Solve.
-A local pond can only hold up to 53 geese. Seven geese are introduced into the pond. Assume that the rate of grow th of the population is $\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0015 ) ( 53 - \mathrm { P } ) \mathrm { P }$ where $t$ is time in weeks. How long will it take for the goose population to be 25 ?
A) 22.26 weeks
B) 46.73 weeks
C) 26.02 weeks
D) 14.23 weeks

Question

Quizplus · Accepted Answer

To solve for the time it takes for the goose population to reach 25, we need to solve the differential equation $\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0015 ) ( 53 - \mathrm { P } ) \mathrm { P }$ with the initial condition that $P(0) = 7$. Separating variables and integrating both sides gives us the solution in terms of $t$. Solving this specific equation with the given initial condition and solving for $t$ when $P = 25$ yields approximately 22.26 weeks.

Solve -A Local Pond Can Only Hold Up to 53 dPdt=(0.0015)(53−P)P\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0015 ) ( 53 - \mathrm { P } ) \mathrm { P }dtdP​=(0.0015)(53−P)P

Solve
-A Local Pond Can Only Hold Up to 53 $\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0015 ) ( 53 - \mathrm { P } ) \mathrm { P }$