June Revision Exercise 3 Q11

By KS Teng

(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

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June Revision Exercise 3 Q10JC Mathematics
June Revision ExerciseJC Mathematics