The number of individuals N in a population of a certain species is given by the following function of time, t:
N(t)= 100/(1+9e^-t), t>=0
a.)how many ppl are in the population initially(at t=0)
b.)determine the size ofa the populations as tgoes towards infiniti?
c. what is the growth rate of the pop. at time t? what is the growth rate at t=0?
A 100/(1+9*1)=10
b lim as t->infinity is 100
c the growth rate at time 0 is 9
To answer these questions, we will evaluate the given function N(t) and analyze its behavior. Let's start by addressing each question separately:
a) To find the number of individuals in the population initially (at t=0), we substitute t=0 into the function N(t):
N(0) = 100 / (1 + 9e^0)
Since any number raised to the power of 0 is 1, we have:
N(0) = 100 / (1 + 9(1))
N(0) = 100 / (1 + 9)
N(0) = 100 / 10
N(0) = 10
Therefore, initially at t=0, there are 10 individuals in the population.
b) To determine the size of the population as t goes towards infinity, we need to analyze the behavior of the function N(t). As t approaches infinity, the exponential term e^(-t) will approach 0. When the exponent approaches negative infinity, the exponential function evaluates to zero, resulting in a denominator of 1 in the function N(t). Hence:
lim (t → ∞) N(t) = 100 / (1 + 9e^(-∞))
Since e^(-∞) approaches 0,
lim (t → ∞) N(t) = 100 / (1 + 9(0))
lim (t → ∞) N(t) = 100 / (1 + 0)
lim (t → ∞) N(t) = 100 / 1
lim (t → ∞) N(t) = 100
Therefore, as t approaches infinity, the population size will stabilize at 100 individuals.
c) The growth rate of the population at any time t is given by the derivative of the function N(t) with respect to t: dN/dt.
To find the growth rate of the population at time t, we differentiate N(t) with respect to t:
N(t) = 100 / (1 + 9e^(-t))
dN/dt = d(100 / (1 + 9e^(-t)))/dt
By applying the chain rule, we have:
dN/dt = (-900e^(-t)) / (1 + 9e^(-t))^2
To calculate the growth rate at a specific time, substitute the given value of t into this derivative equation.
At t=0, the growth rate can be found by evaluating the derivative:
dN/dt | t=0 = (-900e^(0)) / (1 + 9e^(0))^2
dN/dt | t=0 = (-900) / (1 + 9)^2
dN/dt | t=0 = (-900) / (10)^2
dN/dt | t=0 = (-900) / 100
dN/dt | t=0 = -9
Hence, at t=0, the growth rate of the population is -9 individuals per unit of time.