Band 4

Networks, STD2 EQ-Bank 6 MC

The Gantt chart below shows the activities involved in organising a community market day. Critical path activities are shown as solid bars and non-critical activities show available float as dashed extensions.

Activity $D$ is delayed by 7 hours and activity $B$ is delayed by 1 hour. What is the new minimum completion time for the project?

14 hours
15 hours
16 hours
22 hours

Show Answers Only

$C$

Show Worked Solution

$\text{From the Gantt chart:}$

$\text{D has float of 5 hours. Delay of 7 exceeds float by 2.}$

$\text{B has float of 3 hours. Delay of 1 is within float.}$

$\text{Only the delay to D affects completion time.}$

$\text{New minimum} = 14+2 = 16 \text{ hours}$

$\Rightarrow C$

Networks, STD2 EQ-Bank 25

The construction of a new reptile exhibit is a project involving nine activities, $A$ to $I$. The network diagram below shows the activities and their completion times in weeks. Some values are missing.

The Gantt chart below has been created for this project.

Using the Gantt chart, identify the critical path. (1 mark)
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Use the Gantt chart to determine the missing values in the network diagram. (2 marks)
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Activity $E$ is delayed by 8 weeks. Using the Gantt chart, explain whether this will affect the minimum completion time of the project. (2 marks)
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a. $ACDFGI$

b. $\text{B} = 5 \text{ weeks, H} = 7 \text{ weeks}$

c. $\text{From the Gantt chart, activity E has a float of 6 weeks (see the dashed}$

$\text{extension from week 10 to 16).}$

$\text{The delay of 8 weeks exceeds the float of 6 weeks.}$

$\text{The project will be delayed by } 8-6 = 2 \text{ weeks.}$

$\text{New minimum completion time} = 25+2 = 27 \text{ weeks.}$

Show Worked Solution

a. $\text{The critical path is the continuous solid bar on row 1 of the Gantt chart.}$

$\text{Critical path:}\ ACDFGI$

b. $\text{Activities B and H are not labelled in the network diagram.}$

$\text{From the Gantt chart:}$

$\text{B starts at week 0, ends at week 5} \to \text{duration} = 5 \text{ weeks}$

$\text{H starts at week 7, ends at week 14} \to \text{duration} = 7 \text{ weeks}$

c. $\text{From the Gantt chart, activity E has a float of 6 weeks (see the dashed}$

$\text{extension from week 10 to 16).}$

$\text{The delay of 8 weeks exceeds the float of 6 weeks.}$

$\text{The project will be delayed by } 8-6 = 2 \text{ weeks.}$

$\text{New minimum completion time} = 25+2 = 27 \text{ weeks.}$

Networks, STD2 EQ-Bank 18

A Gantt chart for a project with activities $A, B, C, D, E, F, G$ and $H$ has been created. The Gantt chart can be used to complete the missing information on the edges in the network diagram.

Use the Gantt chart to determine the three missing values in the network diagram. (3 marks)
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State the minimum completion time for the project. (1 mark)
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a. $\text{C} = 4 \text{ hours, D} = 5 \text{ hours, E} = 7 \text{ hours}$

b. $26 \text{ hours}$

Show Worked Solution

a. $\text{Using the Gantt chart}$

$\text{C (between A and F):} $

$\Rightarrow\ \text{Starts hour 7, ends hour 11 = 4 hours duration}$

$\text{D (between B and G):}$

$\Rightarrow\ \text{Starts hour 3, ends hour 8 = 5 hours duration}$

$\text{E (between B and H):}$

$\Rightarrow\ \text{Starts hour 3, ends hour 10 = 7 hours duration}$

b. $\text{From the Gantt chart, the project ends at hour 26.}$

$\text{Minimum completion time} = 26 \text{ hours}$

Measurement, STD2 EQ-Bank 30

A 2500-watt air-conditioning system is turned on for 3 hours each day. Electricity is charged at 27 cents per kWh.

What is the cost of electricity for using the air-conditioning system over a seven-day period? (2 marks)

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`$14.18`

Show Worked Solution

`text{Daily usage}`	`=2500 xx 3`
	`= 7500\ text{Wh}`
	`=7.5\ text{kWh (1000 Wh = 1 kWh)}`

`text{Cost}`	`= 7 xx 7.5 xx 0.27`
	`= 14.175=$14.18\ \ text{(nearest cent)}`

♦♦ Mean mark 39%.

Measurement, STD2 EQ-Bank 25

The table compares the fuel costs of a petrol car with an electric car.

\begin{array} {|l|l|l|}
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \textit{Petrol car} & \textit{Electric car}\\
\hline
\rule{0pt}{2.5ex}\text{Fuel consumption}\rule[-1ex]{0pt}{0pt} & \text{8.6 L/100 km} & \text{18 kWh/100 km} \\ \hline \rule{0pt}{2.5ex}\text{Cost of fuel}\rule[-1ex]{0pt}{0pt} & \text{\$1.87/L} & \text{\$0.25/kWh} \\ \hline \end{array}

Jun travels on average 35 000 km per year.

How much will he save on fuel costs in a year by using an electric car? (3 marks)

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$$4053.70$

Show Worked Solution

$\text{Petrol car fuel costs}$

$\text{Litres used}=\dfrac{8.6\times 35\ 000}{100}=3010\text{ L}$

$\text{Cost of fuel}=3010\times $1.87 = $5628.70$

$\text{Electric car power costs}$

$\text{kWh}=\dfrac{18\times 35\ 000}{100}=6300\text{ kWh}$

$\text{Cost of power}=6300\times $0.25 = $1575$

$\text{Saving}=$5628.70-$1575=$4053.70$

Statistics, EXT1 EQ-Bank 6 MC

The lifetime of a certain brand of batteries has a mean lifetime of 20 hours and a standard deviation of 2 hours. A random sample of 40 batteries is selected.

The probability that the mean lifetime of this sample of 40 batteries exceeds 19.5 hours is closest to

0.0571
0.5987
0.8944
0.9429

Show Answers Only

$D$

Show Worked Solution

$\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{CLT applies}$

$\overline{X} \sim N\left( \mu, \dfrac{\sigma^2}{n}\right) \sim N\left( 20, \dfrac{4}{40}\right)$

$\sigma_{\bar{X}} = \dfrac{1}{\sqrt{10}} $

$\text{By the central limit theorem:}$

$Z=\dfrac{\overline{X}-20}{\frac{1}{\sqrt{10}}}\sim N(0,1)$

$Z=\dfrac{19.5-20}{\frac{1}{\sqrt{10}}}=-1.58 \ \text{(2 d.p.)}$

$\text{Using Normal Distribution Table of Values:}$

$\Pr(\overline{X}>19.5)=\Pr(Z>-1.58)$

$\phantom{\Pr(\overline{X}>19.5)}=0.9429$

$\Rightarrow D$

Statistics, EXT1 EQ-Bank 29

The waiting time, $T$ hours, to see a particular doctor at a clinic has a mean of 0.5 hours and a standard deviation of 0.3 hours.

A sample of 35 waiting times is chosen at random.

Use the standard normal distribution table (provided) to find the probability that the average waiting time of the sample of 35 patients is between 0.43 hours and 0.50 hours. Give your answer as a percentage correct to 1 decimal place. (3 marks)

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$\Pr(0.43<\overline{T}<0.50) \approx 41.6\%$

Show Worked Solution

$\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{Central Limit Theorem applies}$

$\text{The sample mean,}\ \overline{T}, \text{for random samples of size 35 is}$

$\text{approximately normally distributed, where:}$

$\mu=0.5\ \ \text{and}\ \ \sigma=\dfrac{0.3}{\sqrt{n}}=\dfrac{0.3}{\sqrt{35}} \approx 0.0507$

$\text{By the central limit theorem:}$

$Z=\dfrac{\overline{T}-0.5}{\frac{0.3}{\sqrt{35}}} \sim N(0,1)$

$\Pr(0.43<\overline{T}<0.50)$	$=\Pr\left(\dfrac{0.43-0.50}{0.0507}<Z<\dfrac{0.50-0.50}{0.0507}\right)$
	$=\Pr(-1.38<Z<0)$
	$=\Pr(0<Z<1.38)$
	$=0.9162-0.5000$
	$=0.4162 = 41.6\%\ \text{(1 d.p.)}$

Statistics, EXT1 EQ-Bank 22

The gestation period of cats has a mean of 66 days and a variance of 9 days$^2$.

A sample of 32 cats is chosen at random.

Using the Normal Distribution Table of Values, determine the probability that the sample has an average gestation period greater than 65 days. (3 marks)

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Show Answers Only

$0.9706$

Show Worked Solution

$\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{CLT applies}$

$\overline{X} \sim N\left( \mu, \dfrac{\sigma^2}{n}\right) \sim N\left( 66, \dfrac{9}{32}\right)$

$\sigma_{\bar{X}} = \dfrac{3}{\sqrt{32}} $

$\text{By the central limit theorem:}$

$Z=\dfrac{\overline{X}-66}{\frac{3}{\sqrt{32}}}\sim N(0,1)$

$Z=\dfrac{65-66}{\frac{3}{\sqrt{32}}}=-1.89 \ \text{(2 d.p.)}$

$\text{Using Normal Distribution Table of Values:}$

$P(\overline{X}>65)$	$=P(Z>-1.89)$
	$=1-P(Z\leq -1.89)$
	$=1-0.0294$
	$=0.9706$

Statistics, EXT1 EQ-Bank 26

A research team is investigating the amount of sleep obtained by Year 12 students. Previous studies indicate that the sleep time of Year 12 students has a population mean of 7.4 hours and a population standard deviation of 2.42 hours.

A random sample of 100 Year 12 students is selected.

Using the normal distribution table (included), determine the probability that the mean sleep time of the sample is less than 7 hours. Give your answer correct to four decimal places. (3 marks)

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$4.94 \%$

Show Worked Solution

$\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{Central Limit Theorem applies}$

$\text{The sample mean,}\ \overline{X}, \text{for random samples of size 100 is}$

$\text{approximately normally distributed, where:}$

$\mu=7.4\ \ \text{and}\ \ \sigma=\dfrac{2.42}{\sqrt{n}}=\dfrac{2.42}{\sqrt{100}}$

$\text{By the central limit theorem:}$

$Z=\dfrac{\overline{X}-7.4}{\frac{2.42}{\sqrt{100}}} \sim N(0,1)$

$Z=\dfrac{7-7.4}{\frac{2.42}{\sqrt{100}}}=-1.65 \ \text{(2 d.p.)}$

$\text{Using Normal Distribution Table of Values:}$

$P(\overline{X} < 7)=P(Z < -1.65)=0.0494=4.94 \%$

Probability, STD2 EQ-Bank 31

A survey of 60 people found the following information about their exercise habits.

35 enjoy hiking $(H)$
28 enjoy cycling $(C)$
8 enjoy neither hiking nor cycling

Draw a Venn diagram to represent this information. (2 marks)

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One person is selected at random. What is the probability that the person enjoys hiking or cycling? (1 mark)

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b. $\dfrac{13}{15}$

Show Worked Solution

a. $\text{Number in either }H\text{ or }C = 60-8 = 52$

$\text{Number in both }H\text{ and }C = 35+28-52 = 11$

$H\text{ only} = 35-11 = 24$

$C\text{ only} = 28-11 = 17$

b. $P(H \text{ or } C) = \dfrac{24+11+17}{60} = \dfrac{52}{60} = \dfrac{13}{15}$

$\text{OR, using the complement:}$

$P(H \text{ or } C) = 1-\dfrac{8}{60} = \dfrac{13}{15}$

Probability, STD2 EQ-Bank 5 MC

The Venn diagram shows information about 40 people surveyed about whether they own a dog $(D)$ or a cat $(C)$.

Which two-way table correctly represents the information in the Venn diagram?

