A local pond can only hold up to 56 geese. Five geese are introduced into the pond. Assume that the rate of growth of the population is $\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0013 ) ( 56 - \mathrm { P } ) \mathrm { P }$ where t is time in weeks. Find a formula for the goose population in terms of t.
A) $\mathrm { P } ( \mathrm { t } ) = \frac { 56 } { 1 + 51 \mathrm { e } ^ { - 0.073 \mathrm { t } } }$
B) $P ( t ) = \frac { 56 } { 1 + 10.2 e ^ { - 56 t } }$
C) $P ( t ) = \frac { 56 } { 1 + 10.2 e ^ { - 0.073 t } }$
D) $P ( t ) = \frac { 56 } { 1 + 51 e ^ { - 56 t } }$

Question

A local pond can only hold up to 56 geese. Five geese are introduced into the pond. Assume that the rate of growth of the population is $\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0013 ) ( 56 - \mathrm { P } ) \mathrm { P }$ where t is time in weeks. Find a formula for the goose population in terms of t. 
A) $\mathrm { P } ( \mathrm { t } ) = \frac { 56 } { 1 + 51 \mathrm { e } ^ { - 0.073 \mathrm { t } } }$
B) $P ( t ) = \frac { 56 } { 1 + 10.2 e ^ { - 56 t } }$
C) $P ( t ) = \frac { 56 } { 1 + 10.2 e ^ { - 0.073 t } }$
D) $P ( t ) = \frac { 56 } { 1 + 51 e ^ { - 56 t } }$

Quizplus · Accepted Answer

We can solve the given differential equation using separation of variables as follows:
$$\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0013 ) ( 56 - \mathrm { P } ) \mathrm { P }$$
$$\frac { \mathrm { dP } } { ( 56 - \mathrm { P } ) \mathrm { P } } = 0.0013\mathrm { dt } $$
Integrating both sides, we get:
$$\ln \left| \frac { \mathrm { P } } { 56 - \mathrm { P } } \right| = 0.0013 \mathrm { t } + \mathrm { C } $$
where C is the constant of integration. To determine the value of C, we use the initial condition that there are five geese initially, which gives us:
$$\ln \left| \frac { 5 } { 51 } \right| = C $$
$$C = -3.9318$$
Substituting this value back into the previous equation, we get:
$$\ln \left| \frac { \mathrm { P } } { 56 - \mathrm { P } } \right| = 0.0013 \mathrm { t } - 3.9318$$
$$\left| \frac { \mathrm { P } } { 56 - \mathrm { P } } \right| = e ^ { 0.0013 \mathrm { t } - 3.9318 } $$
$$\frac { \mathrm { P } } { 56 - \mathrm { P } } = e ^ { 0.0013 \mathrm { t } - 3.9318 } \text {  or  } \frac { \mathrm { P } } { 56 - \mathrm { P } } = - e ^ { 0.0013 \mathrm { t } - 3.9318 } $$
Solving for P in each case, we get:
$$\mathrm { P } ( \mathrm { t } ) = \frac { 56 } { 1 + 10.2 e ^ { - 0.0013 \mathrm { t } } }$$
or
$$\mathrm { P } ( \mathrm { t } ) = \frac { 56 } { 1 + 51 e ^ { - 0.0013 \mathrm { t } } }$$
Since P represents the population of geese and we know that the maximum capacity of the pond is 56 geese, the population cannot exceed 56. Therefore, the correct formula is:
$$\mathrm { P } ( \mathrm { t } ) = \frac { 56 } { 1 + 10.2 e ^ { - 0.0013 \mathrm { t } } }$$
which is choice (C).

A Local Pond Can Only Hold Up to 56 Geese dPdt=(0.0013)(56−P)P\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0013 ) ( 56 - \mathrm { P } ) \mathrm { P }dtdP​=(0.0013)(56−P)P

A Local Pond Can Only Hold Up to 56 Geese $\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0013 ) ( 56 - \mathrm { P } ) \mathrm { P }$