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Practice Application Task

Offshore Wind Farm Engineering

Area of Study:Vectors & Calculus Task format:Application Task Working time:70 minutes Reading time:10 minutes Total marks:50 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 vectors and calculus.

Scenario: Offshore Wind Farm Engineering

Engineers design an offshore wind farm in Bass Strait. They use vectors for turbine positioning and force analysis, and calculus for power curve modelling and structural load calculations. This task involves optimising turbine placement, analysing forces on structures, and predicting power output and structural loads.

Part A: Turbine Positioning

12 marks

Three turbines are positioned at points A(2, 5, 0), B(8, 3, 0), and C(5, 9, 0) (coordinates in km). These positions define the corners of a triangular platform in the xy-plane.

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Question 1 4 marks

a. Find vector $\overrightarrow{AB}$ and calculate its magnitude. [2 marks]

b. Find the midpoint M of AC and the distance from B to M. [2 marks]

Have you genuinely attempted this question on paper first?

a. [2 marks]

\[\overrightarrow{AB} = B - A = (8-2, 3-5, 0-0) = (6, -2, 0)\]

\[|\overrightarrow{AB}| = \sqrt{6^2 + (-2)^2 + 0^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \text{ km}\]

$\overrightarrow{AB} = (6, -2, 0)$, \quad |\overrightarrow{AB}| = 2\sqrt{10}\) km $\approx 6.32$ km

1 mark: correct vector. 1 mark: correct magnitude.

b. [2 marks]

Midpoint M of AC: \[M = \left(\frac{2+5}{2}, \frac{5+9}{2}, \frac{0+0}{2}\right) = (3.5, 7, 0)\]

Distance from B(8, 3, 0) to M(3.5, 7, 0):\[|BM| = \sqrt{(8-3.5)^2 + (3-7)^2 + 0^2} = \sqrt{4.5^2 + (-4)^2} = \sqrt{20.25 + 16} = \sqrt{36.25} = \frac{\sqrt{145}}{2} \text{ km}\]

M = (3.5, 7, 0), \quad |BM| = $\frac{\sqrt{145}}{2}$ km $\approx 6.02$ km

1 mark: correct midpoint. 1 mark: correct distance.

Question 2 4 marks

a. Find the angle between vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ using the dot product formula. [2 marks]

b. Find the scalar projection of $\overrightarrow{AB}$ onto $\overrightarrow{AC}$. [2 marks]

Have you genuinely attempted this question on paper first?

a. [2 marks]

$\overrightarrow{AC} = C - A = (5-2, 9-5, 0) = (3, 4, 0)$

$\overrightarrow{AB} \cdot \overrightarrow{AC} = (6)(3) + (-2)(4) + (0)(0) = 18 - 8 = 10$

$|\overrightarrow{AB}| = 2\sqrt{10}, \quad |\overrightarrow{AC}| = \sqrt{9 + 16} = 5$

\[\cos\theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| \cdot |\overrightarrow{AC}|} = \frac{10}{2\sqrt{10} \cdot 5} = \frac{10}{10\sqrt{10}} = \frac{1}{\sqrt{10}}\]

\[\theta = \cos^{-1}\left(\frac{1}{\sqrt{10}}\right) \approx 71.57°\]

$\theta \approx 71.57°$ or $\arccos\left(\frac{1}{\sqrt{10}}\right)$

1 mark: correct dot product calculation. 1 mark: correct angle.

b. [2 marks]

Scalar projection of $\overrightarrow{AB}$ onto $\overrightarrow{AC}$:\[\text{proj}_{\overrightarrow{AC}} \overrightarrow{AB} = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AC}|} = \frac{10}{5} = 2 \text{ km}\]

Scalar projection = 2 km

1 mark: correct formula. 1 mark: correct calculation.

Question 3 4 marks

a. Find the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$. [2 marks]

b. Hence find the area of triangle ABC and a unit vector perpendicular to the plane containing the three turbines. [2 marks]

Have you genuinely attempted this question on paper first?

a. [2 marks]

\[\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6 & -2 & 0 \\ 3 & 4 & 0 \end{vmatrix}\]

\[= \mathbf{i}((-2)(0) - (0)(4)) - \mathbf{j}((6)(0) - (0)(3)) + \mathbf{k}((6)(4) - (-2)(3))\]

\[= \mathbf{i}(0) - \mathbf{j}(0) + \mathbf{k}(24 + 6) = 30\mathbf{k} = (0, 0, 30)\]

$\overrightarrow{AB} \times \overrightarrow{AC} = (0, 0, 30)$

1 mark: correct setup. 1 mark: correct answer.

b. [2 marks]

Area of triangle ABC: \[\text{Area} = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2} \cdot 30 = 15 \text{ km}^2\]

Unit normal vector: \[\hat{\mathbf{n}} = \frac{(0, 0, 30)}{|(0, 0, 30)|} = \frac{(0, 0, 30)}{30} = (0, 0, 1) = \mathbf{k}\]

Area = 15 km$^2$, Unit perpendicular vector = $(0, 0, 1)$ or $\mathbf{k}$

1 mark: correct area. 1 mark: correct unit normal vector.