A.	$\begin{array}{\|l\|c\|c\|c\|} \hline & D & \text{Not }D & \text{Total} \\ \hline C & 12 & 8 & 20 \\ \hline \text{Not }C & 5 & 15 & 20 \\ \hline \text{Total} & 17 & 23 & 40 \\ \hline \end{array}$	B.	$\begin{array}{\|l\|c\|c\|c\|} \hline & D & \text{Not }D & \text{Total} \\ \hline C & 8 & 12 & 20 \\ \hline \text{Not }C & 15 & 5 & 20 \\ \hline \text{Total} & 23 & 17 & 40 \\ \hline \end{array}$
C.	$\begin{array}{\|l\|c\|c\|c\|} \hline & D & \text{Not }D & \text{Total} \\ \hline C & 15 & 12 & 27 \\ \hline \text{Not }C & 8 & 5 & 13 \\ \hline \text{Total} & 23 & 17 & 40 \\ \hline \end{array}$	D.	$\begin{array}{\|l\|c\|c\|c\|} \hline & D & \text{Not }D & \text{Total} \\ \hline C & 5 & 12 & 17 \\ \hline \text{Not }C & 15 & 8 & 23 \\ \hline \text{Total} & 20 & 20 & 40 \\ \hline \end{array}$

Show Answers Only

$B$

Show Worked Solution

$\text{Reading the Venn diagram:}$

$\text{C and D (intersection)} = 8$

$\text{C only (not D)} = 12$

$\text{D only (not C)} = 15$

$\text{Neither} = 5$

$\text{In the two-way table:}$

$\text{Row }C\text{: }\ D\text{ column} = 8,\ \text{Not }D\text{ column} = 12,\ \text{Total} = 20$

$\text{Row Not }C\text{: }\ D\text{ column} = 15,\ \text{Not }D\text{ column} = 5,\ \text{Total} = 20$

$\Rightarrow B$

Algebra, STD2 A4 EQ-Bank 27

SunPower Solutions is a business that installs solar panels. Fixed costs are $1200. Each panel costs $150.00 to install and generates revenue of $350.00.

The spreadsheet below models the business's costs and revenue for different numbers of panels installed.

Calculate the spreadsheet values for the installation of 4 panels (cells B9, C9, D9) and 6 panels (cells B10, C10, D10). (2 marks)
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Using the spreadsheet, identify the break-even point and explain what it means for the business. (2 marks)
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SunPower Solutions has received a large order that will see them make a profit of $4200. Calculate the number of solar panels $(x)$ they will be installing. (2 marks)
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a. $\text{4 panels: TC (B9) = \$1800.00}$

$\text{Revenue (C9) = \$1400.00, Profit/Loss (D9) = }-\$400.00$

$\text{6 panels: TC (B10) = \$2100.00}$

$\text{Revenue (C10) = \$2100.00, Profit/Loss (D10) = \$0.00}$

b. $\text{Break-even = 6 panels, See worked solution}$

c. $27 \text{ panels}$

Show Worked Solution

a. $\text{4 panels:}$

$\text{Total Cost (B9)} = \$1200+4\times\$150 = \$1800.00$

$\text{Revenue (C9)} = 4\times\$350 = \$1400.00$

$\text{Profit/Loss (D9)} = \$1400.00-\$1800.00 = -\$400.00$

$\text{6 panels:}$

$\text{Total Cost (B10)} = \$1200+6\times\$150 = \$2100.00$

$\text{Revenue (C10)} = 6\times\$350 = \$2100.00$

$\text{Profit/Loss (D10)} = \$2100.00-\$2100.00 = \$0.00$

b. $\text{When 6 panels are installed:}$

$\text{Revenue = Total costs = \$2100.00 (breakeven)}$

$\text{This is the point at which the business covers all of its costs and}$

$\text{begins to make a profit.}$

$\text{OR}$

$\text{If fewer than 6 panels are installed the business will make a loss.}$

c. $\text{Let } x = \text{the number of solar panels to be installed.}$

$\text{Profit}$	$ = \text{Revenue}-\text{Total costs}$
$4200$	$ = 350x-(1200+150x)$
$4200$	$= 200x-1200$
$5400$	$=200x$
$x$	$=27$

$\text{SunPower Solutions will install 27 solar panels.}$

Algebra, STD2 A4 EQ-Bank 16

Harmony Arts Festival is a cultural event with fixed costs of $510. Each ticket costs $8.00 to provide and sells for $25.00.

The spreadsheet below models the festival's costs and revenue for different numbers of tickets sold.

Calculate the values for cells C12 and D12 in the spreadsheet. (2 marks)
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Using the spreadsheet, identify the break-even point and explain what it means for the festival. (2 marks)
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a. $\text{C12} = \$1250.00 \quad \text{D12} = \$340.00$

b. $\text{See worked solution}$

Show Worked Solution

a. $\text{C12: Revenue} = 50 \times \$25.00 = \$1250.00$

$\text{D12: Profit/Loss} = \$1250.00-\$910.00 = \$340.00$

b. $\text{When 30 tickets sold:}$

$\text{Revenue = Total costs = \$750.00 (Breakeven)}$

$\text{This is the point at which the festival covers all of its costs}$

$\text{and begins to make a profit.}$

$\text{OR}$

$\text{If less than 30 tickets are sold the festival will make a loss.}$

Complex Numbers, EXT2 EQ-Bank 15

Prove that for any complex numbers $z_1$ and $z_2$,
$\abs{z_1+z_2} \leqslant \abs{z_1}+\abs{z_2}$ (2 marks)
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Hence, or otherwise, show that if $\abs{z-1}+\abs{z+1} \leqslant 4$ for $z\in C,$
$\abs{z} \leqslant 2$ (2 marks)

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a. $\text{Proof (See Worked Solutions)}$

b. $\text{See Worked Solutions}$

Show Worked Solution

a. $\text {Prove}\ \ \abs{z_1+z_2} \leqslant \abs{z_1}+\abs{z_2}:$

$\abs{z_1+z_2}^2$	$=\left(z_1+z_2\right)\left(\overline{z}_1+\overline{z}_2\right)$
	$=\abs{z_1}^2+\abs{z_2}^2+z_1 \overline{z}_2+\overline{z}_1 z_2$
	$=\abs{z_1}^2+\abs{z_2}^2+2 \operatorname{Re}\left(z_1 \overline{z}_2\right)$

$\text{Since}\ \ \operatorname{Re}(w) \leqslant\abs{w}\ \ \text{for} \ \ w\in C,$

$\abs{z_1+z_2}^2$	$\leqslant\abs{z_1}^2+\abs{z_2}^2+2\abs{z_1 \overline{z}_2}$
	$\leqslant\abs{z_1}^2+2\abs{z_1}\abs{z_2}+\abs{z_2}^2$
	$\leqslant\left(\abs{z_1}+\abs{z_2}\right)^2$

$\therefore\abs{z_1+z_2} \leqslant\abs{z_1}+\abs{z_2}$

b. $|z-1|+|z+1| \leqslant 4 \ \text{(given)}\ …\ (1)$

$\text {Using triangle inequality:}$

$|(z-1)+(z+1)| \leqslant|z-1|+|z+1|$

$\text{Since}\ \ (z-1)(z+1)=2 z:$

$\abs{2z}$	$\leqslant\abs{z-1}+\abs{z+1}$
$\abs{2z}$	$\leqslant 4\ \ \text{(using (1) above)}$
$2\abs{z}$	$\leqslant 4$
$\abs{z}$	$\leqslant 2$

Calculus, EXT2 C1 2020 HSC 13d*

By expanding `(text{cis}\theta + text{cis}(-theta))^4` show that
`cos^4 theta = frac{1}{8} ( cos (4 theta) + 4 cos (2 theta) + 3 )`. (3 marks)

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Hence, or otherwise, find `int_0^(frac{pi}{2}) cos^4 theta\ d theta`. (2 marks)

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a. `text{See Worked Solution}`

b. `frac{3 pi}{16}`

Show Worked Solution

a. `text{cis}\theta + text{cis}(-theta) = 2 cos theta\ \ …\ (1)`

`(text{cis}\theta + text{cis}(-theta))^4= 16 cos^4(4theta)`

`text{Expand LHS:}`

`(text{cis}\theta + text{cis}(-theta))^4`

`= text{cis}(4theta)+4text{cis}(2theta)+6+4text{cis}(-2theta)+text{cis}(4theta)`

`= 2text{cos}(4theta)+8text{cos}(2theta)+6\ \ text{(using (1) above)}`

`text{Equating sides:}`

`16 cos^4 theta`	`= 2 cos (4 theta) + 8 cos (2 theta) + 6`
`cos^4 theta`	`= frac{1}{8} cos(4 theta) + 1/2 cos(2 theta) + 3/8`
`cos^4 theta`	`= frac{1}{8} (cos(4 theta) + 4 cos(2 theta) + 3)`

b.	`int_0^(frac{pi}{2}) cos^4 theta\ d theta`	`= frac{1}{8} int_0^(frac{pi}{2}) cos(4 theta) + 4 cos(2 theta) + 3\ d theta`
		`= frac{1}{8} [ frac{1}{4} sin(4 theta) + 2 sin (2 theta) + 3 theta ]_0^(frac{pi}{2}`
		`= frac{1}{8} [( frac{1}{4} sin (2 pi) + 2 sin pi + frac{3 pi}{2}) – 0 ]`
		`= frac{1}{8} ( frac{3 pi}{2})`
		`= frac{3 pi}{16}`

Mechanics, EXT2 EQ-Bank 36

An experimental rocket is at a height of 5000 m, ascending at a speed of $50\sqrt{2}$ m s$^{-1}$ at an angle of 45° to the horizontal, when its engine stops. The rocket is then subject to gravity and to air resistance proportional to its velocity. Take $g$ = 10 m s$^{-2}$.

The velocity vector of the rocket, $t$ seconds after the engine stops, is

$\mathbf{v}(t) = 50e^{-0.2t}\,\mathbf{i} + (100e^{-0.2t}-50)\mathbf{j}.$ (Do NOT prove this.)

Show that the rocket reaches its greatest height when $t =5\ln 2$ seconds, and calculate its greatest height. (3 marks)

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The pilot can only operate the ejection seat while the rocket is descending at an angle between 45° and 60° to the horizontal. Find the earliest and latest times at which the pilot can eject. (3 marks)

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As the rocket continues to fall, its speed approaches a limiting value. Find this terminal speed, justifying your answer. (1 mark)

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a. $\text{See Worked Solutions}$

b. $\text{Pilot can eject between 5.5 and 6.6 seconds.}$

c. $\text{Terminal speed}=50 \ \text{ms}^{-1}$

Show Worked Solution

a. $\mathbf{v}(t)=50 e^{-0.2 t}\,\mathbf{i}+\left(100 e^{-0.2 t}-50\right)\mathbf{j}$

$\text{Max height occurs when} \ \ \dot{y}=0:$

$100 e^{-0.2 t}-50$	$=0$
$e^{-0.2 t}$	$=\dfrac{1}{2}$
$-0.2 t$	$=-\ln 2$
$t$	$=5 \ln 2$

$\text{Find} \ y \ \text{when}\ \ t=5 \ln 2:$

$y(t)=\displaystyle \int 100 e^{-0.2 t}-50\, d t=-500 e^{-0.2 t}-50 t+c$

$\text{When} \ \ t=0, y=5000:$

$5000=-500 e^{\circ}+c \ \Rightarrow \ c=5500$

$y=5500-500 e^{-0.2 t}-50 t$

$\text{At} \ \ t=5\ln 2:$

$y=5500-500 e^{-\ln 2}-50 \times 5 \ln 2=5076.71 \ldots=5077 \ \text{m}$.

b. $\text {On descent,} \ \ \dot{y}<0.$

$\text{Let} \ \ \theta=\text{angle below the horizontal}$

$\tan \theta=\dfrac{\abs{\dot{y}}}{\dot{x}}$

$\text{Let}\ \ a=e^{-0.2 t}$

$\tan \theta=\dfrac{50-100 a}{50 a}=\dfrac{1}{a}-2 \ \Rightarrow \ \theta=\tan ^{-1}\left(\dfrac{1}{a}-2\right)$

$\text{When} \ 45^{\circ} \ \text {is reached:}$

$\dfrac{1}{a}-2=1 \ \Rightarrow \ a=3$

$e^{-0.2 t}=\dfrac{1}{3} \ \Rightarrow \ t=\dfrac{\ln 3}{0.2} \approx 5.5 \ \text{s (1 d.p.)}$

$\text{When} \ 60^{\circ} \ \text {is reached:}$

$\dfrac{1}{a}-2=\sqrt{3} \ \Rightarrow \ a=\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}$

$e^{-0.2 t}$	$=2-\sqrt{3}$
$-0.2 t$	$=\ln (2-\sqrt{3})$
$t$	$=-5\ln (2-\sqrt{3}) \approx 6.6 \ \text{s (1 d.p.)}$

