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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)=at+b, where a, b.

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)=pt2+qt+r, where p, q, 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.

dVdt=-πR270560h

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 ht=pt2+qt+r.

[1]
b.iii.

Show that dhdt=-R270560h.

[3]
c.

By solving the differential equation dhdt=-R270560h, show that the general solution is given by h=17640c-R2t2, where c.

[5]
d.

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

[4]
e.

Find this new height.

[3]
f.

Show that dHdt0.2514-0.009873t-0.1405H, where 0tT.

[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

ht=-0.134t+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”.


OR

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


OR

unreliable to extrapolate from original data        A1


OR

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

 

[1 mark]

a.iii.

ht=0.002t2-0.174t+3.2        A1

 

[1 mark]

b.i.

0.002t2-0.174t+3.2=0         (M1)

26.4  26.4046        A1

 

[2 marks]

b.ii.

EITHER

0t26.4   t26.4046           A1


OR

0t20 (due to range of original data / interpolation)           A1

 

[1 mark]

b.iii.

V=π12h               (A1)

EITHER

dVdt=πdhdt             M1


OR

attempt to use chain rule             M1

dhdt=dhdV×dVdt


THEN

dhdt=1π×-πR270560h           A1

dhdt=-R270560h           AG

 

[3 marks]

c.

attempt to separate variables             M1

170560hdh=-R2dt           A1

2h70560=-R2t+c           A1A1


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


h=705602c-R2t           A1


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


h=17640c-R2t2          AG

 

[5 marks]

d.

t=0  3.2=17640c2               (M1)

c=0.0134687               (A1)

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

T=cR2=0.01346870.0232

=25.5 (minutes)  25.4606           A1

 

[4 marks]

e.

h=0  c=R2t

c=0.0232×15 =0.007935              (A1)

t=0  h=176400.0232×152               (M1)

h=1.11 (metres)  1.11068           A1

 

[3 marks]

f.

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

dhdt=-35280R20.0134687-R2t               (M1)(A1)

so dHdt=35280R20.0135-R2t-R270560H               M1A1

dHdt=18.66310.0134687-0.000529t-0.00052970560H

dHdt=0.251367-0.0987279-0.140518H

dHdt0.2514-0.009873t-0.1405H           AG

 

[4 marks]

g.i.

evidence of using Euler’s method correctly

e.g. y1=1.05545               (A1)

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

(1.44678 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.

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.ii.

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.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.

c.

Many candidates were able to correctly separate the variables, but many found the integral of 1h to be too difficult. A common incorrect approach was to use logarithms. A surprising number also incorrectly wrote -R2dt=-R33+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.

d.

Generally done well.

e.

Many candidates found this question too difficult.

f.

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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Topic 4—Statistics and probability
Topic 5—Calculus

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