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06_M_GOLD_9004_12_ch05-1

06_M_GOLD_9004_12_ch05-1 - Chapter 5 Applications of the...

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160 Chapter 5 Applications of the Exponential and Natural Logarithm Functions 5.1 Exponential Growth and Decay 1. a. .02 () t Pt Ce = and since (.02)(0) 3( 0 ) PC e C == = , we have .02 () 3 t e = . b. P (0) = 3, so population was 3 million. c. .02 d. (.02)(8) (8) 3 Pe = 3.52 million e. First determine when the population reaches 4 million people. .02 .02 .02 4 4 3 3 ln 4 3 4 ln .02 14.4 years 3. 0 2 tt t e e e =⇒ = ⎛⎞ = ⎜⎟ ⎝⎠ ( ) (.02)(14.4) (14.4) .02 (14.4) .02 3 .08 million per year or about PP e 80,000 per year f. Solve ( ) .02 ( ) P t = for P ( t ) when () .07 .02 () .07 P t .5 million when it is growing at the rate of 70,000 per year. 2. a. P (0) = 10,000 b. .55 t = and since (.55)(0) 10,000 (0) e C = , we have .55 ( ) 10,000 t e = . c. (.55)(5) (5) 10,000 = 156,426 bacteria d. .55 e. ( ) .55 ( ) (.55)(100,000) 55,000 bacteria per hour P t = f. Solve ( ) .55 ( ) P t = for P ( t ) when = 34,000. .55 P ( t ) = 34,000 P ( t ) 61,818 bacteria 3. a. (.2)(0) (0) 5000 5000 , so 5000 cells were present initially. b. k = .2, so a differential equation is = .2 P ( t ). c. .2 .2 10,000 5000 2 ln 2 ln 2 .2 3.5 hours .2 ee = = d. .2 .2 20,000 5000 4 ln 4 ln 4 .2 6.9 hours .2 = = 4. a. (.01)(0) (0) 300 300 ; so 300 cells were present initially. b. k = .01, so a differential equation is = .01 P ( t ). c. .01 .01 600 300 2 ln 2 ln 2 .01 69.3 days .01 = = d. .01 .01 1200 300 4 ln 4 ln 4 .01 138.6 days .01 = = 5. Let P ( t ) be the population after t days, 0 kt = . It is given that P (40) = 2 P (0), so 40 0( ) 40 00 0 40 22 2 ln 2 ln ln 2 .017 40 kk k k P e ek = = 6. Let P ( t ) be the population after t years, 0 kt = . It is given that P (10) = 3 P (0), so 10 (0) 10 0 10 33 3 ln 3 ln ln 3 .11 10 k k P e = = 7. Let P ( t ) be the population after t years. .05 .05 .05 0 3 3 ln 3 ln 3 .05 22 years .05 t P e = =≈ 8. Let P ( t ) be the population after t years. .04 .04 .04 0 2 2 ln 2 ln 2 .04 17.3 years .04 t P e =

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Section 5.1 Exponential Growth and Decay 161 9. Let P ( t ) be the cell population (in millions) after t hours, 0 () kt Pt Pe = . It is given that P (0) = 1 and that P (10) = 9. Thus, (10) 91 k e =⋅ 10 ln 9 ln 9 ln .22 10 k ek =⇒ = . .22 t e = so P (15) = .22(15) 3.3 ee = 27 (million) cells. 10. Let P ( t ) be the population (in billions) t years after January 1, 1993. ( ) 5.51 kt e = since P (0) = 5.51. Solve P (5) = 5.88 for k . (5) (1/5)ln(5.88/5.51) 5.88 5.51 5.88 5 ln 5.51 15 . 8 8 ln ( ) 5.51 55 . 5 1 k t kP t e = ⎛⎞ = ⎜⎟ ⎝⎠ Solve P ( t ) = 7 for t . (1/ 5)ln(5.88/ 5.51) (1/ 5)ln(5.88/ 5.51) 7 5.51 5.88 5.51 5.51 7 7 5.51 . 8 8 7 ln ln 5 5.51 5.51 5ln 18.4 years ln t t e e t t = = = =≈ The world’s population will reach 7 billion in about 18.4 years (the year 2011). 11. Let P ( t ) be the population (in millions) t years after the beginning of 1990. P (0) = 20.2, so ( ) 20.2 kt e = . Solve P (5) = 23 for k . (5) (1/ 5) ln(23/ 20.2) 23 20.2 23 5 ln 20.2 12 3 ln ( ) 20.2 52 0 . 2 k t t e =⇒= = (1/ 5)ln(23/ 20.2)(20) (20) 20.2 34.0 million 12. a. From the graph, P (4) = 9 million. b. From the graph, P ( t ) = 10 when t = 8 years, so the population was 10 million in 1978. c. ( ) .025 ( ) P t = (4) .025 (4) (.025)(9) PP == = .225 million people per year or 225,000 people per year d. Solve .275 = .025 P ( t ); P ( t ) = 11. P ( t ) = 11 when t = 11. The population is growing at the rate of 275,000 people per year in 1981. 13. a. .021 () 8 t e = b. P (0) = 8 g c. .021 d. ( .021)(10) (10) 8 6.5 grams e. ( ) ( .021)(1) .021 =− The sample is disintegrating at a rate of .021 gram per year f. –.105 = –.021 P ( t ) .105 ( ) 5 grams remaining .021 g. 4 grams will remain after 33 years.
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06_M_GOLD_9004_12_ch05-1 - Chapter 5 Applications of the...

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