# hw11 - HW11 Typeset by Farmer § 5.1 2.(a) Since P ′ ( t...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 one-sided 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 ....
View Full Document

## 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.

### Page1 / 10

hw11 - HW11 Typeset by Farmer § 5.1 2.(a) Since P ′ ( t...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online