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Date	November 2021	Marks available	1	Reference code	21N.3.AHL.TZ0.2
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 0
Command term	Interpret	Question number	2	Adapted from	N/A

Question

This question explores models for the height of water in a cylindrical container as water drains out.

The diagram shows a cylindrical water container of height $3.2$ metres and base radius $1$ metre. At the base of the container is a small circular valve, which enables water to drain out.

Eva closes the valve and fills the container with water.

At time $t = 0$ , Eva opens the valve. She records the height, $h$ metres, of water remaining in the container every $5$ minutes.

Eva first tries to model the height using a linear function, $h (t) = a t + b$ , where $a, b \in ℝ$ .

Eva uses the equation of the regression line of $h$ on $t$ , to predict the time it will take for all the water to drain out of the container.

Eva thinks she can improve her model by using a quadratic function, $h (t) = p t^{2} + q t + r$ , where $p, q, r \in ℝ$ .

Eva uses this equation to predict the time it will take for all the water to drain out of the container and obtains an answer of $k$ minutes.

Let $V$ be the volume, in cubic metres, of water in the container at time $t$ minutes.
Let $R$ be the radius, in metres, of the circular valve.

Eva does some research and discovers a formula for the rate of change of $V$ .

$\frac{d V}{d t} = - π R^{2} \sqrt{70 560 h}$

Eva measures the radius of the valve to be $0.023$ metres. Let $T$ be the time, in minutes, it takes for all the water to drain out of the container.

Eva wants to use the container as a timer. She adjusts the initial height of water in the container so that all the water will drain out of the container in $15$ minutes.

Eva has another water container that is identical to the first one. She places one water container above the other one, so that all the water from the highest container will drain into the lowest container. Eva completely fills the highest container, but only fills the lowest container to a height of $1$ metre, as shown in the diagram.

At time $t = 0$ Eva opens both valves. Let $H$ be the height of water, in metres, in the lowest container at time $t$ .

Find the equation of the regression line of $h$ on $t$ .

[2]

a.i.

Interpret the meaning of parameter $a$ in the context of the model.

[1]

a.ii.

Suggest why Eva’s use of the linear regression equation in this way could be unreliable.

[1]

a.iii.

Find the equation of the least squares quadratic regression curve.

[1]

b.i.

Find the value of $k$ .

[2]

b.ii.

Hence, write down a suitable domain for Eva’s function $h (t) = p t^{2} + q t + r$ .

[1]

b.iii.

Show that $\frac{d h}{d t} = - R^{2} \sqrt{70 560 h}$ .

[3]

By solving the differential equation $\frac{d h}{d t} = - R^{2} \sqrt{70 560 h}$ , show that the general solution is given by $h = 17 640 {(c - R^{2} t)}^{2}$ , where $c \in ℝ$ .

[5]

Use the general solution from part (d) and the initial condition $h (0) = 3.2$ to predict the value of $T$ .

[4]

Find this new height.

[3]

Show that $\frac{d H}{d t} \approx 0.2514 - 0.009873 t - 0.1405 \sqrt{H}$ , where $0 \leq t \leq T$ .

[4]

g.i.

Use Euler’s method with a step length of $0.5$ minutes to estimate the maximum value of $H$ .

[3]

g.ii.

Markscheme

$h (t) = - 0.134 t + 3.1$ A1A1

Note: Award A1 for an equation in $h$ and $t$ and A1 for the coefficient $- 0.134$ and constant $3.1$ .

[2 marks]

a.i.

EITHER

the rate of change of height (of water in metres per minute) A1

Note: Accept “rate of decrease” or “rate of increase” in place of “rate of change”.

the (average) amount that the height (of the water) decreases each minute A1

[1 mark]

a.ii.

EITHER

unreliable to use $h$ on $t$ equation to estimate $t$ A1

unreliable to extrapolate from original data A1

rate of change (of height) might not remain constant (as the water drains out) A1

[1 mark]

a.iii.

$h (t) = 0.002 t^{2} - 0.174 t + 3.2$ A1

[1 mark]

b.i.

