Section 9.4 Models for Population Growth

(1)

Section 9.4 Models for Population Growth

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.

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( )| = ^{−(+)} ⇒

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( )| = ^{−(+)} ⇒

1

(2)

SECTION 9.4 MODELS FOR POPULATION GROWTH ¤ 837

ln( ) = ±^{−(+)}. Letting  = 0, we get ln(0) = ±^−, so

ln( ) = ±^{−−}= ±^−^−= ln(0)^− ⇒  = ^ln(⁰^)^− ⇒

 () =  ^{− ln(}⁰^)^−,  6= 0.

(b) lim

→∞ () = lim

→∞ ^{− ln(}⁰^)^−=  ^{− ln(}⁰⁾^·0=  ⁰= 

(c) The graphs look very similar. For the Gompertz function,

 (40) ≈ 732, nearly the same as the logistic function. The Gompertz function reaches  = 900 at  ≈ 617 and its value at  = 80 is about 959, so it doesn’t increase quite as fast as the logistic curve.

(d) 

 =  ln







 =  (ln  − ln  ) ⇒

²

² = 







−1









+ (ln  − ln  )





= 





−1 + ln







= [ ln( ) ][ln( ) − 1] = ² ln( ) [ln( ) − 1]

Since 0    , ⁰⁰= 0 ⇔ ln( ) = 1 ⇔  =  ⇔  = . ⁰⁰ 0for 0    

and ⁰⁰ 0for     , so ⁰is a maximum (and  grows fastest) when  = .

Note: If   , then ln( )  0, so ⁰⁰()  0.

23. (a)  =  cos( − ) ⇒ ( ) =  cos( − )  ⇒ 

( ) = 

cos( − )  ⇒ ln  = () sin( − ) + . (Since this is a growth model,   0 and we can write ln  instead of ln| |.) Since

 (0) = 0, we obtain ln 0= () sin(−) +  = − () sin  +  ⇒  = ln ⁰+ () sin . Thus, ln  = () sin( − ) + ln ⁰+ () sin , which we can rewrite as ln(0) = ()[sin( − ) + sin ] or, after exponentiation,  () = 0()[sin(−)+sin ].

(b) As  increases, the amplitude increases, but the minimum value stays the same.

As  increases, the amplitude and the period decrease.

A change in  produces slight adjustments in the phase shift and amplitude.

 ()oscillates between 0()(1+sin )and 0^()(^{−1+sin )}(the extreme values are attained when  −  is an odd multiple of^₂), so lim

→∞ ()does not exist.

SECTION 9.4 MODELS FOR POPULATION GROWTH ¤ 837

ln( ) = ±^{−(+)}. Letting  = 0, we get ln(0) = ±^−, so

ln( ) = ±^{−−}= ±^−^−= ln(0)^− ⇒  = ^ln(⁰^)^− ⇒

 () =  ^{− ln(}⁰^)^−,  6= 0.

(b) lim

→∞ () = lim

→∞ ^{− ln(}⁰^)^−=  ^{− ln(}⁰⁾^·0=  ⁰= 

(c) The graphs look very similar. For the Gompertz function,

 (40) ≈ 732, nearly the same as the logistic function. The Gompertz function reaches  = 900 at  ≈ 617 and its value at  = 80 is about 959, so it doesn’t increase quite as fast as the logistic curve.

(d) 

 =  ln







 =  (ln  − ln  ) ⇒

²

² = 







−1









+ (ln  − ln  )





= 





−1 + ln







= [ ln( ) ][ln( ) − 1] = ² ln( ) [ln( ) − 1]

Since 0    , ⁰⁰= 0 ⇔ ln( ) = 1 ⇔  =  ⇔  = . ⁰⁰ 0for 0    

and ⁰⁰ 0for     , so ⁰is a maximum (and  grows fastest) when  = .

Note: If   , then ln( )  0, so ⁰⁰()  0.

23. (a)  =  cos( − ) ⇒ ( ) =  cos( − )  ⇒ 

( ) = 

cos( − )  ⇒ ln  = () sin( − ) + . (Since this is a growth model,   0 and we can write ln  instead of ln| |.) Since

 (0) = 0, we obtain ln 0= () sin(−) +  = − () sin  +  ⇒  = ln ⁰+ () sin . Thus, ln  = () sin( − ) + ln ⁰+ () sin , which we can rewrite as ln(0) = ()[sin( − ) + sin ] or, after exponentiation,  () = 0()[sin(−)+sin ].

(b) As  increases, the amplitude increases, but the minimum value stays the same.

As  increases, the amplitude and the period decrease.

A change in  produces slight adjustments in the phase shift and amplitude.

 ()oscillates between 0()(1+sin )and 0^()(^{−1+sin )}(the extreme values are attained when  −  is an odd multiple of^₂), so lim

→∞ ()does not exist.