# Can someone help out with the question below?

Feb 27, 2017

$P ' \left(16\right) = \frac{75}{32}$

#### Explanation:

The idea here is that you can find the rate of change of the pollution with respect to time by taking the first derivative of your function $P \left(t\right)$.

So this is pretty much an exercise in finding the derivative of the function

$P \left(t\right) = {\left({t}^{\frac{1}{4}} + 3\right)}^{3}$

which can be calculated by using the chain rule and the power rule. Keep in mind that you have

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\frac{d}{\mathrm{dt}} \left[u {\left(t\right)}^{n}\right] = n \cdot u {\left(t\right)}^{n - 1} \cdot \frac{d}{\mathrm{dt}} \left[u \left(t\right)\right]}}}$

$\left\{\begin{matrix}u \left(t\right) = {t}^{\frac{1}{4}} + 3 \\ n = 3\end{matrix}\right.$

This means that the derivate of $P \left(t\right)$ will be

${\overbrace{\frac{d}{d} \left(\mathrm{dt}\right) \left[P \left(t\right)\right]}}^{\textcolor{b l u e}{= {P}^{'} \left(t\right)}} = 3 \cdot {\left({t}^{\frac{1}{4}} + 3\right)}^{2} \cdot \frac{d}{\mathrm{dt}} \left({t}^{\frac{1}{4}} + 3\right)$

${P}^{'} \left(t\right) = 3 \cdot {\left({t}^{\frac{1}{4}} + 3\right)}^{2} \cdot \frac{1}{4} \cdot {t}^{\left(\frac{1}{4} - 1\right)}$

${P}^{'} \left(t\right) = \frac{3}{4} \cdot {t}^{- \frac{3}{4}} \cdot {\left({t}^{\frac{1}{4}} + 3\right)}^{2}$

Now all you have to do is plug in $16$ for $t$ to find

${P}^{'} \left(t\right) = \frac{3}{4} \cdot {16}^{- \frac{3}{4}} \cdot {\left({16}^{\frac{1}{4}} + 3\right)}^{2}$

Since you know that

$16 = {2}^{4}$

you can rewrite the equation as

${P}^{'} \left(16\right) = \frac{3}{4} \cdot {2}^{\left[4 \cdot \left(- \frac{3}{4}\right)\right]} \cdot {\left[{2}^{\left(4 \cdot \frac{1}{4}\right)} + 3\right]}^{2}$

${P}^{'} \left(16\right) = \frac{3}{4} \cdot {2}^{- 3} \cdot {\left(2 + 3\right)}^{2}$

${P}^{'} \left(16\right) = \frac{3}{4} \cdot \frac{1}{8} \cdot 25$

${P}^{'} \left(16\right) = \frac{75}{32}$

And there you have it -- the rate at which the pollution changes after $16$ years.