Questions: Section 1.8 Higher-Order Derivatives Question 22, 1.8.49-LS Part 1 of 8 The function p(t)=2000t/(4t+75) gives the population p of deer in an area after t months. a) Find p'(10), p'(50), and p'(100). b) Find p''(10), p''(50), and p''(100). c) Interpret the meaning of your answers to part (a) and (b). What is happening to the pop

Section 1.8 Higher-Order Derivatives
Question 22, 1.8.49-LS
Part 1 of 8

The function p(t)=2000t/(4t+75) gives the population p of deer in an area after t months.
a) Find p'(10), p'(50), and p'(100).
b) Find p''(10), p''(50), and p''(100).
c) Interpret the meaning of your answers to part (a) and (b). What is happening to the pop

Transcript text: Section 1.8 Higher-Order Derivatives Question 22, 1.8.49-LS Part 1 of 8 The function $p(t)=\frac{2000 t}{4 t+75}$ gives the population $p$ of deer in an area after $t$ months. a) Find $p^{\prime}(10), p^{\prime}(50)$, and $p^{\prime}(100)$. b) Find $p^{\prime \prime}(10), p^{\prime \prime}(50)$, and $p^{\prime \prime}(100)$. c) Interpret the meaning of your answers to part (a) and (b). What is happening to the pop

Solution

Solution Steps

Step 1: Find the First Derivative $ p^{\prime}(t) $

To find the first derivative of the population function $ p(t) = \frac{2000t}{4t + 75} $, we apply the quotient rule. The first derivative is given by: \[ p^{\prime}(t) = \frac{(4t + 75)(2000) - (2000t)(4)}{(4t + 75)^2} = \frac{2000(75)}{(4t + 75)^2} - \frac{8000t}{(4t + 75)^2} \] This simplifies to: \[ p^{\prime}(t) = \frac{6000}{4t + 75} - \frac{8000t}{(4t + 75)^2} \]

Step 2: Evaluate $ p^{\prime}(t) $ at Specific Points

Next, we evaluate the first derivative at $ t = 10 $, $ t = 50 $, and $ t = 100 $: p^{\prime}(10) = \frac{6000}{529}

For $ t = 10 $: \[

\] p^{\prime}(50) = \frac{240}{121}

For $ t = 50 $: \[

\] p^{\prime}(100) = \frac{240}{361}

For $ t = 100 $: \[

Step 3: Find the Second Derivative $ p^{\prime \prime}(t) $

To find the second derivative, we differentiate the first derivative $ p^{\prime}(t) $: \[ p^{\prime \prime}(t) = \frac{d}{dt}\left(\frac{6000}{4t + 75} - \frac{8000t}{(4t + 75)^2}\right) \] This results in: \[ p^{\prime \prime}(t) = \frac{64000t}{(4t + 75)^3} - \frac{16000}{(4t + 75)^2} \]

Step 4: Evaluate $ p^{\prime \prime}(t) $ at Specific Points

Finally, we evaluate the second derivative at $ t = 10 $, $ t = 50 $, and $ t = 100 $: p^{\prime \prime}(10) = -\frac{9600}{12167}