Question

1. (8 points) the population of the boom town of suluclac gulch, ca is well modeled by the function $p(t)=\frac{500t}{2t^{2}+9}$, where $p(t)$ is the population in thousands of people, and $t$ is the number of years since january first, 1849. what was the instantaneous rate of change of the population on january first of 1859? make sure you include units in your answer.

Explanation:

Step1: Identify the function and the value of t

The population function is $P(t)=\frac{500t}{2t^{2}+9}$, and $t = 1859 - 1849=10$.

Step2: Use the quotient - rule to find the derivative

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 500t$, so $u^\prime=500$, and $v = 2t^{2}+9$, so $v^\prime = 4t$. Then $P^\prime(t)=\frac{500(2t^{2}+9)-500t(4t)}{(2t^{2}+9)^{2}}=\frac{1000t^{2}+4500 - 2000t^{2}}{(2t^{2}+9)^{2}}=\frac{4500 - 1000t^{2}}{(2t^{2}+9)^{2}}$.

Step3: Evaluate the derivative at t = 10

Substitute $t = 10$ into $P^\prime(t)$:
$P^\prime(10)=\frac{4500-1000\times10^{2}}{(2\times10^{2}+9)^{2}}=\frac{4500 - 100000}{(200 + 9)^{2}}=\frac{- 95500}{209^{2}}=\frac{-95500}{43681}\approx - 2.19$.

Answer:

The instantaneous rate of change of the population on January 1st, 1859 is approximately $-2.19$ thousand people per year.