Section 9.4 Models for Population Growth

Section 9.4 Models for Population Growth

9. The population of the world was about 6.1 billion in 2000. Birth rates around that time ranged from 35 to 40 million per year and death rates ranged from 15 to 20 million per year. Lets assume that the carrying capacity for world population is 20 billion.

(a) Write the logistic differential equation for these data. (Because the initial population is small compared to the carrying capacity, you can take k to be an estimate of the initial relative growth rate.)

(b) Use the logistic model to estimate the world population in the year 2010 and compare with the actual population of 6.9 billion.

(c) Use the logistic model to predict the world population in the years 2100 and 2500.

Solution:

SECTION 9.4 MODELS FOR POPULATION GROWTH ¤ 831 7. Using Equation 7,  = − ⁰

0

=10,000 − 1000

1000 = 9, so  () = 10,000

1 + 9^−.  (1) = 2500 ⇒ 2500 = 10,000

1 + 9^−(1) ⇒ 1 + 9^−= 4 ⇒ 9^−= 3 ⇒ ^−= ¹₃ ⇒ − = ln¹3 ⇒  = ln 3. After another three years,  = 4, and  (4) = 10,000

1 + 9^{−(ln 3)4} = 10,000

1 + 9 (^{ln 3})⁻⁴ = 10,000

1 + 9(3)⁻⁴ =10,000

1 +¹₉ =10,000

10 9

= 9000.

8. (a) From the graph, we estimate the carrying capacity  for the yeast population to be 680.

(b) An estimate of the initial relative growth rate is 1

0



 = 1

18·39 − 18 2 − 0 = 7

12 = 0583.

(c) An exponential model is  () = 18^712. A logistic model is  () = 680

1 + ^−712, where  = ⁶⁸⁰₁₈^{− 18}= ³³¹₉ . (d)

Time in Hours

Observed Values

Exponential Model

Logistic Model

0 18 18 18

2 39 58 55

4 80 186 149

6 171 596 322

8 336 1914 505

10 509 6147 614

12 597 19,739 658

14 640 63,389 673

16 664 203,558 678

18 672 653,679 679

The exponential model is a poor fit for anything beyond the first two observed values. The logistic model varies more for the middle values than it does for the values at either end, but provides a good general fit, as shown in the figure.

(e)  (7) = 680

1 +³³¹₉ ^−7(712) ≈ 420 yeast cells

9. (a) We will assume that the difference in birth and death rates is 20 million/year. Let  = 0 correspond to the year 2000. Thus,

 ≈ 1





 = 1

61billion

20million year



= 1

305, and

 = 

 1 − 





= 1 305

 1 − 

20



with  in billions.

(b)  =  − ⁰

0

= 20 − 61 61 = 139

61 ≈ 22787.  () = 

1 + ^− = 20

1 +¹³⁹₆₁^−305, so

 (10) = 20

1 +¹³⁹₆₁^−10305 ≈ 624 billion, which underestimates the actual 2010 population of 69 billion.

(c) The years 2100 and 2500 correspond to  = 100 and  = 500, respectively.  (100) = 20

1 +¹³⁹₆₁^−100305 ≈ 757 billion and  (500) = 20

1 +¹³⁹₆₁^−500305 ≈ 1387 billion.

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18. Doomsday Equation Let c be a positive number. A differential equation of the form dy

dt = ky^1+c

where k is a positive constant, is called a doomsday equation because the exponent in the expression ky^1+c is larger than the exponent 1 for natural growth.

(a) Determine the solution that satisfies the initial condition y(0) = y0.

(b) Show that there is a finite time t = T (doomsday) such that lim_t→T⁻y(t) = ∞.

(c) An especially prolific breed of rabbits has the growth term ky^1.01. If 2 such rabbits breed initially and the warren has 16 rabbits after three months, then when is doomsday?

Solution:

834 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS

16. Following the hint, we choose  = 0 to correspond to 1960 and subtract 3500from each of the population figures. We then use a calculator to obtain the models and add 3500 to get the exponential function

() = 1809934(10445)^+ 3500and the logistic function

() = 13489650

1 + 62784^{−00721}+ 3500. is a reasonably accurate

accurate model, while is not, since an exponential model would only be used for the first few data points.

17. (a) 

 =  −  = 

 −



. Let  =  −

, so

 =

 and the differential equation becomes

 = .

The solution is  = 0^ ⇒  −

 =

0−





^ ⇒  () = 

 +

0−





^.

(b) Since   0, there will be an exponential expansion ⇔ ⁰−

  0 ⇔   ⁰. (c) The population will be constant if 0−

 = 0 ⇔  = ⁰. It will decline if 0−

  0 ⇔   ⁰. (d) 0= 8,000,000,  =  −  = 0016,  = 210,000 ⇒   ⁰(= 128,000), so by part (c), the population was

declining.

