Question

If a factory continuously pumps pollutants into the air at the rate of the quotient of the square root of t and 15 tons per day, then the amount dumped after 3 days is?

Answer 1

Given

$D(t)=\frac{10t}{1+2t}$

To find how much waste will the pumping station deliver during the 8 hour period.

Explanation:

It is given that,

$\begin{equation*} D(t)=\frac{10t}{1+2t} \end{equation*}$

Then for t=1,

$\begin{gathered} D(1)=\frac{10(1)}{1+2(1)} \\ =\frac{10}{3} \\ =3.33 \end{gathered}$

For t=2,

$\begin{gathered} D(2)=\frac{10(2)}{1+2(2)} \\ =\frac{20}{5} \\ =4 \end{gathered}$

For t=3,

$\begin{gathered} D(3)=\frac{10(3)}{1+2(3)} \\ =\frac{30}{7} \\ =4.29 \end{gathered}$

For t=4,

$\begin{gathered} D(4)=\frac{10(4)}{1+2(4)} \\ =\frac{40}{9} \\ =4.44 \end{gathered}$

For t=5,

$\begin{gathered} D(5)=\frac{10\times5}{1+2(5)} \\ =\frac{50}{11} \\ =4.55 \end{gathered}$

For t=6,

$\begin{gathered} D(6)=\frac{10(6)}{1+2(6)} \\ =\frac{60}{13} \\ =4.62 \end{gathered}$

For t=7,

$\begin{gathered} D(7)=\frac{10(7)}{1+2(7)} \\ =\frac{70}{15} \\ =4.67 \end{gathered}$

For t=8,

$\begin{gathered} D(8)=\frac{10(8)}{1+2(8)} \\ =\frac{80}{17} \\ =4.71 \end{gathered}$

Then,

$\begin{gathered} Total\text{ waste}=3.33+4+4.29+4.44+4.55+4.62+4.67+4.71 \\ =34.61 \end{gathered}$

Hence, the total waste delivered is 34.61.