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Practice Application Task

Murray River Water Management

Area of Study:Functions and Graphs Task format:Application Task Working time:70 minutes Reading time:10 minutes Total marks:52 marks CAS calculator:Permitted

Instructions: Answer all questions in the spaces provided. Unless otherwise stated, an exact answer is required. Where more than one mark is allocated, appropriate working must be shown. Diagrams are not drawn to scale.

This is a practice paper. Your school's SAC may differ in length, marks, and time allocation. The format and difficulty are calibrated to be representative of a typical Unit 3 Application Task.

Scenario: Murray River Water Management

The Murray-Darling Basin Authority (MDBA) monitors river flow at a gauging station near Echuca, Victoria. Water levels fluctuate throughout the year due to seasonal rainfall, upstream dam releases, and irrigation demand. Engineers use mathematical models to predict flow rates and manage water allocation for farming communities along the river.

Part A: Seasonal Flow Model

14 marks

The average monthly flow rate at Echuca, measured in megalitres per day (ML/day), varies with the seasons. Engineers propose a model of the form: \[f(t) = a\sin\!\Bigl(\frac{\pi}{6}(t - c)\Bigr) + d\] where \(t\) is the time in months after January 1 (\(t = 0\) at the start of January) and \(a\), \(c\), \(d\) are positive constants.

Historical records show that the minimum flow of 1600 ML/day occurs in April (\(t = 3\)) and the maximum flow of 3200 ML/day occurs in October (\(t = 9\)). The pattern repeats each year.

Question 14 marks

a. Determine the values of \(a\) and \(d\). [2 marks]

b. Given that the model has a period of 12 months, show that \(b = \dfrac{\pi}{6}\) is already built into the given rule. [1 mark]

c. Using the fact that the maximum occurs at \(t = 9\), find the value of \(c\). [1 mark]

Have you genuinely attempted this question on paper first?

a. [2 marks]

\[d = \frac{3200 + 1600}{2} = 2400 \qquad a = \frac{3200 - 1600}{2} = 800\]

\(a = 800, \quad d = 2400\)

1 mark: correct \(d\). 1 mark: correct \(a\).

b. [1 mark]

Period \(= \frac{2\pi}{b} = \frac{2\pi}{\pi/6} = 12\) months, as required.

c. [1 mark]

Max when \(\sin = 1\), i.e. \(\frac{\pi}{6}(t - c) = \frac{\pi}{2}\). At \(t = 9\): \(9 - c = 3\).

\(c = 6\)

How did you go? (out of 4)

Question 23 marks

Write down the complete rule for \(f(t)\). State the domain and range of \(f\) in the context of a single calendar year (\(0 \leq t \leq 12\)).

\[f(t) = 800\sin\!\Bigl(\frac{\pi}{6}(t - 6)\Bigr) + 2400\] Domain: \([0, 12]\) Range: \([1600, 3200]\)

1 mark: correct rule. 1 mark: domain. 1 mark: range.

How did you go? (out of 3)

Question 34 marks

The MDBA classifies any flow rate above 3000 ML/day as high flow.

a. Solve \(f(t) = 3000\) for \(0 \leq t \leq 12\). Give your answers correct to two decimal places. [3 marks]

b. Hence determine the total number of months per year during which high-flow conditions are expected. Give your answer correct to one decimal place. [1 mark]

a. [3 marks]

\[\sin\!\Bigl(\frac{\pi}{6}(t-6)\Bigr) = \frac{600}{800} = 0.75\] \[\theta = \sin^{-1}(0.75) = 0.8481\ldots \quad \text{or} \quad \theta = \pi - 0.8481 = 2.2935\ldots\] \[t = 6 + \frac{6\theta}{\pi}: \quad t = 7.62 \quad \text{and} \quad t = 10.38\]

\(t = 7.62\) and \(t = 10.38\)

1 mark: correct equation setup. 1 mark: both values of \(\theta\). 1 mark: correct \(t\) values.

b. [1 mark]

Duration \(= 10.38 - 7.62 = 2.8\) months

How did you go? (out of 4)

Question 43 marks

Sketch the graph of \(y = f(t)\) for \(0 \leq t \leq 12\). Label the coordinates of the maximum point, the minimum point, and both endpoints.

Graph space

Key features: Endpoints \((0, 2400)\) and \((12, 2400)\). Minimum \((3, 1600)\). Maximum \((9, 3200)\). Crosses \(y = 2400\) at \(t = 0, 6, 12\). Smooth sinusoidal curve.

1 mark: correct shape. 1 mark: max/min labelled. 1 mark: endpoints labelled.

