It is estimated that t years from now, the population of a certain community will be $p ( t ) = 14 - \frac { 6 } { 3 + t }$ thousand. An environmental study indicates that the average daily level of carbon monoxide in the air will be $C ( p ) = 0.5 \sqrt { p ^ { 2 } + 2 p + 30 }$ units when the population is p thousand. The rate at which the level of carbon monoxide will be changing 3 years from now is
A) -0.078 ppm per thousand people
B) 0.078 ppm per thousand people
C) 1.000 ppm per thousand people
D) -1.000 ppm per thousand people

Question

Quizplus · Accepted Answer

To find the rate at which the level of carbon monoxide is changing 3 years from now, we need to use the chain rule from calculus. First, we find $p(3) = 14 - \frac{6}{3+3} = 14 - 1 = 13$ thousand. Then, we differentiate $C(p)$ with respect to $p$, $C'(p) = \frac{1}{2} \cdot \frac{1}{\sqrt{p^2 + 2p + 30}} \cdot (2p + 2)$. Substituting $p = 13$, we get $C'(13) = \frac{1}{2} \cdot \frac{26 + 2}{\sqrt{13^2 + 2\cdot13 + 30}} = \frac{14}{\sqrt{239}}$. Next, we differentiate $p(t) = 14 - \frac{6}{3+t}$ with respect to $t$, giving $p'(t) = \frac{6}{(3+t)^2}$. Substituting $t = 3$, we get $p'(3) = \frac{6}{(6)^2} = \frac{1}{6}$. Finally, the rate of change of carbon monoxide levels with respect to time is $C'(p) \cdot p'(t) = \frac{14}{\sqrt{239}} \cdot \frac{1}{6} \approx 0.078$ ppm per thousand people.

It Is Estimated That T Years from Now, the Population p(t)=14−63+tp ( t ) = 14 - \frac { 6 } { 3 + t }p(t)=14−3+t6​

It Is Estimated That T Years from Now, the Population $p ( t ) = 14 - \frac { 6 } { 3 + t }$