for use in the formula. For example, if the growth constant of a certain population is given as 1.7%, then k = 0. 017.
7.6 EXPONENTIAL GROWTH AND DECAY ± 415 size of the population. After some time has passed, factors such as limitations on space or food supply, diseases, and so forth, set in and affect the growth rate of the population.

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## This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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SalasSV_07_06 - 7.6 EXPONENTIAL GROWTH AND DECAY 413 7.6...

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