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Statistics, STD2 S1 2011 HSC 25b

The graph below displays data collected at a school on the number of students
in each Year group, who own a mobile phone.
 

2UG 2011 25b
 

  1. Which Year group has the highest percentage of students with mobile phones? (1 mark)

  2. Two students are chosen at random, one from Year 9 and one from Year 10.

     

    Which student is more likely to own a mobile phone?

     

    Justify your answer with suitable calculations. (2 marks)

  3. Identify a trend in the data shown in the graph. (1 mark)

Show Answers Only
  1. `text{Year 12 (100%)}`
  2. `text(Year 10)`
  3. `text(See Worked Solutions for detail)`
Show Worked Solution

i.   `text(Year 12 (100%))`

MARKER’S COMMENT: Many students ignore the instruction to use calculations and lose marks as a result.
 

ii.     `text(% Ownership in Year 9)` `=55/70`
    `=\ text{78.6%  (1d.p.)}`
      `text(% Ownership in Year 10)` `=50/60`
    `=\ text{83.3%  (1d.p.)}`

  
`:.\ text(The Year 10 student is more likely to own a mobile phone.)`

 

iii.   `text(% Ownership increases as students)`

 `text(progress from Year 7 to Year 12.)`

Algebra, STD2 A2 2010 HSC 27c

The graph shows tax payable against taxable income, in thousands of dollars.

2010 27c

  1. Use the graph to find the tax payable on a taxable income of $21 000.  (1 mark)

  2. Use suitable points from the graph to show that the gradient of the section of the graph marked  `A`  is  `1/3`.    (1 mark)

  3. How much of each dollar earned between  $21 000  and  $39 000  is payable in tax?   (1 mark)

  4. Write an equation that could be used to calculate the tax payable, `T`, in terms of the taxable income, `I`, for taxable incomes between  $21 000  and  $39 000.   (2 marks)

Show Answers Only
  1. `$3000\ \ \ text{(from graph)}`
  2. `1/3`
  3. `33 1/3\ text(cents per dollar earned)`
  4. `text(Tax payable on)\ I = 1/3 I\-4000`
Show Worked Solution
i.

 `text(Income on)\ $21\ 000=$3000\ \ \ text{(from graph)}`

 

ii.  `text(Using the points)\ (21,3)\ text(and)\ (39,9)`

♦♦ Mean mark 25%
`text(Gradient at)\ A` `= (y_2\-y_1)/(x_2\ -x_1)`
  `= (9000-3000)/(39\ 000 -21\ 000)`
  `= 6000/(18\ 000)`
  `= 1/3\ \ \ \ \ text(… as required)`

 

iii.  `text(The gradient represents the tax applicable to each dollar)`

♦♦♦ Mean mark 12%!
MARKER’S COMMENT: Interpreting gradients is an examiner favourite, so make sure you are confident in this area.
`text(Tax)` ` = 1/3\ text(of each dollar earned)`
  ` = 33 1/3\ text(cents per dollar earned)`

 

iv.  `text( Tax payable up to $21 000 = $3000)`

`text(Tax payable on income between $21 000 and $39 000)`

♦♦♦ Mean mark 15%.
STRATEGY: The earlier parts of this question direct students to the most efficient way to solve this question. Make sure earlier parts of a question are front and centre of your mind when devising strategy.

` = 1/3 (I\-21\ 000)`

`:.\ text(Tax payable on)\ \ I` `= 3000 + 1/3 (I\-21\ 000)`
  `= 3000 + 1/3 I\-7000`
  `= 1/3 I\-4000`

Statistics, STD2 S1 2010 HSC 26b

A new shopping centre has opened near a primary school. A survey is conducted to determine the number of motor vehicles that pass the school each afternoon between 2.30 pm and 4.00 pm.

The results for 60 days have been recorded in the table and are displayed in the cumulative frequency histogram.
 

2010 26b

  1. Find the value of  Χ  in the table.   (1 mark)

  2. On the cumulative frequency histogram above, draw a cumulative frequency polygon (ogive) for this data.   (1 mark)
  3. Use your graph to determine the median. Show, by drawing lines on your graph, how you arrived at your answer.   (1 mark)
  4. Prior to the opening of the new shopping centre, the median number of motor vehicles passing the school between  2.30 pm  and  4.00 pm  was 57 vehicles per day.

     

    What problem could arise from the change in the median number of motor vehicles passing the school before and after the opening of the new shopping centre?

     

    Briefly recommend a solution to this problem.   (2 marks)

Show Answers Only
  1. `15`
  2.  
  3.  
  4. `text(Problems)`
  5. `text(- increased traffic delays)`
  6. `text(- increased danger to students leaving school)`

  7.  

    `text(Solutions)`

  8. `text(- signpost alternative routes around school)`
  9. `text(- decrease the speed limit in the area)`
Show Worked Solution
i. `X` `= 25\ -10`
    `= 15`

 

♦♦♦ Mean mark 18%
MARKER’S COMMENT: The ogive was poorly drawn with many students incorrectly joining the middle of each column rather than from corner to corner.
ii.
♦♦ Mean mark 25%
MARKER’S COMMENT: Many students did not “show by drawing lines on the graph” as the question asked.

iii.  `text(Median)\ ~~155`

♦ Mean mark 47%
MARKER’S COMMENT: Short answers were often the best. Be concise when you can.
iv. `text(Problems)`
  `text(- increased traffic delays)`
  `text(- increased danger to students leaving school)`
   
  `text(Solutions)`
  `text(- signpost alternative routes around school)`
  `text(- decrease the speed limit in the area)`

Algebra, STD2 A4 2013 HSC 30a

Wind turbines, such as those shown, are used to generate power.

2013 30a

In theory, the power that could be generated by a wind turbine is modelled using the equation

`T = 20\ 000w^3`

where `T` is the theoretical power generated, in watts 
  `w` is the speed of the wind, in metres per second.

 

  1. Using this equation, what is the theoretical power generated by a wind turbine if the wind speed is 7.3 m/s ?  (1 mark)

In practice, the actual power generated by a wind turbine is only 40% of the theoretical power.

  1. If `A` is the actual power generated, in watts, write an equation for `A` in terms of `w`.    (1 mark)

The graph shows both the theoretical power generated and the actual power generated by a particular wind turbine.
 
        2013 30a2

  1. Using the graph, or otherwise, find the difference between the theoretical power and the actual power generated when the wind speed is 9 m/s.   (1 mark)

A particular farm requires at least 4.4 million watts of actual power in order to be self-sufficient.

  1. What is the minimum wind speed required for the farm to be self-sufficient?    (1 mark)

A more accurate formula to calculate the power (`P`) generated by a wind turbine is

`P = 0.61 xx pi xx r^2 × w^3` 

where     `r` is the length of each blade, in metres
  `w` is the speed of the wind, in metres per second. 

 
Each blade of a particular wind turbine has a length of 43 metres.The turbine operates at a wind speed of 8 m/s.

  1. Using the formula above, if the wind speed increased by 10%, what would be the percentage increase in the power generated by this wind turbine?   (3 marks)

Show Answers Only
  1. `7\ 780\ 340\ text(watts)`
  2. `8000w^3`
  3. `8.8\ text(million watts, or 8 748 000 watts)`
  4. `w = 8.2\ text(m/s)\ \ \ text{(1 d.p.)}`
  5. `text(33%)`
Show Worked Solution
i.    `T=20\ 000w^3`
  `text(If)\ \ w = 7.3`
`T` `=20\ 000 xx (7.3)^3`
  `= 7\ 780\ 340\ text(watts)`

  

♦ Mean mark 34%
ii.    `text(We know)\ A = 40% xx T`
`=> A` `=0.4 xx 20\ 000 xx w^3`
  `=8000w^3`

  

♦ Mean mark 38%
iii.    `text(Solution 1)`
  `text(At)\ w=9`
  `A = text(5.8 million watts)\ \ \ text{(from graph)}`
  `T = text(14.6 million watts)\ \ \ text{(from graph)}`
`text(Difference)` `= text(14.6 million)\-text(5.8 million)`
  `= text(8.8 million watts)`

 

`text(Alternative Solution)`
`text(At)\ w=9`
`T` `= 20\ 000 xx 9^3`
  `= 14\ 580\ 000\ text(watts)`
`A` `= 8000 xx 9^3`
  `= 5\ 832\ 000\ text(watts)`
`text(Difference)` `=14\ 580\ 000\-5\ 832\ 000`
  `=8\ 748\ 000\ \ text(watts)`

  

♦♦ Mean mark 25%
COMMENT: Students need to be comfortable in finding the cube roots of values – a calculation that can be required in a number of topic areas and is regularly examined.
iv.    `text(Find)\ w\ text(if)\ A=4.4\ text(million)`
`8000w^3` `= 4\ 400\ 000`
`w^3` `= (4\ 400\ 000)/8000`
  `= 550`
`:. w` `= root(3)(550)`
  `=8.1932…`
  `=8.2\ text(m/s)\ \ \ text{(1 d.p.)}`

 

`:.\ text(The minimum wind speed required is 8.2 m/s)`

 

♦ Mean mark 41%
MARKER’S COMMENT: Students are reminded that a % increase requires them to find the difference in power generated at different wind speeds and divide this result by the original power output, as shown in the Worked Solution.
v.    `text(Find)\ P\ text(when)\ w=8\ text(and)\ r=43`
`P` `= 0.61 xx pi xx r^2 xx w^3`
  `= 0.61 xx pi xx 43^2 xx 8^3`
  `= 1\ 814\ 205.92\ text(watts)`

 

`text(When speed of wind)\ uarr10%`
`w′ = 8 xx 110text(%) = 8.8\ text(m/s)`
 

`text(Find)\ P\ text(when)\ w′ = 8.8`

`P` `=0.61 xx pi xx 43^2 xx 8.8^3`
  `= 2\ 414\ 708.08\ text(watts)`

 

`text(Increase in Power)` `=2\ 414\ 708.08\-1\ 814\ 205.92`
  `= 600\ 502.16`

 

`:.\ text(% Power increase)` `= (600\ 502.16)/(1\ 814\ 205.92)`
  `= 0.331`
  `= text(33%)\ \ \ text{(nearest %)}`

 

Financial Maths, STD2 F1 2011 HSC 23a

Sri has a gross salary of  $56 350. She has tax deductions of $350 for union fees, $2000 in work-related expenses and $250 in donations to charities.