How did you go? (out of 3)

Part B: Dam Release Event

12 marks

When upstream reservoirs release water, the flow rate at Echuca temporarily increases above the seasonal baseline. A planned dam release on day \(d = 0\) produces additional flow modelled by: \[g(d) = 480\, e^{-0.15d}, \quad d \geq 0\] where \(g(d)\) is the additional flow in ML/day and \(d\) is the number of days after the release begins.

Question 51 mark

Evaluate \(g(0)\) and interpret the result in the context of the dam release.

\(g(0) = 480\) ML/day. The dam release initially adds 480 ML/day to the flow at Echuca.

How did you go? (out of 1)

Question 63 marks

The additional flow is considered negligible when it falls below 10 ML/day. Find the number of days after the release until the additional flow becomes negligible. Give your answer correct to one decimal place.

\[480\,e^{-0.15d} = 10 \quad\Rightarrow\quad e^{-0.15d} = \tfrac{1}{48} \quad\Rightarrow\quad d = \frac{\ln(48)}{0.15}\]

\(d \approx 25.8\) days

1 mark: correct equation. 1 mark: correct log work. 1 mark: answer.

How did you go? (out of 3)

Question 74 marks

a. Find the rule for the inverse function \(g^{-1}\). [2 marks]

b. State the domain of \(g^{-1}\). [1 mark]

c. Evaluate \(g^{-1}(100)\) and interpret the result in context. [1 mark]

a. [2 marks]

Swap and solve: \(\frac{d}{480} = e^{-0.15y} \;\Rightarrow\; y = \frac{20}{3}\ln\!\left(\frac{480}{d}\right)\)

\(g^{-1}(d) = \frac{20}{3}\ln\!\left(\frac{480}{d}\right)\)

b. [1 mark]

Domain: \((0,\, 480]\)

c. [1 mark]

\(g^{-1}(100) \approx 10.5\). It takes approximately 10.5 days for additional flow to decrease to 100 ML/day.

How did you go? (out of 4)

Question 84 marks

During a dam release in month \(t\), the total flow at Echuca on day \(d\) of the release is \(T = f(t) + g(d)\).

a. A dam release occurs at the start of August (\(t = 7\)). Find the total flow rate on day 5 of the release. Give your answer correct to the nearest ML/day. [2 marks]

b. The flood warning threshold is 3500 ML/day. Determine which month(s) a dam release of this type would cause the total flow to immediately exceed the flood warning level. Justify your answer. [2 marks]

a. [2 marks]

\(f(7) = 800\sin(\pi/6) + 2400 = 2800\). \(g(5) = 480e^{-0.75} \approx 226.7\).

Total \(\approx 3027\) ML/day

b. [2 marks]

Immediately: \(f(t) + 480 > 3500\), i.e. \(f(t) > 3020\).
\(f(8) \approx 3093 \Rightarrow 3573 > 3500\) (Sep: yes). \(f(9) = 3200 \Rightarrow 3680\) (Oct: yes). \(f(10) \approx 3093 \Rightarrow 3573\) (Nov: yes).
\(f(7) = 2800 \Rightarrow 3280 < 3500\) (Aug: no). \(f(11) = 2800\) (Dec: no).

September, October, and November.

How did you go? (out of 4)

Part C: Water Allocation

14 marks

The volume of water allocated each month to a nearby farming district depends on the number of registered farms. Due to infrastructure limits and sharing agreements, the total monthly allocation \(A\) (in ML) for \(n\) farms is modelled by: \[A(n) = -n^3 + 24n^2 - 80n, \quad 1 \leq n \leq 20\]

Question 93 marks

Factorise \(A(n)\) fully.

\[A(n) = -n(n^2 - 24n + 80) = -n(n - 4)(n - 20)\]

\(A(n) = -n(n-4)(n-20)\)

1 mark: factor out \(-n\). 1 mark: factorise quadratic. 1 mark: fully factorised.

How did you go? (out of 3)

Question 104 marks

a. State the values of \(A(4)\) and \(A(20)\). [1 mark]

b. Using CAS, find the coordinates of any turning points of \(y = A(n)\) for \(1 \leq n \leq 20\). Give coordinates correct to one decimal place. [2 marks]

c. Sketch the graph of \(y = A(n)\) for \(1 \leq n \leq 20\), labelling the turning point(s) and the zeros. [1 mark]

a. [1 mark]

\(A(4) = 0\) and \(A(20) = 0\)

b. [2 marks]

Local min near \((1.9, -72.2)\). Local max near \((14.1, 840.2)\).

1 mark per turning point.

c. [1 mark]

Cubic: negative leading coefficient. Zeros at \(n = 0, 4, 20\). Max near \((14.1, 840)\). Starts negative at \(n = 1\), crosses axis at \(n = 4\), rises to max, falls to zero at \(n = 20\).

