# + 1 5 x2 + 1 = 2 = 3 x 2 x 2 1 2 1 lim in .NET Draw QR Code in .NET + 1 5 x2 + 1 = 2 = 3 x 2 x 2 1 2 1 lim

22 + 1 5 x2 + 1 = 2 = 3 x 2 x 2 1 2 1 lim
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If we used L H pital s rule, we would conclude mistakenly that x2 + 1 2x = lim 1 = 1 lim 2 = lim x 2 x 1 x 2 2x x 2
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EXPONENTIAL GROWTH AND DECAY
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Example (d) above shows that ex grows much faster than any power of x. There are many natural processes, such as bacterial growth or radioactive decay, in which quantities increase or decrease at an exponential rate. De nition: Assume that a quantity y varies with time t. Then y is said to grow or decay exponentially if its instantaneous rate of change ( 19) is proportional to its instantaneous value; that is, dy = Ky dt where K is a constant. Suppose that y satis es (36.1). Let us make the change of variable u = Kt. Then, by the chain rule, Ky = and so, by Problem 35.28, y = Ceu = CeKt (36.2) dy du dy dy = = K dt du dt du or dy =y du (36.1)
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where C is another constant. We can now see why the process y is called exponential. If K > 0, K is called a growth constant, and y increases exponentially with time. If K < 0, K is called a decay constant, and y decreases exponentially with time. Let y0 be the value of y at t = 0. Substituting 0 for t in (36.2), we obtain y0 = Ce0 = C(1) = C so that (36.2) can be rewritten as y = y0 eKt EXAMPLES
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(a) Assume that a culture consisting of 1000 bacteria increases exponentially with a growth constant K = 0.02, where time is measured in hours. Let us nd a formula for the number y of bacteria present after t hours, and let us compute how long it will take until 100 000 bacteria are present in the culture.1 Since y0 = 1000, the desired formula for y is given by (36.3), y = 1000e0.02t
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1Although populations are measured in positive integers, (36.3) seems to be applicable, even though that formula was derived for a quantity measured in real numbers.
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(36.3)
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L H PITAL S RULE; EXPONENTIAL GROWTH AND DECAY
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Now set y = 1000 00 and solve for t, 100 000 = 1000e0.02t 100 = e0.02t ln 100 = ln e0.02t 2 ln 10 = 0.02t t = 100 ln 10 Appendix E gives the approximate value of 2.3026 for ln 10. Thus, t 230.26 hours Note: Sometimes, instead of specifying the growth constant K, say, K = 0.02, one says that the quantity is increasing at the rate of 2 percent per unit time. This is not quite accurate. A rate of increase of r percent per unit time is approximately the same as a value of K = 0.0r when r is relatively small (say, r 3). In fact, with an r percent growth rate, y = y0 (1 + 0.0r) after one unit of time. Since y = y0 eK when t = 1, eK = 1 + 0.0r, and, therefore, K = ln (1 + 0.0r). This is close to 0.0r, since ln (1 + x) x for small x. [For example, ln (1.02) 0.0198 and ln (1.03) 0.02956.] Thus, many textbooks will automatically interpret a rate of increase of r percent per unit time to mean that K = 0.0r. (b) If the decay constant of a given radioactive element is K < 0, compute the time T after which only half of any original quantity remains. At t = T , (36.3) gives 1 y0 = y0 eKT 2 1 = eKT 2 1 ln = KT 2 ln 2 = KT ln 2 =T K ln 1 = ln x x (36.4) [ln 102 = 2 ln 10; ln eu = u]
The number T is called the half-life of the given element. Knowing either the half-life or the decay constant, we can nd the other from (36.4).