The Medicare levy is 1.5% of her taxable income.

Calculate Sri’s Medicare levy.   (3 marks)

Show Answers Only

 `$806.25`

Show Worked Solution
MARKER’S COMMENT: Clear separate steps results in less calculation errors and more marks for working when errors are made.
`text(Deductions)` `=350+2000+250`
  `=$2600`
`text(Taxable)` `=\ text(Gross Income – deduct)`
  `=56\ 350-2600`
  `=$53\ 750`
`text(Medicare Levy)` `=\ text(1.5%)\ xx 53\ 750`
  `=$806.25`

 

`:.\ text(Sri’s Medicare levy is)\ \ $806.25`

Probability, STD2 S2 2011 HSC 15 MC

An unbiased coin is tossed 10 times.

A tail is obtained on each of the first 9 tosses.

What is the probability that a tail is obtained on the 10th toss?

  1. `1/2^10`
  2. `1/2`
  3. `1/10`
  4. `9/10`
Show Answers Only

`B`

Show Worked Solution

`text(Each toss is an independent event and has an even chance)`

`text(of being a head or tail.)`

`=> B`

Probability, STD2 S2 2010 HSC 23c

On Saturday, Jonty recorded the colour of T-shirts worn by the people at his gym. The results are shown in the graph.

 

  1. How many people were at the gym on Saturday? (Assume everyone was wearing a T-shirt).   (1 mark)

  2. What is the probability that a person selected at random at the gym on Saturday, would be wearing either a blue or green T-shirt?   (1 mark)

Show Answers Only
  1. `34`
  2. `15/34`
Show Worked Solution
i.   `text(# People)` `=5+15+10+3+1`
  `=34`

 

ii.   `P (B\ text{or}\ G)` `=P(B)+P(G)`
  `=5/34+10/34`
  `=15/34`

Financial Maths, STD2 F1 2010 HSC 23a

This advertisement appeared in a newspaper

2010 23a 

 What is the maximum possible salary per annum for this civil engineer, correct to the nearest dollar?   (2 marks)

Show Answers Only

`$86\ 396`

Show Worked Solution

`text(Base wage)=1586.70\ text(pw)`

`text(Max weekly base)` `=$1586.70+(text{3.5%}\ xx 1586.70)`
  `=1586.70+55.53`
  `=1642.23\ text((nearest cent))`
`:.\ text(Max annual base)` `=1642.23xx52`
  `=85\ 395.96`
  `=$85\ 396\ text{(nearest dollar)}`

Financial Maths, STD2 F4 2009 HSC 24e

Jay bought a computer for $3600. His friend Julie said that all computers are worth nothing (i.e. the value is $0) after 3 years.

  1. Find the amount that the computer would depreciate each year to be worth nothing after 3 years, if the straight line method of depreciation is used.   (1 mark)

  2. Explain why the computer would never be worth nothing if the declining balance method of depreciation is used, with 30% per annum rate of depreciation. Use suitable calculations to support your answer.    (2 marks)

Show Answers Only
  1. `$1200`
  2. `text(See Worked Solutions.)`
Show Worked Solution
i.    `S` `= V_0-Dn`
  `0` `= 3600-D xx 3`
  `3D` `= 3600`
  `D` `= 3600/3`
    `= 1200`

 
`:.\ text(Annual depreciation = $1200`

 

♦ Mean mark 45%
ii   `text(Using)\ \ S = V_0 (1-r)^n`
  `text(where)\ r = text(30%)\ \ text(and)\ \ V_0 = 3600`

 

`S` `=3600 (1-30/100)^n`  
  `= 3600 (0.7)^n`  

 
`(0.7)^n > 0\ text(for all)\ n`

`:.\ text(Salvage value is always)\ >0`

Statistics, STD2 S1 2009 HSC 24b

Tayvan is an international company that reports its profits in the USA, Belgium and India at the end of each quarter. The profits for 2008 are shown in the area chart.

2009 24b

  1. What was the total profit for Tayvan on June 30?   (1 mark)
  2. What was Tayvan’s profit in Belgium on March 31?     (1 mark)
Show Answers Only
  1. `$8\ 000\ 000\ \ text{(from graph)}`
  2. `$4\ 000\ 000`
Show Worked Solution
MARKER’S COMMENT: Area charts provide cumulative totals. Many students did not know this and incorrectly answered 14 million.
(i)     `$8\ 000\ 000\ \ \ text{(from graph)}`

 

(ii)    `text{Belgium profit (30 March)}`
  `= $5\ 000\ 000-$1\ 000\ 000` 
  `= $4\ 000\ 000`

Statistics, STD2 S1 2009 HSC 24a

The diagram below shows a stem-and-leaf plot for 22 scores. 
 

2UG-2009-24a
 

  1.  What is the mode for this data?   (1 mark)

  2.  What is the median for this data?     (1 mark)

Show Answers Only
  1. `78`
  2. `46`
Show Worked Solution

i.   `text(Mode) = 78`

 

ii.    `22\ text(scores)`

`=>\ text(Median is the average of 11th and 12th scores)`
 

`:.\ text(Median)` `= (45 + 47)/2`
  `= 46`

Financial Maths, STD2 F1 2009 HSC 10 MC

Billy worked for 35 hours at the normal hourly rate of pay and for five hours at double time. He earned $561.60 in total for this work.

What was the normal hourly rate of pay?

  1. $7.02
  2. $12.48
  3. $14.04
  4. $16.05
Show Answers Only

`B`

Show Worked Solution

`text(Let hourly rate) = $X\ text(per hour)`

`35X + 5(2X)` `= 561.60`
`45X` `=561.60`
`X` `= 561.60/45` 
  `= $12.48`

 
`=>  B`

Financial Maths, STD2 F4 2009 HSC 6 MC

A house was purchased in 1984 for $35 000. Assume that the value of the house has increased by 3% per annum since then. 

Which expression gives the value of the house in 2009?  

  1. `35\ 000(1 + 0.03)^25`
  2. `35\ 000(1 + 3)^25` 
  3. `35\ 000 xx 25 xx 0.03`
  4. `35\ 000 xx 25 xx 3`
Show Answers Only

`A`

Show Worked Solution

`r =\ text(3%)\ = 0.03`

`n = 25\ text(years)`

`text(Using)\ \ FV = PV(1 + r)^n`

` :.\ text(Value in 2009) = 35\ 000(1+0.03)^25` 

`=>  A`

Statistics, STD2 S3 2009 HSC 5 MC

Jamie wants to know how many songs were downloaded legally from the internet in the last 12 months by people aged 18–25 years. He has decided to conduct a statistical inquiry.

After he collects the data, which of the following shows the best order for the steps he should take with the data to complete his inquiry?

  1.    Display, organise, conclude, analyse
  2.    Organise, display, conclude, analyse
  3.    Display, organise, analyse, conclude
  4.    Organise, display, analyse, conclude
Show Answers Only

`D`

Show Worked Solution

`text(Process of statistical enquiry requirements)`

`=>  D`

Statistics, STD2 S1 2009 HSC 3 MC

The eye colours of a sample of children were recorded.

When analysing this data, which of the following could be found?

  1. Mean
  2. Median
  3. Mode
  4. Range
Show Answers Only

`C`

Show Worked Solution

`text(Eye colour is categorical data)`

`:.\ text(Only the mode can be found)`

`=>  C`

Statistics, STD2 S1 2009 HSC 2 MC

The step graph shows the charges for a carpark. 
 

2UG-2009-2MC

 
Maria enters the carpark at 10:10 am and exits at 1:30 pm.

How much will she pay in charges?

  1.   `$6` 
  2.   `$12` 
  3.   `$18` 
  4.   `$24` 
Show Answers Only

`C`

Show Worked Solution

`text(Time in carpark = 3h 20m)`

`text(From graph, charge will be $18)`

`=>  C`

Probability, STD2 S2 2009 HSC 1 MC

A newspaper states: ‘It will most probably rain tomorrow.’

Which of the following best represents the probability of an event that will most probably occur?   

  1.   `33 1/3 text(%)` 
  2.   `text(50%)` 
  3.   `text(80%)` 
  4.   `text(100%)` 
Show Answers Only

`C`

Show Worked Solution

`text(Probably) =>\ text(likelihood > 50%)`

`text(However 100% = certainty)`

`:.\ text(80% is the answer)`

`=> C`

Financial Maths, STD2 F1 2010 HSC 11 MC

Which of the following graphs shows the lowest rate of depreciation over the given time period?
 

