# 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?