$\therefore \ \text{Pilot can eject between 5.5 and 6.6 seconds.}$

c. $\text{As} \ \ t \rightarrow \infty:$

$e^{-0.2 t} \rightarrow 0\ \ \Rightarrow\ \ \dot{x} \rightarrow 0, \ \ \dot{y} \rightarrow -50$

$\therefore \ \text{Terminal speed}=\sqrt{0^2+50^2}=50 \ \text{ms}^{-1}$

Mechanics, EXT2 EQ-Bank 34

A particle is projected from the origin with speed, $V$, at an angle of $\alpha$ above the horizontal. It is subject to both gravity and an air resistance proportional to its velocity, so that its horizontal and vertical components of acceleration while it is rising are given by

$\ddot{x}=-k\dot{x}$ and $\ddot{y} = -g-k\dot{y}$

Show that $\dot{x} = V\cos\,\alpha\ e^{-kt}$ and $\dot{y} = \left( \dfrac{g}{k} + V\sin\,\alpha\right)e^{-kt}-\dfrac{g}{k} $ (2 marks)

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Show that when the particle reaches its greatest height, it has travelled a horizontal distance of
$\dfrac{V^2\sin\,2\alpha}{2(g+Vk\,\sin\,\alpha)}$. (3 marks)

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a. $\text{See Worked Solutions}$

b. $\text{See Worked Solutions}$

Show Worked Solution

a. $\ddot{x}=-k \dot{x} \ \Rightarrow \ \dfrac{d \dot{x}}{d t}=-k \dot{x}$

$\text{Solving 1st order differential equation:}$

$\dot{x}=A e^{-k t}$

$\text{Since}\ \ \dot{x}=V \cos \alpha\ \ \text{when} \ \ t=0\ \ \Rightarrow\ \ A=V \cos \alpha$

$\dot{x}=V \cos \alpha e^{-k t}$

$\ddot{y}=-g-k \dot{y}\ \Rightarrow \ \dfrac{d \dot{y}}{d t}+k \dot{y}=-g$

$\text{Solving 1st order differential equation:}$

$\dot{y}=B e^{-k t}-\dfrac{g}{k}$

$\text{Since} \ \ \dot{y}=V \sin \alpha \ \ \text{when} \ \ t=0:$

$V \sin \alpha=B e^{-k t}-\dfrac{g}{k} \ \Rightarrow \ B=\dfrac{g}{k}+V \sin \alpha$

$\dot{y}=\left(V \sin \alpha+\dfrac{g}{k}\right) e^{-k t}-\dfrac{g}{k}$

b. $\text{At max height,} \ \ \dot{y}=0$

$\left(V \sin \alpha+\dfrac{g}{k}\right) e^{-k t}-\dfrac{g}{k}=0 \ \Rightarrow \ e^{-k t}=\dfrac{g}{g+V k \, \sin \alpha}\ \ldots\ (1)$

$\text{Find horizontal distance} \ (x):$

$x$	$=\displaystyle \int_0^t V \cos \alpha\, e^{-k t}\, d t$
	$=-\dfrac{V \cos \alpha}{k}\big[e^{-k t}\big]_0^t$
	$=\dfrac{V \cos \alpha}{k}\left(1-e^{-k t}\right)$
	$=\dfrac{V \cos \alpha}{k}\left(1-\dfrac{g}{g+Vk\, \sin \alpha}\right)\ \ \ \text{(using (1) above)}$
	$=\dfrac{V \cos \alpha}{k}\left(\dfrac{g+Vk\, \sin \alpha-g}{g+Vk\, \sin \alpha}\right)$
	$=\dfrac{V^2 \sin \alpha\, \cos \alpha}{g+Vk\, \sin \alpha}$
	$=\dfrac{V^2 \sin 2 \alpha}{2(g+Vk\, \sin \alpha)}$

Proof, EXT2 EQ-Bank 21

Prove that $\lim\limits_{x \to 0}x^2\, \sin \left(\dfrac{1}{x}\right)=0$. (2 marks)

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$-1 \leqslant \sin \left(\dfrac{1}{x}\right) \leqslant 1$

$-x^2 \leqslant x^2\, \sin \left(\dfrac{1}{x}\right) \leqslant x^2$

$\text{By squeeze theorem:}$

$\text{Since} \ \ \lim\limits_{x \to 0}-x^2=0\ \ \text{and}\ \ \lim\limits _{x \rightarrow 0} x^2=0$

$\Rightarrow \lim\limits _{x \rightarrow 0} x^2\, \sin \left(\dfrac{1}{x}\right)=0$

Show Worked Solution

$-1 \leqslant \sin \left(\dfrac{1}{x}\right) \leqslant 1$

$-x^2 \leqslant x^2\, \sin \left(\dfrac{1}{x}\right) \leqslant x^2$

$\text{By squeeze theorem:}$

$\text{Since} \ \ \lim\limits_{x \to 0}-x^2=0\ \ \text{and}\ \ \lim\limits _{x \rightarrow 0} x^2=0$

$\Rightarrow \lim\limits _{x \rightarrow 0} x^2\, \sin \left(\dfrac{1}{x}\right)=0$

Vectors, EXT2 EQ-Bank 26

Let $S$ be a sphere with equation

\begin{align*}
\left|r-\left(\begin{array}{c} 2 \\ -3 \\ 3
\end{array}\right)\right|=15
\end{align*}

Show the point $P(1,-1,2)$ lies outside the sphere $S$. (2 marks)

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$\text{Centre of circle} \ (C)=(2,-3,3)$

$\text{Radius}=15 \ \text{(given)}$

$\text{Find distance from} \ C \text { to } P(1,-1,2):$

$\operatorname{dist}=\sqrt{(1-2)^2+(-1+3)^2+(2-3)^2}=\sqrt{6}$

$\text{Since \(\ \sqrt{6}<15, P$ lies inside sphere.}\)

Show Worked Solution

$\text{Centre of circle} \ (C)=(2,-3,3)$

$\text{Radius}=15 \ \text{(given)}$

$\text{Find distance from} \ C \text { to } P(1,-1,2):$

$\operatorname{dist}=\sqrt{(1-2)^2+(-1+3)^2+(2-3)^2}=\sqrt{6}$

$\text{Since \(\ \sqrt{6}<15, P$ lies inside sphere.}\)

Proof, EXT2 EQ-Bank 34

Consider a sequence of rectangles with side lengths $a_{ n }$ and $b_{ n }$.

The first rectangle has $a_1=2$ and $b_1=1$.

For integers $n \geq 1,\ \ a_{n+1}=\dfrac{a_n+b_n}{2}$ and $b_{n+1}=\dfrac{2}{a_{n+1}}.$

Show that every rectangle in the sequence has an area of 2 square units. (1 mark)

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Use the relationship between the arithmetic mean and the geometric mean to prove that $a_n \geq \sqrt{2}$ for any integer $n \geq 1$. (2 marks)

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Use mathematical induction to prove that $a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2})$ for any integer $n \geq 1$. (4 marks)

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Use the squeeze theorem to show that the rectangles approach a square as $n$ approaches infinity. (2 marks)

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Show Worked Solution

a. $\text{Since}\ \ a_1 b_1=2\ \ \text{and}\ \ a_{n+1} b_{n+1}=a_{n+1} \times \dfrac{2}{a_{n+1}}=2$

$\Rightarrow\ \text{Each rectangle has area 2.}$

b. $a_1=2\ \ \Rightarrow\ \ a_1 \geq \sqrt{2}\ \ \text{(true for 1st rectangle)}$

$\text{AM/GM inequality:}\ \ \dfrac{a_n+b_n}{2} \geq \sqrt{a_nb_n} $

$a_nb_n=2\ \ \text{(from part (a))}$

$\dfrac{a_n+b_n}{2} \geq \sqrt{2}\ …\ (1)$

$\text{Since}\ \ a_{n+1}=\dfrac{a_n+b_n}{2}:$

$\ a_{n+1} \geq \sqrt{2}\ \ \ \text{(using (1) above)}$

$\therefore a_n \geq \sqrt{2}$

c. $\text{Prove}\ \ a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2}),\ \ \text{for}\ \ n \geq 1$

$\text{If}\ \ n=1:$

$\ a_1-\sqrt{2}=2-\sqrt{2} \leq \dfrac{1}{2^{0}}(2-\sqrt{2})$.

$\Rightarrow\ \text{True for}\ \ n=1.$

$\text{Assume true for}\ \ n=k:$

$a_k-\sqrt{2} \leq \dfrac{1}{2^{k-1}}(2-\sqrt{2})\ …\ (1) $

$\text{Prove true for}\ \ n=k+1:$

$\text{i.e.}\ \ a_{k+1}-\sqrt{2} \leq \dfrac{1}{2^k}(2-\sqrt{2})$

$\text{By definition,} \ \ a_{k+1}=\dfrac{1}{2}\left(a_k+b_k\right)$

$a_{k+1}-\sqrt{2}$	$=\dfrac{1}{2}\left(a_k-\sqrt{2}\right)+\dfrac{1}{2}\left(b_k-\sqrt{2}\right) $
	$\leq \dfrac{1}{2}\left(\dfrac{1}{2^{k-1}}(2-\sqrt{2})\right)+\dfrac{1}{2}\left(b_k-\sqrt{2}\right)\ \ \text{(see (1) above)}\ $

$\text{Since}\ \ a_k \geq \sqrt{2}\ \ \text{and}\ \ a_kb_k=2:$

$\ b_k \leq \sqrt{2}\ \ \text{and}\ \ b_k-\sqrt{2} \leq 0$

$a_{k+1}-\sqrt{2} \leq \dfrac{1}{2^k}(2-\sqrt{2})+\dfrac{1}{2}\left(b_k-\sqrt{2}\right) \leq \dfrac{1}{2^k}(2-\sqrt{2})$

$\Rightarrow\ \text{True for}\ \ n=k+1.$

$\therefore\ \text{Since true for}\ \ n=1,\ \text{by PMI, true for integers}\ \ n \geq 1.$

d. $\text {Combining parts (b) and (c):}$

$0 \leq a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2}).$

$\text{As}\ \ n \rightarrow \infty, \ \dfrac{1}{2^{n-1}} \rightarrow 0$

$\Rightarrow a_n-\sqrt{2} \rightarrow 0\ \ \text{(by squeeze theorem)}$

$\text{Since rectangles have an area = 2:}$

$\text{As}\ \ n \rightarrow \infty,\ b_n \rightarrow \sqrt{2}$

$\text{i.e. rectangles approach a square.}$

Mechanics, EXT2 EQ-Bank 30

Luggage at an airport is delivered to its owners via a ramp that is inclined at 30° to the horizontal. A 20 kg suitcase, initially at rest at the top of the ramp, slides down the ramp against a resistance of `v` newtons per kilogram, where `v\ text(ms)^(-1)` is the speed of the suitcase.

By resolving forces parallel to the ramp, show that the magnitude of the acceleration, `ddot{x}\ text(ms)^(-2)`, of the suitcase down the ramp is given by `ddot{x} = (g-2v)/2`. (2 marks)

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Using 9.8 `text(ms)^(-2)` as the acceleration due to gravity, find the distance `x` metres that the suitcase has slid as a function of `v`. Give your answer in the form `x = bv + c\ log_e(c/(c-v))`, where `b, c in R`. (3 marks)

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a. `text(Proof)\ \ text{(See Worked Solutions)}`

b. `x = -v + 4.9 ln ((4.9)/(4.9-v))`

Show Worked Solution

`sumF`	`=20g sin 30^@-20v`
`m ddot{x}`	`=10g-20v`
`ddot{x}`	`= (g-2v)/2`

b. `text{Using}\ \ ddot{x}=v *(dv)/(dx):`

`(dv)/(dx)`	`= (g-2v)/(2v)`
`(dx)/(dv)`	`= (2v)/(g-2v)`
`(dx)/(dv)`	`= -(2v)/(2v-g)=-((2v-g + g))/(2v-g)= -1-g/(2v-g)`

`text{Find the distance travelled:}`

`x`	`= int_0^v-1-g/(2v-g)\ dv`
	`= int_0^v-1-g/2 (2/(2v-g))\ dv`
	`= [-v-4.9 xx ln\ \|2v-g\|]_0^v`

`text(When)\ \ x=0, v=0:`

`2v-g < 0\ \ =>\ \ |2v-g| = g-2v`

`x`	`= [-v-4.9 ln (g-2v)]_0^v`
	`= -v-4.9 ln (g-2v)-(0-4.9 ln (g))`
	`= -v + 4.9 ln (g/(g-2v))`
	`=-v + 4.9 ln(9.8/(9.8-2v))`
	`= -v + 4.9 ln ((4.9)/(4.9-v))`

Complex Numbers, EXT2 N2 2023 14a*

Let $z$ be the complex number $z=\text{cis}\dfrac{\pi}{6} $ and $w$ be the complex number $w=\text{cis}\dfrac{3\pi}{4} $.