Part B: Force Analysis

12 marks

Support cables and wind loads act on the turbine structure. We model these using vectors in three dimensions.

Question 4 4 marks

a. A support cable runs through point A(2, 5, 0) in the direction $\mathbf{d} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$. Write the vector equation of this cable line. [1 mark]

b. A second cable runs through point B(8, 3, 0) in the direction $\mathbf{e} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}$. Show that these two lines are skew (do not intersect). [3 marks]

Have you genuinely attempted this question on paper first?

a. [1 mark]

\[\mathbf{r} = (2, 5, 0) + t(2, -1, 3) = (2 + 2t, 5 - t, 3t)\]

$\mathbf{r} = (2, 5, 0) + t(2, -1, 3)$ or $\mathbf{r} = (2 + 2t, 5 - t, 3t)$

b. [3 marks]

Second cable: $\mathbf{r} = (8, 3, 0) + s(1, 2, -1) = (8 + s, 3 + 2s, -s)$

For intersection, we need: $2 + 2t = 8 + s$, $5 - t = 3 + 2s$, $3t = -s$

From equation 3: $s = -3t$

Substitute into equation 1: $2 + 2t = 8 - 3t \Rightarrow 5t = 6 \Rightarrow t = 1.2$

Then $s = -3.6$. Check in equation 2: $5 - 1.2 = 3.8$ and $3 + 2(-3.6) = 3 - 7.2 = -4.2$

Since $3.8 \neq -4.2$, the system has no solution.

The lines do not intersect (and are not parallel since directions $(2,-1,3)$ and $(1,2,-1)$ are not scalar multiples). Therefore the lines are skew.

1 mark: second line equation. 1 mark: setting up system. 1 mark: correct conclusion with reasoning.

Question 5 4 marks

a. A wind force $\mathbf{F} = 120\mathbf{i} + 80\mathbf{j} - 40\mathbf{k}$ (in Newtons) acts on a turbine blade oriented along direction $\mathbf{u} = 3\mathbf{i} + 4\mathbf{j}$. Find the component of $\mathbf{F}$ parallel to $\mathbf{u}$ (the vector projection). [2 marks]

b. Find the perpendicular component of $\mathbf{F}$ to $\mathbf{u}$. [2 marks]

Have you genuinely attempted this question on paper first?

a. [2 marks]

\[\mathbf{F} \cdot \mathbf{u} = (120)(3) + (80)(4) + (-40)(0) = 360 + 320 = 680\]

\[|\mathbf{u}| = \sqrt{9 + 16} = 5\]

\[\text{proj}_{\mathbf{u}} \mathbf{F} = \frac{\mathbf{F} \cdot \mathbf{u}}{|\mathbf{u}|^2} \mathbf{u} = \frac{680}{25}(3, 4, 0) = 27.2(3, 4, 0) = (81.6, 108.8, 0)\]

Parallel component: $(81.6, 108.8, 0)$ N or $\dfrac{680}{25}(3, 4, 0)$

1 mark: correct dot product and magnitude. 1 mark: correct vector projection.

b. [2 marks]

\[\text{perp} = \mathbf{F} - \text{proj}_{\mathbf{u}} \mathbf{F} = (120, 80, -40) - (81.6, 108.8, 0)\]

\[= (38.4, -28.8, -40)\]

Perpendicular component: $(38.4, -28.8, -40)$ N

1 mark: correct method. 1 mark: correct calculation.

