A New Farm Pond Was Stocked With 2500 Crappies In 2003
A new farm pond was stocked with 2500 crappies in 2003. The crappie po
A new farm pond was stocked with 2500 crappies in 2003. The crappie population in 2006 was found to be 4320. Let t be the number of years after 2003 (in other words t = 0 corresponds to 2003). Write down the initial crappie population at t = 3. Find the growth function of the form f(t) = y0 b^t that gives the crappie population t years after 2003. Predict the crappie population in 2010. In what year will the crappie population reach 13,000? Give the exact value for t and then use your calculator to approximate t to get the year. Are your answers for (c) and (d) consistent with the given population data in your answer to (b)?
Paper For Above instruction
The growth of biological populations such as crappies can be modeled mathematically using exponential functions, which describe how populations change over time under ideal conditions. The problem provided involves analyzing the growth pattern of crappies in a farm pond, with known initial conditions and a subsequent population measure, leading to the formulation of a growth function, prediction of future population sizes, and determining when the population reaches a specific number.
Introduction
Understanding the dynamics of fish populations in confined environments such as farm ponds is essential for effective management, conservation, and sustainable harvesting. Exponential growth models are particularly relevant in initial growth phases when resources are abundant. This analysis focuses on determining the initial population, deriving the growth function, predicting future population sizes, and identifying the time when the population reaches a specified threshold.
Initial Population and Growth Function
The initial population at t = 0 (which corresponds to the year 2003) is known to be 2500 crappies. The problem states that by 2006 (which corresponds to t = 3), the population increased to 4320 crappies. The exponential growth model is assumed to be of the form f(t) = y0 b^t, where y0 is the initial population, and b is the growth factor per year.
Given y0 = 2500 and f(3) = 4320, we substitute into the model to solve for b:
4320 = 2500 * b^3
b^3 = 4320 / 2500 = 1.728
b = (1.728)^{1/3} ≈ 1.201
Therefore, the growth function is:
f(t) = 2500 * (1.201)^t
Population Prediction in 2010
To predict the population in 2010, note that t = 7 (since 2003 + 7 = 2010). Using the growth function:
f(7) = 2500 * (1.201)^7
Calculating:
(1.201)^7 ≈ 3.829
f(7) ≈ 2500 * 3.829 ≈ 9,573 crappies
Thus, the predicted population in 2010 is approximately 9,573 crappies.
Time when Population Reaches 13,000
We need to find t when f(t) = 13,000:
13,000 = 2500 * (1.201)^t
(1.201)^t = 13,000 / 2,500 = 5.2
Taking natural logarithms on both sides:
t * ln(1.201) = ln(5.2)
t = ln(5.2) / ln(1.201)
Calculating:
ln(5.2) ≈ 1.6487
ln(1.201) ≈ 0.1823
t ≈ 1.6487 / 0.1823 ≈ 9.045
This corresponds to approximately 9.045 years after 2003, which is around mid-2012 (specifically, around June 2012).
Consistency Check with Data
The predictions for 2010 (around 9,573 crappies) and for when the population reaches 13,000 (around 2012) align reasonably well with the exponential growth model derived from the given data. The model assumes continuous, exponential increase without resource limitations, which could become less accurate at higher population levels due to environmental constraints; however, within this timeframe, the estimates are consistent with observed growth patterns.
Conclusion
By modeling the crappie population with an exponential growth function, we obtained a clear mathematical description of its increase over time. The model's predictions for future populations provide valuable insights for managing fish stocks and planning harvests. Moreover, the consistency with observed data supports the application of exponential models in biological population studies, at least within initial growth phases.
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