QCE Specialist Maths: Related rates: melting spherical shell Q170

Q170 · Practice questionTechnology-activeComplex unfamiliar8 marks

QUESTION 170 (8 marks)

A spherical ice shell surrounds a rigid spherical core of radius

3

cm. The outer radius is initially

5

cm. Ice volume is lost at a constant rate of

6\pi

cubic centimetres per minute. Melting occurs only at the outer surface, and the core remains unchanged. Let

r(t)

be the outer radius, in centimetres. The diagram is a central cross-section.

Equal-scale concentric circles representing a rigid core of radius 3 cm and an outer ice surface of radius r. The annular shell is lightly hatched and dimension arrows do not cross labels.

Derive

dr/dt

and determine

r(t)

before the ice has fully melted.

[3 marks]

Determine the time when all the ice has melted and state the physical time domain.

[2 marks]

When the outer radius is

4

cm, what fraction of the original ice remains? Determine the time when exactly half of the original ice remains.

[3 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

8 marks

ANSWER

(a)

dr/dt=-3/(2r^2)

and

r(t)=(125-9t/2)^{1/3}

. (b) All ice melts at

t=196/9

min. (c) At

r=4

, the remaining fraction is

37/98

; half the ice remains at

t=98/9

min.

Worked solution

(a) Ice volume is

V=4\pi(r^3-27)/3

. Differentiate:

\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=-6\pi, \qquad\frac{dr}{dt}=-\frac3{2r^2}.

Separating gives

r^2dr=-3dt/2

, so

r^3=125-9t/2

, using

r(0)=5

(b) All ice has melted when the outer radius reaches the core radius,

r=3

. Thus

27=125-9t/2

, giving

t=196/9

minutes. The physical model is used only for

0\le t\le196/9

; it must not be continued into the rigid core.

125-27=98

. At

r=4

, it is proportional to

64-27=37

, so the remaining fraction is

37/98

, not

1/2

. Because volume is lost at a constant rate, half the ice remains halfway through the melt time, at

t=98/9

. Equivalently set

r^3-27=49

, giving

r^3=76

. A halfway radius decrease is not a halfway volume decrease.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Use the shell volume and differentiate it.

[1 mark]

Part a: Obtain dr/dt=-3/(2r squared).

[1 mark]

Part a: Integrate and apply the initial condition.

[1 mark]

Part b: Use r=3 as the complete-melting condition.

[1 mark]

Part b: Obtain 196/9 min and the correct physical domain.

[1 mark]

Part c: Compare shell volumes and obtain 37/98.

[1 mark]

Part c: Use constant volume loss or r cubed=76 for half volume.

[1 mark]

Part c: Obtain 98/9 min.

[1 mark]

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

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Related rates: melting spherical shell — Question 170

QUESTION 170 (8 marks)

Practice marking scheme