Section 3.8 Exponential Growth and Decay

3. A culture of the bacterium Salmonella enteritidis initially contains 50 cells. When introduced into a nutrient broth,

the culture grows at a rate proportional to its size. After 1.5 hours the population has increased to 975.

(a) Find an expression for the number of bacteria after t hours.

(b) Find the number of bacteria after 3 hours.

(c) Find the rate of growth after 3 hours.

(d) After how many hours will the population reach 250,000?

Solution:

260 ¤ CHAPTER 3 DIFFERENTIATION RULES

(b) = log22

⇒

[ constant] =

12

ln 2

· −2−2 =

2 ln 2· −2 2

= − 1

ln 2

The negative sign indicates that difficulty decreases with increasing width. While the magnitude of the rate of change

decreases with increasing widththat is,

− 1

ln 2

= 1

ln 2decreases as increases

, the rate of change itself

increases (gets closer to zero from the negative side) with increasing values of .

(c) The answers to (a) and (b) agree with intuition. For fixed width, the difficulty of acquiring a target increases, but less and

less so, as the distance to the target increases. Similarly, for a fixed distance to a target, the difficulty of acquiring the target

decreases, but less and less so, as the width of the target increases.

3.8 Exponential Growth and Decay

1. The relative growth rate is 1

= 04159, so

= 04159 and by Theorem 2,

() = (0)04159 = 3804159 million cells. Thus, (2) = 3804159(2) 87 million cells.

2. (a) By Theorem 2, () = (0) = 50. In 20 minutes13hour

, there are 100 cells, so

13

= 503 = 100 ⇒

3 = 2 ⇒ 3 = ln 2 ⇒ = 3 ln 2 = ln(23) = ln 8.

(b) () = 50(ln 8) = 50 · 8

(c) (6) = 50 · 86 = 50 · 218 = 13,107,200 cells

(d)

= ⇒ 0(6) = (6) = (ln 8) (6) ≈ 27,255,656 cellsh

(e) () = 106 ⇔ 50 · 8 = 1,000,000 ⇔ 8 = 20,000 ⇔ ln 8 = ln 20,000 ⇔ =ln 20,000ln 8

≈ 476 h

3. (a) By Theorem 2, () = (0) = 50. Now (15) = 50(15) = 975 ⇒ 15 = 97550 ⇒

15 = ln195 ⇒ = 115ln 195 ≈ 19803. So () ≈ 5019803 cells.

(b) Using 19803 for , we get (3) = 5019803(3) = 19,01385 19,014 cells.

(c)

= ⇒ 0(3) = · (3) = 19803 · 19,014 [from parts (a) and (b)] = 37,6534 37,653 cellsh

(d) () = 5019803 = 250,000 ⇒ 19803 =250,00050

⇒ 19803 = 5000 ⇒ 19803 = ln 5000 ⇒

=ln 5000

19803 430 h

4. (a) () = (0) ⇒ (2) = (0)2 = 400 and (6) = (0)6 = 25,600. Dividing these equations, we get

62 = 25,600400 ⇒ 4 = 64 ⇒ 4 = ln 26 = 6 ln 2 ⇒ = 32ln 2 ≈ 10397, about 104% per hour.

(b) 400 = (0)2 ⇒ (0) = 4002 ⇒ (0) = 4003 ln 2 = 400ln 2

3= 40023 = 50.

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7. Experiments show that if the chemical reaction

N2O5 → 2NO2 +1

2O2

takes place at 45◦C, the rate of reaction of dinitrogen pentoxide is proportional to its concentration as follows:

−d[N2O5]

dt= 0.0005[N2O5]

(a) Find an expression for the concentration [N2O5] after t seconds if the initial concentration is C.

(b) How long will the reaction take to reduce the concentration of N2O5 to 90% of its original value?

Solution:

SECTION 3.8 EXPONENTIAL GROWTH AND DECAY ¤ 243

(b) As in part (a), () = (1960)(−1960) ⇒ (1980) = 150 = 10020 ⇒ 20 = ln 150100

⇒

= 120

ln 32≈ 00203. Thus, (2000) = 10040 = 225 million, which is an overestimate of the actual population

of 214 million.

(c) As in part (a), () = (1980)(−1980) ⇒ (2000) = 214 = 15020 ⇒ 20 = ln 214150

⇒

= 120

ln 214150

≈ 00178. Thus, (2010) = 15030 ≈ 256, which is an overestimate of the actual population of

243 million.

