MATH SOLVE

2 months ago

Q:
# The population of deer in a certain area of Stanly county grows proportional to itself. The population of deer in 2000 was found to be 145, and by 2010 the population had frown to 185. Find the growth constant k (rounded to 6 decimal places), and the expected deer population in 2013.K=?Population = ?

Accepted Solution

A:

The first step is to determine the equation The change in population over time is:[tex]\frac{dP}{dt}= kP\\ \\\frac{dP}{P} = k *dt\\\\[/tex]To clear the constant k you must integrate:
[tex]\int\ {\frac{dP}{P} } \, = \int\ {k} \, dt \\\\ln(P) = k t + c\\\\P = e^{kt} * e^{c} \\ \\e^{c} = C\\\\P = Ce^{kt}[/tex]Then, the exercise talks about the year 2000 was the first data collection, then it is assumed as the year 0[tex]P(0)= 145 \\\\ P (10)= 185 \\\\[/tex]The constant C is cleared[tex]P = Ce^{kt}\\\\ P(0) = C e^{k(0)} \\\\ 145= C e^{k(0)} \\\\ 145=C\\\ P(t) = 145 e^{kt}[/tex]Then the constant k is clear whit the second condition[tex]P(10) = 145 e^{kt} \\\ 185 = 145 e^{k(10)} \\\ ln (\frac{185}{145} ) = k(10)\\\ 0,0243622 = k[/tex]The poblation in 2013 is:[tex]P(13) = 145 e^{kt} \\\ P = 145 e^{k(13)} \\\ P = 199.027389[/tex]Answer:P= 199.027389
k= 0,0243622