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Calculus, 2ADV C4 2009 HSC 2b

  1. Find  `int 5\ dx`.   (1 mark)

  2. Find  `int 3/((x-6)^2)\ dx`.   (2 marks)

  3. Evaluate  `int_1^4 x^2 + sqrtx\ dx`.   (3 marks)

Show Answers Only

a.    `5x + C`

b.    `(-3)/((x-6)) + C`

c.    `77/3`

Show Worked Solution

a.    `int 5\ dx= 5x + C`
  

b.    `int 3/((x-6)^2)\ dx`

`= 3 int (x-6)^(-2)\ dx`

`= 3 xx 1/(-1) xx (x-6)^(-1) + c`

`= (-3)/((x-6)) + c`
  

c.    `int_1^4 x^2 + sqrtx\ \ dx`

`= int_1^4 (x^2 + x^(1/2))\ dx`

`= [1/3 x^3 + 1/(3/2) x^(3/2)]_1^4`

`= [(x^3)/3 + 2/3x^(3/2)]_1^4`

`= [((4^3)/3 + 2/3 xx 4^(3/2))-(1/3 + 2/3)]`

`= [(64/3 + 16/3)-3/3]`

`= [80/3-3/3]= 77/3`

Trigonometry, 2ADV T3 2010 HSC 8c

The graph shown is  `y = A sin bx`.

 

  1. Write down the value of  `A`.   (1 mark)

  2. Find the value of  `b`.   (1 mark)

  3. On the same set of axes, draw the graph  `y = 3 sin x + 1`  for  `0 <= x <= pi`.   (2 marks)

Show Answers Only

a.    `A = 4`

b.    `b = 2`

c.    `text(See Worked Solutions for sketch)`

Show Worked Solution

a.    `A = 4`

b.    `text(S)text(ince the graph passes through)\ \ (pi/4, 4)`

`text(Substituting into)\ \ y = 4 sin bx`

`4 sin (b xx pi/4)` `=4`
`sin (b xx pi/4)` `= 1`
`b xx pi/4` `= pi/2`
`:. b` `= 2`

  

 MARKER’S COMMENT: Graphs are consistently drawn too small by many students. Aim to make your diagrams 1/3 to 1/2 of a page. 
c.   

Linear Functions, 2UA 2010 HSC 3a

In the diagram  `A`,  `B`  and  `C`  are the points  `( –2, –4 ),\ (12,6)`  and  `(6,8)`  respectively.

The point  `N (2,2)`  is the midpoint of   `AC`. The point  `M`  is the midpoint of  `AB`.
 

2010 3a
 

  1. Find the coordinates of  `M`.   (1 mark)
  2. Find the gradient of  `BC`.   (1 mark)
  3. Prove that  `Delta ABC`  is similar to  `Delta AMN`.   (2 marks)
  4. Find the equation of  `MN`.   (2 marks)
  5. Find the exact length of  `BC`.   (1 mark)
  6. Given that the area of  `Delta ABC`  is  `44`  square units, find the perpendicular distance  from  `A`  to  `BC`.    (1 mark)

 

Show Answers Only
  1. `M (5,1)`
  2. `-1/3`
  3. `text{Proof (See Worked Solutions)}`
  4. `x + 3y  – 8 = 0`
  5. `2 sqrt 10 \ text(units)`
  6. `(22 sqrt 10)/5 \ text(units)`
  7.  
Show Worked Solution

(i)   `M \ text(is the midpoint of) \ AB`

`A text{(–2,–4),}\ \ \ \ B(12,6)`

`:.M` `=((x_1+x_2)/2 ‘ (y_1+y_2)/2)`
  `= ((-2 + 12)/2  ,  (-4 + 6)/2)`
  `= (5,1)`

 

(ii)   `B(12,6),     C(6,8)`

`m_(BC)` `= (y_2 – y_1)/(x_2 – x_1)`
  `= (8– 6)/(6– 12)`
  `= -1/3`

 

(iii)  `text(Prove)\ Delta ABC \ text(|||) \ Delta AMN`

`text(Find gradient of) \ NM`

`N(2,2)   M(5,1)`

`m_(NM)` `= (1- 2)/(5– 2)`
  `= -1/3`

`:. NM\ text(||)\ BC \ \ \ text{(gradients are equal)}`

`angle NAM \ text(is common)`

`angle ANM = angle ACB\ \ \ text{(corresponding angles},\ NM\ text(||)\ BC text{)}`

