Porisma

Worked solution

Model reservoir inflow and quadratic outflow

During rainfall, a reservoir contains xx million cubic metres at time tt hours. Inflow is constant at kk million cubic metres per hour, and outflow is proportional to x2x^2.

(i) Given that the water amount stabilises at 0.50.5 million cubic metres, show that dx/dt=k(1−4x2)dx/dt=k(1-4x^2).

(ii) Find xx in terms of k,tk,t if initially there is 11 million cubic metre.

[2017/MJC/II/2 (modified)]

CJC · Discussion · Question 9

Answer

dx/dt=k(1−4x2)dx/dt=k(1-4x^2) and x=1+3e4kt6e4kt−2=12+13e4kt−1x=\dfrac{1+3e^{4kt}}{6e^{4kt}-2}=\dfrac12+\dfrac1{3e^{4kt}-1}.

(i) Use the equilibrium to calibrate outflow

Write dx/dt=k−cx2dx/dt=k-cx^2, with k,c>0k,c>0. At the equilibrium x=0.5x=0.5, 0=k−c(0.5)20=k-c(0.5)^2, so c=4kc=4k and dx/dt=k(1−4x2)dx/dt=k(1-4x^2).

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(ii) Separate and keep the logarithm sign information

For the branch starting at x=1x=1, x>1/2x>1/2. Integration gives 14ln⁡∣1+2x1−2x∣=kt+d\tfrac14\ln\left|\dfrac{1+2x}{1-2x}\right|=kt+d. Thus (1+2x)/(1−2x)=Ae4kt(1+2x)/(1-2x)=Ae^{4kt}, allowing the negative sign to be absorbed into AA. At t=0,x=1t=0,x=1, A=−3A=-3.

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(ii) Solve the resulting linear equation in x

1+2x=−3e4kt(1−2x)1+2x=-3e^{4kt}(1-2x) gives x=1+3e4kt6e4kt−2=12+13e4kt−1x=\dfrac{1+3e^{4kt}}{6e^{4kt}-2}=\dfrac12+\dfrac1{3e^{4kt}-1}. This satisfies x(0)=1x(0)=1 and, for t≥0t\ge0, decreases towards 1/21/2 without crossing it. The equilibrium excluded during separation is not this initial-value branch.

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Common mistakes

The logarithmic ratio is negative on this branch before taking its absolute value; retain that sign after exponentiating.

Read the hint