Question 6 4 marks

a. Three forces act on a junction point: $\mathbf{F}_1 = 500\mathbf{i} + 300\mathbf{j}$ N $\mathbf{F}_2 = -200\mathbf{i} + 400\mathbf{j} + 100\mathbf{k}$ N $\mathbf{F}_3 = -100\mathbf{i} - 500\mathbf{j} + 200\mathbf{k}$ N Find the resultant force $\mathbf{R}$. [1 mark]

b. Find the magnitude of $\mathbf{R}$. [1 mark]

c. Find the angle that $\mathbf{R}$ makes with the horizontal plane (the xy-plane). [2 marks]

Have you genuinely attempted this question on paper first?

a. [1 mark]

\[\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = (500 - 200 - 100, 300 + 400 - 500, 0 + 100 + 200)\]

\[= (200, 200, 300) \text{ N}\]

$\mathbf{R} = (200, 200, 300)$ N or $200\mathbf{i} + 200\mathbf{j} + 300\mathbf{k}$

b. [1 mark]

\[|\mathbf{R}| = \sqrt{200^2 + 200^2 + 300^2} = \sqrt{40000 + 40000 + 90000} = \sqrt{170000} = 100\sqrt{17} \approx 412.3 \text{ N}\]

$|\mathbf{R}| = 100\sqrt{17}$ N $\approx 412.3$ N

c. [2 marks]

The vertical component is 300 N. The horizontal component (in xy-plane) has magnitude: $\sqrt{200^2 + 200^2} = \sqrt{80000} = 200\sqrt{2}$

\[\sin\phi = \frac{300}{100\sqrt{17}} = \frac{3}{\sqrt{17}}\]

\[\phi = \sin^{-1}\left(\frac{3}{\sqrt{17}}\right) \approx 46.4°\]

Angle with horizontal plane: $\phi \approx 46.4°$ or $\sin^{-1}\left(\dfrac{3}{\sqrt{17}}\right)$

1 mark: identifying vertical component. 1 mark: correct angle calculation.

Part C: Power Curve Modelling

13 marks

The power output $P(v)$ (in kilowatts) of a turbine as a function of wind speed $v$ (in m/s) is modelled by: \[P(v) = 800\, \sin^{-1}\left(\frac{v}{25}\right) \quad \text{for } 0 \leq v \leq 25\] Wind speed at time $t$ hours is given by: $v(t) = 20 - 5\cos\left(\frac{\pi t}{12}\right)$

Question 7 4 marks

a. Find $\dfrac{dP}{dv}$. [2 marks]

b. Evaluate $\dfrac{dP}{dv}$ when $v = 15$ m/s and interpret this result. [2 marks]

Have you genuinely attempted this question on paper first?

a. [2 marks]

\[\frac{d}{dv}\left[\sin^{-1}(x)\right] = \frac{1}{\sqrt{1-x^2}}\]

\[\frac{dP}{dv} = 800 \cdot \frac{1}{\sqrt{1 - (v/25)^2}} \cdot \frac{1}{25} = \frac{800}{25\sqrt{1 - v^2/625}} = \frac{32}{\sqrt{1 - v^2/625}}\]

\[= \frac{32}{\sqrt{(625 - v^2)/625}} = \frac{32 \cdot 25}{\sqrt{625 - v^2}} = \frac{800}{\sqrt{625 - v^2}}\]

$\dfrac{dP}{dv} = \dfrac{800}{\sqrt{625 - v^2}}$ kW/(m/s)

1 mark: correct application of inverse trig derivative. 1 mark: correct simplification.

b. [2 marks]

\[\frac{dP}{dv}\bigg|_{v=15} = \frac{800}{\sqrt{625 - 225}} = \frac{800}{\sqrt{400}} = \frac{800}{20} = 40 \text{ kW/(m/s)}\]

$\dfrac{dP}{dv}\bigg|_{v=15} = 40$ kW/(m/s). This means that at a wind speed of 15 m/s, the power output is increasing at a rate of 40 kilowatts for each additional 1 m/s increase in wind speed.

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

Question 8 5 marks

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

b. Using the chain rule, find $\dfrac{dP}{dt}$ when $t = 6$ hours. [2 marks]

c. At what time is the power output changing most rapidly (i.e., $\dfrac{dP}{dt}$ is maximum)? [2 marks]

Have you genuinely attempted this question on paper first?

a. [1 mark]

\[\frac{dv}{dt} = \frac{d}{dt}\left[20 - 5\cos\left(\frac{\pi t}{12}\right)\right] = 0 - 5 \cdot \left(-\sin\left(\frac{\pi t}{12}\right)\right) \cdot \frac{\pi}{12} = \frac{5\pi}{12}\sin\left(\frac{\pi t}{12}\right)\]

$\dfrac{dv}{dt} = \dfrac{5\pi}{12}\sin\left(\dfrac{\pi t}{12}\right)$ m/s per hour

b. [2 marks]

At $t = 6$: $v(6) = 20 - 5\cos(\pi/2) = 20 - 0 = 20$ m/s

$\dfrac{dv}{dt}\bigg|_{t=6} = \dfrac{5\pi}{12}\sin(\pi/2) = \dfrac{5\pi}{12}$ m/s per hour