`:.  Delta ABC \ text(|||) \ Delta AMN \ \ \ text{(equiangular)}`

 

(iv)   `text(Equation of) \ MN \ text(has) \ m=-1/3 \ \ text(through) \ (2,2)`

`text(Using) \ \ \ y- y_1` `= m(x  – x_1)`
`y  – 2` `= -1/3 (x  – 2)`
`3y  – 6` `= -x + 2`
`x +3y  – 8` `= 0`

 

`:. \ text(Equation of) \ MN \ text(is) \ \ x + 3y  – 8 = 0`

 

(v)   `B(12,6)   C(6,8)` 

`BC` `=sqrt{(x_2 – x_1)^2 + (y_2– y_1)^2}`
  `=sqrt{(6– 12)^2 + (8– 6)^2}`
  `=sqrt(36 +4)`
  `=sqrt(40)`
  `= 2 sqrt(10) \ text(units)`

 

(vi)   `text(Given the Area) \  Delta ABC=44\ text(u²)`

`1/2 xx b xx h` `= 44`
`1/2 xx BC xx h`  `= 44`
`1/2 xx 2 sqrt 10 xx h` `= 44`
`:. h` `= 44/ sqrt 10 xx sqrt 10 / sqrt 10`
  `= (44 sqrt 10 )/ 10`
  `= (22 sqrt 10) / 5`

 

`:. _|_ text(distance from) \ A \ text(to) \ BC \ text(is) \ (22 sqrt 10) / 5  \ text(units)`.

Functions, 2ADV F1 2010 HSC 1d

Solve  `| 2x + 3 | = 9`.   (2 marks)

Show Answers Only

 `x = 3\ text(or)\ -6`

Show Worked Solution

`| 2x + 3 | = 9`

`2x + 3` `=9` `\ \ \ \ -(2x + 3)` `=9`
`2x` `=6` `\ \ \ \ -2x\ – 3` `=9`
`x` `=3` `\ \ \ \ -2x` `=12`
     `x` `=-6`

 

`:. x = 3\ \ text(or)\ \ -6`

Statistics, 2ADV 2011 HSC 5b

Kim has three red shirts and two yellow shirts. On each of the three days, Monday, Tuesday and Wednesday, she selects one shirt at random to wear. Kim wears each shirt that she selects only once.

  1. What is the probability that Kim wears a red shirt on Monday?    (1 mark)
  2. What is the probability that Kim wears a shirt of the same colour on all three days?     (1 mark)
  3. What is the probability that Kim does not wear a shirt of the same colour on consecutive days?   (2 marks)
Show Answers Only
  1. `3/5`
  2. `1/10`
  3. `3/10`
Show Worked Solution
i.    `P (R\ text(on Monday) text{)}` `= text(# Red)/text(# Shirts)`
    `= 3/5`

 

MARKER’S COMMENT: Students could also have drawn a probability tree setting out this problem over 3 different days (stages).
ii.    `text(S)text(ince not enough yellow shirts)`
  `P text{(same colour each day)}`

`= P (R,R,R)`

`= 3/5 xx 2/4 xx 1/3`

`= 1/10`

 

iii.   `P text{(not wearing same colour 2 days in a row)}`

`= P (Y,R,Y) + P (R,Y,R)`

♦ Mean mark 47%.
MARKER’S COMMENT: Students should clearly write what probability they are finding in words or symbols before showing calculations.

`=  (2/5 xx 3/4 xx 1/3)+(3/5 xx 2/4 xx 2/3)`

`= 6/60 + 12/60 `

`= 3/10`

Linear Functions, 2UA 2011 HSC 3c

The diagram shows a line  `l_1`, with equation  `3x + 4y - 12 = 0`, which intersects the  `y`-axis at  `B`.

A second line  `l_2`, with equation  `4x - 3y = 0`, passes through the origin  `O`  and intersects `l_1`  at  `E`.  
  

