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The exponential function \(Q(T)=600(1.35)^{T}\) represents the growth of a species of fish in a lake, where \(T\) is measured in 5-year intervals. a. Determine \(Q(1), Q(2),\) and \(Q(3)\). b. Find another function \(q(t),\) where \(t\) is measured in years. c. Determine \(q(5), q(10),\) and \(q(15)\). d. Compare your answers in parts (a) and (c) and describe your results.

Short Answer

Expert verified
Q(1) = 810, Q(2) = 1093.5, Q(3) = 1476.225. q(t) = 600(1.35)^(t/5). q(5) = 810, q(10) = 1093.5, q(15) = 1476.225.

Step by step solution

01

Determine Q(1), Q(2), and Q(3)

To find the number of fish after 5, 10, and 15 years (since T is in 5-year intervals), use the given function:For Q(1): \[ Q(1) = 600(1.35)^1 = 600 \times 1.35 = 810 \]For Q(2): \[ Q(2) = 600(1.35)^2 = 600 \times 1.8225 = 1093.5 \]For Q(3): \[ Q(3) = 600(1.35)^3 = 600 \times 2.460375 = 1476.225 \]
02

Find q(t) where t is in years

Since T is in 5-year intervals, we need to express the function in terms of years (t). Substitute \( t = 5T \): \[ Q(T) = 600(1.35)^T \]becomes \[ q(t) = 600 \times (1.35)^{\frac{t}{5}} \]
03

Determine q(5), q(10), and q(15)

Now, plug in the year values into the new function q(t):For q(5): \[ q(5) = 600 \times (1.35)^{\frac{5}{5}} = 600 \times 1.35 = 810 \]For q(10): \[ q(10) = 600 \times (1.35)^{\frac{10}{5}} = 600 \times (1.35)^2 = 600 \times 1.8225 = 1093.5 \]For q(15): \[ q(15) = 600 \times (1.35)^{\frac{15}{5}} = 600 \times (1.35)^3 = 600 \times 2.460375 = 1476.225 \]
04

Compare Q(T) and q(t)

Notice that the value of Q(T) for \(T=1\), \(T=2\), and \(T=3\) are the same as the values of q(t) for \(t=5\), \(t=10\), and \(t=15\), respectively. This is because \(q(t)\) is simply a rescaling of \(Q(T)\) to yearly intervals instead of 5-year intervals.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Annual Growth Rate
Understanding the annual growth rate in exponential functions is crucial for interpreting the progress of populations over time. The exponential function in this case, \(Q(T)=600(1.35)^{T}\), relies on the growth rate of 1.35, which implies a 35% increase every 5 years. To determine the equivalent annual growth rate, we need to use the concept of geometric transformation.

To find the annual growth rate from the 5-year rate of 1.35, use the following formula: \text{Annual Growth Rate} = (1.35)^{\frac{1}{5}}\. By calculating this, we get an approximate annual growth factor: \[ 1.35^{\frac{1}{5}} \approx 1.062 \] which indicates the population grows by about 6.2% annually. This transformation helps in understanding the growth on a more granular, yearly basis, and is critical for accurate projections and comparisons.

This exponential growth factor is useful for predicting future populations by applying it each year, facilitating better wildlife management and conservation efforts.
Rescaling Intervals
Rescaling intervals is a method used to convert functions measured over larger time spans to smaller intervals, providing a clearer picture of changes over shorter periods. In our exercise, the original function \(Q(T) = 600(1.35)^{T}\) is measured in 5-year intervals.

To convert this model into a yearly model, we substitute \(t = 5T\), where \(T\) represents intervals of 5 years and \(t\) signifies years. This change of variables transforms the original function into \[ q(t) = 600 \times (1.35)^{\frac{t}{5}} \] This means if you want the fish population in any given year, you can use this new function \(q(t)\).

For instance, to find out the population in 3 years, substitute \(t = 3\): \[ q(3) = 600 \times (1.35)^{\frac{3}{5}} \] and for 7 years, it would be \[ q(7) = 600 \times (1.35)^{\frac{7}{5}} \] Rescaling intervals is a powerful tool for transforming models to different time scales, making them more practical for diverse applications.
This technique is commonly used in various fields, such as finance, biology, and environmental science, to adapt models for fine-grained analysis.
Fish Population Modeling
Fish Population Modeling is an essential tool in ecology for studying and managing aquatic ecosystems. The function \(Q(T) = 600(1.35)^{T}\) helps scientists and conservationists track fish populations over time. This mathematical model is based on exponential growth, where the rate of increase is proportional to the current population, implying more fish lead to more offspring.

Properly modeling fish populations can help in:
  • Understanding growth patterns to set sustainable fishing limits.
  • Monitoring the effects of environmental changes or conservation actions.
  • Predicting future population sizes to prevent overfishing and ensure ecosystem balance.

The exercise involved determining populations over specific intervals: 5, 10, and 15 years, by using the function. For example, after 10 years: \[ Q(2) = 600 \times (1.35)^2 = 1093.5 \] Transforming this model to a yearly basis provides more precise insights. This rescaled function \(q(t) = 600 \times (1.35)^{\frac{t}{5}}\) makes it easier to verify population at any given year, showing similar trends as soon observed in 5-year intervals.

Such models are vital for decision-making processes, helping maintain ecological balance while utilizing resources efficiently. Educational exercises like these develop critical thinking and provide practical skills for managing real-world environmental challenges.

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