Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 25

At what rate is the water level rising when the water is 30cm deep?



We know that the volume $V=$ (area of the base)(length). Recall that the Area of trapezoid is $\displaystyle A = \frac{h}{2}(b_1+b_2)$.


$
\begin{equation}
\begin{aligned}
V &= \frac{h}{2} (b_1 + b_2) (10)\\
\\
V &= 5h (b_1 + b_2) && \text{We will take the measurement in meters to be consistent with the units}
\end{aligned}
\end{equation}
$

Notice that $b_1 = h + 0.3m$ so,

$
\begin{equation}
\begin{aligned}
V &= 5h(h + 0.3+0.3)\\
\\
V &= 5h (h + 0.6)\\
\\
V &= 6h^2 + 3h
\end{aligned}
\end{equation}
$


Taking the derivative with respect to time we have,

$
\begin{equation}
\begin{aligned}
\frac{dV}{dt} &= 5 \frac{d}{dh} (h^2) \frac{dh}{dt} + 3 \frac{d}{dh} +\frac{dh}{dt}\\
\\
\frac{dV}{dt} &= 5 (2h) \frac{dh}{dt} + 3(1) \frac{dh}{dt}\\
\\
\frac{dV}{dt} &= 10 h \frac{dh}{dt} + 3 \frac{dh}{dt}\\
\\
\frac{dh}{dt} &= \frac{\frac{dV}{dt}}{10h + 3} && \text{;where } \frac{dV}{dt} = 0.2 \frac{m^3}{\text{min}} \text{ and } h = 0.3m\\
\\
\frac{dh}{dt} &= \frac{0.2}{10(0.3) + 3}\\
\\
\frac{dh}{dt} &= \frac{1}{30} \frac{m}{\text{min}} \text{ or } \frac{1}{30} \frac{\cancel{m}}{\text{min}} \left( \frac{100\text{cm}}{1m}\right) = \frac{10}{3} \frac{m}{\text{min}}
\end{aligned}
\end{equation}
$