To solve the problem of fish population growth using a continuous exponential growth model, we'll follow these steps:
### Part (a)
Given:
- Initial population ([tex]\(y_0\)[/tex]) = 10 fish
- Population after 11 years ([tex]\(y(11)\)[/tex]) = 28 fish
- Time ([tex]\(t\)[/tex]) = 11 years
We know that the exponential growth formula is:
[tex]\[ y = y_0 \cdot e^{kt} \][/tex]
First, we need to determine the growth rate [tex]\(k\)[/tex], so we start by setting up the equation using the given information:
[tex]\[ 28 = 10 \cdot e^{11k} \][/tex]
To isolate [tex]\(e^{11k}\)[/tex], divide both sides by 10:
[tex]\[ \frac{28}{10} = e^{11k} \][/tex]
[tex]\[ 2.8 = e^{11k} \][/tex]
Next, take the natural logarithm of both sides to solve for [tex]\(k\)[/tex]:
[tex]\[ \ln(2.8) = 11k \][/tex]
[tex]\[ k = \frac{\ln(2.8)}{11} \][/tex]
Thus, the growth rate [tex]\(k\)[/tex] is:
[tex]\[ k = \frac{\ln(2.8)}{11} \][/tex]
Now plug this value back into the exponential growth formula:
[tex]\[ y = 10 \cdot e^{\left(\frac{\ln(2.8)}{11}\right)t} \][/tex]
So, the formula relating [tex]\(y\)[/tex] to [tex]\(t\)[/tex] is:
[tex]\[ y = 10 \cdot e^{\left(\frac{\ln(2.8)}{11}\right)t} \][/tex]
### Part (b)
To find the number of fish after 14 years, we use the formula we obtained in part (a) and set [tex]\(t = 14\)[/tex]:
[tex]\[ y(14) = 10 \cdot e^{\left(\frac{\ln(2.8)}{11}\right) \cdot 14} \][/tex]
Using the value for [tex]\(k\)[/tex] which is approximately 0.0936017651982871, we get:
[tex]\[ y(14) = 10 \cdot e^{0.0936017651982871 \cdot 14} \][/tex]
[tex]\[ y(14) = 10 \cdot e^{1.310424712776} \][/tex]
Which results approximately to:
[tex]\[ y(14) \approx 37 \][/tex]
So, the number of fish after 14 years is:
[tex]\[ 37 \text{ fish} \][/tex]
