When scientists talk about populations they often refer to the carrying capacity of species in a particular environment. Carrying capacity is the largest population that an environment can sustain forever.

Suppose the carrying capacity of seals for a particular group of islands is $2900$ and that there are currently $1800$ seals that inhabit the island. The rate of change of the number of seals is jointly proportional to the number of seals and the difference between the number of seals and the carrying capacity. Let $S$ represent the number of seals on the island after $t$ years. 

1. Write a differential equation that models the rate of change in the number of seals.

   Answer (a):

   $\frac { d S } { d t } = k S ( 2900 - S )$

2. Write a general solution for the differential equation.

   Step (b):

   $\frac { d S } { S ( 2900 - S ) } = k d t$

   Hint (b):

   $\frac { 1 } { S ( 2900 - S ) } = \frac { 1 / 2900 } { S } + \frac { 1 / 2900 } { 2900 - S }$

   Another Step (b):

   $\operatorname { ln } | \frac { S } { 2900 - S } | = 2900 k t + C$

3. Recall that $S = 1800$ when $t = 0$ because that is the current population. Suppose that after ten years, 2500 seals inhabit the island. Write a formula for $S$ in terms of $t$.

   Step (c):

   $S = \frac { 2900 ( \frac { 18 } { 11 } ) e ^ { 2900 k } } { 1 + \frac { 18 } { 11 } e ^ { 230014 } }$