Question

A scientist monitored the number of fish in a lake and found that the number of fish increased at a rate proportional to the number of fish present. At time t=0 years, there were 200 fish in the lake. At time t=2 years, there were 1000 fish in the lake. What will be the number of fish in the lake at time t=4 years?

Answer 1

Answer:

5000 fishes

Explanation:

Let the model of the number of fish be in the form of

$P = Ae^{kt}$

where t is the time (in year). So the rate of increasing would be proportional to the number of fishes presented in the lake

At t=0, P = 200

$200 = Ae^0$

$A = 200

At t = 2, P = 1000

[tex] 1000 = 200e^{2k}$

$e^{2k} = 1000 / 200 = 5$

$2k = ln5$

$k = ln5/2 = 0.805$

At t = 4,

$P = 200e^{0.805*4} = 200*25 = 5000$

Answer 2

The number of fish in the lake at time t=4 years will be 5000.

Given data:

The initial time interval is, t = 0 years.

The number of fish in the lake is, n = 200.

The final time interval is, t' = 2 years.

The number of fish at 2 years is, n' = 1000.

Clearly, the problem is forming a linear model. Where number of fish can be given as,

$n=Ae^{kt}$ ...........................................(1)

Here, t is the time (in year). So the rate of increasing would be proportional to the number of fishes presented in the lake.

At t = 0 and n =200,

$200=Ae^{k \times 0}\\200=Ae^{0}\\A = 200$

For t' =2 and n' =1000. Substitute the value of A in equation (1) as,

$n'=Ae^{kt'}\\\\1000=200 \times e^{k \times 2}\\\\5 = e^{2k}\\\\ln5=2K\\k=0.805$

Then number of fish at 4 years is,

$n=Ae^{kt}\\n=200 \times e^{0.805 \times 4}\\n = 5000$

Thus, we can conclude that the number of fish in the lake at time t=4 years will be 5000.

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