A bacteria culture grows with a constant relative growth rate. After 2 hours there are 400 bacteria and after 8 hours the count is 50,000.(a) Find the initial population.P(0) = bacteria(b) Find an expression for the population after t hours.P(t) = (c) Find the number of cells after 5 hours. (Round your answer to the nearest integer.)P(5) = bacteria(d) Find the rate of growth after 5 hours. (Round your answer to the nearest integer.)P'(5) = bacteria/hour(e) When will the population reach 200,000? (Round your answer to one decimal place.)t = hours
Question
A bacteria culture grows with a constant relative growth rate. After 2 hours there are 400 bacteria and after 8 hours the count is 50,000.(a) Find the initial population.P(0) = bacteria(b) Find an expression for the population after t hours.P(t) = (c) Find the number of cells after 5 hours. (Round your answer to the nearest integer.)P(5) = bacteria(d) Find the rate of growth after 5 hours. (Round your answer to the nearest integer.)P'(5) = bacteria/hour(e) When will the population reach 200,000? (Round your answer to one decimal place.)t = hours
Solution
The growth of a bacteria culture with a constant relative growth rate can be modeled by the exponential growth equation P(t) = P0 * e^(kt), where P(t) is the population at time t, P0 is the initial population, k is the relative growth rate, and e is the base of the natural logarithm.
(a) To find the initial population P0, we need to solve the exponential growth equation for P0. We know that P(2) = 400 and P(8) = 50000. We can set up the following system of equations:
400 = P0 * e^(2k) 50000 = P0 * e^(8k)
Dividing the second equation by the first, we get:
125 = e^(6k)
Taking the natural logarithm of both sides, we get:
ln(125) = 6k
Solving for k, we get:
k = ln(125) / 6 ≈ 0.792
Substituting k back into the first equation, we get:
400 = P0 * e^(2 * 0.792)
Solving for P0, we get:
P0 = 400 / e^(2 * 0.792) ≈ 100 bacteria
(b) The expression for the population after t hours is:
P(t) = 100 * e^(0.792t)
(c) To find the number of cells after 5 hours, substitute t = 5 into the equation:
P(5) = 100 * e^(0.792 * 5) ≈ 2980 bacteria (rounded to the nearest integer)
(d) The rate of growth after 5 hours is the derivative of P(t) at t = 5:
P'(5) = 0.792 * 100 * e^(0.792 * 5) ≈ 2363 bacteria/hour (rounded to the nearest integer)
(e) To find when the population will reach 200,000, set P(t) = 200000 and solve for t:
200000 = 100 * e^(0.792t)
Taking the natural logarithm of both sides, we get:
ln(2000) = 0.792t
Solving for t, we get:
t = ln(2000) / 0.792 ≈ 8.7 hours (rounded to one decimal place)
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