Capture4

Capture5

Show Answers Only

`D`

Show Worked Solution

`text(The lowest rate of depreciation will occur when)`

`text(an item retains value for the longest time.)`

`=>  D`

Measurement, STD2 M6 2011 HSC 9 MC

Two trees on level ground, 12 metres apart, are joined by a cable. It is attached 2 metres above the ground to one tree and 11 metres above the ground to the other.

What is the length of the cable between the two trees, correct to the nearest metre? 

  1.  `9\ text(m)`
  2. `12\ text(m)`
  3. `15\ text(m)`
  4. `16\ text(m)`
Show Answers Only

`C`

Show Worked Solution

`text(Using Pythagoras)`

`c^2` `=12^2+9^2`
  `=144+81`
  `=225`
`:.c` `=15,\ \ c>0`

 
`=>C`

Probability, STD2 S2 2012 HSC 26e

The dot plot shows the number of push-ups that 13 members of a fitness class can do in one minute.

2012 26e

  1.  What is the probability that a member selected at random from the class can do more than 38 push-ups in one minute?   (1 mark)

  2.  A new member who can do 32 push-ups in one minute joins the class.

     

    Does the addition of this new member to the class change the probability calculated in part (i)? Justify your answer.    (1 mark)

Show Answers Only
  1. `7/13`
  2. `text{Yes (See Worked Solutions)}`
Show Worked Solution
i.  `P` `= text(# Members > 38 push-ups)/text(Total members)`
  `= 7/13`

 
ii.   `text(Yes.)`

`Ptext{(+ New member)}` `= text(Members > 38 push-ups)/text(Total members)`
  `= 7/14≠ 7/13`
MARKER’S COMMENT: The most successful candidates used the fraction `7/14` in their part (ii) answer rather than relying solely on words.

Financial Maths, STD2 F4 2012 HSC 26b

Jim buys a photocopier for  $22 000.

Its value is depreciated using the declining balance method at the rate of 15% per annum.

What is its value at the end of 3 years? (2 marks)

Show Answers Only

`$13\ 510.75`

Show Worked Solution
`S` `= V_0 (1-r)^n`
  `= 22\ 000 (1-0.15)^3`
  `= 22\ 000 (0.85)^3`
  `= 13\ 510.75`

 
`:.\ text(After 3 years, it is worth)\ $13\ 510.75`

Statistics, STD2 S1 2010 HSC 1 MC

The results of a survey are displayed in the dot plot.

What is the range of this data?
 

2010 1 MC 
 

  1.    7
  2.    8
  3.    9
  4.    10
Show Answers Only

`C`

Show Worked Solution
`text(Range)` `=text(High)-text(Low)`
  `=9-0`
  `=9`

`=>  C`

Probability, STD2 S2 2013 HSC 7 MC

In an experiment, a standard six-sided die was rolled 72 times. The results are shown in the table.
 

Which number on the die was obtained the expected number of times?

  1.    1
  2.    2
  3.    3
  4.    6
Show Answers Only

`B`

Show Worked Solution

`text(Probability of rolling a specific number)=1/6`

`:.\ text(After 72 rolls, a specific number is expected)`

 `1/6xx72=12\ text(times.)`

`=>\ B`

Quadratic, 2UA 2011 HSC 2a

The quadratic equation  `x^2-6x+2=0`  has roots  `alpha`  and  `beta`.

  1. Find  `alpha+beta`.       (1 mark)
  2. Find  `alpha beta`.        (1 mark)
  3. Find  `1/alpha+1/beta`.      (1 mark)
Show Answers Only
  1. `6`
  2. `2`
  3. `3`
Show Worked Solution

(i)  `alpha+beta=-b/a=6`

(ii)  `alpha beta=c/a=2`

(iii)  `1/alpha+1/beta=(alpha+beta)/(alpha beta)=3`

Quadratic, 2UA 2012 HSC 3 MC

The quadratic equation  `x^2+3x -1=0`  has roots  `alpha`  and  `beta`.

What is the value of  `alpha beta+(alpha+beta)` ?

(A)   `4`

(B)   `2`

(C)  `-4`

(D)  `-2`

Show Answers Only

`C`

Show Worked Solution
`alpha beta+(alpha+beta)` `=c/a+(- b/a)`
  `=-1-3`
  `=-4`

Calculus, 2ADV C3 2009 HSC 9b

An oil rig,  `S`,  is 3 km offshore. A power station,  `P`,  is on the shore. A cable is to be laid from `P`  to  `S`.  It costs $1000 per kilometre to lay the cable along the shore and $2600 per kilometre to lay the cable underwater from the shore to  `S`.

The point  `R`  is the point on the shore closest to  `S`,  and the distance  `PR`  is 5 km.

The point  `Q`  is on the shore, at a distance of  `x`  km from  `R`,  as shown in the diagram.


  

  1. Find the total cost of laying the cable in a straight line from  `P`  to  `R`  and then in a straight line from  `R`  to  `S`.   (1 mark)

  2. Find the cost of laying the cable in a straight line from  `P`  to  `S`.  (1 mark)

  3. Let  `$C`  be the total cost of laying the cable in a straight line from  `P`  to  `Q`,  and then in a straight line from `Q`  to  `S`.
     
    Show that  `C=1000(5-x+2.6sqrt(x^2+9))`.    (2 marks)

  4. Find the minimum cost of laying the cable.    (4 marks)

  5. New technology means that the cost of laying the cable underwater can be reduced to $1100 per kilometre.

     

    Determine the path for laying the cable in order to minimise the cost in this case.    (2 marks)

Show Answers Only
  1. `$12\ 800`
  2. `$15\ 160`
  3. `text{Proof (See Worked Solutions)}`
  4. `$12\ 200`
  5. `P\ text(to)\  S\ text(in a straight line.)`
Show Worked Solution
♦♦ Although specific data is unavailable for question parts, mean marks were 35% for Q9 in total.
i.   `text(C)text(ost)` `=(PRxx1000)+(SRxx2600)`
  `=(5xx1000)+(3xx2600)`
  `=12\ 800`

 
`:. text(C)text(ost is)\   $12\ 800`

 

ii.   `text(C)text(ost)=PSxx2600`

`text(Using Pythagoras:)`

`PS^2` `=PR^2+SR^2`
  `=5^2+3^2`
  `=34`
`PS` `=sqrt34`

 

`:.\ text(C)text(ost)` `=sqrt34xx2600`
  `=15\ 160.474…`
  `=$15\ 160\ \ text{(nearest dollar)}`

 

iii.  `text(Show)\ \ C=1000(5-x+2.6sqrt(x^2+9))`

`text(C)text(ost)=(PQxx1000)+(QSxx2600)`

`PQ` `=5-x`
`QS^2` `=QR^2+SR^2`
  `=x^2+3^2`
`QS` `=sqrt(x^2+9)`
`:.C` `=(5-x)1000+sqrt(x^2+9)\ (2600)`
  `=1000(5-x+2.6sqrt(x^2+9))\ \ text(…  as required)`

 

iv.   `text(Find the MIN cost of laying the cable)`

`C` `=1000(5-x+2.6sqrt(x^2+9))`
`(dC)/(dx)` `=1000(–1+2.6xx1/2xx2x(x^2+9)^(–1/2))`
  `=1000(–1+(2.6x)/(sqrt(x^2+9)))`

`text(MAX/MIN when)\ (dC)/(dx)=0`

IMPORTANT: Tougher derivative questions often require students to deal with multiple algebraic constants. See Worked Solutions in part (iv).

`1000(–1+(2.6x)/(sqrt(x^2+9)))=0`

`(2.6x)/sqrt(x^2+9)` `=1`
`2.6x` `=sqrt(x^2+9)`
`(2.6)^2x^2` `=x^2+9`
`x^2(2.6^2-1)` `=9`
`x^2` `=9/5.76`
  `=1.5625`
`x` `=1.25\ \ \ \ (x>0)`
MARKER’S COMMENT: Check the nature of the critical points in these type of questions. If using the first derivative test, make sure some actual values are substituted in.

`text(If)\ \ x=1,\ \ (dC)/(dx)<0`

`text(If)\ \ x=2,\ \ (dC)/(dx)>0`

`:.\ text(MIN when)\ \ x=1.25`

`C` `=1000(5-1.25+2.6sqrt(1.25^2+9))`
  `=1000(122)`
  `=12\ 200` 

 

`:.\ text(MIN cost is)\  $12\ 200\ text(when)\   x=1.25`

 

v.   `text(Underwater cable now costs $1100 per km)`

`=>\ C` `=1000(5-x)+1100sqrt(x^2+9)`
  `=1000(5-x+1.1sqrt(x^2+9))`
`(dC)/(dx)` `=1000(–1+1.1xx1/2xx2x(x^2+9)^(-1/2))`
  `=1000(–1+(1.1x)/sqrt(x^2+9))`

 
`text(MAX/MIN when)\ (dC)/(dx)=0`

`1000(–1+(1.1x)/sqrt(x^2+9))` `=0`
`(1.1x)/sqrt(x^2+9)` `=1`
`1.1x` `=sqrt(x^2+9)`
`1.1^2x^2` `=x^2+9`
`x^2(1.1^2-1)` `=9`
`x^2` `=9/0.21`
`x` `~~6.5\ text{km (to 1 d.p.)}`
  `=>\ text(no solution since)\ x<=5`

 
`text(If we lay cable)\ PR\ text(then)\ RS`

MARKER’S COMMENT: Many students failed to interpret a correct calculation of  `x>5`  as providing no solution.

