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Practice Application Task

Renewable Energy: Solar Farm Analysis

Area of Study:Calculus Task format:Application Task Working time:85 minutes Reading time:10 minutes Total marks:58 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 covering calculus.

Scenario: Renewable Energy - Solar Farm Analysis

A solar energy company operates a photovoltaic farm near Mildura, Victoria. Engineers model the power output of the installation throughout the day, analyse energy production, and compare solar performance with a nearby wind farm. Mathematical models are used to optimise the system and inform decisions about energy storage and grid supply.

Part A: Power Output Modelling

13 marks

The power output $P$ (in kilowatts) of the solar farm $t$ hours after sunrise is modelled by: \[P(t) = 250t\, e^{-0.2t} \quad \text{for } 0 \leq t \leq 14\] The panels also have a temperature function. The panel temperature $\theta$ (in °C) at time $t$ hours after sunrise is given by: \[\theta(t) = 25 + 40\!\left(1 - e^{-0.2t}\right)\]

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Question 1 4 marks

a. Evaluate $P(0)$ and $P(5)$. Give $P(5)$ correct to one decimal place. [2 marks]

b. Using the product rule, show that $P'(t) = 250(1 - 0.2t)\,e^{-0.2t}$. [2 marks]

Have you genuinely attempted this question on paper first?

a. [2 marks]

\[P(0) = 250(0)\,e^{0} = 0\]

\[P(5) = 250(5)\,e^{-1} = 1250\,e^{-1} = 459.9 \text{ kW}\]

$P(0) = 0$ kW, \quad $P(5) = 459.9$ kW

1 mark: $P(0) = 0$. 1 mark: correct evaluation of $P(5)$.

b. [2 marks]

Let $u = 250t$ and $v = e^{-0.2t}$. Then $u' = 250$ and $v' = -0.2\,e^{-0.2t}$.

\[P'(t) = u'v + uv' = 250\,e^{-0.2t} + 250t(-0.2)\,e^{-0.2t}\]

\[= 250\,e^{-0.2t}(1 - 0.2t)\]

$P'(t) = 250(1 - 0.2t)\,e^{-0.2t}$ as required.

1 mark: correct application of product rule. 1 mark: simplification to required form.

Question 2 2 marks

a. Calculate $P'(2)$ correct to one decimal place. Interpret this result in the context of the solar farm. [2 marks]

Have you genuinely attempted this question on paper first?

a. [2 marks]

\[P'(2) = 250(1 - 0.4)\,e^{-0.4} = 250 \times 0.6 \times e^{-0.4} = 100.5 \text{ kW/hr}\]

$P'(2) = 100.5$ kW per hour. Two hours after sunrise, the power output is increasing at a rate of 100.5 kilowatts per hour.

1 mark: correct numerical value. 1 mark: interpretation in context.

Question 3 4 marks

a. Find the exact time at which the power output reaches its maximum. [1 mark]

b. Find the maximum power output, giving your answer correct to the nearest kilowatt. [1 mark]

c. Find $P''(t)$. [1 mark]

d. Use the second derivative to confirm that the stationary point is a maximum. [1 mark]

Have you genuinely attempted this question on paper first?

a. [1 mark]

$P'(t) = 0$ when $1 - 0.2t = 0$, i.e. $t = 5$ hours.

$t = 5$ hours (5 hours after sunrise)

b. [1 mark]

\[P(5) = 250 \times 5 \times e^{-1} = \frac{1250}{e} \approx 459.9 \approx 460 \text{ kW}\]

Maximum power output is approximately 460 kW.

c. [1 mark]

Applying the product rule to $P'(t) = 250(1-0.2t)\,e^{-0.2t}$: \[P''(t) = 250\bigl[(-0.2)\,e^{-0.2t} + (1-0.2t)(-0.2)\,e^{-0.2t}\bigr]\] \[= -50\,e^{-0.2t}(1 - 0.2t)\]

$P''(t) = -50(1 - 0.2t)\,e^{-0.2t}$

d. [1 mark]

\[P''(5) = -50(2 - 1)\,e^{-1} = \frac{-50}{e} \approx -18.4 < 0\]

Since $P''(5) < 0$, the stationary point is a local maximum.