$0.002 t^{2} - 0.174 t + 3.2 = 0$ (M1)

$26.4 (26.4046 \dots)$ A1

[2 marks]

b.ii.

EITHER

$(0 \leq) t \leq 26.4 (t \leq 26.4046 \dots)$ A1

$(0 \leq) t \leq 20$ (due to range of original data / interpolation) A1

[1 mark]

b.iii.

$V = π {(1)}^{2} h$ (A1)

EITHER

$\frac{d V}{d t} = π \frac{d h}{d t}$ M1

attempt to use chain rule M1

$\frac{d h}{d t} = \frac{d h}{d V} \times \frac{d V}{d t}$

THEN

$\frac{d h}{d t} = \frac{1}{π} \times - π R^{2} \sqrt{70 560 h}$ A1

$\frac{d h}{d t} = - R^{2} \sqrt{70 560 h}$ AG

[3 marks]

attempt to separate variables M1

$\int \frac{1}{\sqrt{70 560 h}} d h = \int - R^{2} d t$ A1

$\frac{2 \sqrt{h}}{\sqrt{70 560}} = - R^{2} t + c$ A1A1

Note: Award A1 for each correct side of the equation.

$\sqrt{h} = \frac{\sqrt{70 560}}{2} (c - R^{2} t)$ A1

Note: Award the final A1 for any correct intermediate step that clearly leads to the given equation.

$h = 17 640 {(c - R^{2} t)}^{2}$ AG

[5 marks]

$t = 0 \Rightarrow 3.2 = 17640 c^{2}$ (M1)

$c = 0.0134687 \dots$ (A1)

substituting $h = 0$ and their non-zero value of $c$ (M1)

$T = \frac{c}{R^{2}} = \frac{0.0134687 \dots}{0 . 023^{2}}$

$= 25.5$ (minutes) $(25.4606 \dots)$ A1

[4 marks]

$h = 0 \Rightarrow c = R^{2} t$

$c = 0 . 023^{2} \times 15 (= 0.007935)$ (A1)

$t = 0 \Rightarrow h = 17640 {(0 . 023^{2} \times 15)}^{2}$ (M1)

$h = 1.11$ (metres) $(1.11068 \dots)$ A1

[3 marks]

let $h$ be the height of water in the highest container from parts (d) and (e) we get

$\frac{d h}{d t} = - 35280 R^{2} (0.0134687 \dots - R^{2} t)$ (M1)(A1)

so $\frac{d H}{d t} = 35280 R^{2} (0.0135 - R^{2} t) - R^{2} \sqrt{70560 H}$ M1A1

$(\frac{d H}{d t} = 18.6631 \dots (0.0134687 \dots - 0.000529 t) - 0.000529 \sqrt{70560 H})$

$(\frac{d H}{d t} = 0.251367 \dots - 0.0987279 \dots - 0.140518 \dots \sqrt{H})$

$\frac{d H}{d t} \approx 0.2514 - 0.009873 t - 0.1405 \sqrt{H}$ AG

[4 marks]

g.i.

evidence of using Euler’s method correctly

e.g. $y_{1} = 1.05545 \dots$ (A1)

maximum value of $H = 1.45$ (metres) (at $8.5$ minutes) A2

( $1.44678 \dots$ metres)

[3 marks]

g.ii.

Examiners report

All parts were answered well. In part(a)(i) a few candidates lost a mark from either not writing an equation or not using the variables $h$ and $t$ . In part (a)(ii) some candidates incorrectly stated it was the rate of change of water, instead of the rate of change of the height of the water. A few weaker candidates simply stated it is the gradient of the line. In part (a)(iii) some candidates incorrectly criticized the linear model, instead of addressing the question about why it could be unreliable to use the model to make a prediction about the future.

a.i.

a.ii.

a.iii.

This question was answered well by many candidates. In part (b)(ii) a small number of candidates incorrectly
gave two answers for $k$ , showing a lack of understanding of the context of the model.

b.i.