`=>\ text(C)text(ost)=5xx1100+3xx1000=8500`

`text(If we lay cable directly underwater via)\ PS`

`=>\ text(C)text(ost)=sqrt34xx1100=6414.047…`

`:.\ text{MIN cost is  $6414 by cabling directly from}\ P\ text(to)\ S`.

Mechanics, EXT2* M1 2011 HSC 6b

The diagram shows the trajectory of a ball thrown horizontally, at speed `v` m/s, from the top of a tower `h` metres above the ground level.
 

2011 6b
 

The ball strikes the ground at an angle of  45°, `d` metres from the base of the tower, as shown in the diagram. The equations describing the trajectory of the ball are

`x=vt`   and   `y=h-1/2 g t^2`,    (DO NOT prove this)

where `g` is the acceleration due to gravity, and `t` is time in seconds.

  1. Prove that the ball strikes the ground at time  
     
         `t=sqrt((2h)/(g))`  seconds.    (1 mark)

  2. Hence, or otherwise, show that  `d=2h`.    (2 marks)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}` 
  2. `text{Proof  (See Worked Solutions)}`
Show Worked Solution

i.   `text(Show)\ \ y=0\ \ text(when)\ t=sqrt((2h)/g)\ \ text(seconds:)`

`0` `=h-1/2 g t^2`
`1/2 g t^2` `=h`
`t^2` `=(2h)/g`
`:.t` `=sqrt((2h)/g)\ \ text(seconds,)\ \ t>=0\ \ text(… as required)`

 

ii.   `text(Show that)\ d=2h`

`x=vt\ \ \ \ =>\ \ \ dotx=v`

`y=h-1/2 g t^2\ \ \ \ =>\ \ \ doty=-g t`

`text(At)\ t=sqrt((2h)/g)`,

`doty` `=-gxxsqrt((2h)/g)`
  `=-sqrt(2gh)`

 

`text(S)text(ince the ball strikes the ground at)\ 45^@,\ text(we know)`

♦♦ Mean mark 33%
STRATEGY: The key to unlocking this proof is understanding that `tan theta`, where  `theta` is the angle of trajectory at impact of a projectile, equals  `|\ doty\ |/dotx`.
`tan45^@=` `|\ doty\ |/dotx`
`1=` `sqrt(2gh)/v`
`v=` `sqrt(2gh)`

 
`text(S)text(ince)\ \ x=d=vt\ \ text(when)\ \ t=sqrt((2h)/g)`

`d` `=sqrt(2gh)xxsqrt((2h)/g)`
  `=sqrt((2h)^2)`
  `=2h\ \ text( … as required)`

Calculus in the Physical World, 2UA 2008 HSC 6b

The graph shows the velocity of a particle,  `v`  metres per second, as a function of time,  `t`  seconds.
 


 

  1. What is the initial velocity of the particle?   (1 mark)
  2. When is the velocity of the particle equal to zero?    (1 mark)
  3. When is the acceleration of the particle equal to zero?    (1 mark)
  4. By using Simpson's Rule with five function values, estimate the distance travelled by the particle between  `t=0`  and  `t=8`.   (3 marks)
  5.  
Show Answers Only
  1. `20\ text(m/s)`
  2. `t=10\ text(seconds)`
  3. `t=6\ text(seconds)`
  4. `493 1/3\ text(metres)`
Show Worked Solution

(i)    `text(Find)\   v  \ text(when)  t=0`

`v=20\ \ text(m/s)`

 

(ii)    `text(Particle comes to rest at)\  t=10\ text{seconds  (from graph)}`

 

(iii)  `text(Acceleration is zero when)\ t=6\ text{seconds  (from graph)}`

 

(iv)   

MARKER’S COMMENT: Less errors were made by students using a table and the given formula. Note however, that the formula `A~~(b-a)/6xx` `[f(a)+4f((a+b)/2)+f(b)]` produced the most errors.
`text(Area)` `~~h/3[y_0+y_n+4text{(odds)}+2text{(evens)}]`
  `~~h/3[y_0+y_4+4(y_1+y_3)+2(y_2)]`
  `~~2/3[20+60+4(50+80)+2(70)]`
  `~~2/3[740]`
  `~~493 1/3`

 

`:.\ text{Distance travelled is 493 1/3 m (approx.)}` 

Calculus, 2ADV C4 2009 HSC 7a

The acceleration of a particle is given by

`a=8e^(-2t)+3e^(-t)`,

where  `x`  is the displacement in metres and  `t`  is the time in seconds.

Initially its velocity is  `text(– 6 ms)^(–1)` and its displacement is 5 m.

  1. Show that the displacement of the particle is given by
  2. `qquad  x=2e^(-2t)+3e^-t+t`.   (2 marks) 

  3. Find the time when the particle comes to rest.    (3 marks)

  4. Find the displacement  when the particle comes to rest.    (1 mark)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `ln4\ text(seconds)`
  3. `7/8+ln4\ \ text(units)`
Show Worked Solution

i.    `text(Show)\ \ x=2e^(-2t)+3e^-t+t`

`a=8e^(-2t)+3e^-t\ \ text{(given)}`

`v=int a\ dt=-4e^(-2t)-3e^-t+c_1`

`text(When)\ t=0,  v=-6\ \ text{(given)}`

`-6` `=-4e^0-3e^0+c_1`
`-6` `=-7+c_1`
`c_1` `=1`

 
`:. v=-4e^(-2t)-3e^-t+1`
 

`x` `=int v\ dt`
  `=int(-4e^(-2t)-3e^-t+1)\ dt`
  `=2e^(-2t)+3e^-t+t+c_2`

 
`text(When)\ \ t=0,\ x=5\ \ text{(given)}`

`5` `=2e^0+3e^0+c_2`
`c_2` `=0`

 
`:.\ x=2e^(-2t)+3e^-t+t\ \ text(… as required)`

 

ii.   `text(Particle comes to rest when)\ \ v=0`

`text(i.e.)\ \ -4e^(-2t)-3e^-t+1=0`

`text(Let)\ X=e^-t\ \ \ \ =>X^2=e^(-2t)`

`-4X^2-3X+1` `=0`
`4X^2+3X-1` `=0`
`(4X-1)(X+1)` `=0`

 
 `:.\ \ X=1/4\ \ text(or)\ \ X=-1`

`text(When)\ \ X=1/4:`

`e^-t` `=1/4`
`lne^-t` `=ln(1/4)`
`-t` `=ln(1/4)`
`t` `=-ln(1/4)=ln(1/4)^-1=ln4`

 
`text(When)\ \ X=-1:`

`e^-t=-1\ \ text{(no solution)}`
 

`:.\ text(The particle comes to rest when)\ t=ln4\ text(seconds)`
  

iii.  `text(Find)\ \ x\ \ text(when)\ \ t=ln4 :`

`x=2e^(-2t)+3e^-t+t`

`\ \ =2e^(-2ln4)+3e^-ln4+ln4`

`\ \ =2(e^ln4)^-2+3(e^ln4)^-1+ln4`

`\ \ =2xx4^-2+3xx4^-1+ln4`

`\ \ =2/16+3/4+ln4`

`\ \ =7/8+ln4`

ALGEBRA TIP: Helpful identity  `e^lnx=x`. Easily provable as follows:
`e^ln2=x`
`\ =>lne^ln2=lnx\ `
`\ => ln2=lnx\ `
`\ =>x=2`.

Calculus, EXT1* C1 2010 HSC 7a

The acceleration of a particle is given by

`ddotx=4cos2t`,

where  `x`  is the displacement in metres and  `t`  is the time in seconds. 

Initially the particle is at the origin with a velocity of  `text(1 ms)^(–1)`.

  1. Show that the velocity of the particle is given by

     

      `dotx=2sin2t+1`.    (2 marks)

  2. Find the time when the particle first comes to rest.    (2 marks)

  3. Find the displacement,  `x`,  of the particle in terms of  `t`.    (2 marks)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `(7pi)/12\ text(seconds)`
  3. `x=1-cos2t+t`
Show Worked Solution
i.   `text(Show)\ dotx` `=2sin2t+1`
`dotx` `=intddotx\ dt`
  `=int4cos2t\ dt`
  `=2sin2t+c`

 
`text(When)\ t=0, \ \ dotx=1\ \ text{(given)}`

`1=2sin0+c`

`c=1`

 
`:. dotx=2sin2t+1 \ \ \ text(… as required)`
 

ii.   `text(Find)\ t\ text(when)\ dotx=0 :`

`2sin2t+1` `=0`
`sin2t` `=-1/2`

 
`=>sin theta=1/2\ text(when)\ theta=pi/6`

`text(S)text(ince)\ \ sin theta\ \ text(is negative in 3rd and 4th quadrants)`

`2t` `=pi + pi/6`
`2t` `=(7pi)/6`
`t` `=(7pi)/12`

 
`:.\ text(Particle first comes to rest at)\ t=(7pi)/12\ text(seconds)`
 

iii.    `x` `=intdotx\ dt`
  `=int(2sin2t+1)\ dt`
  `=t-cos2t+c`

 
`text(When)\ t=0,\ x=0\ \ text{(given)}`

`0=0-cos0+c`

`c=1`

 
`:. x=t-cos2t+1`

Calculus, 2ADV C4 2012 HSC 15b

The velocity of a particle is given by

`v=1-2cost`,

where  `x`  is the displacement in metres and  `t`  is the time in seconds. Initially the particle is 3 m to the right of the origin.

