Solving a Related Rate Problem

A rectangular swimming pool \(10 \, {\rm m}\) long and \(5 \,{\rm m}\) wide has a depth of \(3 \,{\rm m}\) at one end and \(1 \,{\rm m}\) at the other end. If water is pumped into the pool at the rate of \(300\) liters per minute (\({\rm liter}/{\rm min}\)), at what rate is the water level rising when it is \(1.5 \,{\rm m}\) deep at the deep end?

Solution

Step 1 We draw a picture of the cross-sectional view of the pool, as shown in Figure 4.

Step 2 The width of the pool is \(5 \,{\rm m}\), the water level (measured at the deep end) is \(h\), the distance from the wall at the deep end to the edge of the water is \(L\), and the volume of water in the pool is \(V\). Each of the variables \(h\), \(L\), and \(V\) varies with respect to time \(t\). \[ {\rm We\;are\;given} \dfrac{\textit{dV}}{\textit{dt}}=300\, {\rm liter}/ {\rm min}\; {\rm and\;are\;asked\;to\;find} \dfrac{\textit{dh}}{\textit{dt}} {\rm when}\; h=1.5\; {\rm m}. \]

Step 3 The volume \(V\) is related to \(L\) and \(h\) by the formula \[ V=\hbox{(Cross-sectional triangular area)(width)}=\left( \dfrac{1}{2} Lh\right) (5) =\dfrac{5}{2}Lh \]

See Figure 4. Using similar triangles, \(L\) and \(h\) are related by the equation \[ \frac{L}{h}=\frac{10}{2} \qquad\hbox{so} \qquad L=5h \]

Now we can write \(V\) as \[ \begin{eqnarray*} V = \frac{5}{2} \textit{Lh} \underset{\underset{{\color{#0066A7}{\hbox{L=5h}}}}{\color{#0066A7}{\uparrow }}} {=} \frac{5}{2}(5h) h=\dfrac{25}{2}h^{2} \nonumber\\[-8.1pt]\tag{1} \end{eqnarray*} \]

Both \(V\) and \(h\) vary with time \(t\).

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Step 4 We differentiate both sides of equation (1) with respect to \(t.\) \[ \dfrac{\textit{dV}}{\textit{dt}}=25h\,\frac{\textit{dh}}{\textit{dt}} \]

Step 5 Substitute \(h=1.5\) m and \(\dfrac{\textit{dV}}{\textit{dt}}=300\ {\rm liter}/{\rm min}= \dfrac{300}{1000}{\rm m}^{3\!\!}/{\rm min}=0.3{\rm m}^{3\!\!}/{\rm min}\). Then \[ \begin{eqnarray*} 0.3 &=&25(1.5) \dfrac{\textit{dh}}{\textit{dt}}\qquad\qquad \color{#0066A7}{{\hbox{\(\dfrac{dV}{dt}=25 h\dfrac{dh}{dt}\)}}} \\[3pt] \frac{\textit{dh}}{\textit{dt}} &=&\frac{0.3}{25(1.5)}=0.008 \end{eqnarray*} \]

When the height of the water is \(1.5\, {\rm m}\), the water level is rising at a rate of \(0.008\, {\rm m}/{\rm min}\).