A culture of bacteria is growing at a rate of 4e^0.5t per hour, with t in hours and 0 ≤ t ≤ 20.
A) How many new bacteria will be in the culture after the first six hours? ( I got 153 bacteria)
B) How many new bacteria are introduced from the end of the fifth hour through the fifteenth hour? (I got 23687 bacteria)
C) At approximately how many hours after t=0 will the culture contain 200 new bacteria? (idk.)
please check my answers, and show your work. Thanks!
So, how did you arrive at these answers? If we disagree, then there's something to check.
and on C, what are the "new" bacteria? Those which have grown during the last hour? The last minute? If you got B right, then C should then be easy to answer.
(A) if you mean how many total bacteria have grown during the first 6 hours, then that is of course ∫[0,6] 4e^(t/2) dt = 8(e^3 - 1) = 152.684... round it is 153
which is right
(B) ∫[5,15] 4e^(t/2) dt = 14366.879...round to get 14367
(C) ∫[0,x] 4e^(t/2) dt = 200
8(e^(x/2) - 1) = 200
e^(x/2) - 1 = 25
e^(x/2) = 26
x/2 = ln26
x = 2 ln26 = 6.516
A) To find the number of new bacteria after the first six hours, we can use the growth rate formula:
N(t) = initial number of bacteria × e^(growth rate × t)
Given that the growth rate is 4e^(0.5t), we can substitute t = 6 into the equation:
N(6) = initial number of bacteria × e^(4e^(0.5 × 6))
To solve this problem, we need to know the initial number of bacteria. Please provide that information so we can proceed with the calculation.
B) To find the number of new bacteria introduced from the end of the fifth hour through the fifteenth hour, we need to calculate the total growth during that period. We can subtract the number of bacteria at the end of the fifth hour from the number of bacteria at the end of the fifteenth hour.
N(15) = initial number of bacteria × e^(4e^(0.5 × 15))
N(5) = initial number of bacteria × e^(4e^(0.5 × 5))
Total new bacteria introduced = N(15) - N(5) = initial number of bacteria × (e^(4e^(0.5 × 15)) - e^(4e^(0.5 × 5)))
To provide an accurate result, please provide the initial number of bacteria.
C) To determine approximately when the culture contains 200 new bacteria, we need to solve the equation N(t) = 200.
N(t) = initial number of bacteria × e^(4e^(0.5t))
200 = initial number of bacteria × e^(4e^(0.5t))
To find the value of t, we need to know the initial number of bacteria.
To find the answers to these questions, we need to use the given growth rate equation and perform the necessary calculations. Let's go through each question together:
A) How many new bacteria will be in the culture after the first six hours?
To find the number of new bacteria after a certain time, we need to integrate the growth rate equation with respect to time. Given the growth rate is 4e^(0.5t) bacteria per hour, we integrate it from 0 to 6 hours:
∫(4e^(0.5t)) dt = [8e^(0.5t)] evaluated from 0 to 6
= 8e^(0.5(6)) - 8e^(0.5(0))
= 8e^(3) - 8(1)
≈ 8(20.0855) - 8
≈ 161.24 - 8
≈ 153 bacteria
Your answer of 153 bacteria is correct for this question.
B) How many new bacteria are introduced from the end of the fifth hour through the fifteenth hour?
To find the number of new bacteria introduced during a specific time interval, we again need to integrate the growth rate equation within that interval. We will integrate from t = 5 to t = 15 hours:
∫(4e^(0.5t)) dt = [8e^(0.5t)] evaluated from 5 to 15
= 8e^(0.5(15)) - 8e^(0.5(5))
= 8e^(7.5) - 8e^(2.5)
≈ 8(1801.83) - 8(12.18)
≈ 14414.64 - 97.44
≈ 14317.2 bacteria
Therefore, the number of new bacteria introduced from the end of the fifth hour through the fifteenth hour is approximately 14317 bacteria, not 23687.
C) At approximately how many hours after t=0 will the culture contain 200 new bacteria?
To find the time when the culture contains a certain number of new bacteria, we need to solve the growth rate equation for t. In this case, the equation is:
4e^(0.5t) = 200
Dividing both sides by 4, we get:
e^(0.5t) = 50
Taking the natural logarithm of both sides:
0.5t = ln(50)
Dividing by 0.5:
t = (2 / 0.5) * ln(50)
t ≈ 4 * 3.912
t ≈ 15.648
So, at approximately 15.648 hours after t=0, the culture will contain 200 new bacteria.
Please note that your answer for question B was incorrect. However, your answers for questions A and C are correct.