(d) (2020) = 150(2020−1980) ≈ 305 million. This estimate will probably be an overestimate since this model gave us an

overestimate in part (c) — indicating that is too large. Creating a model with more recent data would likely result in an

improved estimate.

7. (a) If = [N2O5] then by Theorem 2,

= −00005 ⇒ () = (0)−00005 = −00005.

(b) () = −00005 = 09 ⇒ −00005 = 09 ⇒ −00005 = ln 09 ⇒ = −2000 ln 09 ≈ 211 s

8. (a) The mass remaining after days is () = (0) = 50. Since the half-life is 28 days, (28) = 5028 = 25 ⇒28 = 1

2⇒ 28 = ln 1

2⇒ = −(ln 2)28, so () = 50−(ln 2)28 = 50 · 2−28.

(b) (40) = 50 · 2−4028 ≈ 186mg (d)

(c) () = 2 ⇒ 2 = 50 · 2−28 ⇒ 250

= 2−28 ⇒

(−28) ln 2 = ln 125

⇒ =−28 ln 1

25

ln 2 ≈ 130 days

9. (a) If () is the mass (in mg) remaining after years, then () = (0) = 100.

(30) = 10030 = 12(100) ⇒ 30 = 1

2⇒ = −(ln 2)30 ⇒ () = 100−(ln 2)30 = 100 · 2−30

(b) (100) = 100 · 2−10030 ≈ 992 mg

(c) 100−(ln 2)30 = 1 ⇒ −(ln 2)30 = ln 1100

⇒ = −30 ln 001ln 2

≈ 1993 years

10. (a) If () is the mass after days and (0) = , then () = .

(1) = = 0945 ⇒ = 0945 ⇒ = ln 0945.

Then (ln 0945) = 12 ⇔ ln (ln 0945) = ln 1

2⇔ (ln 0945) = ln 1

2⇔ = − ln 2

ln 0945≈ 1225 years.

(b) (ln 0945) = 020 ⇔ (ln 0945) = ln 15⇔ = − ln 5

ln 0945≈ 2845 years

11. Let () be the level of radioactivity. Thus, () = (0)− and is determined by using the half-life:

(5730) = 12(0) ⇒ (0)−(5730) = 1

2(0) ⇒ −5730 = 1

2⇒ −5730 = ln 1

2⇒ = − ln 1

2

5730=

ln 2

5730.

If 74% of the 14C remains, then we know that () = 074(0) ⇒ 074 = −(ln 2)5730 ⇒ ln 074 = − ln 2

5730⇒

= −5730(ln 074)

ln 2≈ 2489 ≈ 2500 years.

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12. Scientists can determine the age of ancient objects by the method of radiocarbon dating. The bombardment of the

upper atmosphere by cosmic rays converts nitrogen to a radioactive isotope of carbon, 14C, with a half-life of about

5730 years. Vegetation absorbs carbon dioxide through the atmosphere and animal life assimilates 14C through

food chains. When a plant or animal dies, it stops replacing its carbon and the amount of 14C begins to decrease

through radioactive decay. Therefore the level of radioactivity must also decay exponentially.

Dinosaur fossils are too old to be reliably dated using carbon-14. Suppose we had a 68-millionyear- old dinosaur

fossil. What fraction of the living dinosaur�s 14C would be remaining today? Suppose the minimum detectable

amount is 0.1%. What is the maximum age of a fossil that we could date using 14C?

1

Solution:244 ¤ CHAPTER 3 DIFFERENTIATION RULES

12. From Exercise 11, we have the model () = (0)− with = (ln 2)5730. Thus,

(68,000,000) = (0)−68000000 ≈ (0) · 0 = 0. There would be an undetectable amount of 14C remaining for a

68-million-year-old dinosaur.

Now let () = 01% (0), so 0001(0) = (0)− ⇒ 0001 = − ⇒ ln 0001 = − ⇒

=ln 0001

− =ln 0001

−(ln 2)5730≈ 57,104, which is the maximum age of a fossil that we could date using 14C.

