# 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 $\displaystyle \frac{dy}{dt} = – \frac{k}{A} \sqrt{y}$.

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

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