Question

the function ( p(t) = \frac{2000t}{41 + 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) = 2.307 ) deer/month
(type an integer or decimal rounded to three decimal places as needed.)
( p(90) = 0.793 ) deer/month
(type an integer or decimal rounded to three decimal places as needed.)

b) ( p(9) = -0.877 ) deer/month²
(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:

First, we need to find the first derivative \( p'(t) \) of the function \( p(t)=\frac{2000t}{4t + 75} \) (assuming there was a typo and it's \( 4t \) instead of \( 41 \), as \( 41 \) might be incorrect for a logistic - like growth model; if it's indeed \( 41 \), the process is similar).

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

Then \( p'(t)=\frac{2000(4t + 75)-2000t\times4}{(4t + 75)^{2}}=\frac{8000t+150000 - 8000t}{(4t + 75)^{2}}=\frac{150000}{(4t + 75)^{2}} \)

To find the second derivative \( p''(t) \), we differentiate \( p'(t)=\frac{150000}{(4t + 75)^{2}}=150000(4t + 75)^{-2} \)

Using the chain rule: if \( y = a(u)^{-n} \), where \( u = 4t+75 \), \( a = 150000 \), \( n = 2 \), then \( y'=-2a(u)^{-3}\times u' \)

\( u' = 4 \), so \( p''(t)=150000\times(- 2)\times(4t + 75)^{-3}\times4=\frac{-1200000}{(4t + 75)^{3}} \)

Step 1: Calculate \( p''(45) \)

Substitute \( t = 45 \) into \( p''(t) \)
\( p''(45)=\frac{-1200000}{(4\times45 + 75)^{3}} \)
First, calculate the denominator: \( 4\times45+75=180 + 75=255 \)
Then \( 255^{3}=255\times255\times255 = 255\times65025=16581375 \)
\( p''(45)=\frac{-1200000}{16581375}\approx - 0.072 \) (rounded to three decimal places)

Step 2: Calculate \( p''(90) \)

Substitute \( t = 90 \) into \( p''(t) \)
\( p''(90)=\frac{-1200000}{(4\times90+75)^{3}} \)
Calculate the denominator: \( 4\times90 + 75=360+75 = 435 \)
\( 435^{3}=435\times435\times435=435\times189225 = 81312875 \)
\( p''(90)=\frac{-1200000}{81312875}\approx - 0.015 \) (rounded to three decimal places)

(If the original function was \( p(t)=\frac{2000t}{41 + 75}=\frac{2000t}{116} \) (a linear function), then \( p'(t)=\frac{2000}{116}\approx17.241 \), and \( p''(t) = 0 \), but the given \( p'(9) = 12.174 \) suggests that the denominator is \( 4t + 75 \) instead of \( 41+75 \))

Answer:

For \( p''(45) \approx - 0.072 \) deer/month² (if the function is \( p(t)=\frac{2000t}{4t + 75} \))

For \( p''(90)\approx - 0.015 \) deer/month² (if the function is \( p(t)=\frac{2000t}{4t + 75} \))

If the function was \( p(t)=\frac{2000t}{41 + 75}=\frac{2000t}{116} \), then \( p''(t) = 0 \) for all \( t \), so \( p''(45)=0 \) and \( p''(90)=0 \)