Differential equations and related rates — Question 250

Original QCE Vault practice · 12 marks

Q250 · Practice questionTechnology-freeVery complex unfamiliar12 marks

QUESTION 250 (12 marks)

A fountain vessel is formed by rotating the profile

r(h)=\sqrt{h/(1-h)}

about the vertical axis for

0\le h\le3/4

. Here

h

is height and

r

is radius, both measured in metres. Water enters at

\pi/2

^3

/min and leaves at

\pi h

^3

/min. Initially

h=3/4

. Assume the flow model holds for

h>0

A symmetric vessel section with horizontal radius axis r and vertical height axis h, rim at height 3/4, and a dashed generic water level H. The section is formed from r squared equal to h divided by one minus h.

Determine the water volume

V(h)

and

dV/dh

[3 marks]

Derive a differential equation for

h(t)

and determine its equilibrium height.

[2 marks]

Find the exact time at which the height first reaches

3/5

[5 marks]

Explain why the water height approaches

1/2

m but does not reach it in any finite time.

[2 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

12 marks

ANSWER

(a)

V=\pi[-h-\ln(1-h)]

dV/dh=\pi h/(1-h)

. (b)

h^{\prime}=(1/2-h)(1-h)/h

; equilibrium

h=1/2

m. (c)

t=\ln(32/5)

min. (d)

h(t)\downarrow1/2

t\to\infty

, with no finite arrival.

Worked solution

(a)

V(h)=\pi\int_0^h u/(1-u)\,du

. Since

u/(1-u)=-1+1/(1-u)

, integration gives

V=\pi[-h-\ln(1-h)]

. Differentiation or the fundamental theorem gives

dV/dh=\pi h/(1-h)

(b) Net inflow gives

dV/dt=\pi(1/2-h)

. Applying

(dV/dh)h^{\prime}=dV/dt

gives

h^{\prime}=(1/2-h)(1-h)/h

. The only equilibrium in

0<h\le3/4

h=1/2

h>1/2

dt=-h\,dh/[(h-1/2)(1-h)]

. Decompose

h/[(h-1/2)(1-h)]=1/(h-1/2)+2/(1-h)

. An antiderivative is

G(h)=\ln(h-1/2)-2\ln(1-h)

. Thus

t=G(3/4)-G(h)=\ln[4(1-h)^2/(h-1/2)]

. At

h=3/5

this is

\ln(32/5)

(d) For

1/2<h\le3/4

h^{\prime}<0

. The time expression increases without bound as

h\downarrow1/2

, since its denominator tends to zero while its numerator tends to 1. Hence the height decreases towards the equilibrium and reaches it only in the limit.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Form the volume integral with the correct limits.

[1 mark]

Part a: Evaluate it, including the lower limit.

[1 mark]

Part a: Obtain the correct derivative.

[1 mark]

Part b: Use net flow and the chain rule to obtain the equation.

[1 mark]

Part b: Identify the physical equilibrium height.

[1 mark]

Part c: Separate variables with the correct sign.

[1 mark]

Part c: Obtain the two partial fractions.

[1 mark]

Part c: Integrate to the logarithmic expression.

[1 mark]

Part c: Apply the initial condition.

[1 mark]

Part c: Obtain the exact requested time.

[1 mark]

Part d: Use the sign of

h^{\prime}

to establish the decreasing approach.

[1 mark]

Part d: Use the diverging time expression to rule out finite arrival.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.

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Differential equations and related rates — Question 250

QUESTION 250 (12 marks)

Practice marking scheme

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