the small intestine bacteriaa, while inhabiting areas optimal for growth have a doubling time of roughly 10 hours. A normal small intestine starting population would be approximately 10,000 bacteria per ml of fluid
a) write an equation to model this exponential growth, with b(x) representing the number of bacteria per ml and x representing the time in hours
b) how long will it take for there to be 100,000 bacteria per ml (round answer to two decimal places in necessary)
c) determine the average rate of change between 20 hours and 30 hours
d) estimate the instantaneous rate of change at 24 hours
for b), i did it like
100000 = 10000(2)^(x/10)
10 = 2^(x/10)
log 10 = x/10 log 2
log 10/log 2 = x/10
3.3219 = x/10
33.22 = x
would this still be correct?
b(x) = 10000(2)^(x/10)
b) when b(x) = 100000
10000(2)^(x/10) = 100000
2^(x/10) = 10
take log of both sides and use log rules
(x/10)log2 = log 10
x/2 = 1/log2
x = 2/log2 = ..... , don't have my calculator handy
c) find b(30) and b(20)
avg rate of change = (b(30) - b(20))/(30-20) = ....
d) do something like
rate = (b(24.01) - b(24))/(24.01-24)
use your calculator
for b) yes, you are correct.
I have a typo, I did x/2 instead of x/10
Can't explain why I did that, just a typo I guess.
a) The equation to model exponential growth would be: b(x) = 10,000 * 2^(x/10)
b) To calculate the time it takes for there to be 100,000 bacteria per ml, we can set b(x) = 100,000 and solve for x:
10,000 * 2^(x/10) = 100,000
2^(x/10) = 10
x/10 = log2(10)
x = 10 * log2(10)
x ≈ 33.22 hours
c) The average rate of change between 20 hours and 30 hours can be calculated by finding the difference in the number of bacteria over the time period:
Average rate of change = (b(30) - b(20)) / (30 - 20)
Average rate of change = (10,000 * 2^(30/10) - 10,000 * 2^(20/10)) / 10
Average rate of change ≈ (10,000 * 2^3 - 10,000 * 2^2) / 10
Average rate of change ≈ (10,000 * 8 - 10,000 * 4) / 10
Average rate of change ≈ (80,000 - 40,000) / 10
Average rate of change ≈ 40,000 / 10
Average rate of change = 4,000 bacteria per hour
d) To estimate the instantaneous rate of change at 24 hours, we can calculate the slope of the tangent line at that point. Since the doubling time is 10 hours, we can use the derivative to estimate the instantaneous rate of change:
b'(x) = 10,000 * ln(2) * 2^(x/10)
The instantaneous rate of change at 24 hours can be estimated by evaluating the derivative at x = 24:
b'(24) = 10,000 * ln(2) * 2^(24/10)
b'(24) ≈ 10,000 * ln(2) * 2^2.4
b'(24) ≈ 10,000 * 0.693147 * 5.2780316
b'(24) ≈ 3,520,482.31 bacteria per hour
a) The equation to model exponential growth can be written as: b(x) = b0 * 2^(x / t), where b(x) represents the number of bacteria per ml at time x, b0 represents the initial population, x represents the time in hours, and t represents the doubling time.
In this case, the initial population (b0) is 10,000 bacteria per ml, and the doubling time (t) is 10 hours. So the equation becomes:
b(x) = 10,000 * 2^(x / 10)
b) To find out how long it will take for there to be 100,000 bacteria per ml, we need to solve the equation b(x) = 100,000 for x. Let's substitute the values into the equation:
100,000 = 10,000 * 2^(x / 10)
To isolate the exponent, divide both sides by 10,000:
10 = 2^(x / 10)
To get rid of the exponent, take the logarithm of both sides (base 2):
log2(10) = x / 10
Multiply both sides by 10 to solve for x:
x = 10 * log2(10)
Using a calculator, the approximate value of x is 33.22 hours. So it will take approximately 33.22 hours for there to be 100,000 bacteria per ml.
c) The average rate of change between 20 hours and 30 hours can be found by calculating the difference in the number of bacteria per ml at these two times and dividing it by the difference in time.
Let's calculate the average rate of change:
Average rate of change = (b(30) - b(20)) / (30 - 20)
Substituting the values into the equation:
Average rate of change = (10,000 * 2^(30 / 10) - 10,000 * 2^(20 / 10)) / (30 - 20)
Simplifying the equation:
Average rate of change = (10,000 * 2^3 - 10,000 * 2^2) / 10
Using exponent rules, 2^3 equals 8 and 2^2 equals 4:
Average rate of change = (10,000 * 8 - 10,000 * 4) / 10
The calculation gives us an average rate of change of 4,000 bacteria per ml per hour.
d) To estimate the instantaneous rate of change at 24 hours, we can approximate it by finding the average rate of change between 23 hours and 25 hours.
Using the same formula as in part c:
Instantaneous rate of change = (b(25) - b(23)) / (25 - 23)
Substituting the values into the equation:
Instantaneous rate of change = (10,000 * 2^(25 / 10) - 10,000 * 2^(23 / 10)) / (25 - 23)
Simplifying the equation:
Instantaneous rate of change = (10,000 * 2^2.5 - 10,000 * 2^2.3) / 2
Using a calculator, the approximate value of the instantaneous rate of change at 24 hours is 941.37 bacteria per ml per hour.