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Solve Logarithmic Population Model and Average Rate of Change for Rainbow Trout
Mathematics
Grade 11 of Senior High School
Question Content
Starting in 2012, a biologist studied the population of rainbow trout in a particular lake over several years. In 2013 (t=1), there were 390 rainbow trout in the lake. In 2023 (t=10), there were 820 rainbow trout in the lake. The number of rainbow trout in the lake can be modeled by the function L given by L(t) = a + b ln(t), where L(t) is the number of rainbow trout in the lake during year t, and t is the number of years since 2012. (A) (i) Use the given data to write two equations that can be used to find the values for constants a and b in the expression for L(t). (2pts) (ii) Find the values of a and b. (2pts) (B) (i) Use the given data to find the average rate of change of the number of rainbow trout in the lake, in fish per year, from t=1 to t=10. Express your answer as a decimal approximation. Show the computations that lead to your answer. (2pts) (ii) Interpret the meaning of your answer from (i) in the context of the problem. (2pts)
Correct Answer
(A)(i) $390 = a + b\\ln(1)$; $820 = a + b\\ln(10)$ (A)(ii) $a=390$, $b\\approx192.67$ (B)(i) $\\approx47.78$ fish per year (B)(ii) The average number of rainbow trout in the lake increased by approximately 47.78 fish each year between 2013 (t=1) and 2023 (t=10)
Detailed Solution Steps
1
(A)(i): Substitute the given t and L(t) values into the function L(t)=a + b ln(t). When t=1, L(1)=390, so $390 = a + b\\ln(1)$. When t=10, L(10)=820, so $820 = a + b\\ln(10)$.
2
(A)(ii): First, solve $390 = a + b\\ln(1)$. Since $\\ln(1)=0$, this simplifies to $390=a+0$, so $a=390$. Substitute a=390 into the second equation: $820=390 + b\\ln(10)$. Rearrange to solve for b: $b=\\frac{820-390}{\\ln(10)}=\\frac{430}{\\ln(10)}$. Using $\\ln(10)\\approx2.3026$, calculate $b\\approx\\frac{430}{2.3026}\\approx192.67$.
3
(B)(i): Use the average rate of change formula $\\frac{L(t_2)-L(t_1)}{t_2-t_1}$. Here, $t_1=1$, $L(t_1)=390$, $t_2=10$, $L(t_2)=820$. Substitute values: $\\frac{820-390}{10-1}=\\frac{430}{9}\\approx47.78$.
4
(B)(ii): Translate the numerical average rate of change into the context of the problem: it represents the average annual increase in the rainbow trout population over the 9-year period.
Knowledge Points Involved
1
Logarithmic Function Modeling
This involves using a function with a logarithmic term (here $L(t)=a + b\\ln(t)$) to represent real-world population growth or change. It is applied when a quantity changes at a rate that slows over time, fitting the behavior of logarithmic functions. We substitute real data points into the function to create equations for unknown constants.
2
System of Linear Equations with Logarithms
When given a function with two unknown constants, we use paired input-output data to form two equations, creating a system. For logarithmic models, one of the equations may simplify using properties of logarithms (like $\\ln(1)=0$) to solve for constants step-by-step.
3
Average Rate of Change
The average rate of change of a function $f(x)$ from $x_1$ to $x_2$ is calculated by $\\frac{f(x_2)-f(x_1)}{x_2-x_1}$. It measures the average amount the output changes per unit of input over an interval, used to describe average growth or decline in real-world contexts like population change.
4
Logarithm Evaluation
Natural logarithms ($\\ln(x)$) are logarithms with base $e$ (approximately 2.71828). Key values include $\\ln(1)=0$ and $\\ln(10)\\approx2.3026$, which are used to simplify and solve equations involving logarithmic terms in modeling problems.
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