  1. Find the initial velocity of the particle.    (1 mark)

  2. Find the maximum velocity of the particle.    (1 mark)

  3. Find the displacement, `x`,  of the particle in terms of  `t`.    (2 marks)

  4. Find the position of the particle when it is at rest for the first time.    (2 marks)

Show Answers Only
  1. `-1\ text(m/s)`
  2. `3\ text(m/s)`
  3. `x=t-2sint+3`
  4. `pi/3-sqrt3+3`
Show Worked Solution

i.    `text(Find)\ \ v\ \ text(when)\ \ t=0`:

`v` `=1-2cos0`
  `=1-2`
  `=-1`

 
`:.\ text(Initial velocity is)\ -1\ text(m/s.)`

 

ii.  `text(Solution 1)`

`text(Max velocity occurs when)\ \ a=(d v)/(dt)=0`

♦♦ Mean mark 29%
MARKER’S COMMENT: Solution 2 is more efficient here. Using the -1 and +1 limits of trig functions can be very a effective way to calculate max/min values.

`a=2sint`
 

`text(Find)\ \ t\ \ text(when)\ \ a=0 :`

`2sint=0`

`t=0`,  `pi`,  `2pi`, …

`text(At)\ \ t=0,\ \   v=-1\ text(m/s)`

`text(At)\ \ t=pi,\ \ v=1-2(-1)=3\ text(m/s)`

 
`:.\ text(Maximum velocity is 3 m/s)`

 

`text(Solution 2)`

`v=1-2cost`

`text(S)text(ince)\ \ -1` `<cost<1`
`-2` `<2cost<2`
`-1` `<1-2cost<3`

 
`:.\ text(Maximum velocity is 3 m/s)`

 

iii.   `x` `=int v\ dt`
  `=int(1-2cost)\ dt`
  `=t-2sint+c`

 
`text(When)\ \ t=0,\ \ x=3\ \ text{(given)}`

`3=0-2sin0+3`

`c=3`

 
`:. x=t-2sint+3`

 

iv.  `text(Find)\ \ x\ \ text(when)\ \ v=0\ \ text{(first time):}`

♦ Mean mark 50%
MARKER’S COMMENT: Many students found  `t=pi/3`  but failed to gain full marks by omitting to find  `x`. Remember that for calculus, angles are measured in radians, NOT degrees!

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

`0` `=1-2cost`
`cost` `=1/2`
`t` `=cos^-1(1/2)`
  `=pi/3\ \ \ text{(first time)}`

 
`text(Find)\ \ x\ \ text(when)\ \ t=pi/3 :`

`x` `=pi/3-2sin(pi/3)+3`
  `=pi/3-2xxsqrt3/2+3`
  `=pi/3-sqrt3+3\ \ text(units)`

Calculus, EXT1 C1 2010 HSC 2b

The mass `M` of a whale is modelled by

`M=36-35.5e^(-kt)` 

where  `M`  is measured in tonnes,  `t`  is the age of the whale in years and  `k`  is a positive constant.

  1. Show that the rate of growth of the mass of the whale is given by the differential equation
     
    `qquad qquad (dM)/(dt)=k(36-M)`    (1 mark)

  2. When the whale is 10 years old its mass is 20 tonnes.

     

    Find the value of  `k`,  correct to three decimal places.    (2 marks)

  3. According to this model, what is the limiting mass of the whale?    (1 mark)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `0.080`
  3. `36\ text(tonnes)`
Show Worked Solution

i.  `M=36-35.5e^(-kt)`

IMPORTANT: Know this standard proof well and be able to produce it quickly.

`35.5e^(-kt)=36-M`

`:. (dM)/(dt)` `=-kxx-35.5e^(-kt)`
  `=kxx35.5e^(-kt)`
  `=k(36-M)\ \ \ text(… as required)`

 

ii.  `text(Find)\ \ k`

`text(When)\ \ t=10,\ \ M=20`

`M` `=36-35.5e^(-kt)`
`20` `=36-35.5e^(-10k)`
`35.5e^(-10k)` `=16`
`lne^(-10k)` `=ln(16/35.5)`
`-10k` `=ln(16/35.5)`
`:. k` `=-ln(16/35.5)/10`
  `=0.07969…`
  `=0.080\ \ text{(to 3 d.p.)}`

 

iii.  `text(As)\ t->oo,  e^(-kt)=1/e^(kt)\ ->0,\ \ k>0`

`M->36`

`:.\ text(The whale’s limiting mass is 36 tonnes.)`

Calculus, EXT1 C1 2013 HSC 12c

A cup of coffee with an initial temperature of 80°C is placed in a room with a constant temperature of 22°C.

The temperature,  `T`°C,  of the coffee after `t` minutes is given by

`T=A+Be^(-kt)`,

where  `A`,  `B`  and  `k`  are positive constants. The temperature of the coffee drops to 60°C after 10 minutes.

How long does it take for the temperature of the coffee to drop to 40°C?  Give your answer to the nearest minute.    (3 marks)

Show Answers Only

`text(28 minutes)`

Show Worked Solution

`T=A+Be^(-kt)`

`text(S)text(ince constant room temp is)\  22°text(C)“=>\ A=22`

`T=22+Be^(-kt)`
 

`text(At)\ \ t=0,\ T=80`

`80=22+Be^0`

`=>\ B=58`

IMPORTANT: Students should either keep  `k` in exact form (using their calculator’s memory button), or ensure they take `k`  to a sufficient number of decimal places to reach an accurate answer.

`:.\ T=22+58e^(-kt)`
 

`text(At)\ \ t=10,\ T=60`

`60` `=22+58e^(-10k)`
`58e^(-10k)` `=38`
`e^(-10k)` `=38/58`
`lne^(-10k)` `=ln(38/58)`
`-10k` `=ln(38/58)`
`:.\ k` `=-1/10ln(38/58)`

 

`text(Find)\ \ t\ \ text(when)\  T=40°text(C) :`

ALGEBRA TIP: Keep `k`  in the equation right until the last calculation and then substitute in (see Worked Solution).
`40` `=22+58e^(-kt)`
`58e^(-kt)` `=18`
`-kt` `=ln(18/58)`
`t` `=-ln(18/58)/k,\ \ text(where) \ k=–1/10ln(38/58)`
  `=10 xx ln(18/58)/ln(38/58)`
  `=27.6706…`
  `=28\ text{mins  (nearest minute)}`

 
`:.\ text(It takes 28 minutes to cool to 40°C.)`

Calculus, EXT1* C1 2008 HSC 5c

Light intensity is measured in lux. The light intensity at the surface of a lake is 6000 lux. The light intensity,  `I`  lux, a distance  `s`  metres below the surface of the lake is given by

`I=Ae^(-ks)`

where  `A`,  and  `k`  are constants.

  1. Write down the value of  `A`.   (2 marks)

  2. The light intensity 6 metres below the surface of the lake is 1000 lux. Find the value of  `k`.    (2 marks)

  3. At what rate, in lux per metre, is the light intensity decreasing 6 metres below the surface of the lake?    (2 marks)

Show Answers Only
  1. `6000`
  2. `- 1/6 ln(1/6)\ text(or)\ 0.29863`
  3. `299`
Show Worked Solution

i.   `I=Ae^(-ks)`

`text(Find)\ A,\ text(given)\   I=6000\ text(at)\  s=0`

`6000` `=Ae^0`
`:.\ A` `=6000`

 

ii.   `text(Find)\ k\ text(given)\  I=1000\ text(at)  s=6`

MARKER’S COMMENT: Many students used the “log” function on their calculator rather than the `log_e` function. BE CAREFUL!
`1000` `=6000e^(-6xxk)`
`e^(-6k)` `=1/6`
`lne^(-6k)` `=ln(1/6)`
`-6k` `=ln(1/6)`
`k` `=- 1/6 ln(1/6)`
  `=0.2986…`
  `=0.30\ \ \ text{(2 d.p.)}`

 

iii.  `text(Find)\ \ (dI)/(ds)\ \ text(at)\ \ s=6`

ALGEBRA TIP: Tidy your working by calculating `(dI)/(ds)` using `k` and then only substituting for `k` in part (ii) at the final stage.
`I` `=6000e^(-ks)`
`:.(dI)/(ds)` `=-6000ke^(-ks)`

 
`text(At)\ s=6,`

`(dI)/(ds)` `=-6000ke^(-6k),\ \ \ text(where)\ k=- 1/6 ln(1/6)`
  `=-298.623…`
  `=-299\ \ text{(nearest whole number)}`

 
`:. text(At)\ s=6,\ text(the light intensity is decreasing)`

`text(at 299 lux per metre.)`

Calculus, EXT1* C1 2009 HSC 6b

Radium decays at a rate proportional to the amount of radium present. That is, if  `Q(t)`  is the amount of radium present at time  `t`,  then  `Q=Ae^(-kt)`,  where  `k`  is a positive constant and  `A`  is the amount present at  `t=0`. It takes 1600 years for an amount of radium to reduce by half.

