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: Calculate 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

The average rate of change of a function \(y = f(x)\) over \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\). Here, \(a = 0\), \(b = 2.5\), \(P(0)=100+30\times0 + 4\times0^{2}=100\), \(P(2.5) = 200\).
The average rate of change is \(\frac{P(2.5)-P(0)}{2.5-0}=\frac{200 - 100}{2.5}=\frac{100}{2.5}=40\).

Step3: Estimate instantaneous rate of change

First, find the derivative of \(P(t)\) using the power - rule. If \(P(t)=100 + 30t+4t^{2}\), then \(P^\prime(t)=30 + 8t\).
Substitute \(t = 2.5\) into \(P^\prime(t)\): \(P^\prime(2.5)=30+8\times2.5=30 + 20=50\).

Step4: Explain differences

Part a) gives the population number at a specific time. Part b) gives the average rate of change over an interval, which is a measure of how the population changes on average over that time span. Part c) gives the instantaneous rate of change at a single point in time. The average rate of change over an interval is a kind of 'overall' measure for that interval, while the instantaneous rate of change is the rate at a precise moment.

Answer:

a) 200
b) 40
c) 50
d) Part a) gives population value, part b) gives average rate over an interval and part c) gives instantaneous rate at a point.