Question

an ecologist began studying a certain type of plant species in a wetlands area in 2013. in 2015 (t = 2), there were 59 plants. in 2021 (t = 8), there were 118 plants.
the number of plants of this species can be modeled by the function ( p(t) = ab^t ), where ( p(t) ) is the number of plants during year ( t ), and ( t ) is the number of years since 2013.

part b
(i) use the given data to find the average rate of change of the number of plants, in plants per year, from ( t = 2 ) to ( t = 8 ) years. express your answer as a decimal approximation. show the computations that lead to your answer.
(ii) use the average rate of change found in part b (i) to estimate the number of plants for ( t = 10 ) years. show the work that leads to your answer.
(iii) the average rate of change found in part b (i) can be used to estimate the number of plants during year ( t ), for ( t > 10 ) years. are these estimates, found using the average rate of change, less than or greater than the number of plants predicted by the model ( p ) during year ( t ), for ( t > 10 ) years? explain your reasoning. your explanation should include a reference to the graph of ( p ) and its relationship to the estimates found using the average rate of change.

Explanation:

Step1: Find the average rate of change from \( t = 2 \) to \( t = 8 \)

The formula for the average rate of change of a function \( P(t) \) from \( t = a \) to \( t = b \) is \( \frac{P(b) - P(a)}{b - a} \). We know \( P(2) = 59 \) (2015, \( t = 2 \)) and \( P(8) = 118 \) (2021, \( t = 8 \)). So the average rate of change is \( \frac{118 - 59}{8 - 2} \).

$$ \frac{118 - 59}{8 - 2}=\frac{59}{6}\approx9.83 $$

Step2: Estimate \( P(10) \) using the average rate of change

First, we know the average rate of change from \( t = 2 \) to \( t = 8 \) is approximately \( 9.83 \). The time from \( t = 8 \) to \( t = 10 \) is \( 10 - 8 = 2 \) years. We can use the formula \( P(10)=P(8)+ \text{average rate of change} \times (10 - 8) \). Substituting the values, \( P(10)=118+9.83\times2 \).

$$ P(10)=118 + 19.66 = 137.66 $$

Step3: Analyze the average rate of change for \( t>10 \)

The function \( P(t)=ab^t \) is an exponential function. For an exponential function \( y = ab^t \) with \( b>1 \) (since the number of plants is increasing), the rate of change (derivative or average rate of change over intervals) is increasing as \( t \) increases. The average rate of change we calculated was for \( t = 2 \) to \( t = 8 \). As \( t \) increases beyond 10, the exponential function will have a steeper slope (since exponential growth has an increasing rate of change), so the average rate of change for \( t>10 \) will be greater than the average rate of change we found (from \( t = 2 \) to \( t = 8 \)). This means the estimates using the average rate of change (which is a constant approximation for the exponential growth) will be less than the actual number of plants predicted by the exponential model \( P(t) \), because the actual rate of change is increasing, so the exponential function will grow faster than a linear approximation (using the average rate of change) for \( t>10 \).

Answer:

(i) The average rate of change is \(\frac{59}{6}\approx9.83\) plants per year. (ii) \(P(10)\approx137.66\) plants. (iii) The estimates using the average rate of change will be less than the number of plants predicted by \(P(t)\) for \(t > 10\) because \(P(t)\) is an exponential growth function with an increasing rate of change, while the average rate of change is a constant approximation, so the exponential function grows faster than the linear - like approximation for \(t>10\).