By first writing $z$ and $w$ in Cartesian form, or otherwise, show that
$|z+w|^2=\dfrac{4-\sqrt{6}+\sqrt{2}}{2}$. (3 marks)

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The complex numbers $z, w$ and $z+w$ are represented in the complex plane by the vectors $\overrightarrow{O A},\overrightarrow{O B}$ and $\overrightarrow{O C}$ respectively, where $O$ is the origin.
Show that $\angle A O C=\dfrac{7 \pi}{24}$. (2 marks)

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Deduce that $\cos \dfrac{7 \pi}{24}=\dfrac{\sqrt{8-2 \sqrt{6}+2 \sqrt{2}}}{4}$. (1 mark)

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i. $\text{See Worked Solutions}$

ii. $\text{See Worked Solutions}$

iii. $\text{See Worked Solutions}$

Show Worked Solution

i. $z= \cos\,\dfrac{\pi}{6} + i \,\sin\,\dfrac{\pi}{6} = \dfrac{\sqrt3}{2} + \dfrac{1}{2}i $

$w= \cos\,\dfrac{3\pi}{4} + i \,\sin\,\dfrac{3\pi}{4} = -\dfrac{1}{\sqrt2} + \dfrac{i}{\sqrt2} $

$\|z+w\|^2$	$=\Bigg{\|} \dfrac{\sqrt3}{2}+\dfrac{1}{2}i-\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2} \Bigg{\|}$
	$=\Bigg{\|} \Bigg{(}\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2} \Bigg{)} +\Bigg{(}\dfrac{1}{2}+\dfrac{1}{\sqrt2}\Bigg{)}\,i \Bigg{\|}$
	$=\Bigg{\|} \dfrac{\sqrt6-2}{2\sqrt2}+\dfrac{\sqrt2+2}{2\sqrt2}\,i \Bigg{\|}$
	$= \dfrac{(\sqrt6-2)^2+(\sqrt2+2)^2}{(2\sqrt2)^2}$
	$= \dfrac{6-4\sqrt6+4+2+4\sqrt2+4}{8}$
	$=\dfrac{16-4\sqrt6+4\sqrt2}{8} $
	$=\dfrac{4-\sqrt6+\sqrt2}{2} $

ii.

$\angle AOB= \arg(w)-\arg(z)=\dfrac{3\pi}{4}-\dfrac{\pi}{6}=\dfrac{7\pi}{12} $

$ |z|=|w|=1\ \Rightarrow AOBC\ \text{is a rhombus.} $

$\overrightarrow{OC}\ \text{is a diagonal of rhombus}\ AOBC $

$\Rightarrow \overrightarrow{OC}\ \text{bisects}\ \angle AOB $

$\therefore \angle AOC= \dfrac{1}{2} \times \dfrac{7\pi}{12}=\dfrac{7\pi}{24} $

iii. $\text{In}\ \triangle AOC: $

$ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA} = \overrightarrow{OB} $

$\Rightarrow \overrightarrow{OB}\ \text{is represented by}\ w. $

$\text{Using the cos rule in}\ \triangle AOC: $

$\cos\,\dfrac{7\pi}{24}$	$=\dfrac{\|z\|^2+\|z+w\|^2-\|w\|^2}{2\|z\|\|z+w\|}$
	$=\dfrac{ 1+\frac{4-\sqrt6+\sqrt2}{2}-1}{2 \times 1 \sqrt{\frac{4-\sqrt6+\sqrt2}{2}}} $
	$=\dfrac{\sqrt{\frac{4-\sqrt6+\sqrt2}{2}} \times 2} {2 \times 2} $
	$=\dfrac{\sqrt{4( \frac{4-\sqrt6+\sqrt2}{2})}} {4} $
	$=\dfrac{8-2\sqrt6+2\sqrt2}{4} $

♦♦ Mean mark (iii) 26%.

Mechanics, EXT2 EQ-Bank 19

A light inextensible string passes over a smooth pulley, as shown below, with particles of mass 1 kg and $m$ kg attached to the ends of the string.

The acceleration due to gravity is 9.8 m s$^{-2}$.

If the acceleration of the 1 kg particle is 4.9 ms$^{-2}$ upwards, then determine the value of $m$. (2 marks)

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$m=3\ \text{kg}$

Show Worked Solution

$\text{Resolving the forces:}$

$\text{Using}\ \ \Sigma F = m \ddot{x},\ \text{consider forces on 1 kg mass:}$

$T-(9.8 \times 1) = 4.9 \times 1\ \ \Rightarrow\ \ T=14.7$

$\text{Consider forces on}\ m\ \text{kg mass:}$

$m \times 9.8-T$	$=m \times 4.9$
$4.9m$	$= 14.7$
$:. m$	$= \dfrac{14.7}{4.9}=3\ \text{kg}$

Mechanics, EXT2 EQ-Bank 16

Particles of mass 3 kg and 5 kg are attached to the ends of a light inextensible string that passes over a fixed smooth pulley, as shown above. The system is released from rest and with acceleration due to gravity equal to 9.8 m s$^{-2}$.

Assuming the system remains connected, determine the speed of the 5 kg mass after two seconds. (3 marks)

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$v= 4.9\ \text{m s}^{-1}$

Show Worked Solution

$\Sigma F=5g-3g=2 \times 9.8 = 19.6\ \text{N}$

$\text{Find acceleration, using}\ \ \Sigma F=m \ddot{x}:$

$19.6$	$=(5+3) \ddot{x}$
$\ddot{x}$	$=\dfrac{19.6}{8}=2.45\ \text{m s}^{-2}$

$v= \displaystyle \int \ddot{x}\,dt=\int 2.45\,dt=2.45t+c$

$v=0\ \ \text{at}\ \ t=0\ \ \Rightarrow\ \ c=0$

$\text{At}\ \ t=2:$

$v=2 \times 2.45 = 4.9\ \text{m s}^{-1}$

Mechanics, EXT2 2019 SPEC1 9

A light inextensible string is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs in equilibrium from a smooth ring on the string, as shown in the diagram below. The string makes an angle `alpha` with the ceiling.

Express the tension, `T` newtons, in the string in terms of `m`, `g` and `alpha`. (1 mark)

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A different light inextensible sting is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs from a smooth ring on the string. A horizontal force of `F` newtons is applied to the ring. The tension in the sting has a constant magnitude and the system is in equilibrium. At one end the string makes an angle `beta` with the ceiling and at the other end the string makes an angle `2beta` with the ceiling, as shown in the diagram below.

Show that `F = mg((1-cos(beta))/(sin(beta)))`. (3 marks)

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a. `T = (mg)/(2sinalpha)`

b. `text(See Worked Solutions)`

Show Worked Solution

`2 xx Tsinalpha`	`= mg`
`:.T`	`= (mg)/(2sinalpha)`

`text(Resolving forces vertically:)`

`Tsin(beta) + Tsin(2beta)`	`= mg`
`T`	`= (mg)/(sin(beta) + sin(2beta))`

`text(Resolving forces horizontally:)`

`F + Tcos(2beta)`	`= Tcos(beta)`
`F`	`= Tcos(beta)-Tcos(2beta)`
	`= T(cos(beta)-cos(2beta))`
	`= T[cos(beta)-(2cos^2beta-1)]`
	`= T(−2cos^2(beta) + cos(beta) + 1)`
	`= T(−2cos(beta)-1)(cos(beta)-1)`
	`= (mg(1 -cos(beta))(2cosbeta + 1))/(sin(beta) + 2sin(beta)cos(beta))`
	`= (mg(1-cos(beta))(2cos(beta) + 1))/(sin(beta)(1+2cos(beta)))`
	`= mg((1-cos(beta))/(sin(beta)))`

Mechanics, EXT2 EQ-Bank 25

The diagram below shows objects of mass 5 kg and $M$ kg attached to the ends of a light, inextensible string that passes over a smooth pulley.

The 5 kg object is accelerating upwards at a rate of 4.9 m/s$^2$. Let the tension in the string be $T$ newtons.

Using 9.8 m s$^{-2}$ as the acceleration due to gravity, find the value of $T$ and hence determine the value of $M$. (3 marks)

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$M=15\ \text{kg}$

Show Worked Solution

$\text{Using} \ \ F=m \ddot{x}:$

$\text{Consider the 5 kg mass:}$

$T-5 g$	$=m \ddot{x}$
$T-5(9.8)$	$=5(4.9)$
$T$	$=73.5 \ N$

$\text{Consider the \(M$ kg mass:}\)

$\text{Mass is accelerating downward at 4.9 ms\(^{-2}$}\)

$M \times 9.8-73.5$	$=M \times 4.9$
$4.9 M$	$=73.5$
$M$	$=\dfrac{73-5}{49}=15 \ \text{kg}$

Mechanics, EXT2 EQ-Bank 20

A 6-kilogram mass is placed on a frictionless plane inclined 30° to the horizontal.

The mass is connected to another 10-kilogram mass by a light, inextensible string via a pulley, as shown in the diagram

The 6-kilogram mass is released from rest and slides up the plane.

Given the acceleration due to gravity is $g$ m s$^{-2}$, express the velocity, $v$, of the 6 kilogram mass in terms of $t$ as it moves up the plane. (3 marks)

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$v=\dfrac{7}{16}gt$

Show Worked Solution

$\text{Resolving forces on the 6 kg mass:}$

$\text{Let} \ \ F_d=\text{force down slope:}$

$\sin 30^{\circ}$	$=\dfrac{F_d}{6 g}$
$F_d$	$=6 g\, \sin 30=3 g$

$\text{Using} \ \ \sum F=m a$:

$T-3 g=6 a\ \ldots\ (1)$

$\text{Consider 10 kg mass:}$

$10 g-T=10 a\ \ldots\ (2)$

$\text{Add (1) + (2):}$

$7 g=16 a \ \Rightarrow \ a=\dfrac{7}{16} g$

$v=\displaystyle \int \dfrac{7}{16}g \, dt=\dfrac{7}{16} g t+c$

$\text{When} \ \ t=0 \quad v=0 \ \Rightarrow \ c=0$

$\therefore v=\dfrac{7}{16}gt$

Mechanics, EXT2 EQ-Bank 22

A 12 kilogram object is suspended from a horizontal ceiling by two light, inextensible strings at angles of 30° and 45°, as shown in the diagram.

The acceleration due to gravity is $g$ m s$^{-2}$ and the tensions in the strings are $T_1$ newtons and $T_2$ newtons.

Show that $T_2=\sqrt{\dfrac{3}{2}} T_1$ (2 marks)

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Determine the tensions, in newtons, of $T_1$ and $T_2.$ (2 marks)

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a. $\text{See Worked Solutions}$

b. $T_1=12(\sqrt{3}-1)g$

$T_2=6 \sqrt{6}(\sqrt{3}-1)g\ \ \ (=6 \sqrt{2}(3-\sqrt{3}))$

Show Worked Solution

a. $\text{Resolve forces into horizontal / vertical components:}$

$\text{Horizontal forces are equal.}$

$T_1\cos 30^{\circ}$	$=T_2 \cos 45^{\circ}$
$T_1 \times \dfrac{\sqrt{3}}{2}$	$=T_2 \times \dfrac{1}{\sqrt{2}}$
$T_2$	$=\dfrac{\sqrt{3} \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \, T_1=\sqrt{\dfrac{3}{2}}\, T_1$

b. $\text{Vertical forces are equal.}$

$T_1 \sin 30+T_2 \sin 45$	$=12 g$
$T_1 \times \dfrac{1}{2}+T_2 \times \dfrac{1}{\sqrt{2}}$	$=12 g$

$\text{Substitute} \ \ T_2=\sqrt{\dfrac{3}{2}}\, T_1:$

$T_1 \times \dfrac{1}{2}+T_1 \times \sqrt{\dfrac{3}{2}} \times \dfrac{1}{\sqrt{2}}$	$=12 g$
$T_1\left(\dfrac{\sqrt{3}+1}{2}\right)$	$=12 g$

$T_1=\dfrac{24g}{\sqrt{3}+1} \times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}=12(\sqrt{3}-1)g$

$T_2$	$=\sqrt{\dfrac{3}{2}} \times 12(\sqrt{3}-1)g$
	$=6 \sqrt{6}(\sqrt{3}-1)g\ \ \ (=6 \sqrt{2}(3-\sqrt{3}))$

Mechanics, EXT2 EQ-Bank 17

Two light inextensible strings are attached to a horizontal surface and suspended a 10-kilogram object as shown in the diagram below

The tension in the strings are $T_1$ newtons and $T_2$ newtons.

