Question

1. suppose that the level of carbon monoxide in a city’s air may be modeled by the formula co(p) = √(.5p + 15) parts per million when the population is p thousand people. it is estimated that t years after today, the city’s population will be p(t) = 100 + .04t² thousand people. at what rate will the carbon monoxide level be changing with respect to time 5 years from now?

Explanation:

Step1: Use the chain - rule

The chain - rule states that if $y = f(u)$ and $u = g(t)$, then $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. Here, $CO(p)=\sqrt{0.5p + 15}=(0.5p + 15)^{\frac{1}{2}}$ and $p(t)=100 + 0.04t^{2}$. First, find $\frac{d(CO)}{dp}$ and $\frac{dp}{dt}$.

Step2: Differentiate $CO(p)$ with respect to $p$

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, if $CO(p)=(0.5p + 15)^{\frac{1}{2}}$, then $\frac{d(CO)}{dp}=\frac{1}{2}(0.5p + 15)^{-\frac{1}{2}}\cdot0.5=\frac{0.5}{2\sqrt{0.5p + 15}}=\frac{1}{4\sqrt{0.5p + 15}}$.

Step3: Differentiate $p(t)$ with respect to $t$

If $p(t)=100 + 0.04t^{2}$, then $\frac{dp}{dt}=0.08t$.

Step4: Apply the chain - rule

By the chain - rule, $\frac{d(CO)}{dt}=\frac{d(CO)}{dp}\cdot\frac{dp}{dt}=\frac{1}{4\sqrt{0.5p + 15}}\cdot0.08t=\frac{0.08t}{4\sqrt{0.5p + 15}}=\frac{0.02t}{\sqrt{0.5p + 15}}$.

Step5: Find $p$ when $t = 5$

When $t = 5$, $p(5)=100+0.04\times5^{2}=100 + 0.04\times25=100 + 1=101$ (thousand people).

Step6: Substitute $t = 5$ and $p = 101$ into $\frac{d(CO)}{dt}$

$\frac{d(CO)}{dt}\big|_{t = 5}=\frac{0.02\times5}{\sqrt{0.5\times101+15}}=\frac{0.1}{\sqrt{50.5 + 15}}=\frac{0.1}{\sqrt{65.5}}\approx\frac{0.1}{8.093}\approx0.0124$ parts per million per year.

Answer:

Approximately $0.0124$ parts per million per year.