This question was answered well by many candidates. In part (b)(ii) a small number of candidates incorrectly
gave two answers for $k$ , showing a lack of understanding of the context of the model.

b.ii.

This question was answered well by many candidates. In part (b)(ii) a small number of candidates incorrectly
gave two answers for $k$ , showing a lack of understanding of the context of the model.

b.iii.

Many candidates recognized the need to use related rates of change, but could not present coherent working to reach the given answer. Often candidates either did not appreciate the need to use the equation for the volume of a cylinder or did not simplify their equation using $r = 1$ . Many candidates wrote nonsense arguments trying to cancel the factor of $π$ . In these long paper 3 questions, the purpose of “show that” parts is often to enable candidates to re-enter a question if they are unable to do a previous part.

Many candidates were able to correctly separate the variables, but many found the integral of $\frac{1}{\sqrt{h}}$ to be too difficult. A common incorrect approach was to use logarithms. A surprising number also incorrectly wrote $\int - R^{2} d t = - \frac{R^{3}}{3} + c$ , showing a lack of understanding of the difference between a parameter and a variable. Given that most questions in this course will be set in context, it is important that candidates learn to distinguish these differences.

Generally done well.

Many candidates found this question too difficult.

Part (g)(i) was often left blank and was the worst answered question on the paper. Part (g)(ii) was answered correctly by a number of candidates, who made use of the given answer from part (g)(i).

g.i.

Part (g)(i) was often left blank and was the worst answered question on the paper. Part (g)(ii) was answered correctly by a number of candidates, who made use of the given answer from part (g)(i).

g.ii.

Syllabus sections

Topic 4—Statistics and probability » SL 4.4—Pearsons, scatter diagrams, eqn of y on x

