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 \(h\) .
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.
If the glass is filled completely, how long will it take for all the water to evaporate?
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]
\(\frac{{{\text{d}}V}}{{{\text{d}}t}} = - 3 \times \) 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]
\({V_0} = 5000\pi \) (\( = 15700\) 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]
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.
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.
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.