13. Let measure time since a dinosaur died in millions of years, and let () be the amount of 40K in the dinosaur’s bones at

time . Then () = (0)− and is determined by the half-life: (1250) = 12(0) ⇒ (0)−(1250) = 1

2(0) ⇒

−1250 = 12⇒ −1250 = ln 1

2⇒ = − ln 1

2

1250=

ln 2

1250. To determine if a dinosaur dating of 68 million years is

possible, we find that (68) = (0)−(68) ≈ 0963(0), indicating that about 96% of the 40K is remaining, which is

clearly detectable. To determine the maximum age of a fossil by using 40K, we solve () = 01%(0) for .

(0)− = 0001(0) ⇔ − = 0001 ⇔ − = ln 0001 ⇔ =ln 0001

−(ln 2)1250≈ 12,457 million, or

12457 billion years.

14. From the information given, we know that

= 2 ⇒ = 2 by Theorem 2. To calculate we use the point (0 5):

5 = 2(0) ⇒ = 5. Thus, the equation of the curve is = 52.

15. (a) Using Newton’s Law of Cooling,

= ( − ), we have

= ( − 22). Now let = − 22, so

(0) = (0)− 22 = 85− 22 = 63, so is a solution of the initial-value problem = with (0) = 63 and by

Theorem 2 we have () = (0) = 63.

(30) = 6330 = 65− 22 ⇒ 30 = 4363

⇒ = 130

ln 4363, so () = 63

130 ln

4363

and

(45) = 634530

ln

4363

≈ 36◦C. Thus, (45) ≈ 36 + 22 = 58◦C.

(b) () = 40 ⇒ () = 18. () = 63130 ln

4363

= 18 ⇒

130 ln

4363

= 18

63= 2

7⇒ 1

30 ln 43

63= ln 2

7⇒

=30 ln 2

7

ln 4363

≈ 98 min.

16. Let () be the temperature of the body hours after 1:30 PM. Then (0) = 325 and (1) = 303. Using Newton’s Law of

Cooling,

= ( − ), we have

= ( − 20). Now let = − 20, so (0) = (0)− 20 = 325− 20 = 125,

so is a solution to the initial value problem = with (0) = 125 and by Theorem 2 we have

() = (0) = 125.

(1) = 303− 20 ⇒ 103 = 125(1) ⇒ = 10312.5 ⇒ = ln 103

12.5 . The murder occurred when

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13. Dinosaur fossils are often dated by using an element other than carbon, such as potassium-40, that has a longer

half-life (in this case, approximately 1.25 billion years). Suppose the minimum detectable amount is 0.1% and a

dinosaur is dated with 14K to be 68 million years old. Is such a dating possible? In other words, what is the

maximum age of a fossil that we could date using 14K?

Solution:

244 ¤ CHAPTER 3 DIFFERENTIATION RULES

12. From Exercise 11, we have the model () = (0)− with = (ln 2)5730. Thus,

(68,000,000) = (0)−68000000 ≈ (0) · 0 = 0. There would be an undetectable amount of 14C remaining for a

68-million-year-old dinosaur.

Now let () = 01% (0), so 0001(0) = (0)− ⇒ 0001 = − ⇒ ln 0001 = − ⇒

=ln 0001

− =ln 0001

−(ln 2)5730≈ 57,104, which is the maximum age of a fossil that we could date using 14C.

13. Let measure time since a dinosaur died in millions of years, and let () be the amount of 40K in the dinosaur’s bones at

time . Then () = (0)− and is determined by the half-life: (1250) = 12(0) ⇒ (0)−(1250) = 1

2(0) ⇒

−1250 = 12⇒ −1250 = ln 1

2⇒ = − ln 1

2

1250=

ln 2

1250. To determine if a dinosaur dating of 68 million years is

possible, we find that (68) = (0)−(68) ≈ 0963(0), indicating that about 96% of the 40K is remaining, which is

clearly detectable. To determine the maximum age of a fossil by using 40K, we solve () = 01%(0) for .

(0)− = 0001(0) ⇔ − = 0001 ⇔ − = ln 0001 ⇔ =ln 0001

−(ln 2)1250≈ 12,457 million, or

12457 billion years.

14. From the information given, we know that

= 2 ⇒ = 2 by Theorem 2. To calculate we use the point (0 5):

5 = 2(0) ⇒ = 5. Thus, the equation of the curve is = 52.

15. (a) Using Newton’s Law of Cooling,

= ( − ), we have

= ( − 22). Now let = − 22, so

(0) = (0)− 22 = 85− 22 = 63, so is a solution of the initial-value problem = with (0) = 63 and by

Theorem 2 we have () = (0) = 63.