$\dfrac{dP}{dv}\bigg|_{v=20} = \dfrac{800}{\sqrt{625 - 400}} = \dfrac{800}{15} = \dfrac{160}{3}$ kW/(m/s)

\[\frac{dP}{dt}\bigg|_{t=6} = \frac{dP}{dv} \cdot \frac{dv}{dt} = \frac{160}{3} \cdot \frac{5\pi}{12} = \frac{800\pi}{36} = \frac{200\pi}{9} \approx 69.8 \text{ kW/hr}\]

$\dfrac{dP}{dt}\bigg|_{t=6} = \dfrac{200\pi}{9}$ kW/hr $\approx 69.8$ kW/hr

1 mark: correct calculation of $v(6)$ and $dv/dt$ at $t=6$. 1 mark: correct chain rule application and final answer.

c. [2 marks]

\[\frac{dP}{dt} = \frac{dP}{dv} \cdot \frac{dv}{dt} = \frac{800}{\sqrt{625 - v^2}} \cdot \frac{5\pi}{12}\sin\left(\frac{\pi t}{12}\right)\]

Since $\dfrac{dP}{dv} > 0$ for all valid $v$, $\dfrac{dP}{dt}$ is maximum when $\sin\left(\dfrac{\pi t}{12}\right) = 1$, i.e., when $\dfrac{\pi t}{12} = \dfrac{\pi}{2}$

\[t = 6 \text{ hours}\]

Power output changes most rapidly at $t = 6$ hours (6 hours after the start of the day).

1 mark: recognising that maximum occurs when $\sin\left(\dfrac{\pi t}{12}\right) = 1$. 1 mark: correct time.

Question 9 4 marks

a. The blade pitch angle $\theta$ and power coefficient $C$ are related implicitly by: \[C^3 + 2C\theta - \theta^2 = 1\] Use implicit differentiation to find $\dfrac{dC}{d\theta}$ in terms of $C$ and $\theta$. [2 marks]

b. Given that when $\theta = 1$, $C = 1$, find the value of $\dfrac{dC}{d\theta}$ at this point. [1 mark]

c. Interpret the sign of $\dfrac{dC}{d\theta}$ at the point $(\theta, C) = (1, 1)$. [1 mark]

Have you genuinely attempted this question on paper first?

a. [2 marks]

\[\frac{d}{d\theta}\left[C^3 + 2C\theta - \theta^2\right] = \frac{d}{d\theta}[1]\]

\[3C^2\frac{dC}{d\theta} + 2C + 2\theta\frac{dC}{d\theta} - 2\theta = 0\]

\[3C^2\frac{dC}{d\theta} + 2\theta\frac{dC}{d\theta} = 2\theta - 2C\]

\[\frac{dC}{d\theta}(3C^2 + 2\theta) = 2\theta - 2C\]

$\dfrac{dC}{d\theta} = \dfrac{2\theta - 2C}{3C^2 + 2\theta} = \dfrac{2(\theta - C)}{3C^2 + 2\theta}$

1 mark: correct differentiation of implicit equation. 1 mark: correct rearrangement.

b. [1 mark]

\[\frac{dC}{d\theta}\bigg|_{(\theta,C)=(1,1)} = \frac{2(1 - 1)}{3(1)^2 + 2(1)} = \frac{0}{5} = 0\]

$\dfrac{dC}{d\theta}\bigg|_{(\theta,C)=(1,1)} = 0$

c. [1 mark]

The derivative is zero, meaning that the power coefficient is at a stationary point (local maximum or minimum) with respect to pitch angle. A small change in pitch angle has no first-order effect on the power coefficient at this operating point.

1 mark: correct interpretation of zero derivative in context.

Part D: Structural Load Analysis

13 marks

The structural loads and stress distribution on the turbine foundations require integration to find total quantities.

Question 10 4 marks

a. The stress on a support beam at distance $x$ from the base is: $\sigma(x) = \dfrac{600x}{x^2 + 9}$ Using the substitution $u = x^2 + 9$, find $\int \sigma(x)\, dx$. [2 marks]

b. Find the total stress from $x = 0$ to $x = 4$. [2 marks]

Have you genuinely attempted this question on paper first?

a. [2 marks]

Let $u = x^2 + 9$, then $du = 2x\, dx$, so $x\, dx = \dfrac{1}{2}du$

\[\int \frac{600x}{x^2 + 9}\, dx = \int \frac{600}{u} \cdot \frac{1}{2}\, du = 300 \int \frac{1}{u}\, du = 300\ln|u| + C\]

\[= 300\ln(x^2 + 9) + C\]

$\int \sigma(x)\, dx = 300\ln(x^2 + 9) + C$

1 mark: correct substitution and setup. 1 mark: correct integration.

