Answered

An initial population of 610 fish is introduced into a lake. This fish population grows according to a continuous exponential growth model. There are 915 fish in the lake after 14 years.

(a) Let [tex] t [/tex] be the time (in years) since the initial population is introduced, and let [tex] y [/tex] be the number of fish 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]\[ y = 610e^{kt} \][/tex]
[tex]\[ 915 = 610e^{14k} \][/tex]

Solve for [tex] k [/tex] and then write the complete formula for [tex] y [/tex]:
[tex]\[ k = \frac{\ln(\frac{915}{610})}{14} \][/tex]
[tex]\[ y = 610e^{\left(\frac{\ln(\frac{915}{610})}{14}\right)t} \][/tex]

(b) How many fish are there 17 years after the initial population is introduced? Do not round any intermediate computations, and round your answer to the nearest whole number.
[tex]\[ y = 610e^{\left(\frac{\ln(\frac{915}{610})}{14}\right)17} \][/tex]

Calculate the exact number of fish:
[tex]\[ y \approx \square \text{ fish} \][/tex]



Answer :

GinnyAnswer
To address the problem step-by-step, follow along as we derive the solution.

Part (a): Formula Relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex]

We know that the population follows a continuous exponential growth model. This model can be described 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 growth rate,
- [tex]\( t \)[/tex] is the time in years since the initial population is introduced,
- [tex]\( y \)[/tex] is the population at time [tex]\( t \)[/tex].

Given:
- Initial population, [tex]\( y_0 = 610 \)[/tex] fish,
- Population after 14 years, [tex]\( y_{14} = 915 \)[/tex] fish,
- Time, [tex]\( t = 14 \)[/tex] years.

We use these data points to determine the growth rate [tex]\( k \)[/tex].

Using the population after 14 years, we can set up the equation:
[tex]\[ y_{14} = y_0 \cdot e^{k \cdot 14} \][/tex]
Substitute the given values:
[tex]\[ 915 = 610 \cdot e^{14k} \][/tex]
Solve for [tex]\( k \)[/tex] as follows:
[tex]\[ e^{14k} = \frac{915}{610} \][/tex]
[tex]\[ 14k = \ln\left(\frac{915}{610}\right) \][/tex]
[tex]\[ k = \frac{1}{14} \cdot \ln\left(\frac{915}{610}\right) \][/tex]

So, the formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is:
[tex]\[ y = 610 \cdot e^{\left(\frac{1}{14} \ln\left(\frac{915}{610}\right)\right) t} \][/tex]

Part (b): Number of Fish After 17 Years

We use the derived formula to calculate the number of fish at [tex]\( t = 17 \)[/tex] years:
[tex]\[ y = 610 \cdot e^{kt} \][/tex]
Substitute [tex]\( t = 17 \)[/tex] and the previously computed [tex]\( k \)[/tex]:
[tex]\[ y = 610 \cdot e^{\left(\frac{1}{14} \ln\left(\frac{915}{610}\right)\right) \cdot 17} \][/tex]

Evaluate the expression:
[tex]\[ y = 610 \cdot e^{0.028961793436297456 \cdot 17} \][/tex]

This gives:
[tex]\[ y \approx 998 \][/tex]

Therefore, the number of fish after 17 years is approximately 998.

Answers:
(a) The formula is:
[tex]\[ y = 610 \cdot e^{\left(\frac{1}{14} \ln \left(\frac{915}{610}\right) \right) t} \][/tex]

(b) The number of fish after 17 years is:
[tex]\[ \boxed{998} \][/tex]

Other Questions