Problem 40 - Calculus Early Transcendentals

Question

Starting with an initial value of \(P(0)=55,\) the population of a prairie dog community grows at a rate of \(P^{\prime}(t)=20-t / 5\) (prairie dogs/month), for \(0 \leq t \leq 200,\) where \(t\) is measured in months. a. What is the population 6 months later? b. Find the population \(P(t),\) for \(0 \leq t \leq 200\)

Step-by-Step Solution

Answer

Based on the step-by-step solution provided: a. The population of the prairie dog community after 6 months is 102 prairie dogs. b. The general population function P(t) for the given time interval, 0 ≤ t ≤ 200, is \(P(t) = 20t - \frac{1}{10}t^2 + 55\).

Step 1: Integration of the population growth function

We are given the rate of population change as \(P'(t) = 20 - \frac{t}{5}\). Firstly, we must find P(t) by integrating P'(t) with respect to time t: $$ P(t) = \int P'(t) \, dt = \int (20 - \frac{t}{5}) \, dt $$

Step 2: Applying the integration

Now, integrate each term separately: $$ P(t) = 20 \int dt - \frac{1}{5} \int t \, dt = 20t - \frac{1}{5} \cdot \frac{t^2}{2} + C $$ where C is the integration constant.

Step 3: Using the initial condition

We are given the initial population as P(0) = 55. Apply this condition to find the value of C: $$ 55 = 20(0) - \frac{1}{5} \cdot \frac{0^2}{2} + C \Rightarrow C = 55 $$ Now we have the expression for P(t) as: $$ P(t) = 20t - \frac{1}{10}t^2 + 55 $$ #a#

Step 4: Finding the population after 6 months

To find the population after 6 months, simply substitute t=6 into the derived equation for P(t): $$ P(6) = 20\cdot 6 - \frac{1}{10}(6)^2 + 55 $$ This becomes: $$ P(6) = 120 - \frac{36}{10} + 55 = 120 - 18 + 55 $$ So, after 6 months, the population is: $$ P(6) = 102 \text{ (prairie dogs)} $$ #b#

Step 5: Population function for 0 ≤ t ≤ 200

Now that we have found the population after 6 months, we can provide the general population function for the given time interval. Since we already derived it in Step 3: The population function P(t) for 0 ≤ t ≤ 200 is: $$ P(t) = 20t - \frac{1}{10}t^2 + 55 $$