  1. Find the value of  `k`.   (2 marks)

  2. A factory site is contaminated with radium. The amount of radium on site is currently three times the safe level.

     

    How many years will it be before the amount of radium reaches the safe level.    (2 marks)

Show Answers Only
  1. `(-ln(1/2))/1600\ \ text(or 0.000433)`
  2. `2536\ text(years)`
Show Worked Solution

i.   `Q=Ae^(-kt)`

MARKER’S COMMENT: Students must be familiar with “half-life” and the algebra required. i.e. using `Q=1/2 A` within their calculations.

`text(When)\ \ t=0,\ \ Q=A`

`text(When)\ \ t=1600,\ \ Q=1/2 A`

`:.1/2 A` `=A e^(-1600xxk)`
`e^(-1600xxk)` `=1/2`
`lne^(-1600xxk)` `=ln(1/2)`
`-1600k` `=ln(1/2)`
`k` `=(-ln(1/2))/1600`
  `=0.0004332\ \ text{(to 4 sig. figures)}`

 

ii.   `text(Find)\ \ t\ \ text(when)\ \ Q=1/3 A :`

IMPORTANT: Know how to use the memory function on your calculator to store the exact value of `k` found in part (i) to save time.
`1/3 A` `=A e^(-kt)`
`e^(-kt)` `=1/3`
`lne^(-kt)` `=ln(1/3)`
`-kt` `=ln(1/3)`
`:.t` `=(-ln(1/3))/k,\ \ \ text(where)\ \ \ k=(-ln(1/2))/1600`
  `=(ln(1/3) xx1600)/ln(1/2)`
  `=2535.940…`

 
`:.\ text(It will take  2536  years.)`

Calculus, EXT1* C1 2012 HSC 14c

Professor Smith has a colony of bacteria. Initially there are 1000 bacteria. The number of bacteria,  `N(t)`,  after  `t`  minutes is given by

`N(t)=1000e^(kt)`. 

  1. After 20 minutes there are 2000 bacteria.

     

    Show that `k=0.0347`  correct to four decimal places.   (1 mark)

  2. How many bacteria are there when  `t=120`?    (1 mark)

  3. What is the rate of change of the number of bacteria per minute, when  `t=120`?     (1 mark)

  4. How long does it take for the number of bacteria to increase from 1000 to 100 000?    (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions).}`
  2. `64\ 328`
  3. `2232`
  4. `133\ text(minutes)` 
Show Worked Solution

i.   `N=1000e^(kt)\ text(and given)\ N=2000\ text(when)\  t=20`

`2000` `=1000e^(20xxk)`
`e^(20k)` `=2`
`lne^(20k)` `=ln2`
`20k` `=ln2`
`:.k` `=ln2/20`
  `=0.0347\ \ text{(to 4 d.p.)  … as required}`

 

 ii.  `text(Find)\ \ N\ \ text(when)\  t=120`

NOTE: Students could have used the exact value of  `k=ln2/20`  in parts (ii), (iii) and (iv), which would yield the answers (ii) 64,000, (iii) 2,218, and (iv) 133 minutes.
`N` `=1000e^(120xx0.0347)`
  `=64\ 328.321..`
  `=64\ 328\ \ text{(nearest whole number)}`

 
`:.\ text(There are)\  64\ 328\ text(bacteria when t = 120.)`
 

iii.  `text(Find)\ (dN)/(dt)\ text(when)\  t=120`

MARKER’S COMMENT: This part proved challenging for many students. Differentiation is required to find the rate of change in these type of questions.
`(dN)/(dt)` `=0.0347xx1000e^(0.0347t)`
  `=34.7e^(0.0347t)`

 
`text(When)\ \ t=120`

`(dN)/(dt)` `=34.7e^(0.0347xx120)`
  `=2232.1927…`
  `=2232\ \ text{(nearest whole)}`

 
`:.\ (dN)/(dt)=2232\ text(bacteria per minute at)\  t=120`

 

iv.  `text(Find)\ \ t\ \ text(such that)\  N=100,000`

`=>100\ 000` `=1000e^(0.0347t)`
`e^(0.0347t)` `=100`
`lne^(0.0347t)` `=ln100`
`0.0347t` `=ln100`
`t` `=ln100/0.0347`
  `=132.7138…`
  `=133\ \ text{(nearest minute)}`

 
`:.\ N=100\ 000\ text(when)\  t=133\ text(minutes.)`

Calculus, EXT1* C1 2013 HSC 16b

Trout and carp are types of fish. A lake contains a number of trout. At a certain time, 10 carp are introduced into the lake and start eating the trout. As a consequence, the number of trout,  `N`,  decreases according to

`N=375-e^(0.04t)`,

where `t` is the time in months after the carp are introduced.

The population of carp,  `P`,  increases according to  `(dP)/(dt)=0.02P`.

  1. How many trout were in the lake when the carp were introduced?    (1 mark)

  2. When will the population of trout be zero?    (1 mark)

  3. Sketch the number of trout as a function of time.     (1 marks)

  4. When is the rate of increase of carp equal to the rate of decrease of trout?    (3 marks)

  5. When is the number of carp equal to the number of trout?    (2 marks)

Show Answers Only
  1. `text(374 trout)`
  2. `text{148 months (nearest month)}`
  3.  
     
    2UA 2013 HSC 16b Answer
  4. `text{After 80 months (nearest month)}`
  5. `text{After 135 months (nearest month)}`
Show Worked Solution

i.    `text(Carp introduced at)\ \ t=0`

MARKER’S COMMENT: A number of students did not equate `e^0=1` in this part.

`N=375-e^0=374`

`:.\ text(There was 374 trout when carp were introduced.)`

 

ii.    `text(Trout population will be zero when)`

NOTE: The last line of the solution isn’t necessary but is included as good practice as a check that the answer matches the exact question asked.
`N` `=375-e^(0.04t)=0`
`e^(0.04t)` `=375`
`0.04t` `=ln375`
`t` `=ln375/0.04`
  `=148.173 …`
  `=148\ text{months (nearest month)}`

 
`:.\ text(After 148 months, the trout population will be zero.)`
 

iii.   2UA 2013 HSC 16b Answer

♦♦ Mean mark 33% for part (iii)

 

iv.   `text(We need) \ |(dN)/(dt)|=(dP)/(dt)`

`text(Given)\ N=375-e^(0.04t)`

`(dN)/(dt)=-0.04e^(0.04t)`
  

`text(Find)\ P\ text(in terms of)\  t`

`text(Given)\ (dP)/(dt)=0.02P`

`=> P=Ae^(0.02t)`

♦♦♦ Mean mark 20%
COMMENT: Students who progressed to `0.2e^(0.02t)“=0.04e^(0.04t)` received 2 full marks in this part. Show your working!

 `text(Find)\ A\ \ =>text(when)\  t=0,\ P=10`

`10` `=Ae^0`
`:.A` `=10`
`=>P` `=10xx0.02e^(0.02t)`
  `=0.2e^(0.02t)`

 
`text(Given that)\ \ (dP)/(dt)=|(dN)/(dt)|`

`0.2e^(0.02t)` `=0.04e^(0.04t)`
`5e^(0.02t)` `=e^(0.04t)`
`e^(0.04t)/e^(0.02t)` `=5`
`e^(0.04t-0.02t)` `=5`
`lne^(0.02t)` `=ln5`
`0.02t` `=ln5`
`t` `=ln5/0.02`
  `=80.4719…`
  `=80\ text{months (nearest month)}`

 

v.   `text(Find)\ t\ text(when)\ N=P`

`text(i.e.)\ \ 375-e^(0.04t)` `=10e^(0.02t)`
`e^(0.04t)+10e^(0.02t)-375` `=0`

`text(Let)\ X=e^(0.02t),\ text(noting)\ \ X^2=(e^(0.02t))^2=e^(0.04t)`

♦♦♦ Mean mark 4%!
MARKER’S COMMENT: Correctly applying substitution to exponentials to form a solvable quadratic proved very difficult for almost all students.
`:.\ X^2+10X-375` `=0`
`(X-15)(X+25)` `=0`

`X=15\ \ text(or)\ \ –25`

 
`text(S)text(ince)\  X=e^(0.02t)`

`e^(0.02t)` `=15\ \ \ \ (e^(0.02t)>0)`
`lne^(0.02t)` `=ln15`
`0.02t` `=ln15`
`t` `=ln15/0.02`
  `=135.4025…`
  `=135\ text(months)`

Calculus, 2ADV C3 2011 HSC 7b

The velocity of a particle moving along the `x`-axis is given by

`v=8-8e^(-2t)`,

where `t` is the time in seconds and `x` is the displacement in metres.

