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).
A) $B(t)=500 e^{7.6 t}$
B) $B(t)=500 e^{0.38 t}$
C) $B(t)=500\left(1-e^{0.38 t}\right)$
D) $B(t)=500\left(20-e^{7.6 t}\right)$

Quizplus · Accepted Answer

The differential equation is $\frac{dB}{dt} = kB$, where $k$ is the growth rate. Given that 100,000 is 10% of the population in 1899, the population was 1,000,000 at that time. Solving $\frac{dB}{dt} = kB$ with initial condition $B(0) = 500$ and $B(20) = 1,000,000$, we find $k = 0.38$. Thus, $B(t) = 500 e^{0.38 t}$.

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