b. [2 marks]

\[\int_0^4 \frac{600x}{x^2 + 9}\, dx = \left[300\ln(x^2 + 9)\right]_0^4\]

\[= 300\ln(16 + 9) - 300\ln(0 + 9) = 300\ln(25) - 300\ln(9)\]

\[= 300\ln\left(\frac{25}{9}\right) = 300\ln\left(\frac{25}{9}\right) \approx 300 \times 1.022 = 306.5\]

Total stress from $x = 0$ to $x = 4$: $300\ln\left(\dfrac{25}{9}\right) \approx 306.5$

1 mark: correct limits and anti-derivative evaluation. 1 mark: correct numerical answer or exact answer.

Question 11 5 marks

a. The fatigue load is modelled by $F(t) = t e^{-0.5t}$ (in kN). Using integration by parts, find $\int t e^{-0.5t}\, dt$. [3 marks]

b. Find the total fatigue load over the interval $[0, 6]$. [2 marks]

Have you genuinely attempted this question on paper first?

a. [3 marks]

Use integration by parts: $\int u\, dv = uv - \int v\, du$

Let $u = t$, $dv = e^{-0.5t}\, dt$

Then $du = dt$, $v = \dfrac{e^{-0.5t}}{-0.5} = -2e^{-0.5t}$

\[\int t e^{-0.5t}\, dt = t \cdot (-2e^{-0.5t}) - \int (-2e^{-0.5t})\, dt\]

\[= -2te^{-0.5t} + 2\int e^{-0.5t}\, dt = -2te^{-0.5t} + 2 \cdot \frac{e^{-0.5t}}{-0.5} + C\]

\[= -2te^{-0.5t} - 4e^{-0.5t} + C = -2e^{-0.5t}(t + 2) + C\]

$\int t e^{-0.5t}\, dt = -2e^{-0.5t}(t + 2) + C$

1 mark: correct choice of $u$ and $dv$. 1 mark: correct evaluation of $uv$. 1 mark: completing the integration by parts.

b. [2 marks]

\[\int_0^6 t e^{-0.5t}\, dt = \left[-2e^{-0.5t}(t + 2)\right]_0^6\]

\[= -2e^{-3}(6 + 2) - \left(-2e^0(0 + 2)\right) = -16e^{-3} + 4\]

\[= 4 - 16e^{-3} = 4 - \frac{16}{e^3} \approx 4 - 0.798 = 3.202 \text{ kN}\]

Total fatigue load over $[0,6]$: $4 - 16e^{-3}$ kN $\approx 3.20$ kN

1 mark: correct evaluation at limits. 1 mark: correct simplification and numerical answer.

Question 12 4 marks

a. Flow rate through a cooling pipe is: $Q(x) = \dfrac{24}{x^2 + 5x + 6}$ Express $Q(x)$ in partial fractions. [2 marks]

b. Find $\displaystyle\int_1^4 Q(x)\, dx$, giving an exact answer. [2 marks]

Have you genuinely attempted this question on paper first?

a. [2 marks]

\[x^2 + 5x + 6 = (x + 2)(x + 3)\]

\[\frac{24}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}\]

Multiply by $(x+2)(x+3)$: $24 = A(x+3) + B(x+2)$

When $x = -2$: $24 = A(1) \Rightarrow A = 24$

When $x = -3$: $24 = B(-1) \Rightarrow B = -24$

$Q(x) = \dfrac{24}{x+2} - \dfrac{24}{x+3}$

1 mark: correct factorisation. 1 mark: correct partial fractions.

b. [2 marks]

\[\int_1^4 Q(x)\, dx = \int_1^4 \left(\frac{24}{x+2} - \frac{24}{x+3}\right)\, dx\]

\[= \left[24\ln(x+2) - 24\ln(x+3)\right]_1^4\]

\[= 24\ln(6) - 24\ln(7) - (24\ln(3) - 24\ln(4))\]

\[= 24\left[\ln(6) - \ln(7) - \ln(3) + \ln(4)\right] = 24\ln\left(\frac{6 \cdot 4}{7 \cdot 3}\right) = 24\ln\left(\frac{24}{21}\right) = 24\ln\left(\frac{8}{7}\right)\]

$\displaystyle\int_1^4 Q(x)\, dx = 24\ln\left(\dfrac{8}{7}\right)$

1 mark: correct anti-derivatives. 1 mark: correct evaluation and simplification.

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Practice Application Task

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Practice Application Task

Part A: Turbine Positioning

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Part B: Force Analysis

Part C: Power Curve Modelling

Part D: Structural Load Analysis

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