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?
B) How many new bacteria are introduced from the end of the fifth hour through the fifteenth hour?
C) At approximately how many hours after t=0 will the culture contain 200 new bacteria?
(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)
(B) ∫[8,15] 4e^(t/2) dt
(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) Well, if I try to imagine bacteria growing, all I see is a bunch of tiny Elvis Presleys popping out. Anyway, to calculate the number of new bacteria after six hours, we can use the formula N(t) = 4e^(0.5t), where t is the number of hours. So, plugging in t = 6, we get N(6) = 4e^(0.5*6). I'll do the math for you. Give me a sec... Ah, the magic of mathematics! After six hours, you should have approximately 480 new bacteria in your culture.
B) Now let's calculate the number of new bacteria from the end of the fifth hour through the fifteenth hour, shall we? We'll need to find the difference between N(15) and N(5). So, let's summon our mathematical powers once again. N(15) - N(5) equals 4e^(0.5*15) - 4e^(0.5*5). Drumroll, please... And the answer is approximately 1570 new bacteria that popped into existence during that time frame.
C) Lastly, we have to determine approximately when the culture will contain 200 new bacteria. To solve this, we need to find the value of t when N(t) = 200. So, let's put on our detective hats and solve the equation 4e^(0.5t) = 200. Let's see...I'm doing some calculations here, and the approximate answer is t ≈ 5.36 hours. So, after about 5.36 hours, your culture will have 200 new bacteria. Just make sure they don't throw a wild party while you're not looking!
To find the number of new bacteria in the culture at a given time, we can integrate the given growth rate function with respect to time. Let's go step by step to solve each question:
A) To find the number of new bacteria after the first six hours, we need to integrate the growth rate function from 0 to 6.
∫4e^(0.5t) dt
Applying the power rule of integration, we have:
= 8e^(0.5t) evaluated from 0 to 6
= 8e^(0.5 * 6) - 8e^(0.5 * 0)
= 8e^3 - 8e^0
≈ 8e^3 - 8
Therefore, there will be approximately 484.95 new bacteria in the culture after the first six hours.
B) To find the number of new bacteria introduced between the end of the fifth hour and the fifteenth hour, we need to integrate the growth rate function from 5 to 15.
∫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)
≈ 2990.87 - 166.16
≈ 2824.71
Therefore, approximately 2,824.71 new bacteria are introduced from the end of the fifth hour through the fifteenth hour.
C) To find approximately how many hours after t=0 the culture contains 200 new bacteria, we need to set up an equation and solve for t.
4e^(0.5t) = 200
Divide both sides by 4:
e^(0.5t) = 50
Take the natural logarithm of both sides:
ln(e^(0.5t)) = ln(50)
Using logarithm properties:
0.5t * ln(e) = ln(50)
Since ln(e) = 1:
0.5t = ln(50)
Solve for t:
t ≈ (ln(50))/(0.5)
t ≈ 4.61 hours
Therefore, at approximately 4.61 hours after t=0, the culture will contain 200 new bacteria.
To answer these questions, we can use the given growth rate equation: 4e^0.5t per hour.
A) To find the number of new bacteria after the first six hours, we need to calculate the integral of the growth rate equation over the interval from 0 to 6.
The integral of 4e^0.5t with respect to t from 0 to 6 is:
∫[0 to 6] 4e^0.5t dt
To solve this integral, we can use the formula for the integral of e^kt:
∫ e^kt dt = (1/k)e^kt + C
Applying this formula with k = 0.5, we can find the integral:
= (4/0.5)e^0.5t |[0 to 6]
= 8e^0.5(6) - 8e^0
= 8e^3 - 8
So, there will be approximately 8e^3 - 8 new bacteria in the culture after the first six hours.
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 integral of the growth rate equation over the interval from 5 to 15.
The integral of 4e^0.5t with respect to t from 5 to 15 is:
∫[5 to 15] 4e^0.5t dt
Using the same formula for the integral of e^kt, we have:
= (4/0.5)e^0.5t |[5 to 15]
= 8e^0.5(15) - 8e^0.5(5)
= 8e^(15/2) - 8e^(5/2)
So, there will be approximately 8e^(15/2) - 8e^(5/2) new bacteria introduced from the end of the fifth hour through the fifteenth hour.
C) To find the number of hours after t=0 when the culture contains 200 new bacteria, we need to solve the equation:
4e^0.5t = 200
Dividing both sides by 4:
e^0.5t = 50
Taking the natural logarithm of both sides:
0.5t = ln(50)
Solving for t:
t ≈ 2ln(50)/0.5
Calculating this value:
t ≈ 2 * 3.912 / 0.5 ≈ 15.648
So, at approximately t ≈ 15.648 hours, the culture will contain 200 new bacteria.
dB/dt = 4 e^ (t/2)
B = 8 e^(t/2) + C
at t = 0, say there are -8 so B is only the new ones
B = 8 e^(t/2)
A) B = 8 e^3 = 160
B) B = 8e^7.5 - 8 e^2.5 = 8 (1808 - 12) = 14,368
C) 200 = 8 e^t/2
e^t/2 = 25
ln e^t/2 = t/2 = ln 25
so t = 2 ln 25 = 6.43 hours