On January 1, 1879, records show that 500 of a fish called Atlantic striped bass were introduced into the San Francisco Bay. In 1899, the first year fishing for bass was allowed, 100,000 of these bass were caught, representing 10% of the population at the start of 1899. Owing to reproduction, at any moment in time the bass population is growing at a rate proportional to the population at that moment. Write a differential equation satisfied by B(t), the number of Atlantic striped bass a time t, where t is in years since January 1, 1879 and 0 $\le$ t

Question

On January 1, 1879, records show that 500 of a fish called Atlantic striped bass were introduced into the San Francisco Bay. In 1899, the first year fishing for bass was allowed, 100,000 of these bass were caught, representing 10% of the population at the start of 1899. Owing to reproduction, at any moment in time the bass population is growing at a rate proportional to the population at that moment. Write a differential equation satisfied by B(t), the number of Atlantic striped bass a time t, where t is in years since January 1, 1879 and 0 $\le$ t < 20 and solve it for B(t).

Quizplus · Accepted Answer

Let B(t) be the number of Atlantic striped bass at time t, in years since January 1, 1879. We are told that the population is growing at a rate proportional to the population at any given moment. This means that the rate of change of B(t) is proportional to B(t), or:

d B(t)/dt = k B(t)

where k is the constant of proportionality. To solve for B(t), we can use separation of variables and integrate both sides:

d B(t)/B(t) = k dt

ln(B(t)) = kt + C

where C is the constant of integration. We can exponentiate both sides to solve for B(t):

B(t) = e^(kt+C) = e^C e^(kt)

To find the value of C, we can use the initial condition given in the problem:

B(0) = 500

Plugging this into our equation, we get:

500 = e^C e^(k*0) = e^C

Therefore, C = ln(500). Our final equation for B(t) is:

B(t) = 500 e^(kt)

To solve for k, we can use the fact that in 1899, the population was 10 times what it was in 1879, or:

B(20) = 10 B(0)

Substituting in our equation for B(t), we get:

500 e^(20k) = 10(500)

Simplifying, we get:

e^(20k) = 10

Taking the natural logarithm of both sides, we get:

20k = ln(10)

k = ln(10)/20

Therefore, our final equation for B(t) is:

B(t) = 500 e^(ln(10)/20 * t)

which simplifies to:

B(t) = 500^(1+(t/20))

On January 1, 1879, Records Show That 500 of a Fish