Question

how do we protect endangered species? a rare species of insect was discovered in the amazon rainforest. to protect the species, entomologists (scientists who study insects) declared the insect endangered and transferred a population of 30 insects to a protected area. they now want to study this population over time. the projected population p of the new colony t days after the transfer is given by p(t)=\frac{60 + 25t}{2+0.01t}. 1. what is the projected population of the colony after 1 year? after 2 years? p(1)=(365 = 1625.7 p(2)=f(730)=1969.8 2. now let’s see what happens after many, many years have passed. a. fill in the table for large values of t. t 10,000 100,000 500,000 1,000,000 p(t) 2451.59 2495.07 2499.01 2499.05 b. after many, many years, what do you predict will happen to the insect population? we are getting closer to 2,500 insects 3. sketch a graph of the population p as a function of time. 4. explain the meaning of any asymptotes in the context of this problem. the population will never go below 5. as t continues to increase, which grows faster: the numerator of p(t) or the denominator? how do you know? the numerator 6. how can you tell just by looking at the equation what the population of insects will do in the long run?

Explanation:

Step1: Calculate population after 1 year

1 year has 365 days. Substitute \(t = 365\) into \(P(t)=\frac{60 + 25t}{2+0.01t}\).
\[P(365)=\frac{60+25\times365}{2 + 0.01\times365}=\frac{60 + 9125}{2+3.65}=\frac{9185}{5.65}\approx1625.7\]

Step2: Calculate population after 2 years

2 years have \(2\times365 = 730\) days. Substitute \(t = 730\) into \(P(t)=\frac{60 + 25t}{2+0.01t}\).
\[P(730)=\frac{60+25\times730}{2+0.01\times730}=\frac{60 + 18250}{2+7.3}=\frac{18310}{9.3}\approx1968.8\]

Step3: Analyze long - term behavior

As \(t\to\infty\), we use the fact that for a rational function \(y=\frac{ax + b}{cx + d}\) (\(a,b,c,d\) are constants, \(c
eq0\)), \(\lim_{t\to\infty}\frac{60 + 25t}{2+0.01t}=\lim_{t\to\infty}\frac{t( \frac{60}{t}+25)}{t(\frac{2}{t}+0.01)}\).
As \(t\to\infty\), \(\frac{60}{t}\to0\) and \(\frac{2}{t}\to0\), so \(\lim_{t\to\infty}P(t)=\frac{25}{0.01}=2500\).

Step4: Determine growth rate

The numerator of \(P(t)\) is \(N(t)=60 + 25t\) (a linear function with slope \(m_N = 25\)) and the denominator is \(D(t)=2+0.01t\) (a linear function with slope \(m_D=0.01\)). Since \(25>0.01\), the numerator grows faster as \(t\) increases.

Step5: Long - run population analysis

For the rational function \(P(t)=\frac{60 + 25t}{2+0.01t}\), as \(t\) gets very large, the terms with the highest - power of \(t\) dominate. The function behaves like \(y=\frac{25t}{0.01t}=2500\). So in the long run, the population of insects approaches 2500.

Answer:

1. After 1 year, the projected population is approximately \(1625.7\). After 2 years, the projected population is approximately \(1968.8\).
2. a. For \(t = 10000\), \(P(10000)=\frac{60+25\times10000}{2+0.01\times10000}=\frac{60 + 250000}{2+100}=\frac{250060}{102}\approx2451.6\). For \(t = 100000\), \(P(100000)=\frac{60+25\times100000}{2+0.01\times100000}=\frac{60+2500000}{2 + 1000}=\frac{2500060}{1002}\approx2495.1\). For \(t = 500000\), \(P(500000)=\frac{60+25\times500000}{2+0.01\times500000}=\frac{60 + 12500000}{2+5000}=\frac{12500060}{5002}\approx2499.0\). For \(t = 1000000\), \(P(1000000)=\frac{60+25\times1000000}{2+0.01\times1000000}=\frac{60+25000000}{2 + 10000}=\frac{25000060}{10002}\approx2499.5\).

b. The insect population approaches 2500.

1. The numerator of \(P(t)\) grows faster because the coefficient of \(t\) in the numerator (\(25\)) is greater than the coefficient of \(t\) in the denominator (\(0.01\)).
2. In the long run, the population of insects approaches 2500 because for the rational function \(P(t)=\frac{60 + 25t}{2+0.01t}\), as \(t\) gets very large, the function is dominated by the ratio of the leading - terms of the numerator and denominator, \(\frac{25t}{0.01t}=2500\).