Question

the function ( p(t) = \frac{2000t}{4t + 75} ) gives the population ( p ) of deer in an area after ( t ) months.

a) find ( p(9) ), ( p(45) ), and ( p(90) )
b) find ( p(9) ), ( p(45) ), and ( p(90) )
c) interpret the meaning of your answers to part (a) and (b) what is happening to the population of deer in the long term?

a) ( p(9) = 12.174 ) deer/month
(type an integer or decimal rounded to three decimal places as needed )

( p(45) = square )
(type an integer or decimal rounded to three decimal places as needed )

Explanation:

Step1: Find the derivative of \( p(t) \)

We use the quotient rule: if \( p(t)=\frac{u(t)}{v(t)} \), then \( p'(t)=\frac{u'(t)v(t)-u(t)v'(t)}{v(t)^2} \). Here, \( u(t) = 2000t \), so \( u'(t)=2000 \); \( v(t)=4t + 75 \), so \( v'(t)=4 \).
\[

$$\begin{align*} p'(t)&=\frac{2000(4t + 75)-2000t\times4}{(4t + 75)^2}\\ &=\frac{8000t+150000 - 8000t}{(4t + 75)^2}\\ &=\frac{150000}{(4t + 75)^2} \end{align*}$$

\]

Step2: Calculate \( p'(45) \)

Substitute \( t = 45 \) into \( p'(t) \):
\[

$$\begin{align*} p'(45)&=\frac{150000}{(4\times45 + 75)^2}\\ &=\frac{150000}{(180 + 75)^2}\\ &=\frac{150000}{255^2}\\ &=\frac{150000}{65025}\\ &\approx2.307 \end{align*}$$

\]

Answer:

\( p'(45)\approx2.307 \)