Question

A population of deer inside a park has a carrying capacity of 100 and a growth rate of 1.8%. If the initial population is 75 deer, what is the population of deer at any given time?

Answer 1

Answer:

The population for the first year is 75.35.

Step-by-step explanation:

Solution:

Given Data;

Carrying capacity (K) = 100

Growth rate (r) = 1.8% = 0.018

Initial population (Po) = 75

time = first year

Population at any given time (pt) = ?

Using the population growth rate formula, we have

dp/dt = rp (1-p/t)---------------------------1

Where r is the growth rate, t is the given time and p is the population growth

Equation 1 is differentiated and is obtained as;

Pt = K/ [1 + (K-Po/Po)e^-rt]------------2

Substituting into equation 2, we have

Pt = 100/[ 1+ (100-75/75) *e^-0.018*1]

Pt = 100/[1+ (0.333 * e^-0.018]

   = 100/[1 +(0.333*0.982)]

   = 100/[1 + 0.327]

   = 100/1.327

  =75.35