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Date May 2011 Marks available 3 Reference code 11M.2.hl.TZ1.14
Level HL only Paper 2 Time zone TZ1
Command term Find Question number 14 Adapted from N/A

Question

An open glass is created by rotating the curve \(y = {x^2}\) , defined in the domain \(x \in [0,10]\), \(2\pi \) radians about the y-axis. Units on the coordinate axes are defined to be in centimetres.

When the glass contains water to a height \(h\) cm, find the volume \(V\) of water in terms of \(h\) .

[3]
a.

If the water in the glass evaporates at the rate of 3 cm3 per hour for each cm2 of exposed surface area of the water, show that,

\(\frac{{{\text{d}}V}}{{{\text{d}}t}} = - 3\sqrt {2\pi V} \) , where \(t\) is measured in hours.

[6]
b.

If the glass is filled completely, how long will it take for all the water to evaporate?

[7]
c.

Markscheme

volume \( = \pi \int_0^h {{x^2}{\text{d}}y} \)     (M1)

\(\pi \int_0^h {y{\text{d}}y} \)     M1

\(\pi \left[ {\frac{{{y^2}}}{2}} \right]_0^h = \frac{{\pi {h^2}}}{2}\)     A1

[3 marks]

a.

\(\frac{{{\text{d}}V}}{{{\text{d}}t}} = - 3 \times \) surface area     A1

surface area \( = \pi {x^2}\)     (M1)

\( = \pi h\)     A1

\(V = \frac{{\pi {h^2}}}{2} \Rightarrow h\sqrt {\frac{{2V}}{\pi }} \)     M1A1

\(\frac{{{\text{d}}V}}{{{\text{d}}t}} = - 3\pi \sqrt {\frac{{2V}}{\pi }} \)     A1

\(\frac{{{\text{d}}V}}{{{\text{d}}t}} = - 3\sqrt {2\pi V} \)     AG

Note: Assuming that \(\frac{{{\text{d}}h}}{{{\text{d}}t}} = - 3\) without justification gains no marks.

[6 marks]

b.

\({V_0} = 5000\pi \) (\( = 15700\) cm3)     A1

\(\frac{{{\text{d}}V}}{{{\text{d}}t}} =  - 3\sqrt {2\pi V} \)

attempting to separate variables     M1

EITHER

\(\int {\frac{{{\text{d}}V}}{{\sqrt V }}}  =  - 3\sqrt {2\pi } \int {{\text{d}}t} \)     A1

\(2\sqrt V  =  - 3\sqrt {2\pi t}  + c\)     A1

\(c = 2\sqrt {5000\pi } \)     A1

\(V = 0\)     M1

\( \Rightarrow t = \frac{2}{3}\sqrt {\frac{{5000\pi }}{{2\pi }}}  = 33\frac{1}{3}\) hours     A1

 

OR

\(\int_{5000\pi }^0 {\frac{{{\text{d}}V}}{{\sqrt V }}}  =  - 3\sqrt {2\pi } \int_0^T {{\text{d}}t} \)     M1A1A1

Note: Award M1 for attempt to use definite integrals, A1 for correct limits and A1 for correct integrands.

\(\left[ {2\sqrt V } \right]_{5000\pi }^0 = 3\sqrt {2\pi } T\)     A1

\(T = \frac{2}{3}\sqrt {\frac{{5000\pi }}{{2\pi }}}  = 33\frac{1}{3}\) hours     A1

 

[7 marks]

c.

Examiners report

This question was found to be challenging by many candidates and there were very few completely correct solutions. Many candidates did not seem able to find the volume of revolution when taken about the y-axis in (a). Candidates did not always recognize that part (b) did not involve related rates. Those candidates who attempted the question made some progress by separating the variables and integrating in (c) but very few were able to identify successfully the values necessary to find the correct answer.

a.

This question was found to be challenging by many candidates and there were very few completely correct solutions. Many candidates did not seem able to find the volume of revolution when taken about the y-axis in (a). Candidates did not always recognize that part (b) did not involve related rates. Those candidates who attempted the question made some progress by separating the variables and integrating in (c) but very few were able to identify successfully the values necessary to find the correct answer.

b.

This question was found to be challenging by many candidates and there were very few completely correct solutions. Many candidates did not seem able to find the volume of revolution when taken about the y-axis in (a). Candidates did not always recognize that part (b) did not involve related rates. Those candidates who attempted the question made some progress by separating the variables and integrating in (c) but very few were able to identify successfully the values necessary to find the correct answer.

c.

Syllabus sections

Topic 6 - Core: Calculus » 6.5 » Volumes of revolution about the \(x\)-axis or \(y\)-axis.
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