QCE Specialist Maths: Rates of change and differential equations: conical tank Q162

Q162 · Practice questionTechnology-activeComplex unfamiliar9 marks

QUESTION 162 (9 marks)

An inverted conical tank has height

6

m and top radius

3

m. Water depth above the vertex is

h

metres, and the radius of its surface is

r

metres. Water enters at

6\pi

cubic metres per minute and leaves at

2\pi h

cubic metres per minute. Initially

h(0)=4

. Assume the model applies for

0<h<6

Schematic section of an inverted cone, full height 6 m and top radius 3 m. Water surface radius r and depth h are labelled inside. Dashed centreline and dimension arrows are separate from the tank outline.

Show that

dh/dt=8(3-h)/h^2

[3 marks]

Determine the equilibrium depth and explain, using the model, whether the water can ever reach a depth of

2

[3 marks]

Determine the time taken for the depth to fall from

4

m to

3.5

m. Give an exact expression and a decimal value to four decimal places.

[3 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

9 marks

ANSWER

(a)

dh/dt=8(3-h)/h^2

. (b) Equilibrium depth 3 m; the water never reaches 2 m. (c)

t=27/64+(9/8)\ln2\approx1.2017

min.

Worked solution

(a) Similar triangles give

r=h/2

. Thus

V=\frac13\pi r^2h=\frac{\pi h^3}{12},\qquad \frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}.

Net inflow is

6\pi-2\pi h=2\pi(3-h)

. Equating rates gives

dh/dt=8(3-h)/h^2

(b) The equilibrium is

h=3

. Above 3 the depth decreases, and below 3 it increases. Starting at 4, separation shows that the elapsed time to a depth

h

with

3<h\le4

t=\frac18\int_h^4\frac{u^2}{u-3}\,du.

Since

u^2/(u-3)=u+3+9/(u-3)

, this integral diverges as

h\to3^+

. The depth approaches 3 from above but cannot reach or pass it in finite time; it therefore never reaches 2.

h=7/2

t=\frac18\left[\frac{u^2}{2}+3u+9\ln(u-3)\right]_{7/2}^{4} =\frac{27}{64}+\frac98\ln2\approx1.2017\ \mathrm{min}.

Both limits exceed the equilibrium, so the logarithm arguments are positive. The changing cross-sectional area is essential; volume loss cannot be converted to depth loss by a constant factor.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Use similarity to obtain r=h/2 and V=pi h cubed /12.

[1 mark]

Part a: Differentiate V and form the net flow.

[1 mark]

Part a: Obtain the required depth equation.

[1 mark]

Part b: Identify equilibrium h=3 and the sign of the rate above it.

[1 mark]

Part b: Use separation, uniqueness, or another valid argument to rule out crossing equilibrium.

[1 mark]

Part b: Conclude that depth 2 is never attained from depth 4.

[1 mark]

Part c: Separate variables with limits 3.5 and 4 and divide the rational expression.

[1 mark]

Part c: Integrate and obtain 27/64+(9/8)ln2.

[1 mark]

Part c: Give 1.2017 minutes.

[1 mark]

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

View the QCAA syllabus

How many marks did you earn?

Compare your working with the guide above.

Rates of change and differential equations: conical tank — Question 162

QUESTION 162 (9 marks)

Practice marking scheme