(30) = 6330 = 65− 22 ⇒ 30 = 4363

⇒ = 130

ln 4363, so () = 63

130 ln

4363

and

(45) = 634530

ln

4363

≈ 36◦C. Thus, (45) ≈ 36 + 22 = 58◦C.

(b) () = 40 ⇒ () = 18. () = 63130 ln

4363

= 18 ⇒

130 ln

4363

= 18

63= 2

7⇒ 1

30 ln 43

63= ln 2

7⇒

=30 ln 2

7

ln 4363

≈ 98 min.

16. Let () be the temperature of the body hours after 1:30 PM. Then (0) = 325 and (1) = 303. Using Newton’s Law of

Cooling,

= ( − ), we have

= ( − 20). Now let = − 20, so (0) = (0)− 20 = 325− 20 = 125,

so is a solution to the initial value problem = with (0) = 125 and by Theorem 2 we have

() = (0) = 125.

(1) = 303− 20 ⇒ 103 = 125(1) ⇒ = 10312.5 ⇒ = ln 103

12.5 . The murder occurred when

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14. A curve passes through the point (0, 5) and has the property that the slope of the curve at every point P is twice

the y-coordinate of P . What is the equation of the curve?

Solution:

244 ¤ CHAPTER 3 DIFFERENTIATION RULES

12. From Exercise 11, we have the model () = (0)− with = (ln 2)5730. Thus,

(68,000,000) = (0)−68000000 ≈ (0) · 0 = 0. There would be an undetectable amount of 14C remaining for a

68-million-year-old dinosaur.

Now let () = 01% (0), so 0001(0) = (0)− ⇒ 0001 = − ⇒ ln 0001 = − ⇒

=ln 0001

− =ln 0001

−(ln 2)5730≈ 57,104, which is the maximum age of a fossil that we could date using 14C.

13. Let measure time since a dinosaur died in millions of years, and let () be the amount of 40K in the dinosaur’s bones at

time . Then () = (0)− and is determined by the half-life: (1250) = 12(0) ⇒ (0)−(1250) = 1

2(0) ⇒

−1250 = 12⇒ −1250 = ln 1

2⇒ = − ln 1

2

1250=

ln 2

1250. To determine if a dinosaur dating of 68 million years is

possible, we find that (68) = (0)−(68) ≈ 0963(0), indicating that about 96% of the 40K is remaining, which is

clearly detectable. To determine the maximum age of a fossil by using 40K, we solve () = 01%(0) for .

(0)− = 0001(0) ⇔ − = 0001 ⇔ − = ln 0001 ⇔ =ln 0001

−(ln 2)1250≈ 12,457 million, or

12457 billion years.

14. From the information given, we know that

= 2 ⇒ = 2 by Theorem 2. To calculate we use the point (0 5):

5 = 2(0) ⇒ = 5. Thus, the equation of the curve is = 52.

15. (a) Using Newton’s Law of Cooling,

= ( − ), we have

= ( − 22). Now let = − 22, so

(0) = (0)− 22 = 85− 22 = 63, so is a solution of the initial-value problem = with (0) = 63 and by

Theorem 2 we have () = (0) = 63.

(30) = 6330 = 65− 22 ⇒ 30 = 4363

⇒ = 130

ln 4363, so () = 63

130 ln

4363

and

(45) = 634530

ln

4363

≈ 36◦C. Thus, (45) ≈ 36 + 22 = 58◦C.

(b) () = 40 ⇒ () = 18. () = 63130 ln

4363

= 18 ⇒

130 ln

4363

= 18

63= 2

7⇒ 1

30 ln 43

63= ln 2

7⇒

=30 ln 2

7

ln 4363

≈ 98 min.

16. Let () be the temperature of the body hours after 1:30 PM. Then (0) = 325 and (1) = 303. Using Newton’s Law of

Cooling,

= ( − ), we have

= ( − 20). Now let = − 20, so (0) = (0)− 20 = 325− 20 = 125,

so is a solution to the initial value problem = with (0) = 125 and by Theorem 2 we have

() = (0) = 125.

(1) = 303− 20 ⇒ 103 = 125(1) ⇒ = 10312.5 ⇒ = ln 103

12.5 . The murder occurred when

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2