18. (a) 

 = ^1+ ⇒ ^−1− =   ⇒ ^−

− =  + . Since (0) = 0, we have  = ^−₀

−. Thus,

^−

− =  +^−₀

−, or ^−= ^−₀ − . So ^= 1

₀^−−  = ^₀

1 − 0^and () = 0

(1 − 0^)^1. (b) () → ∞ as 1 − ^0 → 0, that is, as  → 1

^₀. Define  = 1

₀^. Then lim

→⁻() = ∞.

(c) According to the data given, we have  = 001, (0) = 2, and (3) = 16, where the time  is given in months. Thus,

0= 2and 16 = (3) = 0

(1 − ^0 · 3)^1. Since  = 1

^₀, we will solve for 0^. 16 = 2

(1 − 3^0)¹⁰⁰ ⇒ 1 − 3^0 =₁

8

001

= 8^−001 ⇒ ^0 = ¹₃

1 − 8^−001

. Thus, doomsday occurs when

 =  = 1

₀^ = 3

1 − 8^−001 ≈ 14577 months or 1215 years.

19. (a) The term −15 represents a harvesting of fish at a constant rate—in this case, 15 fishweek. This is the rate at which fish are caught.

(b) (c) From the graph in part (b), it appears that  () = 250 and  () = 750 are the equilibrium solutions. We confirm this analytically by solving the equation  = 0 as follows: 008 (1 − 1000) − 15 = 0 ⇒ 008 − 000008²− 15 = 0 ⇒

−000008(²− 1000 + 187,500) = 0 ⇒ ( − 250)( − 750) = 0 ⇒  = 250 or 750.

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1

21. There is considerable evidence to support the theory that for some species there is a minimum population m such that the species will become extinct if the size of the population falls below m. This condition can be incorporated into the logistic equation by introducing the factor (1 − m/P ). Thus the modified logistic model is given by the differential equation

dP dt = kP

 1 − P

M

 1 −m

P



(a) Use the differential equation to show that any solution is increasing if m < P < M and decreasing if 0 < P < m.

(b) For the case where k = 0.08, M = 1000, and m = 200, draw a direction field and use it to sketch several solution curves. Describe what happens to the population for various initial populations. What are the equilibrium solutions?

(c) Solve the differential equation explicitly, either by using partial fractions or with a computer algebra system.

Use the initial population P0.

(d) Use the solution in part (c) to show that if P0< m, then the species will become extinct. [Hint: Show that the numerator in your expression for P (t) is 0 for some value of t.]

Solution:

836 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS (c) 

 = 008 − 000008²− . 

 = 0 ⇔  =−008 ±

(008)²− 4(−000008)(−)

2(−000008) , which has at least one solution when the discriminant is nonnegative ⇒ 00064 − 000032 ≥ 0 ⇔  ≤ 20. For 0 ≤  ≤ 20, there is at least one value of  such that  = 0 and hence, at least one equilibrium solution. For   20,   0 and the population always dies out.

(d) The weekly catch should be less than 20 fish per week.

21. (a) 

 = ( )

 1 − 



1 −



. If     , then  = (+)(+)(+) = + ⇒  is increasing.

If 0    , then  = (+)(+)(−) = − ⇒  is decreasing.

(b)  = 008,  = 1000, and  = 200 ⇒



 = 008

 1 − 

1000



1 −200





For 0  0 200, the population dies out. For 0= 200, the population is steady. For 200  0 1000, the population increases and approaches 1000. For 0 1000, the population decreases and approaches 1000.

The equilibrium solutions are  () = 200 and  () = 1000.

(c) 

 = 

 1 − 





1 −





= 

 − 



 − 





= 

( −  )( − ) ⇔

 

( −  )( − ) =

 

 . By partial fractions, 1

( −  )( − ) = 

 −  + 

 − , so

( − ) + ( −  ) = 1.

If  = ,  = 1

 − ; if  = ,  = 1

 − , so 1

 − 

  1

 −  + 1

 − 



 =

 

  ⇒ 1

 − (− ln | −  | + ln | − |) = 

 +  ⇒ 1

 − ln



 − 

 − 



 = 

 +  ⇒ ln



 − 

 − 



 = ( − )



 + 1 ⇔  − 

 −  = ^{(−)()} [ = ±^1].

Let  = 0: 0− 

 − ⁰ = . So  − 

 −  = 0− 

 − ⁰^{(−)()}. Solving for  , we get  () =( − ⁰) +  (0− )^{(−)()}

 − ⁰+ (0− )^{(−)()} .

(d) If 0  , then 0−   0. Let () be the numerator of the expression for  () in part (c). Then

 (0) = 0( − )  0, and ⁰−   0 ⇔ lim

→∞ (0− )^{(−)()}= −∞ ⇒ lim

→∞ () = −∞.

Since  is continuous, there is a number  such that () = 0 and thus  () = 0. So the species will become extinct.

22. (a) 

 =  ln







 ⇒

 

 ln( ) =



 . Let  = ln







= ln  − ln  ⇒  = −

 ⇒



−

 =  +  ⇒ ln|| = − −  ⇒ || = ^{−(+)} ⇒ |ln( )| = ^{−(+)} ⇒

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