Express $T_1$ in terms of $T_2$. (2 marks)

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If the acceleration due to gravity is 9.8 ms$^{-2}$, determine the exact values of $T_1$ and $T_2$, in newtons. (2 marks)

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a. $\text{See Worked Solutions}$

b. $T_1=49 \sqrt{3}, T_2=49$

Show Worked Solution

a. $\text{Resolve forces into horizontal/vertical components:}$

$\text{Horizontal forces are equal.}$

$T_1 \cos 60^{\circ}$	$=T_2 \cos 30^{\circ}$
$T_1 \times \dfrac{1}{2}$	$=T_2 \times \dfrac{\sqrt{3}}{2}$
$T_1$	$=\sqrt{3}\, T_2$

b. $\text{Vertical forces are equal.}$

$T_1 \sin 60^{\circ}+T_2 \sin 30^{\circ}$	$=10 \times 9.8$
$T_1 \times \dfrac{\sqrt{3}}{2}+T_2 \times \frac{1}{2}$	$=98$

$\text {Substitute} \ \ T_1=\sqrt{3}\, T_2 :$

$\sqrt{3}\, T_2 \times \dfrac{\sqrt{3}}{2}+T_2 \times \dfrac{1}{2}$	$=98$
$2\, T_2$	$=98$
$T_2$	$=49 \ \text{newtons}$

$\therefore T_1=49 \sqrt{3}, \ T_2=49$

Mechanics, EXT2 EQ-Bank 30

In a circus act, an 8 kg cannon ball is projected from the origin into the air with an initial velocity of 26 m s$^{-1}$ and at an angle of 67.4° to the horizontal. The ball is caught at the top of its trajectory by a performer who is at the position $(A, B)$.

The velocity vector, $\mathbf{v} (t)$, of the ball at time $t$ seconds after launch is given by

$\mathbf{v}(t)=10 e^{-0.8 t} \mathbf{i} +\left[36.5 e^{-0.8 t}-12.5\right] \mathbf{j}$. (Do NOT Prove this.)

Show that the ball reaches the performer at $t=1.339$ (to three decimal places). (2 marks)

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Find the values of $A$ and $B$. (3 marks)

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What is the speed of the ball when it reaches the performer? (1 mark)

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What is the magnitude of the force on the ball when it reaches the performer? (2 marks)

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a. $t=1.339 \ \text{s}$

b. $A=8.217\ \text{m},\ \ B=13.257\ \text{m}$

c. $\text{Speed}\ = 3.426\ \text{m s}^{-1}$

d. $82.951 \ \text{N}$

Show Worked Solution

a. $\text{At top of trajectory:}$

$36.5 e^{-0.8 t}-12.5$	$=0$
$36.5 e^{-0.8 t}$	$=12.5$
$e^{-0.8 t}$	$=\dfrac{12.5}{36.5}$
$-0.8 t$	$=\ln \dfrac{12.5}{36.5}$
$ t$	$=\dfrac{\ln\frac{12.5}{36.5}}{-0.8}=1.339 \ \text{s (3 d.p.)}$

b. $\text{Horizontal velocity}\ =10 e^{-0.8 t}$

$x=\displaystyle \int 10 e^{-0.8 t}\,d t=\dfrac{10}{-0.8} e^{-0.8 t}+c_1=-12.5 e^{-0.8 t}+c_1$

$\text{When} \ \ t=0, \ x=0:$

$0=-12.5 e^0+c_1\ \ \Rightarrow\ \ c_1=12.5$

$x=-12.5 e^{-0.8 t}+12.5$

$\text{When } t=1.339, \ x=A:$

$A=-12.5 e^{-0.8 \times 1.339}+12.5=8.217\ \text{m (3 d.p.)}$

$\text {Vertical velocity}\ =36.5 e^{-0.8 t}-12.5$

$y=\displaystyle \int\left(36.5 e^{-0.8 t}-12.5\right)\,d t=\dfrac{36.5}{-0.8} e^{-0.8 t}-12.5 t+c_2=-45.625 e^{-0.8 t}-12.5 t+c_2$

$\text{When} \ \ t=0, \ y=0:$

$0=-45.625+c_2\ \ \Rightarrow\ \ c_2=45.625$

$y=-45.625 e^{-0.8 t}-12.5 t+45.625$

$\text{When}\ \ t=1.339, \ y=B:$

$B=-45.625 e^{-0.8 \times 1.339}-12.5 \times 1.339+45.625=13.257\ \text{m (3 d.p.)}$

c. $\text{At top of trajectory,}\ \mathbf{j} \text{-component of velocity = 0.}$

$\Rightarrow\ \text{Speed at top is the}\ \mathbf{i} \text{-component of velocity at}\ t=1.339:$

$\text{Speed}\ =10e^{-0.8 \times 1.339} = 3.426\ \text{m s}^{-1}\ \text{(3 d.p.)}$

d. $\text{Using}\ \ F=m \ddot{x}:$

$\mathbf{v}(t)=10 e^{-0.8 t}\mathbf{i} +\left(36.5 e^{-0.8 t}-12.5\right) \mathbf{j}$

$\mathbf{a} =\dfrac{d v }{d t}=-8 e^{-0.8 t} \mathbf{i} -29.2 e^{-0.8 t} \mathbf{j}$

$\text{When} \ \ t=1.339:$

$\mathbf{a}=-8 e^{-0.8 \times 1.339}\mathbf{i}-29.2 e^{-0.8 \times 1.339}\mathbf{j}=-2.74 1\,\mathbf{i} -10\, \mathbf{j}$

$\text{Magnitude of acceleration}$

$=\sqrt{(2.741)^2+(10)^2}=10.36885 \ldots\ \text{ms}^{-2}$

$\therefore \ \text{Magnitude of force}$

$=8 \times 10.36885 \ldots =82.951 \ \text{N (3 d.p.)}$

Mechanics, EXT2 EQ-Bank 18

An object of mass 5 kg is on a slope that is inclined at an angle of 60° to the horizontal. The acceleration due to gravity is $g \ \text{ms} ^{-2}$ and the velocity of the object down the slope is $v \ \text{ms} ^{-1}$.

As well as the force due to gravity, the object is acted on by two forces, one of magnitude $2 v$ newtons and one of magnitude $2 v^2$ newtons, both acting up the slope.

Show that the resultant force down the slope is
$\dfrac{5 \sqrt{3}}{2} g-2 v-2 v^2 $ newtons. (2 marks)

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There is one value of $v$ such that the object will slide down the slope at a constant speed.
Find this value of $v$ in $\text{ms}^{-1}$, correct to 1 decimal place, given that $g=10$. (2 marks)

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a. $\text{See Worked Solution}$

b. $v=4.2 \ \text{ms}^{-1} $

Show Worked Solution

$\Sigma F\ \text{(down slope) }$	$=5 g\ \cos 30^{\circ}-2 v-2 v^2$
	$=\dfrac{5 \sqrt{3}}{2} g-2 v-2 v^2 \ \ \text{newtons }$

b. $\text{Constant speed} \ \Rightarrow \ \ \Sigma F=0$

$\dfrac{5 \sqrt{3}}{2} \times 10$	$=2 v+2 v^2$
$0$	$=2 v^2+2 v-25 \sqrt{3}$

$v=\dfrac{-2 \pm \sqrt{4+4(2)(25 \sqrt{3})}}{4} =4.1798 \ldots \text { or }-5.1798 \ldots$

$\text{Since object is moving down the slope,}\ \ v \gt 0.$

$\therefore v=4.1798 = 4.2 \ \text{ms}^{-1} \ \ \text{(1 d.p.)}$

Vectors, EXT2 EQ-Bank 21

Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in the plane, where $\mathbf{v}\,\neq\, \mathbf{0}$.

For every real number $t$, let $P(t)=\abs{\mathbf{u} -t \mathbf{v}}^2$.

Show that $P(t)=\abs{\mathbf{u}}^2-2 t( \mathbf{u} \cdot \mathbf{v} )+t^2 \abs{\mathbf{v}}^2$. (1 mark)

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Show that $P(t)$ has a minimum value at $t=\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}$. (2 marks)

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Hence prove the Cauchy-Schwarz inequality, $\abs{\mathbf{u}\cdot\mathbf{v}} \leq \abs{\mathbf{u}}\abs{\mathbf{v}}$. (3 marks)

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a. $\text{See Worked Solutions}$

b. $\text{See Worked Solutions}$

c. $\text{See Worked Solutions}$

Show Worked Solution

a.	$P(t)$	$=\abs{\mathbf{u} -t \mathbf{v}}^2$
		$=( \mathbf{u} -t \mathbf{v} ) \cdot( \mathbf{u} -t\mathbf{v} )$
		$= \mathbf{u} \cdot \mathbf{u} + \mathbf{u} \cdot(-t \mathbf{v} )+(-t \mathbf{v}) \cdot \mathbf{u} +t^2(\mathbf{v} \cdot \mathbf{v} )$
		$=\abs{\mathbf{u}}^2-2 t( \mathbf{u} \cdot \mathbf{v} )+t^2\abs{\mathbf{v}}^2$

b. $\text{Find}\ t\ \text{when}\ \ \dfrac{d P}{d t}=0:$

$2 t\abs{\mathbf{v}}^2-2 \mathbf{u} \cdot \mathbf{v}$	$=0$
$2 t\abs{\mathbf{v}}^2$	$=2 \mathbf{u} \cdot \mathbf{v}$
$t$	$=\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}$

$\dfrac{d^2 P}{d t^2}=2\abs{\mathbf{v}}^2>0$

$\therefore\ \text{Min SP at}\ \ t=\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}$

c. $P(t)=\abs{\mathbf{u} -t \mathbf{v}}^2\ \ \Rightarrow\ \ P(t) \geq 0$

$P\left(\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\right)$	$=\left(\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\right)^2\abs{\mathbf{v}}^2-2\left(\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\right) \mathbf{u} \cdot \mathbf{v} +\abs{\mathbf{u}}^2$
	$=-\dfrac{( \mathbf{u} \cdot \mathbf{v} )^2}{\abs{\mathbf{v}}^2}+\abs{\mathbf{u}}^2$

$0$	$ \leq -\dfrac{( \mathbf{u} \cdot \mathbf{v} )^2}{\abs{\mathbf{v}}^2}+\abs{\mathbf{u}}^2$
$(\mathbf{u} \cdot \mathbf{v} )^2$	$\leq \abs{\mathbf{u}}^2\abs{\mathbf{v}}^2$
$\abs{ \mathbf{u} \cdot \mathbf{v} }$	$\leq\abs{\mathbf{u}}\abs{\mathbf{v}}\ \ \text{(i.e. the Cauchy-Schwarz inequality)}$

Vectors, EXT2 EQ-Bank 27

Line 1 is given by the equations $x=-1+2 s, \ y=1-2 s$ and $z=1+2 s$, where $s$ is a parameter.

Line 2 is given by the equations $x=1+2 t, \ y=-1-t$ and $z=4+3 t$, where $t$ is a parameter.

