Answered

At the beginning of a study, a certain culture of bacteria has a population of 730. The population grows according to a continuous exponential growth model. After 6 days, there are 949 bacteria.

(a) Let [tex]$t$[/tex] be the time (in days) since the beginning of the study, and let [tex]$y$[/tex] be the number of bacteria at time [tex]$t$[/tex]. Write a formula relating [tex]$y$[/tex] to [tex]$t$[/tex]. Use exact expressions to fill in the missing parts of the formula. Do not use approximations.
[tex]\[ y = \square e^{\square t} \][/tex]

(b) How many bacteria are there 16 days after the beginning of the study? Do not round any intermediate computations, and round your answer to the nearest whole number.
[tex]\[ \square \text{ bacteria} \][/tex]



Answer :

GinnyAnswer
To solve this problem, we will use the continuous exponential growth model for the population of bacteria. Let's follow these steps:

### Part (a)

1. Model the initial conditions:
- Initial population at [tex]\( t = 0 \)[/tex] days: [tex]\( y_0 = 730 \)[/tex] bacteria.
- Population after [tex]\( t = 6 \)[/tex] days: [tex]\( y(6) = 949 \)[/tex] bacteria.

2. Form the exponential growth equation:
The general form of the continuous exponential growth model is given by:
[tex]\[ y = y_0 e^{kt} \][/tex]
where:
- [tex]\( y \)[/tex] is the number of bacteria at time [tex]\( t \)[/tex].
- [tex]\( y_0 \)[/tex] is the initial number of bacteria.
- [tex]\( k \)[/tex] is the growth rate constant.
- [tex]\( t \)[/tex] is the time in days.

3. Determine the growth rate constant [tex]\( k \)[/tex]:
We know:
[tex]\[ y(t) = 949 \quad \text{when} \quad t = 6 \][/tex]
Substitute these values into the exponential growth equation:
[tex]\[ 949 = 730 e^{6k} \][/tex]

4. Solve for [tex]\( k \)[/tex]:
[tex]\[ e^{6k} = \frac{949}{730} \][/tex]
Take the natural logarithm of both sides:
[tex]\[ 6k = \ln\left(\frac{949}{730}\right) \][/tex]
Solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{1}{6} \ln\left(\frac{949}{730}\right) \][/tex]

Thus, the formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is:
[tex]\[ y = 730 e^{\left(\frac{1}{6} \ln\left(\frac{949}{730}\right) t\right)} \][/tex]

### Part (b)

1. Find the population at [tex]\( t = 16 \)[/tex] days:
Using the formula derived in part (a):
[tex]\[ y = 730 e^{kt} \][/tex]
where [tex]\( k = \frac{1}{6} \ln\left(\frac{949}{730}\right) \)[/tex].

2. Substitute [tex]\( t = 16 \)[/tex] into the equation:
[tex]\[ y(16) = 730 e^{\left(\frac{1}{6} \ln\left(\frac{949}{730}\right) \cdot 16\right)} \][/tex]

3. Compute the value:
Evaluate the expression inside the exponent:
[tex]\[ \frac{1}{6} \ln\left(\frac{949}{730}\right) \cdot 16 \approx 0.04372737741124851 \cdot 16 \][/tex]

4. Determine the population at 16 days:
[tex]\[ y(16) = 730 e^{(0.04372737741124851 \cdot 16)} \][/tex]

Upon evaluating this:
[tex]\[ y(16) \approx 1469.507475164839 \][/tex]

After rounding to the nearest whole number, the population of bacteria 16 days after the beginning of the study is approximately:
[tex]\[ 1469 \text{ bacteria} \][/tex]

Therefore, the straight answer for both parts are:

### Part (a):
[tex]\[ y = 730 e^{\left(\frac{1}{6} \ln\left(\frac{949}{730}\right) t\right)} \][/tex]

### Part (b):
[tex]\[ 1469 \text{ bacteria} \][/tex]

Other Questions