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=
Solution
- 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.
- 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.
- 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)
- 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.
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