(i)

$\frac{dV}{dt} = 300 - kV, k > 0$

$\int \frac{1}{300 - kV} dV = \int 1 dt$

$-\frac{1}{k} \mathrm{ln}|300 - kV| = t + C$

$300 - kV = Ae^{-kt}$

When $t = 0, V = 0 \Rightarrow A = 300$

$\therefore V=\frac{300(1-e^{-kt})}{k}$

When $t = 20, V=4500 \Rightarrow 4500 = \frac{300(1-e^{-20t})}{k}$

Using GC, $k = 0.030293$

When $V = 6000, t = 30.7$

$\Rightarrow 30.7- 20 = 10.7$ time interval between the first and second alarm.

(iii)
As $t \rightarrow \infty, V \rightarrow 9903 m^3$ which is impossible as the canal has only a fixed capacity of $6000 m^3$. Thus, the model is not valid for large values of $t$

(iv)
Assume that the weather condition remain unchanged.

Back to June Revision Exercise 3