Find the amount of $H_2 O$ that flows from the tank during the first 10min. Suppose that the $H_2 O$ flows from the bottom of a storage tank at a rate of $r(t) = 200 - 4t$ $\displaystyle \frac{\text{liters}}{\text{min}}$, where $0 \leq t \leq 50$.
Using Net Change Theorem
$
\begin{equation}
\begin{aligned}
\int^b_a r(t) dt &= \int^{10}_0 (200 -4t) dt\\
\\
\int^b_a r(t) dt &= \left[ 200t - 2t^2 \right]^{10}_0\\
\\
\int^b_a r(t) dt &= 200(10) - 2(10)^2 - \left[ 200(0) - 2(0)^2 \right]\\
\\
\int^b_a r(t) dt &= 2000 - 200 - 0\\
\\
\int^b_a r(t) dt &= 1800 \text{ liters}
\end{aligned}
\end{equation}
$
This means that during the first 10 mins. the amount of water that flows inside the tank is 1800 liters.
Wednesday, April 24, 2019
Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 60
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