Question 4 3 marks

Panel efficiency $E$ (as a percentage) depends on temperature according to:

\[E(\theta) = 100 - 0.4(\theta - 25)\]

a. Find $\dfrac{d\theta}{dt}$. [1 mark]

b. Using the chain rule, find $\dfrac{dE}{dt}$ when $t = 2$. Give your answer correct to two decimal places and interpret the result. [2 marks]

Have you genuinely attempted this question on paper first?

a. [1 mark]

\[\frac{d\theta}{dt} = 40 \times 0.2\,e^{-0.2t} = 12\,e^{-0.2t}\]

$\dfrac{d\theta}{dt} = 12\,e^{-0.2t}$

b. [2 marks]

$\dfrac{dE}{d\theta} = -0.4$. By the chain rule: $\dfrac{dE}{dt} = \dfrac{dE}{d\theta} \times \dfrac{d\theta}{dt} = -0.4 \times 8\,e^{-0.2t} = -3.2\,e^{-0.2t}$

At $t = 2$: $\dfrac{dE}{dt} = -3.2\,e^{-0.4} = -2.14$% per hour.

$\dfrac{dE}{dt}\big|_{t=2} = -2.14$% per hour. Two hours after sunrise, panel efficiency is decreasing at a rate of 2.14 percentage points per hour due to rising panel temperature.

1 mark: correct chain rule setup and calculation. 1 mark: interpretation.

Part B: Daily Energy Production

17 marks

The rate of energy production (in kWh per hour) follows the same model as Part A: \[R(t) = 250t\,e^{-0.2t} \quad \text{for } 0 \leq t \leq 14\] where $t = 0$ is sunrise and the effective daylight period lasts 14 hours.

Question 5 5 marks

a. Using CAS or otherwise, find $\displaystyle\int 250t\,e^{-0.2t}\,dt$. [2 marks]

b. Verify your result by differentiating your answer. [2 marks]

c. Evaluate the total energy produced from sunrise to $t = 5$ hours. Give your answer correct to the nearest kWh. [1 mark]

Have you genuinely attempted this question on paper first?

a. [2 marks]

Using integration by parts: $u = t$, $dv = 250e^{-0.2t}dt$, so $du = dt$, $v = -1250e^{-0.2t}$. \[\int 250t\,e^{-0.2t}\,dt = -1250te^{-0.2t} + 1250\int e^{-0.2t}\,dt = -1250te^{-0.2t} - 6250e^{-0.2t}\] \[= -1250(t+5)\,e^{-0.2t} + C\]

$\displaystyle\int 250t\,e^{-0.2t}\,dt = -1250(t + 5)\,e^{-0.2t} + C$

1 mark: correct anti-derivative (integration by parts). 1 mark: correct simplification to factored form.

b. [2 marks]

\[\frac{d}{dt}\!\left[-1250(t+5)\,e^{-0.2t}\right] = -1250\bigl[e^{-0.2t} + (t+5)(-0.2)\,e^{-0.2t}\bigr]\] \[= -1250\,e^{-0.2t}[1 - 0.2t - 1] = -1250\,e^{-0.2t}(-0.2t) = 250t\,e^{-0.2t} \; \checkmark\]

1 mark: correct differentiation using product rule. 1 mark: simplification confirms original integrand.

c. [1 mark]

\[\int_0^5 250t\,e^{-0.2t}\,dt = \left[-1250(t+5)\,e^{-0.2t}\right]_0^5\] \[= -1250(10)\,e^{-1} - (-1250(5)\,e^0) = -\frac{12500}{e} + 6250 \approx -4598.5 + 6250 = 1651.5\]

Total energy produced in the first 5 hours is approximately 1652 kWh.

Question 6 5 marks

a. Calculate the total energy produced over the full 14-hour daylight period. Give your answer correct to the nearest kWh. [2 marks]

b. Find the average power output over the 14-hour period. Give your answer correct to one decimal place and interpret this result. [3 marks]

Have you genuinely attempted this question on paper first?

a. [2 marks]

\[\int_0^{14} 250t\,e^{-0.2t}\,dt = \left[-1250(t+5)\,e^{-0.2t}\right]_0^{14}\] At $t = 14$: $-1250(19)\,e^{-2.8} = -23750 \times 0.0608 \approx -1443.6$
At $t = 0$: $-1250(5) = -6250$

Total energy $= -1443.6 - (-6250) = 4806$ kWh

1 mark: correct setup of definite integral. 1 mark: correct evaluation and rounding.

b. [3 marks]

\[\text{Average power} = \frac{1}{14}\int_0^{14} R(t)\,dt = \frac{4806}{14} = 343.3 \text{ kW}\]

The average power output over the 14-hour period is 343.3 kW.

