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 $u = g(x)$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Here, $CO(p)=\sqrt{0.5p + 15}=(0.5p + 15)^{\frac{1}{2}}$ and $p(t)=100 + 0.04t^{2}$. So, $\frac{dCO}{dt}=\frac{dCO}{dp}\cdot\frac{dp}{dt}$.

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

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

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

Using the power - rule, $\frac{dp}{dt}=0.08t$.

Step4: Find $p$ when $t = 5$

Substitute $t = 5$ into $p(t)$: $p(5)=100+0.04\times5^{2}=100 + 0.04\times25=100 + 1=101$.

Step5: Calculate $\frac{dCO}{dt}$ at $t = 5$

First, find $\frac{dCO}{dp}$ at $p = 101$: $\frac{dCO}{dp}\big|_{p = 101}=\frac{0.5}{2\sqrt{0.5\times101+15}}=\frac{0.5}{2\sqrt{50.5 + 15}}=\frac{0.5}{2\sqrt{65.5}}$.
Second, find $\frac{dp}{dt}$ at $t = 5$: $\frac{dp}{dt}\big|_{t = 5}=0.08\times5 = 0.4$.
Then, $\frac{dCO}{dt}\big|_{t = 5}=\frac{dCO}{dp}\big|_{p = 101}\cdot\frac{dp}{dt}\big|_{t = 5}=\frac{0.5}{2\sqrt{65.5}}\times0.4=\frac{0.2}{2\sqrt{65.5}}=\frac{0.1}{\sqrt{65.5}}\approx\frac{0.1}{8.093}\approx0.0123$ parts per million per year.

Answer:

Approximately $0.0123$ parts per million per year.