009A Sample Midterm 1, Problem 2 Detailed Solution

Suppose the size of a population at time  $t$  is given by

$N(t)={\frac {1000t}{5+t}},~t\geq 0.$

(a) Determine the size of the population as  $t\rightarrow \infty .$  We call this the limiting population size.

(b) Show that at time  $t=5,$  the size of the population is half its limiting size.

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Background Information:
Recall that
$\lim _{x\rightarrow \infty }{\frac {a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{0}}{b_{n}x^{n}+b_{n-1}x^{n-1}+\cdots +b_{0}}}={\frac {a_{n}}{b_{n}}}$
provided  $a_{n}\neq 0$  and  $b_{n}\neq 0.$

Solution:

(a)

Step 1:
We have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{t\rightarrow \infty }N(t)=\lim _{t\rightarrow \infty }{\frac {1000t}{5+t}}.}

Step 2:
Using the Background Information, we have
${\begin{array}{rcl}\displaystyle {\lim _{t\rightarrow \infty }N(t)}&=&\displaystyle {\frac {1000}{1}}\\&&\\&=&\displaystyle {1000.}\end{array}}$

(b)
We have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {N(5)}&=&\displaystyle {\frac {1000(5)}{5+5}}\\&&\\&=&\displaystyle {\frac {1000(5)}{10}}\\&&\\&=&\displaystyle {100(5)}\\&&\\&=&\displaystyle {500}\\&&\\&=&\displaystyle {{\frac {1000}{2}}.}\end{array}}}

Final Answer:
(a)     $1000$
(b)     Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle N(5)=500}

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