1 mark: correct average value formula setup. 1 mark: correct substitution and evaluation. 1 mark: correct answer with units and interpretation.

Question 7 3 marks

On a cloudy day, the power output is reduced to $R_c(t) = 150t\,e^{-0.2t}$ for $0 \leq t \leq 14$.

a. Without recalculating the integral from scratch, explain how the total energy on a cloudy day relates to the clear-day total found in Question 6. [1 mark]

b. The company needs at least 3500 kWh per day to meet its grid supply commitments. Determine whether each day type (clear and cloudy) meets this requirement. [2 marks]

Have you genuinely attempted this question on paper first?

a. [1 mark]

Since $R_c(t) = \frac{150}{250} \times R(t) = 0.6 \times R(t)$, the cloudy-day total is 60% of the clear-day total.

Cloudy total $= 0.6 \times 4806 \approx 2884$ kWh

b. [2 marks]

Clear day: 4806 kWh > 3500 kWh. The requirement is met.
Cloudy day: 2884 kWh < 3500 kWh. The requirement is not met.

1 mark: correct cloudy day total. 1 mark: comparison with requirement for both.

Question 8 4 marks

The company installs a battery storage system. The battery charges whenever the power output exceeds the farm's own consumption of 300 kW. The net charging rate is:

\[B(t) = R(t) - 300 \quad \text{when } R(t) \geq 300\]

a. Find the values of $t$ (correct to two decimal places) for which $R(t) = 300$. [2 marks]

b. Calculate the total energy stored in the battery during the period when $R(t) \geq 200$. Give your answer correct to the nearest kWh. [2 marks]

Have you genuinely attempted this question on paper first?

a. [2 marks]

$250t\,e^{-0.2t} = 300$, i.e. $t\,e^{-0.2t} = 1.2$. Using CAS:

$t \approx 1.69$ and $t \approx 11.12$

1 mark each for each correct intersection.

b. [2 marks]

\[\text{Energy stored} = \int_{1.69}^{11.12}(250t\,e^{-0.2t} - 300)\,dt\] \[= \left[-1250(t+5)\,e^{-0.2t} - 300t\right]_{1.69}^{11.12}\] Using CAS to evaluate:

Total energy stored $\approx 951$ kWh

1 mark: correct integral setup. 1 mark: correct evaluation.

Part C: System Optimisation

14 marks

The company is expanding operations. Engineers need to optimise the physical layout and economic performance of the farm.

Question 9 5 marks

A new rectangular solar panel array must have a total panel area of 1200 m². A security fence will surround the array, with a 3-metre gap between the panels and the fence on all sides. Let the panel array have dimensions $x$ metres by $y$ metres.

a. Show that $y = \dfrac{1200}{x}$. [1 mark]

b. Show that the total area enclosed by the fence is $A(x) = 1236 + 6x + \dfrac{7200}{x}$. [1 mark]

c. Find the value of $x$ that minimises the total fenced area, and find this minimum area correct to the nearest m². [3 marks]

Have you genuinely attempted this question on paper first?

a. [1 mark]

Panel area: $xy = 1200$, so $y = \dfrac{1200}{x}$.

b. [1 mark]

Fenced dimensions: $(x+6)$ by $(y+6)$. \[A = (x+6)(y+6) = xy + 6x + 6y + 36 = 1200 + 6x + \frac{7200}{x} + 36 = 1236 + 6x + \frac{7200}{x}\]

c. [3 marks]

\[A'(x) = 6 - \frac{7200}{x^2} = 0 \quad \Rightarrow \quad x^2 = 1200 \quad \Rightarrow \quad x = 20\sqrt{3} \approx 34.64 \text{ m}\]

\[A = 1236 + 6(20\sqrt{3}) + \frac{7200}{20\sqrt{3}} = 1236 + 120\sqrt{3} + \frac{360}{\sqrt{3}} = 1236 + 120\sqrt{3} + 120\sqrt{3} = 1236 + 240\sqrt{3}\]

The value of $x$ that minimises total fenced area is $x = 20\sqrt{3}$ metres. The minimum area is approximately 1652 m².

