Knowee
Questions
Features
Study Tools

A bacteria culture P grows according to dPdt =kP. The bacteria count was 1296 after 2 hours and 531441 after 6 hours. Answer the following questions.1. What is the relative growth rate?Answer: k= 2. What was the initial size of the culture?Answer: P0= 3. Find an expression for the number of bacteria after t hours.Answer: P(t)= 4. When will the population reach 1000000?Answer (in hours): t=

Question

A bacteria culture P grows according to dPdt =kP. The bacteria count was 1296 after 2 hours and 531441 after 6 hours. Answer the following questions.1. What is the relative growth rate?Answer: k= 2. What was the initial size of the culture?Answer: P0= 3. Find an expression for the number of bacteria after t hours.Answer: P(t)= 4. When will the population reach 1000000?Answer (in hours): t=

🧐 Not the exact question you are looking for?Go ask a question

Solution

  1. What is the relative growth rate?

The given differential equation is dP/dt = kP, which is the equation for exponential growth. The solution to this equation is P(t) = P0 * e^(kt), where P0 is the initial population, k is the relative growth rate, and t is time.

We know that P(2) = 1296 and P(6) = 531441. We can use these two equations to solve for k:

1296 = P0 * e^(2k) 531441 = P0 * e^(6k)

Dividing the second equation by the first, we get:

531441 / 1296 = e^(4k) k = ln(531441 / 1296) / 4 ≈ 1.5

So, the relative growth rate is k = 1.5.

  1. What was the initial size of the culture?

Substitute k = 1.5 into the first equation to solve for P0:

1296 = P0 * e^(2 * 1.5) P0 = 1296 / e^3 ≈ 64

So, the initial size of the culture was P0 = 64.

  1. Find an expression for the number of bacteria after t hours.

Substitute P0 = 64 and k = 1.5 into the equation P(t) = P0 * e^(kt) to get:

P(t) = 64 * e^(1.5t)

  1. When will the population reach 1000000?

Set P(t) = 1000000 and solve for t:

1000000 = 64 * e^(1.5t) t = ln(1000000 / 64) / 1.5 ≈ 8.53 hours

So, the population will reach 1000000 after approximately 8.53 hours.

This problem has been solved

Similar Questions

A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420.(a) Find an expression for the number of bacteria after t hours.P(t) = 100·4.2t (b) Find the number of bacteria after 4 hours. (Round your answer to the nearest whole number.)P(4) = bacteria(c) Find the rate of growth after 4 hours. (Round your answer to the nearest whole number.)P'(4) = bacteria per hour(d) When will the population reach 10,000? (Round your answer to one decimal place.)t = hr

Find the rate of growth after 4 hours. (Round your answer to the nearest whole number.)P'(4) = bacteria per hour

A bacteria population is 7000 at time t = 0 and its rate of growth is 1000 · 2t bacteria per hour after t hours. What is the population after one hour? (Round your answer to the nearest whole number.)

a) A population of bacteria has an initial size 200. After one hour, the population has reached 220. The population grows in such a way that the rate of growth doubles each hour.(2 marks)i. What is the population after five hours? ii. Write a recurrence relation for growth of bacteria, be, after n hours. (3 marks)

A culture of bacteria has an initial population of 9100 bacteria and doubles every 4 hours. Using the formula P, start subscript, t, end subscript, equals, P, start subscript, 0, end subscript, dot, 2, start superscript, start fraction, t, divided by, d, end fraction, end superscriptP t​ =P 0​ ⋅2 dt​ , where P, start subscript, t, end subscriptP t​ is the population after t hours, P, start subscript, 0, end subscriptP 0​ is the initial population, t is the time in hours and d is the doubling time, what is the population of bacteria in the culture after 5 hours, to the nearest whole number?

1/3

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.