  1. Show that the particle is initially at rest.     (1 mark)

  2. Show that the acceleration of the particle is always positive.     (1 mark)

  3. Explain why the particle is moving in the positive direction for all  `t>0`.     (2 marks)

  4. As  `t->oo`, the velocity of the particle approaches a constant.

     

    Find the value of this constant.     (1 mark) 

  5. Sketch the graph of the particle's velocity as a function of time.     (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `text(See Worked Solutions.)`
  4. `8\ text(m/s)`
  5. `text(See sketch in Worked Solutions)`
Show Worked Solution

i.   `text(Initial velocity when)\ \ t=0`

`v` `=8-8e^0`
  `=0\ text(m/s)`
 
`:.\ text(Particle is initially at rest.)`

 

 

MARKER’S COMMENT: Students whose working showed `e^(-2t)` as `1/e^(2t)`, tended to score highly in this question.

ii.   `a=d/(dt) (v)=-2xx-8e^(-2t)=16e^(-2t)`

`text(S)text(ince)\  e^(-2t)=1/e^(2t)>0\ text(for all)\  t`.

`=>\ a=16e^(-2t)=16/e^(2t)>0\ text(for all)\  t`.
 

`:.\ text(Acceleration is positive for all)\ \ t>0`.
 

iii.  `text{S}text{ince the particle is initially at rest, and ALWAYS}`

♦♦♦ Mean mark 22%
COMMENT: Students found part (iii) the most challenging part of this question by far.

`text{has a positive acceleration.`
 

`:.\ text(It moves in a positive direction for all)\ t`.
 

iv.   `text(As)\ t->oo`,  `e^(-2t)=1/e^(2t)->0`

`=>8/e^(2t)->0\  text(and)`

`=>v=8-8/e^(2t)->8\ text(m/s)`
 

`:.\ text(As)\ \ t->oo,\ text(velocity approaches 8 m/s.)`

 

IMPORTANT: Use previous parts to inform this diagram. i.e. clearly show velocity was zero at  `t=0`  and the asymptote at  `v=8`. 
v.   

Calculus in the Physical World, 2UA 2011 HSC 7b Answer

Financial Maths, 2ADV M1 2008 HSC 4b

The zoom function in a software package multiplies the dimensions of an image by 1.2.  In an image, the height of a building is 50 mm. After the zoom function is applied once, the height of the building in the image is 60 mm. After the second application, it is 72 mm.

  1. Calculate the height of the building in the image after the zoom function has been applied eight times. Give your answer to the nearest mm.     (2 marks)

  2. The height of the building in the image is required to be more than 400 mm. Starting from the original image, what is the least number of times the zoom function must be applied?     (2 marks)

Show Answers Only
  1. `text(215 mm)`
  2. `12`
Show Worked Solutions
i.    `T_1` `=a=50`
  `T_2` `=ar^1=50(1.2)=60`
  `T_3` `=ar^2=50(1.2)^2=72`

 
`=>\ text(GP where)\ \ a=50,\ \ r=1.2`

`\ \ vdots` 

MARKER’S COMMENT: Within this GP, note that `T_9` is the term where the zoom has been applied 8 times.
`T_9` `=50(1.2)^8`
  `=214.99`

 

`:.\ text{Height will be 215 mm  (nearest mm)}`

 

ii.    `T_n=ar^(n-1)` `>400`
  `:.\ 50(1.2)^(n-1)` `>400`
  `1.2^(n-1)` `>8`
  `ln 1.2^(n-1)` `>ln8`
  `n-1` `>ln8/ln1.2`
  `n` `>12.405`

 

`:.\ text(The height of the building in the 13th image)`

`text(will be higher than 400 mm, which is the 12th)`

`text(time the zoom would be applied.)`

Financial Maths, 2ADV M1 2009 HSC 4a

A tree grows from ground level to a height of 1.2 metres in one year. In each subsequent year, it grows `9/10` as much as it did in the previous year.

Find the limiting height of the tree.     (2 marks)

 

Show Answer Only

`12\ text(m)`

Show Worked Solution

`a=1.2, \ \ \ r=9/10`

`text(S)text(ince)\ \ |\ r\ |<1,`

`S_oo` `=a/(1-r)`
  `=1.2/(1-(9/10))`
  `=12\ text(m)`

 

`:.\ text(Limiting height of tree is 12 m.)`

Calculus, EXT1* C1 2013 HSC 14a

The velocity of a particle moving along the `x`-axis is given by  `dotx=10-2t`, where `x` is the displacement from the origin in metres and `t` is the time in seconds. Initially the particle is 5 metres to the right of the origin.

  1. Show that the acceleration of the particle is constant.  (1 mark)

  2. Find the time when the particle is at rest.  (1 mark)

  3. Show that the position of the particle after 7 seconds is 26 metres to the right of the origin.  (2 marks)

  4. Find the distance travelled by the particle during the first 7 seconds.   (2 marks)

Show Answers Only
  1.  `ddotx=-2\ \ text{(constant)}`
  2. `t=5\ text(seconds)`
  3. `text{Proof (See Worked Solutions)}`
  4. `29\ text(metres)`
Show Worked Solutions

i.   `dotx=10-2t`

`text(Acceleration)=ddotx=d/(dx)dotx=-2`

 
`:.\ text(The acceleration is constant.)`

 

 ii.  `text(Particle comes to rest when)\  dotx=0`

`10-2t=0\ text(when)\ t=5`

 
`:.\ text(Particle comes to rest after 5 seconds.)`

 

 iii. `text(Show)\  x=26\ text(when)\ t=7`

`x` `=int dotx\ dt`
  `=int (10-2t)\ dt`
  `=10t -t^2+c`

 

`text(Given)\  t=0\ text(when)\ x=5`

`=> 5` `=10(0)-0^2+c`
`c` `=5`

 
`:. x=10t-t^2+5`
 

`text(At)\ \ t=7`

`x` `=10(7)-7^2+5`
  `=70-49+5`
  `=26`

 
`:.\ text(The particle is 26 metres to the right of the origin)`

`text(after 7 seconds … as required)`

 

 iv.   `text(Find the distance travelled in the first 7 seconds:)`

♦ Mean mark 37%.
IMPORTANT: Students can also draw a number line that shows where a particle starts from, changes direction and finishes in order to reduce errors.

`text(At)\  t=5`,

`x` `=10(5)-5^2+5`
  `=50-25+5`
  `=30\ text(m)`

 
`=>\ text{The particle travels 25 m to the right}\ (x=5\ text{to}\ 30)`

`text{then 4 m to the left}\  (x=30\ text{to}\ 26)`
 

`:.\ text(The total distance travelled in 7 seconds)`

`=25+4`

`=29\ text(m)`

Calculus, 2ADV C2 2012 HSC 11d

Differentiate    `(3+e^(2x))^5`.    (2 marks) 

Show Answer Only

`10e^(2x)(3+e^(2x))^4`

Show Worked Solutions

`y=(3+e^(2x))^5`

`(dy)/dx` `=5(3+e^(2x))^4 xx  d/(dx)(3+e^(2x))`
  `=5(3+e^(2x))^4 xx 2e^(2x)`
  `=10e^(2x)(3+e^(2x))^4`

 

Calculus, 2ADV C2 2013 HSC 11d

Differentiate  `x^2e^x`    (2 marks)

Show Answer Only

 `xe^x(x+2)`

Show Worked Solutions

`text{Using the product rule}`

`text(Let)\ \ u=x^2,` `\ \ \ \ \ \ u^{\prime}=2x`
`text(Let)\ \ v=e^x,` `\ \ \ \ \ \ v^{\prime}=e^x`
`{d(uv)}/dx` `=u prime v+v prime u`
  `=2x e^x +x^2 e^x `
  `=xe^x(x+2)`

Financial Maths, 2ADV M1 2009 HSC 8b

One year ago Daniel borrowed $350 000 to buy a house. The interest rate was 9% per annum, compounded monthly. He agreed to repay the loan in 25 years with equal monthly repayments of $2937.

  1. Calculate how much Daniel owed after his first monthly repayment.    (1 mark)

Daniel has just made his 12th monthly repayment. He now owes $346 095. The interest rate now decreases to 6% per annum, compounded monthly.

 

The amount  `$A_n`, owing on the loan after the `n`th monthly repayment is now calculated using the formula  
 
`qquad qquad A_n=346,095xx1.005^n-1.005^(n-1)M-\ ... -1.005M-M`
  
where `$M` is the monthly repayment, and `n=1,2,\ ...,288`.   (DO NOT prove this formula.)

  1. Calculate the monthly repayment if the loan is to be repaid over the remaining 24 years (288 months).    (3 marks)

  2. Daniel chooses to keep his monthly repayments at $2937. Use the formula in part (ii) to calculate how long it will take him to repay the $346 095.   (3 marks)

  3. How much will Daniel save over the term of the loan by keeping his monthly repayments at $2937, rather than reducing his repayments to the amount calculated in part (ii)?   (1 mark)

Show Answers Only
  1. `$349\ 688`
  2. `$2270.31\ \ text{(nearest cent)}`
  3. `178.37\ text(months)`
  4. `$129\ 976.59\ \ text{(nearest cent)}`
Show Worked Solutions

i.   `text(Let)\ L_n= text(the amount owing after)\ n\ text(months)`

`text(Repayment)\ =M=$2937\ \ text(and)\ \ r=text(9%)/12=0.0075\ text(/month)`

`:.\ L_1` `=350\ 000(1+r)-M`
  `=350\ 000(1.0075)-2937`
  `=349\ 688`

 

`:.\ text(After 1 month, the amount owing is)\  $349\ 688`

 

ii.   `text(After 12 repayments, Daniel owes)\ $346\ 095,\  text(and)\ r darr 6%`

`:.\ r=(6%)/12=0.005`

`text(Loan is repaid over the next 24 years. i.e.)\ $A_n=0\ text(when)\  n=288`

`A_n` `=346\ 095(1.005^n)-1.005^(n-1)M-\ ..\ -1.005M-M`
  `=346\ 095(1.005^n)-M(1+1.005+..+1.005^(n-1))`
`A_288` `=346\ 095(1.005^288)-M(1+1.005+..+1.005^287)=0`

`=>\ GP\ text(where)\ a=1,\ text(and)\  r=1.005`

MARKER’S COMMENT: Careless setting out and poor handwriting, especially where indexes were involved, was a major contributor to errors in this question.

