Question

1. a population of raccoons moves into a wooded area. at t months, the number of raccoons, p(t), can be modelled using the equation p(t)=100 + 30t+4t². a) determine the population of raccoons at 2.5 months. b) determine the average rate of change in the raccoon population over the interval from 0 months to 2.5 months. c) estimate the rate of change in the raccoon population at exactly 2.5 months. d) explain why your answers for parts a), b), and c) are different.

Explanation:

Step1: Find population at 2.5 months

Substitute \(t = 2.5\) into \(P(t)=100 + 30t+4t^{2}\).
\[

$$\begin{align*} P(2.5)&=100+30\times2.5 + 4\times(2.5)^{2}\\ &=100 + 75+4\times6.25\\ &=100+75 + 25\\ &=200 \end{align*}$$

\]

Step2: Calculate average rate of change from 0 to 2.5 months

The average rate of change formula is \(\frac{P(2.5)-P(0)}{2.5 - 0}\). First, find \(P(0)=100+30\times0 + 4\times0^{2}=100\). Then \(\frac{P(2.5)-P(0)}{2.5 - 0}=\frac{200 - 100}{2.5}=\frac{100}{2.5}=40\).

Step3: Find derivative of \(P(t)\)

Differentiate \(P(t)=100 + 30t+4t^{2}\) with respect to \(t\). Using the power - rule \((x^n)^\prime=nx^{n - 1}\), \(P^\prime(t)=30 + 8t\).

Step4: Evaluate derivative at \(t = 2.5\)

Substitute \(t = 2.5\) into \(P^\prime(t)\). \(P^\prime(2.5)=30+8\times2.5=30 + 20=50\).

Step5: Explain differences

Part a) gives the population at a point. Part b) gives the average rate of change over an interval, which is a measure of the overall change in population divided by the change in time. Part c) gives the instantaneous rate of change at a specific time. The average rate of change over an interval may not equal the instantaneous rate of change at the end - point of the interval because the population is changing non - linearly.

Answer:

a) 200
b) 40
c) 50
d) Part a) is population value, part b) is average rate of change over an interval and part c) is instantaneous rate of change at a point. The non - linear nature of population growth causes the differences.