Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 44

Suppose that a cylindrical tank holds a 100,000 gallons of water, which can be drained from the bottom
of the tank in an hour, then Torricelli's Law gives the volume $V$ of water remaining in the tank after
$t$ minutes as



$\quad \displaystyle V(t) = 100,000 \left( 1 - \frac{t}{60} \right)^2 \qquad 0 \leq t \leq 60$



Determine the rate at which the water is flowing out of the tank for times $t =$ 0, 10, 20, 30, 40, 50 and 60 $min$.
Find the flow rate and the amount of water remaining in the tank. Also, determine the time at which the flow rate is
least and greatest.


Based from the definition,
$f'(a) = \lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h}$


$
\begin{equation}
\begin{aligned}
\nu'(t) & = \lim\limits_{h \to 0} \frac{\nu ( t + h ) + \nu (t)}{h}\\
\nu'(t) & = \lim\limits_{h \to 0} \frac{100,000 \left( 1 - \frac{(t+h)}{60}\right)^2 - \left[ 100,000 \left( 1 - \frac{t}{60} \right)^2\right]}{h}\\
\nu'(t) & = \lim\limits_{h \to 0} \frac{100,000 \left( \frac{60-t-h}{60}\right)^2 - 100,000 \left( \frac{60-t}{60} \right)^2}{h}\\
\nu'(t) & = \lim\limits_{h \to 0} \frac{\frac{100,000}{60^2}[(60-t-h)^2 - (60-t)^2]}{h}\\
\nu'(t) & = \lim\limits_{h \to 0} \frac{250[\cancel{60^2}-\cancel{60t}-60h-\cancel{60t}+\cancel{t^2}
+th-60h+th+h^2-\cancel{60^2}+\cancel{60t}+\cancel{60t}-\cancel{t^2}]}{9h}\\
\nu'(t) & = \lim\limits_{h \to 0} \frac{250[-120h+2th+h^2]}{9h}\\
\nu'(t) & = \lim\limits_{h \to 0} \frac{250\cancel{h}[-120+2t+h]}{9\cancel{h}}\\
\nu'(t) & = \lim\limits_{h \to 0} \left[ \frac{-10000}{3} + \frac{500t}{9}+h\right]\\
\nu'(t) & = \frac{-10000}{3} + \frac{500t}{9} + 0\\
\nu'(t) & = \frac{-10000}{3} + \frac{500t}{9} \frac{\text{volume}}{\text{min}}
\end{aligned}
\end{equation}
$



$\displaystyle V(t) = 100,000 \left( 1-\frac{t}{60}\right)^2$



Flow rate, $\displaystyle \nu(t) = \frac{10,000}{3} + \frac{500t}{9}$



$
\begin{array}{|c|c|c|}
\hline
t\text{(min)} & V(t) & \text{Flow rate, } \nu'(t)\\
\hline\\
0 & 100,000 & -3333.\overline{33}\\
\\
10 & 69444.\overline{44} & -2777.78\\
\\
20 & 44444.\overline{44} & -2222.\overline{22}\\
\\
30 & 25000 & -1666.67\\
\\
40 & 11111.\overline{11} & -1111.11\\
\\
50 & 2777.78 & -555.56\\
\\
60 & 0 & 0\\
\\
\hline
\end{array}
$



The amount of water remaining in the tank can be computed by substituting the values of $t$
to the equation $V(t)$. Also, flow rates are obtained by substituting the values of $t$ in the equation of $\nu'(t)$.
The negative values represent how fast the water is flowing out of the tank until such time that all water is drained from the tank.
Flow rate is greatest at $t=0 $ while it is least at $t=60$.

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