Chapter 5.1, Problem 13E

To determine

The lower estimate and upper estimate for the total amount of oil that leaked out.

Answer to Problem 13E

The lower estimate for the total amount of oil leaked is 63.2L.

The upper estimate for the total amount of oil leaked is 70 L.

Explanation of Solution

Given:

The function r(t) is an decreasing function.

The number of intervals is $n=5$.

Each interval value is $\Delta t=2\text{h}$.

Calculation:

The distance approximation $R_5$ provides the lower estimate for the total amount of oil leaked from the tank and $L_5$ provides the upper estimate for the total amount of oil leaked from the tank.

The expression to find the lower estimate for the total amount of oil leaked $\left(R{n}_{}\right)$ is shown below:

$R_n=r\left(t_1\right)\Delta t+r\left(t_2\right)\Delta t+\mathrm{...}+r\left(t_n\right)\Delta t$ (1)

Here, for the time $t_1$, the rate is $r\left(t_1\right)$, the time interval is $\Delta t$, for the time $t_2$, the rate is $r\left(t_2\right)$, and for the time $t_n$, the rate is $r\left(t_n\right)$.

Substitute 5 for n in Equation (1).

$R_5=r\left(t_1\right)\Delta t+r\left(t_2\right)\Delta t+r\left(t_3\right)\Delta t+r\left(t_4\right)\Delta t+r\left(t_5\right)\Delta t$ (2)

Substitute 7.6L/h for $r\left(t_1\right)$, 2h for $\Delta t$, 6.8L/h for $r\left(t_2\right)$, 6.2L/h for $r\left(t_3\right)$, 5.7L/h for $r\left(t_4\right)$, and 5.3L/h for $r\left(t_5\right)$ in Equation (2).

$\begin{array}{c}{R}_5=\left(7.6\times 2\right)+\left(6.8\times 2\right)+\left(6.2\times 2\right)+\left(5.7\times 2\right)+\left(5.3\times 2\right)\\ =15.2+13.6+12.4+11.4+10.6\\ =63.2\text{L}\text{.}\end{array}$

Therefore, the lower estimate for the total amount of oil leakage from tank is 63.2 L.

The expression to find the upper estimate for the total amount of oil leaked $\left(L_n\right)$ is shown below:

$L_n=r\left(t_0\right)\Delta t+r\left(t_1\right)\Delta t+\mathrm{...}+r\left(t_{n-1}\right)\Delta t$ (3)

Here, for the time $t_0$, the rate is $r\left(t_0\right)$ and for the time $t_{n-1}$, the rate is $r\left(t_{n-1}\right)$.

Substitute 5 for n in Equation (3).

$L_5=r\left(t_0\right)\Delta t+r\left(t_1\right)\Delta t+r\left(t_2\right)\Delta t+r\left(t_3\right)\Delta t+r\left(t_4\right)\Delta t$ (4)

Substitute 8.7L/h for $r\left(t_0\right)$, 2h for $\Delta t$, 7.6L/h for $r\left(t_1\right)$, 6.8 L/h for $r\left(t_2\right)$, 6.2L/h for $r\left(t_3\right)$, and 5.7L/h for $r\left(t_4\right)$ in Equation (4).

$\begin{array}{c}{L}_5=\left(8.7\times 2\right)+\left(7.6\times 2\right)+\left(6.8\times 2\right)+\left(6.2\times 2\right)+\left(5.7\times 2\right)\\ =17.4+15.2+13.6+12.4+11.4\\ =70\text{L}\text{.}\end{array}$

Therefore, the upper estimate for the total amount of oil leakage from tank is 70 L.