Given that the number of bacteria in a culture decreases according to a continuous exponential decay model, we can solve the problem step-by-step as follows:
### Part (a)
We need to write a formula relating the number of bacteria [tex]\( y \)[/tex] at time [tex]\( t \)[/tex] in minutes.
The model for exponential decay can be expressed by the formula:
[tex]\[ y = y_0 \cdot e^{kt} \][/tex]
where:
- [tex]\( y_0 \)[/tex] is the initial population
- [tex]\( k \)[/tex] is the decay constant
- [tex]\( t \)[/tex] is the time elapsed
Given:
- [tex]\( y_0 = 3100 \)[/tex] (initial population)
- [tex]\( y = 868 \)[/tex] (population after 5 minutes)
- [tex]\( t = 5 \)[/tex] minutes
First, we need to determine the decay constant [tex]\( k \)[/tex]. We use the data for the population at 5 minutes to find [tex]\( k \)[/tex]:
[tex]\[
868 = 3100 \cdot e^{5k}
\][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[
\frac{868}{3100} = e^{5k}
\][/tex]
[tex]\[
e^{5k} = \frac{868}{3100}
\][/tex]
[tex]\[
5k = \ln\left(\frac{868}{3100}\right)
\][/tex]
[tex]\[
k = \frac{1}{5} \ln\left(\frac{868}{3100}\right)
\][/tex]
So the general formula for the number of bacteria at any time [tex]\( t \)[/tex] is:
[tex]\[
y = 3100 \cdot e^{\left(\frac{1}{5} \ln\left(\frac{868}{3100}\right) \cdot t \right)}
\][/tex]
### Part (b)
We need to find the number of bacteria 17 minutes after the beginning of the study.
Using the formula we derived:
[tex]\[
y = 3100 \cdot e^{\left(\frac{1}{5} \ln\left(\frac{868}{3100}\right) \cdot 17 \right)}
\][/tex]
To simplify the expression, remember that we already determined the decay constant [tex]\( k \)[/tex]:
[tex]\[
k = \frac{1}{5} \ln\left(\frac{868}{3100}\right) \approx -0.2545931351625775
\][/tex]
Then substituting [tex]\( k \)[/tex] and [tex]\( t = 17 \)[/tex] into the equation:
[tex]\[
y = 3100 \cdot e^{(-0.2545931351625775 \cdot 17)}
\][/tex]
[tex]\[
y \approx 41
\][/tex]
Therefore, the number of bacteria 17 minutes after the beginning of the study is approximately 41 bacteria.
