Question

Find the function for the percentage of carbon dioxide present in the room at time\(t\). The amount of carbon dioxide present is\(0.15\% \), the percentage of fresh air is\(0.05\% \)and the rate of mixed in the mixed out carbon dioxide is\(2\;{{\rm{m}}^3}/{\rm{min}}\). And find the limit of the function that we found

Short answer

The function for the percentage is \(p(t) = 0.05 + 0.01{e^{\frac{{ - t}}{{90}}}}\).

The percentage of carbon dioxide level goes to \(0.05\)when there is a long run.

Step-by-step solution

the rate of change of carbon dioxide with respect to time\({\rm{t}}\).

Initial the amount of carbon dioxide present \( = 0.15\% \)

Percentage of fresh air\( = 0.05\% \)

The rate of mixed in and mixed out of carbon dioxide\( = 2\;{{\rm{m}}^3}/{\rm{min}}\).

Let\(y\)be the amount of carbon dioxide.

At the initial stage carbon dioxide level is\(y(0) = 015\% \left( {180{m^3}} \right)\)at the time zero.

\(\begin{aligned}y &= 0.0015(180)\\ &= 0.27{m^3}\end{aligned}\)

Here\(180\;{{\rm{m}}^2}\)is the volume carbon dioxide constantly.

Then the percentage of carbon dioxide is\(\frac{y}{{180}}(100)\)at the time\(t\).

The rate of change of carbon dioxide with respect to the time\(t\),

\(\frac{{dy}}{{dt}} = \frac{{0.05}}{{100}}(2) - \frac{y}{{180}}(2)\)

Here\(\frac{y}{{180}}(2)\)be the carbon dioxide flows inside.

\(\begin{aligned}\frac{{dy}}{{dt}} &= 0.001 - \frac{y}{{90}}\\\frac{{dy}}{{dt}} &= \frac{{0.09 - y}}{{90}}\end{aligned}\)

Multiply\(100\)on both sides,

\(\frac{{dy}}{{dt}} = \frac{{9 - 100y}}{{9000}}\)

Separate the variables,

\(\frac{{dy}}{{9 - 100y}} = \frac{{dt}}{{9000}}\)

Integrating on both sides,

\(\begin{aligned}\int {\frac{{dy}}{{9 - 100y}}} &= \int {\frac{{dt}}{{9000}}} \\ - \frac{1}{{100}}\ln (9 - 100y) &= \frac{1}{{9000}}t + C\end{aligned}\) …… (1)

Substitute the condition \({\rm{y(0) = 0}}{\rm{.27}}\) in equation (1).

Substitute the initial condition \({\rm{y(0) = 0}}{\rm{.27}}\)in equation (1),

\(\begin{aligned} - \frac{1}{{100}}\ln (9 - 100(0.27)) &= \frac{1}{{9000}}(0) + C\\ - \frac{1}{{100}}\ln (18) &= C\end{aligned}\)

Substitute\(c\)value in equation (1),

\(\begin{aligned} - \frac{1}{{100}}\ln (9 - 100y) &= \frac{1}{{9000}}t - \frac{1}{{100}}\ln (18)\\ - \ln (9 - 100y) &= \frac{t}{{90}} - \ln (18)\\In(9 - 100y) &= - \frac{t}{{90}} + \ln (18)\end{aligned}\)

Take exponentials on both sides,

\(\ln (9 - 100y) = - \frac{t}{{90}} + \ln (18)\)

\({\rm{9 - 100 y}}\)is negative. So that it can be written as follows,

\(\begin{aligned}(100y - 9) &= {e^{ - \frac{t}{{90}} + \ln (18)}}\\(100y - 9) &= {e^{ - \frac{t}{{90}}}}18\\100y &= {e^{ - \frac{t}{{90}}}}18 + 9\\y &= \frac{{{e^{ - \frac{t}{{90}}}}18 + 9}}{{100}}\end{aligned}\)

Therefore, the carbon dioxide level after \(t\) minutes is \(y = \frac{{{e^{ - \frac{s}{{90}}}}18 + 9}}{{100}}\).

Consider the percentage of carbon dioxide level,

\(p(t) = \frac{y}{{180}}(100)\) …… (2)

Substitute\(y\)value in equation (2),

\(\begin{aligned}p(t) &= \frac{{\frac{{{e^{ - \frac{t}{{90}}}}18 + 9}}{{100}}}}{{180}}(100)\\ &= \frac{{{e^{ - \frac{t}{{90}}}}0.18 + 0.09}}{{180}}(100)\\ &= \left( {0.0005 + 0.001{e^{\frac{{ - t}}{{90}}}}} \right)(100)\\ &= 0.05 + 0.01{e^{\frac{{ - t}}{{90}}}}\end{aligned}\)

Obtain the percentage of the long run as follows,

Take limit on both sides,

\(\mathop {\lim }\limits_{t \to \infty } p(t) = 0.05 + 0.1{e^{\frac{{ - \pi }}{{90}}}}\)

\( = 0.05 + 0.1(0)\;\;\;\left( {{e^{ - \infty }} = 0} \right)\)

\( = 0.05\)

Thus, the percentage of carbon dioxide level goes to \(0.05\)when there is a long run, that means \(0.05\% \).