Question

Help please

The number of bacteria in a culture is given by the function
n(t)=945e^0.2t
where t is measured in hours.
(a) What is the relative rate of growth of this bacterium population?
Your answer is ---------- percent

(b) What is the initial population of the culture (at t=0)?
Your answer is ------------

(c) How many bacteria will the culture contain at time t=5?
Your answer is --------------

Answer 1

a) 20%

b) 945

c) 2571

Explanation:

(a) To find the relative rate of growth of the bacterium population, we need to find the derivative of the given function with respect to time t:

n(t) = 945e^0.2t

Taking the derivative of both sides with respect to t, we get:

dn/dt = 945(0.2)e^0.2t

Simplifying this expression, we get:

dn/dt = 189e^0.2t

Now, we can find the relative rate of growth by dividing the derivative by the original function:

Relative rate of growth = (dn/dt) / n(t)

Substituting the expressions we found earlier, we get:

Relative rate of growth = 189e^0.2t / 945e^0.2t = 0.2

Therefore, the relative rate of growth is 0.2 or 20%.

(b) To find the initial population of the culture, we simply need to substitute t = 0 in the given function:

n(0) = 945e^0.2(0) = 945e^0 = 945(1) = 945

Therefore, the initial population of the culture is 945.

(c) To find the number of bacteria in the culture at time t=5, we simply need to substitute t = 5 in the given function:

n(5) = 945e^0.2(5) = 945e = 945(2.71828) = 2571.2266

Therefore, the culture will contain approximately 2571 bacteria at time t=5.