chapter 5

chapter 5 - Chapter 5 Exercises 5.1 1 a d.02t P(t = Ce and...

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189 Chapter 5 Exercises 5.1 1. a. and since , we have . P ( t ) = Ce .02 t 3 = P (0) = Ce (.02)(0) = C P ( t ) = 3 e .02 t b. P (0) = 3, so population was 3 million. c. .02 d. 3.52 million P (8) = 3 e (.02)(8) e. .065 million per year or 65,000 per year P (4) = .02 P = .02(3 e (.02)(4) ) f. Solve for P ( t ) when , .02 P ( t ) = .07 P ( t ) 3.5 million when it is growing at the rate of 70,000 per year. P ( t ) = P ( t ) P ( t ) = .07 2. a. and since , we have . P ( t ) = Ce .55 t 10,000 = P = Ce (.55)(0) = C P ( t ) = 10,000 e .55 t b. P (0) = 10,000 c. 156,426 bacteria P (5) = 10,000 e (.55)(5) d. .55 e. = .55 P ( t ) = (.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 P ( t ) P ( t ) 3. a. , so 5000 cells were present initially. P = 5000 e (.2)(0) = 5000 b. k = .2, so a differential equation is = .2 P ( t ). P ( t ) c. 10,000 = 5000 e .2 t 2 = e .2 t ln 2 = .2 t t = ln2 .2 3.5 hours d. 20,000 = 5000 e .2 t 4 = e .2 t ln 4 = .2 t t = ln 4 6.9 hours 4. a. ; so 300 cells were present initially. P = 300 e (.01)(0) = 300 b. k = .01, so a differential equation is P ( t ) = .01 P ( t ). c. 600 = 300 e .01 t 2 = e .01 t ln 2 = .01 t t = .01 69.3 days d. 1200 = 300 e .01 t 4 = e .01 t ln 4 = .01 t t = ln 4 138.6 days 5. Let P ( t ) be the population after t days, P ( t ) = P 0 e kt . It is given that P (40) = 2 P (0), i.e., P 0 e 40 k = 2 P 0 e 0( k ) = 2 P 0 ; e 40 k = 2; ln e 40 k = ln2; k = 40 .017 6. Let P ( t ) be the population after t years, P ( t ) = P 0 e kt . It is given that P (10) = 3 P (0), i.e., P 0 e 10 k = 3 P 0 e k = 3 P 0; e 10 k = 3; ln e 10 k = ln3; k = ln 3 10 .11. 7. Let P ( t ) be the population after t years. P ( t ) = P 0 e .05 t 3 P 0 = P 0 e .05 t 3 = e .05 t ln 3 = .05 t t = ln3 .05 22 years

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Chapter 5: Applications of Exponential & Natural Logarithm Functions ISM: Calculus & Its Applications, 11e 190 8. Let P ( t ) be the population after t years. P ( t ) = P 0 e .04 t 2 P 0 = P 0 e .04 t 2 = e .04 t ln 2 = .04 t t = ln2 .04 17.3 years 9. Let P ( t ) be the cell population (in millions) after t hours, P ( t ) = P 0 e kt . It is given that P (0) = 1 and that P (10) = 9. Thus, 9 = 1 e k (10) ; ln 9 = ln e 10 k k = ln9 10 .22. so P (15) = e 27 (million) P ( t ) = e .22 t .22(15) = e 3.3 10. Let P ( t ) be the population (in billions) t years after January 1, 1993. since P (0) = 5.51. Solve P (5) = 5.88 for k . P ( t ) = 5.51 e kt 5.51 e k (5) = 5.88 5 k = ln 5.88 5.51 k = 1 5 ln 5.88 5.51 Solve P ( t ) = 7 for t . P ( t ) = 5.51 e (1/5)ln(5.88/5.51) t 5.51 e (1/5) ln(5.88/5.51) t = 7 e (1/5) ln(5.88/5.51) t = 7 5.51 1 5 ln 5.88 5.51 t = ln 7 5.51 t = 5ln 7 5.51 () ln 5.88 5.51 18.4 years The world’s population will reach 7 billion in the year 2011. 11. Let P ( t ) be the population (in millions) t years after the beginning of 1990. P (0) = 20.2, so P ( t ) = 20.2 e kt . Solve P (5) = 23 for k . 20.2 e k = 23 5 k = ln 23 20.2 k = 1 5 ln 23 20.2 P ( t ) = 20.2 e (1/5)ln(23/20.2) t P (20) = 20.2 e (1/5)ln(23/20.2)(20) 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. P ( t ) = .025 P ( t ) P (4) = .025 P (4) = (.025)(9) = .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. P ( t ) = 8 e .021 t b. P (0) = 8 g c. .021 d. P (10) = 8 e ( .021)(10) 6.5 grams e. P ( t ) = ( .021)(1) =− .021 Disintegrating at a rate of .021 gram per year f. –.105 = –.021 P ( t ) P ( t ) = .105 .021 = 5 grams remaining g. 4 grams will remain after 33 years.
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chapter 5 - Chapter 5 Exercises 5.1 1 a d.02t P(t = Ce and...

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