Show that Line 1 and Line 2 are skew. (3 marks)

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Show Worked Solution

$\text{Skew lines do not intersect and are not parallel.}$

$\text{Solving for \(s$ and $t$ in $x$ and $y$:}\)

$\text{From} \ x: \quad$	$-1+2 s$	$=1+2 t\ \ldots\ (1)$
$\text{From} \ y: \quad$	$1-2 s$	$=-1-t\ \ldots\ (2)$
$(1)+(2)$	$0$	$=t$

$\text{Substitute}\ \ t=0\ \ \text{into (1):}$

$-1+2 s=1\ \ \Rightarrow\ \ s=1$

$\text{Substitute}\ \ s=1\ \ \text{and}\ \ t=0\ \ \text{into}\ z:$

$z=1+2=3\ \text{(line 1)}, \ z=4\ \text{(line 2)}$

$\Rightarrow\ \text{Lines 1 and 2 do not intersect.}$

$\text{Direction vectors:}$

$\text{Line 1 = }\left( \begin{array}{r}2 \\ -2 \\ 2\end{array}\right),\ \ \text{Line 2 = } \left(\begin{array}{r}2 \\ -1 \\ 3\end{array}\right).$

$\left(\begin{array}{r}2 \\ -2 \\ 2\end{array}\right) \neq k\left(\begin{array}{r}2 \\ -1 \\ 3\end{array}\right) \ \text{for any}\ \ k \in R$

$\Rightarrow\ \text{Lines 1 and 2 are not parallel.}$

$\therefore\ \text{Lines 1 and 2 are skew as they do not intersect and are not parallel.}$

Calculus, EXT1 EQ-Bank 4 MC

Which slope field best matches the differential equation $\dfrac{d y}{d x}=-y^2\left(1- y ^2\right)$ ?

Show Answers Only

$C$

Show Worked Solution

$\text{By elimination:}$

$\text{At}\ \ y=0, \ \dfrac{dy}{dx}=0\ \ \text{(eliminate B)}$

$\text{At}\ \ y=1, \ \dfrac{dy}{dx}=-1(1-1)=0\ \ \text{(eliminate D)}$

$\text{At}\ \ y=-\dfrac{3}{2}, \ \dfrac{dy}{dx}=-\dfrac{9}{4}\left(1-\dfrac{9}{4}\right) \gt 0\ \ \text{(eliminate A)}$

$\Rightarrow C$

Vectors, EXT2 EQ-Bank 18

Consider the triangle with vertices $A(4,5,1), B(8,1,2)$ and the origin $O(0,0,0)$. The triangle has three medians.

The median through $B$ has vector equation

$\lambda \, \overrightarrow{O M}+(1-\lambda) \overrightarrow{O B}=\left(\begin{array}{l}8 \\ 1 \\ 2\end{array}\right)+\lambda\left(\begin{array}{c}-6 \\ 1.5 \\ -1.5\end{array}\right)$ (Do NOT prove this.)

for a parameter $\lambda \in[0,1]$ and where $M$ is the midpoint of $O A$.

Write down the median through $A$ as a vector equation. (2 marks)

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The three medians of any triangle meet at a point known as the centroid.
Find the value of $\lambda$ corresponding to the centroid. (2 marks)

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Into what ratio does the centroid divide the median $B M$? (1 mark)

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a. $\left(\begin{array}{c}4 \\5 \\1\end{array}\right) + \mu \left(\begin{array}{c}0 \\-4.5 \\0\end{array}\right)\ \ \text{for}\ \ \mu \in[0,1]$

b. $\lambda=\dfrac{2}{3}.$

c. $2:1.$

Show Worked Solution

a. $\text{Let}\ N = \text{midpoint of}\ OB:$

$N \equiv \left(\dfrac{8+0}{2}, \dfrac{1+0}{2}, \dfrac{2+0}{2}\right) \equiv (4,0.5,1).$

$\text{Direction vector of the median through} \ A \ \text {is}\left(\begin{array}{c}4-4 \\0.5-5 \\1-1\end{array}\right) = \left(\begin{array}{c}0 \\-4.5 \\0\end{array}\right).$

$\text{Equation of median from}\ A:$

$\left(\begin{array}{c}4 \\5 \\1\end{array}\right) + \mu \left(\begin{array}{c}0 \\-4.5 \\0\end{array}\right)\ \ \text{for}\ \ \mu \in[0,1]$

b. $\text{Equation of median from}\ B:$

$\left(\begin{array}{c}8 \\1 \\2\end{array}\right) + \lambda \left(\begin{array}{c}-6 \\1.6 \\-1.5\end{array}\right)\ \ \text{for}\ \ \lambda \in[0,1]$

$\text{Medians intersect at centroid}$

$x\text{-coordinate of median through}\ B = 8-6\lambda$

$x\text{-coordinate of median through}\ A = 4$

$\text{Equating}\ x\text{-coordinates:}$

$8-6 \lambda=4\ \ \Rightarrow\ \ \lambda=\dfrac{2}{3}$

$\therefore\ \text{Point of intersection occurs at } \lambda=\dfrac{2}{3}.$

c. $\text{The centroid divides the median in the ratio of} \ 2:1.$

Mechanics, EXT2 EQ-Bank 24

A 2 kg mass is initially at rest on a smooth horizontal surface. The mass is then acted on by two constant forces that cause the mass to move horizontally. One force has magnitude 10 N and acts in a direction 60° upwards from the horizontal, and the other force has magnitude 5 N and acts in a direction 30° upwards from the horizontal, as shown in the diagram below.

Find the normal reaction force, in newtons, that the surface exerts on the mass. (2 marks)
Find the acceleration of the mass, in ms⁻², after it begins to move. (2 marks)

Show Answers Only

a. `R = 2g-5/2-5 sqrt 3\ text(N)`

b. `ddot{x} = 5/2-(5 sqrt 3)/2\ text(ms)^(-2)`

Show Worked Solution

`text{Let}\ R =\ text{normal force}`

`text(Resolving forces vertically:)`

`2g`	`= 5 sin 30 + 10 sin 60 + R`
`2g`	`= 5/2 + 5 sqrt 3 + R`
`R`	`= 2g-5/2-5 sqrt 3\ text(N)`

b. `text{Using}\ Sigma F=m ddotx :`

	`2ddot{x}`	`= 10 cos 60-5 cos 30`
	`2ddot{x}`	`= 5-(5 sqrt 3)/2`
	`:.ddot{x}`	`= 5/2-(5 sqrt 3)/4\ text(ms)^(-2)`

Mechanics, EXT2 2020 SPEC2 18 MC

A particle of mass `m` kilogram hangs from a string that is attached to a fixed point. The particle is acted on by a horizontal force of magnitude `F` newtons. The system is in equilibrium when the string makes an angle `alpha` to the horizontal, as shown in the diagram below. The tension in the string has magnitude `T` newtons.

The value of `tan\ alpha` is

`(mg)/T`
`T/(mg)`
`F/(mg)`
`(mg)/F`

Show Answers Only

`D`

Show Worked Solution

`text(Resolving forces vertically:)`

`mg = Tsin(alpha)`

`text(Resolving forces horizontally:)`

`F = Tcos(alpha)`

`:. tan\ alpha = (mg)/F`

`=>D`

Mechanics, EXT2 EQ-Bank 21

A mass of `m_1` kilograms is placed on a plane inclined at 30° to the horizontal. It is connected by a light inextensible string to a second mass of `m_2` kilograms that hangs below a frictionless pulley situated at the top end of the incline, over which the string passes.

Given that the inclined plane is smooth, find the relationship between `m_1` and `m_2` if the mass `m_1` moves down the plane at constant speed. (3 marks)

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`m_1=2m_2`

Show Worked Solution

`m_1g\ sin30-m_2g = (m_1 + m_2)a`

`text(S)text(ince)\ m_1\ text(moves at constant speed,)\ a = 0`

`m_1 g · 1/2-m_2 g`	`= 0`
`m_1`	`= 2m_2`

Financial Maths, STD2 EQ-Bank 29

Shown below is part of the output from a spreadsheet used to model a 25-year reducing balance loan with equal monthly repayments and a constant interest rate, $r$, expressed as a decimal.

Note that many rows and values have been removed from the output shown.

Show that the amount owing at the end of month 120 is $407 860.99. (4 marks)

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$\text{See worked solution}$

Show Worked Solution

$\text{Find } r:$

$\text{Interest}$	$= r \times \text{amount owing}$
$3500$	$= r \times 500\,000$
$r$	$= 0.007$

$\text{Find monthly repayment using PVIF table:}$

$\text{At } N = 300\ \text{ and }\ r = 0.007,\ \text{ PVIF} = 125.23492$

$\text{Monthly repayment}= \dfrac{500\,000}{125.23492}= \$3992.50$

$\text{Amount owing at start of month 120:}$

$\text{Amount owing at end of month 119} = \$408\,990.56$

$\therefore\ \text{Interest charged in month 120}$	$= 0.007 \times \$408\,990.56$
	$= \$2862.93$

$\text{Amount owing at end of month 120:}$

$= \$408\,990.56+\$2862.93-\$3992.50$

$= \$407\,860.99 \quad \checkmark$

Trigonometry, EXT1 T1 EQ-Bank 28

On the number plane provided below, draw $y=\cos ^{-1} x+\sin ^{-1} x$ by first drawing $y=\cos ^{-1} x$ and $y=\sin ^{-1} x$. (3 marks)

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Show Worked Solution

Statistics, EXT1 EQ-Bank 12

An office has 7 printers and 5 photocopiers. On average, each printer is used 73% of the time and each photocopier is used 46% of the time.

Write an expression for the probability that, at a particular time, at least one printer is in use. (1 mark)

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Write an expression for the probability that, at a particular time, at least one printer and exactly three photocopiers are in use. (2 marks)

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a. $1-{ }^7 C_0(0.27)^7$

b. $\left[1-(0.27)^7\right] \times { }^5 C_3(0.46)^3(0.54)^2$

Show Worked Solution

a. $P\text{(printer in use)} = 0.73, \ \ P\text{(printer not in use)} = 0.27$

$P\text{(at least 1 printer in use)}$	$=1-P\text{(no printer in use)}$
	$=1-{ }^7 C_0(0.27)^7$

b. $P\text{(copier in use)} = 0.46, \ \ P\text{(copier not in use)} = 0.54$

$P\text{(at least 1 printer and exactly 3 copiers in use)}$

$=\left[1-(0.27)^7\right] \times { }^5 C_3(0.46)^3(0.54)^2$

Statistics, EXT1 EQ-Bank 25

In a large school, the average amount of money spent per student per day at the canteen is $8 with a standard deviation of 6.5 .

At the end of each day, 50 randomly chosen students are asked how much they spent at the canteen on that day.

Use the standard normal distribution table (included) to find the probability that the sample mean on a particular day is greater than $10. (3 marks)

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$1.46 \%$

Show Worked Solution

$\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{Central Limit Theorem applies}$

$\text{The sample mean,}\ \overline{X}, \text{for random samples of size 50 is}$

$\text{approximately normally distributed, where:}$

$\mu=8\ \ \text{and}\ \ \sigma=\dfrac{6.5}{\sqrt{n}}=\dfrac{6.5}{\sqrt{50}}$

$\text{By the central limit theorem:}$

$Z=\dfrac{\overline{X}-8}{\frac{6.5}{\sqrt{50}}} \sim N(0,1)$

$Z=\dfrac{10-8}{\frac{6.5}{\sqrt{50}}}=2.18 \ \text{(2 d.p.)}$

$\text{Using Normal Distribution Table of Values:}$

$P(\overline{X} \geq 10)$	$=1-P(Z \leq 2.18)$
	$=1-0.9854$
	$=1.46 \%$

Vectors, EXT1 EQ-Bank 26

The position vector of a particle that is moving along a curve at time `t` is given by

`\mathbf{r}(t) = 3 cos (t) \mathbf{i} + 4 sin (t) \mathbf{j}, \ t >= 0`.

Determine the first time when the speed of the particle is a minimum. (3 marks)

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`t_1 = pi/2`

Show Worked Solution

`underset ~v(t)= -3 sin(t) underset ~i + 4 cos (t) underset ~j`

`text{Since speed =}\ |underset ~v(t)|:`

`\|underset ~v(t)\|`	`= sqrt (9 sin^2(t) + 16 cos^2(t))`
	`= sqrt (9 sin^2(t) + 9 cos^2(t) + 7 cos^2(t))`
	`= sqrt (9 + 7 cos^2(t))`

`text(Minimised speed occurs when)\ cos(t) = 0:`

`:. t_1 = pi/2`

COMMENT:
Undertilde notation in answers is allowed even when boldface vector notation appears in the question.

Vectors, EXT1 EQ-Bank 4 MC

The position of a body is given by `\mathbf{r} = 3\mathbf{i} + \mathbf{j} ` metres at a particular time. The body moves with constant velocity and two seconds later its displacement is `−\mathbf{i} + 5\mathbf{j} ` metres.