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Draw the regression line on the scatter diagram.
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EXN.1.SL.TZ0.8c.i:
Explain why your line should not be used to estimate the value of $m$ at which the temperature is $10.0$ $° C$ .
EXN.1.SL.TZ0.8d:
State a more appropriate model for the water temperature in the lake over an extended period of time. You are not expected to calculate any parameters.
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Use your line to find an estimate for the water temperature on the first day of May.
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Explain in context why your line should not be used to predict the value for December (month $12$ ).
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Use your answer to part (b) to write down the value of $n$ to the nearest integer.
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Label the point ${\text{M}}(\bar x,{\text{ }}\bar C)$ on the scatter diagram.
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Use the given equation of the regression line to estimate the number of IB Diploma points that this girl obtained.
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Use the equation of the regression line to estimate the least number of folders that the factory needs to sell in a month to exceed its production cost for that month.
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Calculate $x$ , the mean wind speed.
20N.2.SL.TZ0.T_1d.i:
Calculate $r$ , Pearson’s product–moment correlation coefficient.
20N.2.SL.TZ0.T_1d.ii:
Describe the correlation between the wind speed and the time to fully charge the robot.
20N.2.SL.TZ0.T_1e.i:
Write down the equation of the regression line $y$ on $x$ , in the form $y = m x + c$ .
19M.2.SL.TZ2.S_1a.i:
Write down the value of $a$ and of $b$ .
16N.2.SL.TZ0.T_1c:
Plot the point $(\bar x,{\text{ }}\bar y)$ on your scatter diagram and label this point M.
20N.2.SL.TZ0.T_1e.ii:
Draw this regression line on your scatter diagram.
19N.3.AHL.TZ0.Hsp_1b:
Peter uses the regression line of $y$ on $x$ as $y = 0.248x + 83.0$ and calculates that a student with a Mathematics test score of 73 will have a running time of 101 seconds. Comment on the validity of his calculation.
17M.2.SL.TZ1.T_1e:
Find the percentage error in your estimate in part (d).
20N.2.SL.TZ0.T_1e.iii:
Hence or otherwise estimate the charging time when the wind speed is $27 {km h}^{- 1}$ .
17M.2.SL.TZ1.T_1c:
Write down the equation of the regression line $y$ on $x$ , in the form $y = mx + c$ .
17M.2.SL.TZ2.T_3a:
Draw a scatter diagram for this data. Use a scale of 2 cm for 5000 folders on the horizontal axis and 2 cm for 10 000 Euros on the vertical axis.
21M.3.AHL.TZ2.1e.i:
Use Juliet’s data to find the value of $a$ and of $b$ .
21M.2.SL.TZ1.1f:
Find the percentage score calculated by Jason.
21M.2.SL.TZ1.1g:
State whether it is valid to use the regression line $y$ on $x$ for Jason’s estimate. Give a reason for your answer.
21M.2.SL.TZ1.1e:
Describe the correlation.
19N.1.SL.TZ0.T_6a:
Use your graphic display calculator to find the equation of the regression line $y$ on $x$ .
16N.2.SL.TZ0.T_1e:
Write down the equation of the regression line $y$ on $x$ for these eight male students.
19N.1.SL.TZ0.T_6b:
Use your regression equation to estimate the cost of a flight from Hong Kong to Tokyo with Galois Airways.
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Interpret, referring to income and happiness, what the value of $a$ represents.
17M.2.SL.TZ2.T_3f:
Use your graphic display calculator to find the equation of the regression line $C$ on $x$ .
20N.2.SL.TZ0.T_1c:
Plot and label the point $M$ on your scatter diagram.
EXN.1.SL.TZ0.8a:
Assuming the data follows a linear model for this period, find the regression line of $T$ on $m$ for the remaining data.
EXM.1.AHL.TZ0.15c.i:
find the value of $a$ and of $b$ .
EXN.1.AHL.TZ0.12b:
Find the equation of the least squares regression line of $\log_{10} P$ against $\log_{10} d$ .
EXN.1.AHL.TZ0.12c.ii:
Find an expression for $P$ in terms of $d$ .
21N.1.SL.TZ0.1b:
Comment on your answer to part (a), using the information that Eduardo found.
21N.1.AHL.TZ0.12a:
Use the data in the second table to find the value of $m$ and the value of $b$ for the regression line, $\ln x = m (\ln d) + b$ .
21N.1.AHL.TZ0.12b:
Assuming that the model found in part (a) remains valid, estimate the percentage of trees in stock when $d = 25$ .
21N.1.SL.TZ0.1c:
Write down the equation of the regression line of $t$ on $x$ , in the form $t = a x + b$ .
21N.1.SL.TZ0.1d:
A 57-year-old male also ran in the $5000 m$ race.

Use the equation of the regression line to estimate the time he took to complete the $5000 m$ race.
21N.3.AHL.TZ0.2a.i:
Find the equation of the regression line of $h$ on $t$ .
21N.3.AHL.TZ0.2a.iii:
Suggest why Eva’s use of the linear regression equation in this way could be unreliable.
21N.3.AHL.TZ0.2b.i:
Find the equation of the least squares quadratic regression curve.
21N.3.AHL.TZ0.2b.ii:
Find the value of $k$ .
21N.3.AHL.TZ0.2b.iii:
Hence, write down a suitable domain for Eva’s function $h (t) = p t^{2} + q t + r$ .
21N.3.AHL.TZ0.2c:
Show that $\frac{d h}{d t} = - R^{2} \sqrt{70 560 h}$ .
21N.3.AHL.TZ0.2d:
By solving the differential equation $\frac{d h}{d t} = - R^{2} \sqrt{70 560 h}$ , show that the general solution is given by $h = 17 640 {(c - R^{2} t)}^{2}$ , where $c \in ℝ$ .
21N.3.AHL.TZ0.2e:
Use the general solution from part (d) and the initial condition $h (0) = 3.2$ to predict the value of $T$ .
21N.3.AHL.TZ0.2f:
Find this new height.
21N.3.AHL.TZ0.2g.i:
Show that $\frac{d H}{d t} \approx 0.2514 - 0.009873 t - 0.1405 \sqrt{H}$ , where $0 \leq t \leq T$ .
21N.3.AHL.TZ0.2g.ii:
Use Euler’s method with a step length of $0.5$ minutes to estimate the maximum value of $H$ .

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