Question

An oil storage tank ruptures at time\({\rm{t = 0}}\)and oil leaks from the tank at a rate of\({\rm{r(t) = 100}}{{\rm{e}}^{{\rm{ - 0}}{\rm{.01t}}}}\)litters per minute. How much oil leaks out during the first hour?

Short answer

\({\rm{4512liters}}\)Oil leaks out during the first hour.

Step-by-step solution

Given data.

Time\({\rm{t = 0}}\).

Oil leaks from the tank at a rate of \({\rm{r(t) = 100}}{{\rm{e}}^{{\rm{ - 0}}{\rm{.01t}}}}\)litters per minute.

 Volume leaked during the first hour.

The volume of the total amount spilled is determined by the integral of the leak rate. Because the rate is in liters per minute, multiply from \({\rm{0}}\) to \({\rm{60}}\) to calculate the volume lost in the first hour.

\({\rm{V = }}\int_{\rm{0}}^{{\rm{60}}} {\rm{1}} {\rm{00}}{{\rm{e}}^{{\rm{ - 0}}{\rm{.01t}}}}{\rm{dt}}\)

Limits of\({\rm{u}}\),

\(\begin{aligned}{c}{\rm{u = - 0}}{\rm{.01 t }}\\{\rm{d u = - 0}}{\rm{.01 d t }}\\{\rm{ - }}\frac{{\rm{1}}}{{{\rm{0}}{\rm{.01}}}}{\rm{du = dt}}\\{\rm{ - 100du = dt}}\end{aligned}\)

Values are,

\(\begin{aligned}{c}{\rm{a = - 0}}{\rm{.01}}\\{\rm{t = - 0}}{\rm{.01(0)}}\\{\rm{ = 0}}\\{\rm{b = - 0}}{\rm{.01(60)}}\\{\rm{ = - 0}}{\rm{.6}}\end{aligned}\)

Therefore,

\(\begin{aligned}{c}{\rm{V = }}\int_{\rm{0}}^{{\rm{60}}} {\rm{1}} {\rm{00}}{{\rm{e}}^{{\rm{ - 0}}{\rm{.01t}}}}{\rm{dt}}\\{\rm{ = }}\int_{\rm{0}}^{{\rm{ - 0}}{\rm{.6}}} {\rm{1}} {\rm{00( - 100)}}{{\rm{e}}^{\rm{u}}}{\rm{du}}\\{\rm{ = - 10000}}\left( {{{\rm{e}}^{\rm{u}}}} \right)_{\scriptstyle{\rm{0}}\atop\scriptstyle}^{{\rm{ - 0}}{\rm{.6}}}\\{\rm{ = - 10000}}\left( {{{\rm{e}}^{{\rm{ - 0}}{\rm{.6}}}}{\rm{ - 1}}} \right)\\ \approx {\rm{4512 liters}}\end{aligned}\)

So, the oil leaked in first hour is approximately\({\rm{4512liters}}\).