Question

Nana has a water purifier that filters \dfrac13 3 1 ​ start fraction, 1, divided by, 3, end fraction of the contaminants each hour. She used it to purify water that had \dfrac12 2 1 ​ start fraction, 1, divided by, 2, end fraction kilogram of contaminants. Write a function that gives the remaining amount of contaminants in kilograms, C(t)C(t)C, left parenthesis, t, right parenthesis, ttt hours after Nana started purifying the water.

Answer 1

Given:

Nana has a water purifier that filters $\dfrac{1}{3}$ of the contaminants each hour.

Water has contaminants = $\dfrac{1}{2}$

To find:

The function that gives the remaining amount of contaminants in kilograms, C(t), t hours after Nana started purifying the water.

Solution:

Let C(t) be the remaining amount of contaminants in kilograms after t hours.

Initial amount of contaminants = $\dfrac{1}{2}$

Decreasing rate is $\dfrac{1}{3}$ .

Using the exponential decay model:

$C(t)=C_0(1-r)^t$

where, $C_0$ is initial amount of contaminants, r is the decreasing rate and t is time in hours.

Substituting the values, we get

$C(t)=\dfrac{1}{2}(1-\dfrac{1}{3})^t$

$C(t)=\dfrac{1}{2}(\dfrac{2}{3})^t$

Therefore, the required function is $C(t)=\dfrac{1}{2}(\dfrac{2}{3})^t$.