1 mark for differentiation. 1 mark for equating and solving. 1 mark for evaluating the minimum area.

Question 10 5 marks

Each panel in the array generates annual revenue of $\$(2400 - 2n)$, where $n$ is the total number of panels installed. (As more panels are installed, increased shading reduces the per-panel output.)

The annual operating cost for the array is $\$(100n + 5000)$.

a. Show that the annual profit is $\Pi(n) = -2n^2 + 2300n - 5000$. [1 mark]

b. Find the number of panels that maximises annual profit. [1 mark]

c. Find the maximum annual profit. [1 mark]

d. The company requires at least \$500,000 annual profit. Find the range of values of $n$ for which this is achieved. [2 marks]

Have you genuinely attempted this question on paper first?

a. [1 mark]

Revenue $= n(2400 - 2n) = 2400n - 2n^2$. \[\Pi = 2400n - 2n^2 - 100n - 5000 = -2n^2 + 2300n - 5000\]

b. [1 mark]

$\Pi'(n) = -4n + 2300 = 0 \;\Rightarrow\; n = 575$

575 panels maximises profit.

c. [1 mark]

\[\Pi(575) = -2(575)^2 + 2300(575) - 5000 = -661\,250 + 1\,322\,500 - 5000 = 656\,250\]

Maximum annual profit is \$656,250.

d. [2 marks]

$-2n^2 + 2300n - 5000 \geq 500\,000$ \[-2n^2 + 2300n - 505\,000 \geq 0 \quad \Rightarrow \quad 2n^2 - 2300n + 505\,000 \leq 0\] \[n = \frac{2300 \pm \sqrt{2300^2 - 4(2)(505\,000)}}{2(2)} = \frac{2300 \pm \sqrt{5\,290\,000 - 4\,040\,000}}{4} = \frac{2300 \pm \sqrt{1\,250\,000}}{4}\] \[\sqrt{1\,250\,000} = 500\sqrt{5} \approx 1118.0\] \[n = \frac{2300 \pm 1118.0}{4}\]

$n \approx 295.5$ or $n \approx 854.5$, so $296 \leq n \leq 854$.

1 mark: correct inequality setup. 1 mark: correct range.

Question 11 4 marks

As the solar farm ages, the annual maintenance cost (in thousands of dollars) in year $t$ after installation is modelled by:

\[M(t) = 5 + 3e^{0.12t}, \quad \text{for } t \geq 0\]

a. Find the total maintenance cost over the first 10 years after installation. Give your answer correct to the nearest thousand dollars. [2 marks]

b. Find the average annual maintenance cost over the first 10 years. Give your answer correct to the nearest thousand dollars. [2 marks]

Have you genuinely attempted this question on paper first?

a. [2 marks]

\[\int_0^{10} (5 + 3e^{0.12t})\,dt = \left[5t + \frac{3}{0.12}e^{0.12t}\right]_0^{10} = \left[5t + 25e^{0.12t}\right]_0^{10}\]

\[= (50 + 25e^{1.2}) - (0 + 25e^0) = 50 + 25(3.3201) - 25 = 50 + 83.0 - 25 = 108\]

Total maintenance cost over first 10 years is approximately $108,000 or $108 thousand.

1 mark: correct anti-derivative. 1 mark: correct evaluation.

b. [2 marks]

\[\text{Average annual cost} = \frac{1}{10}\int_0^{10} M(t)\,dt = \frac{108}{10} = 10.8 \text{ thousand dollars}\]

Average annual maintenance cost over the first 10 years is approximately $10,800 or $10.8 thousand per year.

1 mark: correct average value formula. 1 mark: correct evaluation.

