Question

1. A population of fish in a lake follow a logistic growth model. With an initial growth rate of 10%, an initial population of 50, and a carrying capacity of 500, how many fish will be in the pond after 5 years?
   1. Solve and graph the solution to this equation.
   2. Explain why this differential equation does or does not serve as a realistic model.

Answer 1

This is a simple problem related to modeling an expression for growth of population which is based on the logistic model.

In a logistic model,

If a population is growing in a constrained environment with carrying capacity K, and absent constraint and would grow exponentially with growth rate r, then the population behavior can be described by the logistic growth model:

Where n is the time period .

Now from our question ,

The initial population(Po) =50

Carrying capacity (K)=500

Growth rate (r) =10% or 0.10

Keeping the values we can get population as per logistic growth for each year

Let us find the population after 1st year

P1 =50+0.10*(1-50/500)*5

P1 =54.5

Similarly we can find P2,P3, ....

The given data is

Rate	0.1
Carry capacity	500

The population after each year is shown in the table below.

Year	Population
0	50
1	54.50
2	59.36
3	64.59
4	70.21
5	76.25
6	82.71
7	89.61
8	96.97
9	104.78
10	113.06
11	121.81
12	131.03
13	140.70
14	150.81
15	161.34
16	172.27
17	183.56
18	195.18
19	207.08
20	219.21
21	231.52
22	243.95
23	256.44
24	268.93
25	281.36
26	293.67
27	305.78
28	317.66
29	329.25
30	340.49
31	351.35
32	361.80
33	371.80
34	381.33
35	390.38
36	398.94
37	407.00
38	414.57
39	421.66
40	428.26
41	434.41
42	440.11
43	445.38
44	450.24
45	454.72
46	458.84
47	462.62
48	466.08
49	469.24
50	472.13
51	474.76
52	477.16
53	479.34
54	481.32
55	483.11
56	484.75
57	486.23
58	487.56
59	488.78
60	489.87
61	490.87
62	491.76
63	492.57
64	493.30
65	493.97
66	494.56
67	495.10
68	495.58
69	496.02
70	496.42
71	496.77
72	497.09
73	497.38
74	497.64
75	497.88
76	498.09
77	498.28
78	498.45
79	498.61
80	498.74
81	498.87
82	498.98
83	499.08
84	499.18
85	499.26
86	499.33
87	499.40
88	499.46
89	499.51
90	499.56
91	499.61
92	499.64
93	499.68
94	499.71
95	499.74
96	499.77
97	499.79
98	499.81
99	499.83
100	499.85
101	499.86
102	499.88
103	499.89
104	499.90
105	499.91
106	499.92
107	499.93
108	499.93
109	499.94
110	499.95
111	499.95
112	499.96
113	499.96
114	499.96
115	499.97
116	499.97
117	499.97
118	499.98
119	499.98
120	499.98
121	499.98
122	499.98
123	499.99
124	499.99
125	499.99
126	499.99
127	499.99
128	499.99
129	499.99
130	499.99
131	499.99
132	499.99
133	500.00

Thus we see that after 5 years ,the fish population will be roughly equal to 76

and it will attain its carrying capacity is about 133 years.

b). The exponential growth is practically un-realistic because it does not consider the constraints due to environmental limits that have very high consequences on the growth of the population of the sample under consideration .This is because according to this model, a population can grow with no limits.Thus the growth can be bottomless .

Now reflect on the logistic growth model.

It is a growth model which takes into account the constraints or limiting capacity of the environment it is within.

Such being the case, the logistic growth model is a more realistic model because it considers those environmental limits that are limiting or detrimental to the population growth. It tells us that the population has a limit because of those environmental factors.

For example,in our example shared above, the fish tank capacity is the limiting condition and hence it tells us that the eventual population cant be limitless . It has to stop at 500 .