Question

ecologists estimate that, when the population of a certain city is x thousand persons, the average level l of carbon monoxide in the air above the city will be l ppm (parts per million), where l = 90 + 0.4x+0.006x². the population of the city is estimated to be x = 752+23t + 0.5t² thousand persons t years from the present. find the rate of change of carbon monoxide with respect to the population when t = 2. how fast (with respect to time) is the carbon monoxide level changing at time t = 2? what is the rate of change of carbon monoxide with respect to the population of the city? (type an expression using x as the variable.)

Explanation:

Step1: Recall the derivative formula for rate - of - change

The rate of change of a function $y = f(x)$ with respect to $x$ is given by $y^\prime=\frac{dy}{dx}$. We want to find the rate of change of $L$ with respect to $x$, where $L = 90+0.4x + 0.006x^{2}$.

Step2: Differentiate $L$ with respect to $x$

Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, for the constant term $90$, $\frac{d}{dx}(90)=0$; for the term $0.4x$, $\frac{d}{dx}(0.4x)=0.4$; for the term $0.006x^{2}$, $\frac{d}{dx}(0.006x^{2})=2\times0.006x=0.012x$. So, $\frac{dL}{dx}=0.4 + 0.012x$.

Answer:

$0.4 + 0.012x$