Part D: Comparing Energy Sources

14 marks

The solar farm's output is compared with a nearby wind farm. The wind farm's power output $W$ (in kW) over the same 14-hour daylight period is modelled by: \[W(t) = 180 + 60\cos\!\left(\frac{\pi t}{7}\right) \quad \text{for } 0 \leq t \leq 14\]

Question 12 4 marks

a. Find the power output of the wind farm at $t = 0$, $t = 7$, and $t = 14$. [2 marks]

b. Using CAS, find the values of $t$ in $[0, 14]$ for which $S(t) = W(t)$, where $S(t) = 250t\,e^{-0.2t}$. Give your answers correct to two decimal places. [2 marks]

Have you genuinely attempted this question on paper first?

a. [2 marks]

$W(0) = 180 + 60\cos(0) = 240$ kW

$W(7) = 180 + 60\cos(\pi) = 180 - 60 = 120$ kW

$W(14) = 180 + 60\cos(2\pi) = 240$ kW

$W(0) = 240$, $W(7) = 120$, $W(14) = 240$ kW.

b. [2 marks]

Solve $250t\,e^{-0.2t} = 180 + 60\cos(\pi t/7)$ using CAS:

$t \approx 1.17$ and $t \approx 13.15$

1 mark each for each intersection point.

Question 13 5 marks

a. For what time interval does the solar farm produce more power than the wind farm? [1 mark]

b. Calculate the total amount of energy (in kWh) by which the solar farm exceeds the wind farm during the interval found in part (a). Give your answer correct to the nearest kWh. [2 marks]

c. Calculate the total energy produced by the wind farm over $[0, 14]$. [2 marks]

Have you genuinely attempted this question on paper first?

a. [1 mark]

Solar exceeds wind for $1.17 < t < 13.15$ (approximately 11.98 hours).

b. [2 marks]

\[\int_{1.17}^{13.15}\!\bigl[S(t) - W(t)\bigr]\,dt = \int_{1.17}^{13.15}\!\bigl[250t\,e^{-0.2t} - 180 - 60\cos(\pi t/7)\bigr]\,dt\] Using CAS to evaluate:

Solar excess $\approx 2250$ kWh

1 mark: correct integral expression. 1 mark: correct evaluation.

c. [2 marks]

\[\int_0^{14}\!\bigl[180 + 60\cos(\pi t/7)\bigr]\,dt = \left[180t + \frac{420}{\pi}\sin(\pi t/7)\right]_0^{14}\] \[= 180(14) + \frac{420}{\pi}\sin(2\pi) - 0 = 2520 + 0\]

Total wind energy = 2520 kWh.

1 mark: correct anti-derivative. 1 mark: correct evaluation.

Question 14 5 marks

The company proposes a hybrid system that always uses whichever source produces more power at any given time. Let $H(t) = \max\!\bigl(S(t),\, W(t)\bigr)$.

a. Write a piecewise definition for $H(t)$ over $[0,\, 14]$, using the intersection values from Question 12. [1 mark]

b. Calculate the total energy produced by the hybrid system over $[0, 14]$. Give your answer correct to the nearest kWh. [2 marks]

c. How much more energy (as a percentage) does the hybrid system produce compared to using only the wind farm? Give your answer correct to one decimal place. [2 marks]

Have you genuinely attempted this question on paper first?

a. [1 mark]

\[H(t) = \begin{cases} W(t) & 0 \leq t \leq 1.17 \\ S(t) & 1.17 < t < 13.15 \\ W(t) & 13.15 \leq t \leq 14 \end{cases}\]

b. [2 marks]

The hybrid system uses wind when wind is higher, and solar when solar is higher. \[\int_0^{14} H(t)\,dt = \int_0^{14} W(t)\,dt + \int_{1.17}^{13.15}\bigl[S(t) - W(t)\bigr]\,dt\] \[= 2520 + 2250 = 4770 \text{ kWh}\]

Total hybrid energy = 4770 kWh.

1 mark: recognising method (wind total + solar excess). 1 mark: correct answer.

c. [2 marks]

\[\text{Extra energy} = 4770 - 2520 = 2250 \text{ kWh}\] \[\text{Percentage increase} = \frac{2250}{2520} \times 100 = 89.3\%\]

The hybrid system produces 89.3% more energy than the wind farm alone.

1 mark: correct calculation. 1 mark: expressing as percentage.

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Practice Application Task

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Practice Application Task

Part A: Power Output Modelling

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Part B: Daily Energy Production

Part C: System Optimisation

Part D: Comparing Energy Sources

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