# 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} \\