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Unformatted text preview: HW11 Typeset by Farmer § 5.1 2.(a) Since P ′ ( t ) = . 55 P ( t ) for all t ≥ 0, by 4.3(3), P ( t ) = Ce . 55 t , for some constant C . (See the * note following this problem for a technical detail.) Since P (0) = 10 4 , the number of bacteria initially present (at t = 0), 10 4 = P (0) = Ce . 55(0) = Ce = C. So P ( t ) = 10 4 e . 55 t . (b) As discussed in (a), P (0) = 10 4 . (c) At t = 5, i.e., after 5 hours, there are P (5) = 10 4 e . 55(5) = 10 4 e 2 . 75 bacteria present. (d) The growth constant of P ( t ) is . 55. (e) When P ( t ) = 10 5 , P ′ ( t ) = . 55 P ( t ) = . 55(10 5 ) = 55 , 000, so at that time, the population is increasing by 55,000 per hour. (f) When P ′ ( t ) = 34 , 000, P ( t ) = P ′ ( t ) / ( . 55) = 34 , 000 / ( . 55) = 20(34000) / 11 = 680000 / 11 = 61818 2 11 ≈ 67818 . *This note refers to 2(a). Since the domain of P ( t ) is t ≥ 0, “ P ′ (0) = . 55 P (0)” refers to the onesided derivative at t = 0. 4.3(3) applies in this situation too; i.e., the only functions defined on the interval t ≥ 0, satisfying the given differential equation, are of the form Ce . 55 t . 4(a) We assume “initially” means at t = 0. So there were P (0) = 300 e . 01(0) = 300 e = 300 insects present. (b) Differentiating P ( t ), P ′ ( t ) = 300 e . 01 t ( . 01) = . 01 P ( t ) by the chain rule. So we have the differential equation P ′ ( t ) = . 01 P ( t ). (On page 277 the general form of this is mentioned, so you could just quote that.) (c) The population doubles once P ( t ) = 600, so 600 = 300 e . 01 t 2 = e . 01 t We can take logs since this is a positive number: ln2 = ln( e . 01 t ) = . 01 t by 4.4(2). So 100ln 2 = t. 1 ln2 is around . 693, so it will take about 69 days. (d) The population will equal 1200 when 1200 = 300 e . 01 t 4 = e . 01 t ln4 = . 01 t 2ln 2 = . 01 t (by § 4.6), so 200ln 2 = t ≈ 138 . 6 ≈ 139 . Notice that it took the same length of time for the population to double from 300 to 600 as to double from 600 to 1200. 6(a)Let P ( t ) denote the size of the population t years after some starting time (it doesn’t matter what starting point we choose). Since P ( t ) is growing at a rate proportional to its size, we have P ′ ( t ) = kP ( t ), where k is the growth constant. So by 4.3(3), P ( t ) = Ce kt , where C is the size of the population at our chosen starting time. Since P ( t ) triples every 10 years, Ce k (10) = P (10) = 3 P (0) = 3 C. So e 10 k = 3 ln e 10 k = ln3 10 k = ln3 k = (ln3) / 10 . 10. We assume that by “world’s population” the questioners are referring to humans. Let H ( t ) be the number of humans on Earth, in billions, t years after Jan 1, 1993. Let P ( t ) be the actual number of humans, so P ( t ) = 10 9 H ( t ). We’re assuming the population is growing at a rate proportional to its size, so P ′ ( t ) = kP ( t ) for some constant k ....
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This note was uploaded on 04/03/2008 for the course MATH 16A taught by Professor Stankova during the Spring '07 term at University of California, Berkeley.
 Spring '07
 Stankova
 Math

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