The population of a slow growing bacterial culture can be represented by the function P(t)=t^2-0.7t+10, where t is measured in seconds.
a) Determine the populations of bacteria at 2.5 seconds.
b) Determine the average rate of change in the bacteria population over the interval from 0 to 2.5.
c) Estimate the instantaneous rate of change in the bacteria population at exactly 2.5 seconds. (use 0.1 centered interval)
a) Well, let's find out how many bacteria we have after 2.5 seconds. Just plug in t=2.5 into the function:
P(2.5) = (2.5)^2 - 0.7(2.5) + 10
Doing the math, we get P(2.5) = 6.25 - 1.75 + 10 = 14.5
So, the population of bacteria at 2.5 seconds is approximately 14.5.
b) To find the average rate of change, we need to calculate the change in population over the interval from 0 to 2.5 seconds, and then divide it by the length of the interval.
The change in population is given by P(2.5) - P(0), so:
Change = P(2.5) - P(0) = (2.5)^2 - 0.7(2.5) + 10 - (0)^2 - 0.7(0) + 10 = 14.5 - 0 + 10 = 24.5
The length of the interval is 2.5 seconds, so the average rate of change is:
Average rate of change = Change / Interval = 24.5 / 2.5 = 9.8
So, the average rate of change in the bacteria population over the interval from 0 to 2.5 seconds is approximately 9.8 bacteria per second.
c) To estimate the instantaneous rate of change at exactly 2.5 seconds, we can calculate the average rate of change over a small interval centered around 2.5 seconds. Let's use a 0.1-second interval.
The change in population over this interval is given by P(2.6) - P(2.4), so:
Change = P(2.6) - P(2.4) = (2.6)^2 - 0.7(2.6) + 10 - (2.4)^2 - 0.7(2.4) + 10
Doing the math, we get Change = 15.96 - 3.64 + 10 - 14.44 + 3.36 + 10 = 22.28
The length of the interval is 0.1 seconds, so the estimated instantaneous rate of change is:
Estimated instantaneous rate of change = Change / Interval = 22.28 / 0.1 = 222.8
So, the estimated instantaneous rate of change in the bacteria population at exactly 2.5 seconds, using a 0.1 centered interval, is approximately 222.8 bacteria per second. Don't worry, I won't give you any tickles!
a) To determine the population of bacteria at 2.5 seconds, plug in t = 2.5 into the function P(t):
P(2.5) = (2.5)^2 - 0.7(2.5) + 10
= 6.25 - 1.75 + 10
= 14.5
Therefore, the population of bacteria at 2.5 seconds is 14.5.
b) The average rate of change in the bacteria population over the interval from 0 to 2.5 can be found by calculating the slope of the line connecting the two points (0, P(0)) and (2.5, P(2.5)). In other words, we need to find the change in population divided by the change in time:
Average rate of change = (P(2.5) - P(0)) / (2.5 - 0)
Using the values we found before, we can calculate:
Average rate of change = (14.5 - P(0)) / 2.5
To determine P(0), plug in t = 0 into the function P(t):
P(0) = (0)^2 - 0.7(0) + 10
= 10
Therefore:
Average rate of change = (14.5 - 10) / 2.5
= 4.5 / 2.5
= 1.8
The average rate of change in the bacteria population over the interval from 0 to 2.5 seconds is 1.8.
c) To estimate the instantaneous rate of change in the bacteria population at exactly 2.5 seconds, we can use the concept of the derivative. The derivative of the function P(t) will give us the rate of change of the population with respect to time at each point.
We can find the derivative of P(t) with respect to t:
P'(t) = 2t - 0.7
To estimate the instantaneous rate of change at exactly 2.5 seconds, we need to find the value of P'(t) at t = 2.5.
P'(2.5) = 2(2.5) - 0.7
= 5 - 0.7
= 4.3
Therefore, the estimated instantaneous rate of change in the bacteria population at exactly 2.5 seconds is 4.3.
To determine the populations of bacteria at 2.5 seconds, we need to substitute t = 2.5 into the function P(t) = t^2 - 0.7t + 10.
a) P(2.5) = (2.5)^2 - 0.7(2.5) + 10
= 6.25 - 1.75 + 10
= 14.5
Therefore, the population of bacteria at 2.5 seconds is 14.5.
To determine the average rate of change in the bacteria population over the interval from 0 to 2.5, we need to calculate the change in population over this time interval and divide it by the duration of the interval.
b) Average rate of change = (P(2.5) - P(0)) / (2.5 - 0)
P(0) = (0)^2 - 0.7(0) + 10
= 10
Average rate of change = (14.5 - 10) / 2.5
= 4.5 / 2.5
= 1.8
Therefore, the average rate of change in the bacteria population over the interval from 0 to 2.5 is 1.8.
To estimate the instantaneous rate of change in the bacteria population at exactly 2.5 seconds using a 0.1 centered interval, we need to calculate the average rate of change over a very small time interval around 2.5 seconds.
c) Instantaneous rate of change ≈ (P(2.5 + 0.1) - P(2.5 - 0.1)) / (0.1 + 0.1)
P(2.6) = (2.6)^2 - 0.7(2.6) + 10
= 6.76 - 1.82 + 10
= 15.94
P(2.4) = (2.4)^2 - 0.7(2.4) + 10
= 5.76 - 1.68 + 10
= 14.08
Instantaneous rate of change ≈ (15.94 - 14.08) / 0.2
≈ 9.3
Therefore, the estimated instantaneous rate of change in the bacteria population at exactly 2.5 seconds is approximately 9.3.