calculus word problem

Question

A lake initially contains 5000 fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by 9% each month. However, factoring in all causes, 100 fish are lost each month. How many fish will be in the pond after 8 months? (Don't round until the very end.)

Answer 1

Solution:

Present population of fish P_o=5000, Percent increase in fish is 9% per month, Death rate of fish is 100 per month. t = 8 month

Use the formula P(t) = P₀ e^kt Here k is the rate of population increase or decrease positive for increase and negative for decrease. In this problem you have both. The population amount increases by 9% and decreases by a hundred

k can be written as +0.09 - (100/5000) where 100/5000 is the percent decrease

k = +0.09-0.02=+0.07

P(t) = 5000 e^(+0.07)(8) = 8753.36 Round to 8753 fish by the end of the 8th month

Lets test this a bit:

P(1) = 5000 + 0.09(5000) -100 = 5350 , P(1) = 5000 e^(+0.07)(1) = 5362.5= 5363

P(2) = 5350 + 0.09(5350) -100 = 5731.5=5732, P(2) = 5000 e^(+0.07)(2) =5751.3=5751

So it varies a bit due to exponential function but the procedure should remain the same because what if you want to find the population after 100 years, you certainly cannot calculate this in this way.

P(100 yrs) = 5000 e^{(+0.07)(12)(100)} = 1.5125 x 10⁴⁰