The velocity, in m s⁻¹, of the body is

`2\mathbf{i} + 6\mathbf{j}`
`−2\mathbf{i} + 2\mathbf{j}`
`−4\mathbf{i} + 4\mathbf{j}`
`4\mathbf{i}-4\mathbf{j}`

Show Answers Only

`B`

Show Worked Solution

`Delta \mathbf{r}= (−1-3)\mathbf{i} + (5-1)\mathbf{j}= −4\mathbf{i} + 4\mathbf{j}`

`\mathbf{v} = (Delta \mathbf{r})/(Delta \mathbf{t})= (−4\mathbf{i} + 4\mathbf{j})/(2-0)=-2\mathbf{i}+\mathbf{j}`

`=> B`

Vectors, EXT1 V1 EQ-Bank 6 MC

The position of a particle at time `t` is given by `\mathbf{r}(t) = (sqrt(t-2))\mathbf{i} + (2t)\mathbf{j}` for `t >= 2`.

The cartesian equation of the path of the particle is

`y = 2x^2 + 4, \ \ \ \ \ x >= 2`
`y = 2x^2 + 2, \ \ \ \ \ x >= 2`
`y = sqrt((x-4)/2),\ \ \ x >= 2`
`y = 2x^2 + 2,\ \ \ \ \ x >= 0`

Show Answers Only

`A`

Show Worked Solution

`underset ~r(t) = (sqrt(t-2)\underset ~i + (2t) underset ~j)`

`x = sqrt(t-2)`

`text(Given)\ \ t >= 2\ \ =>\ \ x >=0`

`y = 2t\ \ =>\ \ t = y/2`

`:. x`	`= sqrt(y/2-2)`
`x^2`	`= y/2-2`
`y/2`	`= x^2 + 2`
`y`	`= 2x^2 + 4`

`=> A`

Vectors, EXT1 EQ-Bank 5 MC

The acceleration vector of a particle that starts from rest is given by

`underset ~a(t) = −4 sin(2t) underset ~i + 20 cos (2t) underset ~j`, where `t >= 0`.

The velocity vector of the particle, `underset ~v(t)`, is given by

`−8 cos(2t) underset ~i-40 sin(2t) underset ~j`
`2 cos(2t) underset ~i + 10 sin(2t) underset ~j`
`(8-8 cos(2t)) underset ~i-40 sin(2t) underset ~j`
`(2 cos(2t)-2) underset ~i + 10 sin(2t) underset ~j`

Show Answers Only

`D`

Show Worked Solution

`underset ~v(t)= int underset ~a (t)\ dt= (2 cos (2t) + c_0) underset ~i + (10 sin (2t) + c_1) underset ~j`

`text(S)text(ince)\ \ v=0\ \ text(when)\ \ t=0:`

`0= (2 cos (0) + c_0) underset ~i + (10 sin (0) + c_1) underset ~j`

`0=(2 + c_0) underset ~i + c_1 underset ~j`

`=> c_0 = -2, \ \ c_1 = 0`

`:. underset ~v(t) = (2 cos (2t)-2) underset ~i + 10 sin(2t) underset ~j`

`=> D`

Vectors, EXT1 EQ-Bank 8 MC

If $\underset{\sim}{u}=2 \underset{\sim}{i}-2 j+\underset{\sim}{k}$ and $\underset{\sim}{v}=3 \underset{\sim}{i}-6 j+2 \underset{\sim}{k}$, the projection of $\underset{\sim}{v}$ onto $\underset{\sim}{u}$ is

$\dfrac{20}{49}(3 \underset{\sim}{i}-6 \underset{\sim}{j}+2 \underset{\sim}{k})$
$\dfrac{20}{3}(2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k})$
$\dfrac{20}{7}(3 \underset{\sim}{i}-6 \underset{\sim}{j}+2 \underset{\sim}{k})$
$\dfrac{20}{9}(2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k})$

Show Answers Only

$D$

Show Worked Solution

$\underset{\sim}{u}=\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right), \ \ \underset{\sim}{v}=\left(\begin{array}{c}3 \\ -6 \\ 2\end{array}\right)$

$\underset{\sim}{u} \cdot \underset{\sim}{v}=6+12+2=20$

$\abs{\underset{\sim}{u}}^2=2^2+(-2)^2+1^2=9$

$\operatorname{proj}_{\underset{\sim}{u}} \underset{\sim}{v}=\left(\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{\abs{\underset{\sim}{u}}^2}\right) \underset{\sim}{u}=\dfrac{20}{9}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)$

$\Rightarrow D$

Vectors, EXT1 EQ-Bank 23

An object is travelling around a circular path of radius 5 m with position vector

$\textbf{r} (t)=5 \cos \left(t^2\right)\textbf{i} +5 \sin \left(t^2\right)\textbf{j} \quad t \geq 0$

where $t$ is the time in seconds.

Find an expression for the speed of the object in terms of $t$. (2 marks)

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$\abs{\textbf{v} (t)}=10 t \ \text{ms}^{-1}$

Show Worked Solution

$\text {The velocity is given by}$

$\textbf{v}(t)=\dfrac{d}{d t} \textbf{r}(t)=-10 t\, \sin \left(t^2\right) \textbf{i} +10 t\, \cos \left(t^2\right) \textbf{j}$

$\text {The speed is the magnitude of the velocity:}$

$\abs{\textbf{v} (t)}$	$=\sqrt{100 t^2\, \sin ^2\left(t^2\right)+100 t^2\, \cos ^2\left(t^2\right)}$
	$=10 t \sqrt{\sin ^2\left(t^2\right)+\cos ^2\left(t^2\right)}$
	$=10 t \ \text{m s}^{-1}$

Vectors, EXT1 EQ-Bank 21

Given the vectors $\textbf{a} = \textbf{i}+3\textbf{j}$ and $\textbf{b} =4\textbf{i} +2\textbf{j}$, find the projection of $\textbf{a}$ onto $\textbf{b}$. (2 marks)

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$\operatorname{proj}_{\textbf{b}}\textbf{a}=\left(\dfrac{\textbf{a} \cdot \textbf{b}}{\abs{\textbf{b}}^2}\right) \textbf{b}=\left(\dfrac{10}{20}\right) \textbf{b}=\dfrac{1}{2} \displaystyle \binom{4}{2}=\binom{2}{1}$

Show Worked Solution

$\displaystyle \textbf{a}=\binom{1}{3}, \ \ \textbf{b}=\binom{4}{2}$

$\textbf{a}\cdot \textbf{b}=1 \times 4+3 \times 2=10$

$\abs{\textbf{b}}^2=4^2+2^2=20$

Vectors, EXT1 EQ-Bank 38

Let $\underset{\sim}{a}=2 \underset{\sim}{i}-3 j+\underset{\sim}{k}$ and $\underset{\sim}{b}=\underset{\sim}{i}+m j-\underset{\sim}{k}$, where $m$ is an integer.

The vector resolute of $\underset{\sim}{a}$ in the direction of $\underset{\sim}{b}$ is $-\dfrac{11}{18}(\underset{\sim}{i}+m\underset{\sim}{j}-\underset{\sim}{k})$.

Find the value of $m$. (3 marks)

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Find the component of $\underset{\sim}{a}$ that is perpendicular to $\underset{\sim}{b}$. (1 mark)

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a. $m=4$

b. $\left(\begin{array}{c}2 \frac{11}{18} \\ -\frac{5}{9} \\ \frac{7}{18}\end{array}\right)$

Show Worked Solution

a. $\underset{\sim}{a}=\left(\begin{array}{c}2 \\ -3 \\ 1\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}1 \\ m \\ -1\end{array}\right)$

$\underset{\sim}{b} \cdot \underset{\sim}{a}=2-3 m-1=1-3 m$

$\abs{\underset{\sim}{b}}=\sqrt{1^2+m^2+(-1)^2}=\sqrt{2+m^2}$

$\operatorname{proj}_{\underset{\sim}{b}}\underset{\sim}{a}=\left(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{b}^2}\right) \underset{\sim}{b}=\dfrac{1-3 m}{2+m^2}\, \underset{\sim}{b}$

$\text{Equating projection vectors:}$

$\dfrac{1-3 m}{m^2+2}$	$=-\dfrac{11}{18}$
$18-54 m$	$=-11 m^2-22$

$11 m^2-54 m+40$	$=0$
$(11 m-10)(m-4)$	$=0$

$\therefore m=4\ \left(m \neq \frac{10}{11}, m \in Z\right)$

b. $\text{Component of \(\underset{\sim}{a}$ perpendicular to $\underset{\sim}{b}$:}\)

$\underset{\sim}{a}-\operatorname{proj}_{\underset{\sim}{b}} \underset{\sim}{a}=\left(\begin{array}{c}2 \\ -3 \\ 1\end{array}\right)+\dfrac{11}{18}\left(\begin{array}{c}1 \\ 4 \\ -1\end{array}\right)=\left(\begin{array}{c}2 \frac{11}{18} \\ -\frac{5}{9} \\ \frac{7}{18}\end{array}\right)$

Vectors, EXT1 EQ-Bank 33

Let `underset ~a = 3 underset ~i-2 underset ~j + m underset ~k` and `underset ~b = 2 underset ~i-underset ~j + 3 underset ~k`, where `m in R`.

Find the value(s) of `m` such that the projection of `underset ~a` onto `underset ~b` has magnitude `sqrt 14`. (3 marks)

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`m = -22/3, 2`

Show Worked Solution

$\underset{\sim}{a}=\left(\begin{array}{c}3 \\ -2 \\ m\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)$

$\underset{\sim}{a} \cdot \underset{\sim}{b}=6+2+3 m=8+3 m$

$\abs{\underset{\sim}{b}}=\sqrt{2^2+(-1)^2+3^2}=\sqrt{14}$

$\text{Since magnitude of projection}=\sqrt{14}$:

$\abs{\operatorname{proj}_{\underset{\sim}{b}} \underset{\sim}{a}}=\dfrac{\abs{\underset{\sim}{a} \cdot \underset{\sim}{b}}}{\abs{\underset{\sim}{b}}}=\dfrac{\abs{8+3 m}}{\sqrt{14}}=\sqrt{14}$

$\abs{8+3 m}$	$=14$
$8+3 m$	$= \pm 14$
$3 m$	$=-8 \pm 14$
$m$	$=-\dfrac{22}{3}, 2$

Vectors, EXT1 2017 NHT 10

Consider the vectors `underset ~a =-underset ~i-2 underset ~j + 3 underset ~k` and `underset ~b = 2 underset ~i + c underset ~j + underset ~k`.

Find the value of `c` if the angle between `underset ~a` and `underset ~b` is `pi/3`. (4 marks)

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`c = -3`

Show Worked Solution

`underset ~a ⋅ underset ~b`	`= -1 xx 2 + (-2) xx c + 3 xx 1`
	`= -2-2c + 3`
	`= 1-2c`

`1-2c`	`= sqrt((-1)^2 + (-2)^2 + 3^3) *sqrt(2^2 + c^2 + 1^2) xx cos (pi/3)`
`1-2c`	`= 1/2(sqrt 14 ⋅ sqrt(5 + c^2))`
`2-4c`	`= sqrt(14(5 + c^2))`
`(2-4c)^2`	`= 14(5 + c^2)`

`4-16c + 16c^2`	`= 70 + 14c^2`
`2c^2-16c-66`	`= 0`
`c^2-8c-33`	`= 0`
`(c-11)(c + 3)`	`= 0`

`c = 11 or c = -3`

`text(S)text(ince)\ \ 2-4c = sqrt(15(5 + c^2))`

`2-4c > 0\ \ =>\ \ c<2`

`:. c = -3`

Vectors, EXT1 2024 SPEC1 4

Consider the vectors $\underset{\sim}{ a }=3 \underset{\sim}{ j }+3 \underset{\sim}{ k }$ and $\underset{\sim}{ b }=2 \underset{\sim}{ i }-\underset{\sim}{ j }-2 \underset{\sim}{ k }$.

