# Oil flows into a tank according to the rate F(t) = \frac{(t^2 +1)}{(1 + t)}, and at the same time...

## Question:

Oil flows into a tank according to the rate {eq}F(t) = \frac{(t^2 +1)}{(1 + t)} {/eq}, and at the same time empties out at the rate {eq}E(t) = \frac{\ln(t + 7)}{(t + 1)} {/eq}, with both {eq}F(t) {/eq} and {eq}E(t) {/eq} measured in gallons per minute. How much oil, to the nearest gallon, is in the tank at time {eq}t = 12 {/eq} minutes.

## Integration:

A mathematical quantity that represents the merge of the derivative of the function is known as integration. It is a sub-part of calculus analysis. It used in engineering application to solve the system problem.

Given Data:

• The inlet rate of oil is: {eq}F\left( t \right) = \dfrac{{\left( {{t^2} + 1} \right)}}{{1 + t}} {/eq}
• The outlet rate of oil is: {eq}E\left( t \right) = \dfrac{{\ln \left( {t + 7} \right)}}{{1 + t}} {/eq}
• {eq}t = 12\;{\rm{minutes}} {/eq}

The expression for oil to nearest gallon is

{eq}\begin{align*} g\left( t \right) &= F\left( t \right) - E\left( t \right)\\ &= \dfrac{{\left( {{t^2} + 1} \right)}}{{1 + t}} - \dfrac{{\ln \left( {t + 7} \right)}}{{1 + t}}\\ &= \dfrac{{\left( {{t^2} + 1} \right) - \ln \left( {t + 7} \right)}}{{1 + t}} \end{align*} {/eq}

Integrate the above expression for limit {eq}0 \le t \le 12 {/eq}

{eq}\begin{align*} g\left( t \right) &= \int_0^{12} {\left[ {\left( {t - 1} \right) + \dfrac{2}{{1 + t}} - \dfrac{{\ln \left( {1 + t} \right)}}{{1 + t}}} \right]} dt\\ &= \int_0^{12} t dt - \int_0^{12} 1 dt + 2\int_0^{12} {\dfrac{1}{{1 + t}}dt - \int_0^{12} {\dfrac{{\ln \left( {1 + t} \right)}}{{1 + t}}} } dt\\ &= \left[ {\dfrac{{{t^2}}}{2}} \right]_0^{12} - \left[ t \right]_0^{12} + 2\ln \left( {1 + t} \right)_0^{12} - {I_o}\\ &= \left[ {\dfrac{{{{\left( {12} \right)}^2}}}{2} - 0} \right] - \left[ {12 + 0} \right] + 2\left[ {\ln \left( {1 + 12} \right) - \ln \left( {0 + 1} \right)} \right] - {I_o}\\ &= 72 - 12 + 2.197 - {I_o}\\ &= 62.197 - {I_o} \cdots\cdots\cdots{\rm(I)} \end{align*} {/eq}

Let {eq}\ln \left( {1 + t} \right) = V \cdots\cdots\cdots{\rm(II)} {/eq}

Differentiate the above expression

{eq}\begin{align*} \dfrac{{d\left( {\ln \left( {1 + t} \right)} \right)}}{{dt}} &= \dfrac{{dV}}{{dt}}\\ \dfrac{1}{{1 + t}} &= \dfrac{{dV}}{{dt}}\\ dt &= \left( {1 + t} \right)dV \end{align*} {/eq}

Substitute the value and solve the expression (II) at {eq}t=0 {/eq}

{eq}\begin{align*} \ln \left( {1 + 0} \right) &= V\\ V &= 0 \end{align*} {/eq}

Substitute the value and solve the expression (II) at {eq}t=12 {/eq}

{eq}\begin{align*} \ln \left( {1 + 12} \right) &= V\\ V &= \ln \left( {13} \right) \end{align*} {/eq}

The domain of variable {eq}V {/eq} is

{eq}0 \le V \le \ln 13 {/eq}

Substitute the value and solve the integration {eq}{I_o} {/eq}

{eq}\begin{align*} {I_o} &= \int_0^{\ln 13} {\dfrac{V}{{1 + t}}\left( {\left( {1 + t} \right)dV} \right)} \\ &= \int_0^{\ln 13} {VdV} \\ &= \left[ {\dfrac{{{V^2}}}{2}} \right]_0^{\ln 13}\\ &= \dfrac{1}{2}\left[ {{{\left( {\ln 13} \right)}^2} - 0} \right]\\ &= \dfrac{1}{2}\left[ {2.56} \right]\\ &= 1.28 \end{align*} {/eq}

Substitute the value and solve the expression (I)

{eq}\begin{align*} g\left( t \right) &= 62.197 - 1.28\\ &= 60.917 \end{align*} {/eq}

Thus the oil to nearest gallon is {eq}60.917\;{\rm{gallons}} {/eq}