How did you go? (out of 4)

Question 113 marks

The allocation per farm is given by \(\displaystyle P(n) = \frac{A(n)}{n}\).

a. Show that \(P(n) = -n^2 + 24n - 80\). [1 mark]

b. By completing the square or otherwise, find the number of farms that maximises the allocation per farm. State this maximum allocation per farm. [2 marks]

a. [1 mark]

\(P(n) = \frac{-n(n-4)(n-20)}{n} = -(n-4)(n-20) = -n^2 + 24n - 80\)

b. [2 marks]

\(P(n) = -(n^2 - 24n) - 80 = -(n-12)^2 + 144 - 80 = -(n-12)^2 + 64\)

Maximum at \(n = 12\) farms, giving 64 ML per farm.

How did you go? (out of 3)

Question 124 marks

The farming district requires a minimum total allocation of 200 ML per month to remain economically viable.

a. Solve \(A(n) \geq 200\) for \(1 \leq n \leq 20\). Give endpoints correct to one decimal place. [3 marks]

b. Interpret your answer in context, noting that \(n\) must be a whole number. [1 mark]

a. [3 marks]

Solve \(-n^3 + 24n^2 - 80n - 200 = 0\) using CAS. Roots at \(n \approx 6.3\) and \(n \approx 19.3\).

\(A(n) \geq 200\) for approximately \(6.3 \leq n \leq 19.3\)

b. [1 mark]

The district needs between 7 and 19 farms (inclusive) for at least 200 ML per month.

How did you go? (out of 4)

Part D: Model Adjustments and Connections

12 marks

Climate models and additional monitoring stations provide further data for the MDBA's planning.

Question 134 marks

A climate-adjusted flow model is given by:

\(\displaystyle h(t) = 640\sin\!\Bigl(\frac{\pi}{6}(t - 8.5)\Bigr) + 2200\)

Describe a sequence of transformations that maps the graph of \(y = \sin(t)\) onto the graph of \(y = h(t)\).

1. Dilation by factor 640 from the \(t\)-axis.
2. Dilation by factor \(\frac{6}{\pi}\) from the \(y\)-axis.
3. Translation 8.5 units in the positive \(t\)-direction.
4. Translation 2200 units in the positive \(y\)-direction.

1 mark per correct transformation (dilations before translations).

How did you go? (out of 4)

Question 143 marks

At a monitoring station upstream near Albury, the flow rate is modelled by:

\(r(t) = f(t + 1.5) + 350\)

a. Interpret the "\(+1.5\)" and the "\(+350\)" in the context of river flow between Albury and Echuca. [2 marks]

b. Find the maximum flow rate at Albury and the month in which it occurs. [1 mark]

a. [2 marks]

The \(+1.5\) means Albury's seasonal pattern leads Echuca by 1.5 months (upstream receives rainfall first). The \(+350\) means Albury's flow is 350 ML/day higher (water lost to irrigation between stations).

b. [1 mark]

Max at \(t = 7.5\) (mid-August): \(3200 + 350 = 3550\) ML/day.

How did you go? (out of 3)

Question 155 marks

A downstream water treatment plant has processing efficiency (as a percentage) that depends on the flow rate. When the flow rate at Echuca is \(r\) hundreds of ML/day, the efficiency is:

\(E(r) = 25\log_{10}(r) + 15, \quad r > 0\)

a. When the flow rate at Echuca is 2400 ML/day (\(r = 24\)), find the processing efficiency. Give your answer correct to one decimal place. [1 mark]

b. Find the flow rate (in ML/day) at which the plant operates at exactly 50% efficiency. Give your answer correct to the nearest whole number. [2 marks]

c. Write down the rule for the composite function \(\displaystyle E\!\left(\frac{f(t)}{100}\right)\) and evaluate the efficiency at \(t = 9\) (October). [2 marks]

a. [1 mark]

\(E(24) = 25\log_{10}(24) + 15 \approx 49.5\%\)

b. [2 marks]

\(25\log_{10}(r) + 15 = 50 \;\Rightarrow\; \log_{10}(r) = 1.4 \;\Rightarrow\; r = 10^{1.4} \approx 25.12\)

Flow rate \(\approx 2512\) ML/day

c. [2 marks]

\(E\!\left(\frac{f(t)}{100}\right) = 25\log_{10}\!\left(\frac{800\sin(\frac{\pi}{6}(t-6)) + 2400}{100}\right) + 15\)

At \(t = 9\): \(f(9) = 3200\), \(r = 32\). \(E(32) \approx 52.6\%\)

How did you go? (out of 5)

~ End of Application Task ~

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Before You Start

Practice Application Task

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Practice Application Task

Part A: Seasonal Flow Model

Part B: Dam Release Event

Part C: Water Allocation

Part D: Model Adjustments and Connections

Your Performance Breakdown