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?

Login or Register to Leave Comments. It's Free!