Rates of Change involving Integration


Rates of Change for finding height and time can be solved by integration. Once integrated, find the integral constant using the given conditions.

Worked Examples for Rates of Change involving Integration

The diagram represents a vertical cylindrical water cooler of constant cross-sectional area \(A\). Water drains through a hole at the bottom of the cooler. From physical principles, it is known that the volume \(V\) of water decreases at a rate given by
$$ \displaystyle \frac{dV}{dt} = -k \sqrt{y}, $$
where \(k\) is a positive constant and \(y\) is the depth of water.
Initially the cooler is full and it takes \(T\) seconds to drain. Thus \(y = y_0 \) when \(t=0\), and \(y=0\) when \(t=T\).

(a)    Show that \(\displaystyle \frac{dy}{dt} = – \frac{k}{A} \sqrt{y} \).

\( \begin{aligned} \displaystyle \require{color}
V &= \text{cross-sectional area } \times \text{ height} \\
&= Ah \\
\frac{dy}{dt} &= \frac{dy}{dV} \times \frac{dV}{dt} \\
&= \frac{1}{A} \times -k \sqrt{y} \\
\therefore \frac{dy}{dt} &= – \frac{k}{A} \sqrt{y} \\
\end{aligned} \\ \)

(b)    Show that \( \displaystyle y = y_0 \bigg(1-\frac{t}{T}\bigg)^2 \text{ for } 0 \le t \le T \).

\( \begin{aligned} \displaystyle \require{color}
\frac{dt}{dy} &= – \frac{A}{k} \frac{1}{\sqrt{y}} &\color{red} \text{by } (a) \\
t &= – \frac{A}{k} \int \frac{1}{\sqrt{y}} dy \\
&= – \frac{2A}{k} \sqrt{y} + C \\
T &= – \frac{2A}{k} \sqrt{0} + C &\color{red} y=0,t=T \\
T=C \\
t &= – \frac{2A}{k} \sqrt{y} + T \\
\frac{2A}{k} \sqrt{y} &= T – t \color{red} \cdots (1) \\
\frac{2A}{k} \sqrt{y_0} &= T – 0 \color{red} \cdots (2) &\color{red} y=y_0,t=0 \\
\frac{\sqrt{y}}{\sqrt{y_0}} &= \frac{T-t}{T} &\color{red} (1) \div (2) \\
&= 1- \frac{t}{T} \\
\frac{y}{y_0} &= \bigg(1 – \frac{t}{T} \bigg)^2 \\
\therefore y &= y_0 \bigg(1 – \frac{t}{T} \bigg)^2 \\
\end{aligned} \\ \)

(c)    Suppose it takes 10 seconds for half the water to drain out. How long does it take to empty the full cooler?

\( \begin{aligned} \displaystyle \require{color}
\frac{1}{2}y_0 &= y_0\bigg(1 – \frac{10}{T}\bigg)^2 &\color{red} t=10,y= \frac{1}{2} y_0 \\
\frac{1}{2} &= \bigg(1 – \frac{10}{T}\bigg)^2 \\
\frac{1}{\sqrt{2}} &= 1 – \frac{10}{T} \\
\frac{10}{T} &= 1 – \frac{1}{\sqrt{2}} \\
\frac{10}{T} &= \frac{\sqrt{2}-1}{\sqrt{2}} \\
\frac{T}{10} &= \frac{\sqrt{2}}{\sqrt{2}-1} \\
T &= \frac{10\sqrt{2}}{\sqrt{2}-1} \\
\therefore T &\approx 34.1 \\
\end{aligned} \\ \)

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