`M((1(1.005^288-1))/(1.005-1))=346\ 095(1.005^288)`

`M` `=(1\ 455\ 529.832)/641.1158`
  `=2270.31`

 

`:.\ text{Monthly repayment is $2270.31  (nearest cent)}` 

 

iii.  `text(Given)\ $M\ text(remains at $2937, find)\  n\ text(such that)`

`$A_n=0\ text{(i.e. loan fully paid off)}`

`:. 346\ 095(1.005^n)-2937((1(1.005^n-1))/(1.005-1))` `=0`
`346\ 095(1.005^n)-587\ 400(1.005^n-1)` `=0`
`(346\ 095-587\ 400)(1.005^n)+587\ 400` `=0`
♦♦ A poorly answered question.
MARKER’S COMMENT: Many students struggled to handle the exponential and logarithm calculations in this question.
ALGEBRA TIP: Dividing by `(1.005-1)` in part (iii) is equivalent to multiplying by 200, and cleans up working calculations (see Worked Solutions).
 

`241\ 305(1.005^n)` `=587\ 400`
`ln1.005^n` `=ln((587\ 400)/(241\ 305))`
`n` `=ln2.43426/ln1.005`
  `=178.37..`

 

`:.\ text{He will pay off the loan in 179 months (note the}`

`text{last payment will be a partial payment).}`

 

iv.  `text(Total paid at $2937 per month)`

`= 2937xx178.37=$523\ 872.69`

`text(Total paid at $2270.31 per month)`

`=2270.31xx288=$653,849.28`

 

`:.\ text(The amount saved)`

`=653\ 849.28-523\ 872.69`

`=$129\ 976.59\ \ text{(nearest cent)}`

Financial Maths, 2ADV M1 2010 HSC 4a

Susanna is training for a fun run by running every week for 26 weeks. She runs 1 km  in the first week and each week after that she runs 750 m more than the previous week, until she reaches 10 km in a week. She then continues to run 10 km each week.

  1. How far does Susannah run in the 9th week?     (1 mark)

  2. In which week does she first run 10 km?     (1 mark)

  3. What is the total distance that Susannah runs in 26 weeks?     (2 marks)

Show Answers Only
  1. `7\ text(km)`
  2. `13 text(th week)`
  3. `201.5\ text(km)`
Show Worked Solutions

i.    `T_1=a=1`

`T_2=a+d=1.75`

`T_3=a+2d=2.50`

`=>\ text(AP where)\  a=1  \ \ d=0.75`

`\ \ vdots`

`T_9` `=a+8d`
  `=1+8(0.75)`
  `=7`

 

`:.\ text(Susannah runs 7 km in the 9th week.)`

 

ii.  `text(Find)\ n\ text(such that)\ T_n=10\ text(km)`

`text(Using)\ T_n=a+(n-1)d`

MARKER’S COMMENT: Better responses wrote the formula for the `nth` term before clearly substituting in known values `a` and `d`.
`1+(n-1)(0.75)` `=10`
`0.75n-0.75` `=9`
`n` `=9.75/0.75`
  `=13`

 

`:.\ text(Susannah runs 10 km for the first time in the 13th Week.)`

 

iii.  `text{Let D = the total distance Susannah runs in 26 weeks}`

MARKER’S COMMENT: Many students incorrectly calculated `S_26`, not taking into account the AP stopped at the 13th term.
`text(D)` `=S_13+13(10)`
  `=n/2[2a+(n-1)d]+13(10)`
  `=13/2[2(1)+(13-1)(0.75)]+130`
  `=13/2(2+9)+130`
  `=201.5`

 

`:.\ text(Susannah runs a total of 201.5 km in 26 weeks.)`

Financial Maths, 2ADV M1 2011 HSC 3a

A skyscraper of 110 floors is to be built. The first floor to be built will cost $3 million. The cost of building each subsequent floor will be $0.5 million more than the floor immediately below.

  1. What will be the cost of building the 25th floor?     (2 marks)

  2. What will be the cost of building all 110 floors of the skyscraper?     (2 marks)

Show Answers Only
  1. `$15\ text(million)`
  2. `$3,327.5\  text(million)`
Show Worked Solutions
i.     `T_1` `=a=3`
`T_2` `=a+d=3.5`
`T_3` `=a+2d=4`

 
`=>\ text(AP where)\ \ a=3\ \ d=0.5` 

MARKER’S COMMENT: Better responses listed the sequence of terms in the series as illustrated.
`T_25` `=a+24d`
  `=3+24(0.5)`
  `=15`

 
 `:.\ text(The 25th floor costs  $15,000,000.)`

MARKER’S COMMENT: Better responses stated the formula BEFORE any calculations were performed. This allows markers to allocate part marks to students who had errors in calculation.

 

ii.     `S_110` `=\ text(Total cost of 110 floors)`
  `=n/2(2a+(n-1)d)` 
  `=110/2(2xx3\ 000\ 000+(110-1)500\ 000)`
  `=55(6\ 000\ 000+49\ 500\ 000)`
  `=$3327.5\  text(million)`

 

`:.\ text{The total cost of 110 floors is $3327.5 million}`

Financial Maths, 2ADV M1 2012 HSC 15a

Rectangles of the same height are cut from a strip and arranged in a row. The first rectangle has width 10cm. The width of each subsequent rectangle is 96% of the width of the previous rectangle.
 

2012 15a
 

  1. Find the length of the strip required to make the first ten rectangles.     (2 marks)

  2. Explain why a strip of 3m is sufficient to make any number of rectangles.     (1 mark)

Show Answers Only
  1. `83.8\ text{cm  (1 d.p.)}`
  2. `S_oo=2.5\ text(m)\ \ =>\ \ text(sufficient.)`
Show Worked Solutions
i.    `T_1` `=a=10`
`T_2` `=ar=10xx0.96=9.6`
`T_3` `=ar^2=10xx0.96^2=9.216`

 
`=>\ text(GP where)\ \ a=10\ \ text(and)\ \ r=0.96`

MARKER’S COMMENT: A common error was to find `T_10` instead of `S_10`
`S_10` `=\ text(Length of strip for 10 rectangles)`
  `=(a(1-r^n))/(1-r)`
  `=10((1-0.96^10)/(1-0.96))`
  `=83.8\ text{cm   (to 1 d.p.)}`

 

ii.   `text(S)text(ince)\ |\ r\ |<\ 1`

`S_oo` `=a/(1-r)`
  `=10/(1-0.96)`
  `=250\ text(cm)`

 

`:.\ text(S)text(ince  3 m > 2.5 m, it is sufficient.)`

Probability, 2ADV S1 2013 HSC 15d

Pat and Chandra are playing a game. They take turns throwing two dice. The game is won by the first player to throw a double six. Pat starts the game.

  1. Find the probability that Pat wins the game on the first throw.     (1 mark)

  2. What is the probability that Pat wins the game on the first or on the second throw?     (2 marks)

  3. Find the probability that Pat eventually wins the game.     (2 marks)

Show Answers Only
  1. `1/36`
  2. `(2521)/(46\ 656)\ \ text(or)\ \ 0.054`
  3. `36/71`
Show Worked Solutions

i.   `P\ text{(Pat wins on 1st throw)}=P(W)`

`P(W)` `=P\ text{(Pat throws 2 sixes)}`
  `=1/6 xx 1/6`
  `=1/36`

 

ii.  `text(Let)\ P(L)=P text{(loss for either player on a throw)}=35/36`

`P text{(Pat wins on 1st or 2nd throw)}` 

♦♦ Mean mark 33%
MARKER’S COMMENT: Many students did not account for Chandra having to lose when Pat wins on the 2nd attempt.

`=P(W) + P(LL W)`

`=1/36\ + \ (35/36)xx(35/36)xx(1/36)`

`=(2521)/(46\ 656)`

`=0.054\ \ \ text{(to 3 d.p.)}`

 

iii.  `P\ text{(Pat wins eventually)}`

`=P(W) + P(LL\ W)+P(LL\ LL\ W)+ … `

`=1/36\ +\ (35/36)^2 (1/36)\ +\ (35/36)^2 (35/36)^2 (1/36)\ +…`

 
`=>\ text(GP where)\ \ a=1/36,\ \ r=(35/36)^2=(1225)/(1296)`

♦♦♦ Mean mark 8%!
 COMMENT: Be aware that diminishing probabilities and `S_oo` within the Series and Applications are a natural cross-topic combination.

 
`text(S)text(ince)\ |\ r\ |<\ 1:`

`S_oo` `=a/(1-r)`
  `=(1/36)/(1-(1225/1296))`
  `=1/36 xx 1296/71`
  `=36/71`

 

`:.\ text(Pat’s chances to win eventually are)\  36/71`.