Find the angle between $\underset{\sim}{ a }$ and $\underset{\sim}{ b }$. (2 marks)

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$\theta=\dfrac{3 \pi}{4}\left(\text{or} \ 135^{\circ}\right)$

Show Worked Solution

$\underset{\sim}{a}=\left(\begin{array}{l}0 \\ 3 \\ 3\end{array}\right) \Rightarrow \abs{\underset{\sim}{a}}=\sqrt{18}=3 \sqrt{2}$

$\underset{\sim}{b}=\left(\begin{array}{c}2 \\ -1 \\ -2\end{array}\right) \Rightarrow\abs{\underset{\sim}{b}}=\sqrt{9}=3$

$\cos \theta=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}} \cdot \abs{\underset{\sim}{b}}}=\dfrac{-3-6}{3 \sqrt{2} \times 3}=-\dfrac{1}{\sqrt{2}}$

$\therefore \theta=\cos ^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)=\dfrac{3 \pi}{4}\left(\text{or} \ 135^{\circ}\right)$

Vectors, EXT1 2014 SPEC1 1

Consider the vector `underset ~a = sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k`, where `underset ~i, underset ~j` and `underset ~k` are unit vectors in the positive directions of the `x, y` and `z` axes respectively.

Find the unit vector in the direction of `underset ~a`. (1 mark)

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Find the acute angle that `underset ~a` makes with the positive direction of the `x`-axis. (2 marks)

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The vector `underset ~b = 2 sqrt 3 underset ~i + m underset ~j-5 underset ~k`.
Given that `underset ~b` is perpendicular to `underset ~a,` find the value of `m`. (2 marks)

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a. `1/sqrt 6 (sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k)`

b. `theta = 45^@`

c. `m = 6 + 5 sqrt 2`

Show Worked Solution

a. `|underset ~a|= sqrt((sqrt 3)^2 + (-1)^2 + (-sqrt 2)^2)= sqrt 6`

`hat underset ~a= underset ~a/|underset ~a|= 1/sqrt 6 (sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k)`

b. `x text{-axis vectors include}\ (1,0,0).`

`underset ~a ⋅ underset ~i = ((\sqrt3),(-1),(-\sqrt2))((1),(0),(0))=\sqrt3`

	`underset ~a ⋅ underset ~i`	`= \|underset ~a\|\|underset ~i\| cos theta= sqrt 6 cos theta`
	`sqrt 3`	`= sqrt 6 cos theta`
	`cos theta`	`=1/sqrt 2`
	`:. theta`	`= 45^@`

c. `underset ~a ⋅ underset ~b = sqrt 3 (2 sqrt 3) + (-1)(m) + (-sqrt 2)(-5) = 0`

`6-m + 5 sqrt 2`	`=0`
`:. m`	`=6 + 5 sqrt 2`

Calculus, EXT1 EQ-Bank 22

Solve the differential equation $\dfrac{d y}{d x}=20 e^{-5 y}$. (3 marks)

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$y=\dfrac{1}{5} \ln \abs{100 x+c}$

Show Worked Solution

$\dfrac{d y}{d x}$	$=20 e^{-5 y}$
$\displaystyle \int e^{5y}\,d y$	$=\displaystyle \int 20\, d x$
$\dfrac{1}{5} e^{5 y}$	$=20 x+c$
$e^{5 y}$	$=100 x+c$
$5 y$	$=\ln \abs{100 x+c}$
$ y$	$=\dfrac{1}{5} \ln \abs{100 x+c}$

Vectors, EXT1* V1 2025 HSC 11d

Force ${\underset{\sim}{F}}_1$ has magnitude 12 newtons in the direction of vector $2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}$.
Show that ${\underset{\sim}{F}}_1=8 \underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}$. (1 mark)

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Force ${\underset{\sim}{F}}_1$ from part (i) and a second force, ${\underset{\sim}{F}}_2=-6 \underset{\sim}{i}+12 \underset{\sim}{j}+4 \underset{\sim}{k}$, both act upon a particle.
Show that the resultant force acting on the particle is given by:
${\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}.$ (1 mark)

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Calculate ${\underset{\sim}{F}}_3 \cdot \underset{\sim}{d}$, where ${\underset{\sim}{F}}_3$ is the resultant force from part (ii) and $\underset{\sim}{d}=\underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k}$. (1 mark)

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i. $\text{Unit vector of the direction vector:}$

$\dfrac{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}{\abs{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}} = \dfrac{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}{\sqrt{2^2+(-2)^2 + 1^2}} = \dfrac{1}{3} \left( 2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k} \right)$

$\text{Since \({\underset{\sim}{F}}_1$ has magnitude 12:}\)

${\underset{\sim}{F}}_1=12 \times \dfrac{1}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)$

${\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}$

ii. ${\underset{\sim}{F}}_3={\underset{\sim}{F}}_1+{\underset{\sim}{F}}_2=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)+\left(\begin{array}{c}-6 \\ 12 \\ 4\end{array}\right)=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)$

${\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}$

iii. ${\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22$

Show Worked Solution

i. $\text{Unit vector of the direction vector:}$

$\text{Since \({\underset{\sim}{F}}_1$ has magnitude 12:}\)

${\underset{\sim}{F}}_1=12 \times \dfrac{1}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)$

${\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}$

${\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}$

iii. ${\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22$

Vectors, EXT1* V1 2024 HSC 11c

Find the angle between the two vectors $\underset{\sim}{u}=\left(\begin{array}{c}1 \\ 2 \\ -2\end{array}\right)$ and $\underset{\sim}{v}=\left(\begin{array}{c}4 \\ -4 \\ 7\end{array}\right)$, giving your answer in radians, correct to 1 decimal place. (2 marks)

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$\theta=2.3^c \ \ \text{(1 d.p.)}$

Show Worked Solution

$\underset{\sim}{u}=\left(\begin{array}{c}1 \\ 2 \\ -2\end{array}\right),\abs{\underset{\sim}{u}}=\sqrt{1+4+4}=3$

$\underset{\sim}{v}=\left(\begin{array}{c}4 \\ -4 \\ 7\end{array}\right),\abs{\underset{\sim}{v}}=\sqrt{16+16+49}=9$

$\cos \theta=\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{|\underset{\sim}{u}||\underset{\sim}{v}|}=\dfrac{1 \times 4-2 \times 4-2 \times 7}{3 \times 9}=-\dfrac{2}{3}$

$\theta=\cos ^{-1}\left(-\dfrac{2}{3}\right)=2.30 \ldots=2.3^c \ \ \text{(1 d.p.)}$

Vectors, EXT1* V1 2022 HSC 11d

A triangle is formed in three-dimensional space with vertices `A(1,-1,2)`, `B(0,2,-1)` and `C(2,1,1)`.

Find the size of `/_ABC`, giving your answer to the nearest degree. (3 marks)

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`33°`

Show Worked Solution

`vec(BA)=((1),(-1),(2))-((0),(2),(-1))=((1),(-3),(3))`

`abs(vec(BA))=sqrt(1^2+3^2+3^2)=sqrt19`

`vec(BC)=((2),(1),(1))-((0),(2),(-1))=((2),(-1),(2))`

`abs(vec(BC))=sqrt(2^2+1^2+2^2)=sqrt9=3`

`vec(BA)*vec(BC)=1xx2+ -3xx-1+3xx2=11`

`cos/_ABC=(vec(BA)*vec(BC))/(abs{vec(BA)}abs{vec(BC)})=11/(3sqrt19)`

`:./_ABC=cos^(-1)(11/(3sqrt19))=32.733…=33°\ \ text{(nearest degree)}`

Vectors, EXT1* V1 2024 HSC 12a

The vector $\underset{\sim}{a}$ is $\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$ and the vector $\underset{\sim}{b}$ is $\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)$.

Find $\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}$. (1 mark)

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Show that $\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}$ is perpendicular to $\underset{\sim}{b}$. (2 marks)

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i. $\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)$

ii. $\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)$

$ \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0$

$\therefore\ \text {Vectors are perpendicular.}$

Show Worked Solution

i. $\underset{\sim}{a}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)$

$\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\dfrac{2+0-12}{4+0+16}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=-\dfrac{1}{2}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)$

$ \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\,\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0$

$\therefore\ \text{Vectors are perpendicular.}$

Vectors, EXT1 EQ-Bank 20

Two vectors are given by `underset ~a = 4 underset ~i + m underset ~j - 3 underset ~k` and `underset ~b = −2 underset ~i + n underset ~j - underset ~k`, where `m`, `n in R^+`.

If `|\ underset ~a\ | = 10` and `underset ~a` is perpendicular to `underset ~b`, determine the exact values of `m` and `n`. (3 marks)

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`m=5sqrt3, \ n=\sqrt{3}/3`

Show Worked Solution

`text(Using)\ \ |\ underset ~a\ | = 10:`

`10`	`= sqrt(4^2 + m^2 + (-3)^2)`
`100`	`=m^2+75`
`m^2`	`= 25`
`m`	`=5sqrt3\ \ (m in R^+)`

`text(S)text(ince)\ \ underset ~a _|_ underset ~b\ \ =>\ \ underset ~a xx underset ~b=0`

`0`	`=4 xx (−2) + mn + (−3) xx (−1)`
`0`	`=n xx 5sqrt3-5`
`n`	`=5/(5\sqrt{3})`
`n`	`=1/\sqrt{3}=\sqrt{3}/3`

\(\Pr(0.43<\overline{T}<0.50)\)	\(=\Pr\left(\dfrac{0.43-0.50}{0.0507}<Z<\dfrac{0.50-0.50}{0.0507}\right)\)
	\(=\Pr(-1.38<Z<0)\)
	\(=\Pr(0<Z<1.38)\)
	\(=0.9162-0.5000\)
	\(=0.4162 = 41.6\%\ \text{(1 d.p.)}\)

\(P(\overline{X}>65)\)	\(=P(Z>-1.89)\)
	\(=1-P(Z\leq -1.89)\)
	\(=1-0.0294\)
	\(=0.9706\)

\(\text{Profit}\)	\( = \text{Revenue}-\text{Total costs}\)
\(4200\)	\( = 350x-(1200+150x)\)
\(4200\)	\(= 200x-1200\)
\(5400\)	\(=200x\)
\(x\)	\(=27\)

\(\abs{z_1+z_2}^2\)	\(=\left(z_1+z_2\right)\left(\overline{z}_1+\overline{z}_2\right)\)
	\(=\abs{z_1}^2+\abs{z_2}^2+z_1 \overline{z}_2+\overline{z}_1 z_2\)
	\(=\abs{z_1}^2+\abs{z_2}^2+2 \operatorname{Re}\left(z_1 \overline{z}_2\right)\)

\(\abs{z_1+z_2}^2\)	\(\leqslant\abs{z_1}^2+\abs{z_2}^2+2\abs{z_1 \overline{z}_2}\)
	\(\leqslant\abs{z_1}^2+2\abs{z_1}\abs{z_2}+\abs{z_2}^2\)
	\(\leqslant\left(\abs{z_1}+\abs{z_2}\right)^2\)

\(\abs{2z}\)	\(\leqslant\abs{z-1}+\abs{z+1}\)
\(\abs{2z}\)	\(\leqslant 4\ \ \text{(using (1) above)}\)
\(2\abs{z}\)	\(\leqslant 4\)
\(\abs{z}\)	\(\leqslant 2\)

\(100 e^{-0.2 t}-50\)	\(=0\)
\(e^{-0.2 t}\)	\(=\dfrac{1}{2}\)
\(-0.2 t\)	\(=-\ln 2\)
\(t\)	\(=5 \ln 2\)

\(e^{-0.2 t}\)	\(=2-\sqrt{3}\)
\(-0.2 t\)	\(=\ln (2-\sqrt{3})\)
\(t\)	\(=-5\ln (2-\sqrt{3}) \approx 6.6 \ \text{s (1 d.p.)}\)

\(\|z+w\|^2\)	\(=\Bigg{\|} \dfrac{\sqrt3}{2}+\dfrac{1}{2}i-\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2} \Bigg{\|}\)
	\(=\Bigg{\|} \Bigg{(}\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2} \Bigg{)} +\Bigg{(}\dfrac{1}{2}+\dfrac{1}{\sqrt2}\Bigg{)}\,i \Bigg{\|}\)
	\(=\Bigg{\|} \dfrac{\sqrt6-2}{2\sqrt2}+\dfrac{\sqrt2+2}{2\sqrt2}\,i \Bigg{\|}\)
	\(= \dfrac{(\sqrt6-2)^2+(\sqrt2+2)^2}{(2\sqrt2)^2}\)
	\(= \dfrac{6-4\sqrt6+4+2+4\sqrt2+4}{8}\)
	\(=\dfrac{16-4\sqrt6+4\sqrt2}{8} \)
	\(=\dfrac{4-\sqrt6+\sqrt2}{2} \)