2011 3c
 

  1. Show that the coordinates of  `B`  are  `(0, 3)`.    (1 mark)
  2. Show that  `l_1`  is perpendicular to  `l_2`.   (2 marks)
  3. Show that the perpendicular distance from  `O`  to  `l_1`  is  `12/5`.   (1 mark) 
  4. Using Pythagoras’ theorem, or otherwise, find the length of the interval  `BE`.   (1 mark)
  5. Hence, or otherwise, find the area of  `Delta BOE`.  (1 mark)
Show Answers Only
  1. `B(0,3)`
  2. `text(Proof)  text{(See Worked Solutions)}`
  3. `text(Proof)  text{(See Worked Solutions)}`
  4. `BE = 9/5\ \ text(or)\ \ 1.8\ text(units)`
  5. `54/25\ \ text(or)\ \ 2.16\ text(u²)`
Show Worked Solution
(i)    `B\ text(is)\ y text(-intercept of)\ l_1`
  `text(When)\ x = 0`
`(3 xx 0) + 4y\ – 12` `= 0`
`4y` `=12`
`y` `=3`

`:.\ B\ text(is)\ (0,3)`

 

MARKER’S COMMENT: Better responses involved turning the equations into the “gradient intercept” form as shown in the worked solutions. Many students who tried to use the general formula `m=- b/a` made errors in the process. 
(ii)    `l_1\ text(is)\ 3x + 4y -12`  `= 0`
  `4y` `= -3x + 12`
  `y` `= -3/4x + 3`
  `=> m\ text(of)\  l_1` `= -3/4`
`l_2\ text(is)\ 4x -3y` `= 0`
`3y` `= 4x`
`y` `= 4/3 x`
`=> m\  text(of)\ l_2` `= 4/3`
`m (l_1) xx m (l_2)` `= -3/4 xx 4/3`
  `= -1`

`:.\ l_1\ text(and)\ l_2\ text(are perpendicular)`

 

(iii)   `text(Show)\ _|_\ text(distance)\ O\ (0,0)\ text(to)\ l_1 = 12/5`
`_|_\ text(dist)` `= | (ax_1 + by_1 + c)/sqrt(a^2 + b^2) |`
  `= | (0 + 0\ – 12)/sqrt(3^2 + 4^2) |`
  ` = | -12/5 |`
  `= 12/5\ text(units    … as required)`

 

(iv)   `text(S)text(ince)\ Delta OBE\ text(is right angled)`
`OB^2` `= OE^2 + BE^2`
`3^2` `= (12/5)^2 + BE^2`
`BE^2` `= 9\ – 144/25`
  `= 81/25`
`:.\ BE` `= 9/5`
  `=1.8\ text(units)`

 

(v)    `text(Area)\ Delta BOE` `= 1/2 bh`
    `= 1/2 xx 9/5 xx 12/5`
    `= 108/50`
    `= 54/25`
    `=2.16\  text(u²)`

Probability, 2UA 2011 HSC 1g

A batch of 800 items is examined. The probability that an item from this batch is defective is 0.02.

How many items from this batch are defective?   (1 mark)

Show Answers Only

 `16`

Show Worked Solution
`text(# Items defective)` `= P text{(defective)} xx text(# Items)`
  `= 0.02 xx 800`
  `= 16`

 

`:.\ text(16 Items are defective.)`

Linear Functions, 2UA 2013 HSC 12b

The points  `A(–2, –1)`,  `B(–2, 24)`,  `C(22, 42)` and  `D(22, 17)` form a parallelogram as shown. The point  `E(18, 39)` lies on  `BC`. The point  `F`  is the midpoint of  `AD`.
 

2013 12b
 

  1. Show that the equation of the line through  `A`  and  `D`  is  `3x- 4y + 2 = 0`.    (2 marks)
  2. Show that the perpendicular distance from  `B`  to the line through  `A`  and  `D`  is  `20`  units.   (1 mark)
  3. Find the length of  `EC`.    (1 mark)
  4. Find the area of the trapezium  `EFDC`.    (2 marks)
Show Answers Only
  1. `3x-4y+2=0`
  2. `text(Proof)\  text{(See Worked Solutions)}`
  3. `text(5 units)`
  4. `200\  text(u²)`
Show Worked Solution
(i)    `text(Need to find the equation of)\ AD`
`m_(AD)` `= (y_2\-y_1)/(x_2\-x_1)`
  `= (17+1)/(22+2)`
  `= 18/24`
  `= 3/4`

`text(Equation of)\ AD\ text(has)\ m=3/4 text(, through)\ A text{(–2,–1)}`

`text(Using)\ y\-y_1` `= m (x\-x_1)`
`y+1` `=3/4 (x +2)`
`4y+4` `=3x+6`
`3x-4y+2` `=0`
MARKER’S COMMENT: Students must know the perpendicular distance formula which was again the cause of many errors such that it was flagged by markers.
 

(ii)    `B(–2,24)`
  `AD\ text(is)\  \ 3x -4y+2 =0`
`_|_ text(dist)` `= | (ax_1 + by_1 + c)/sqrt(a^2 + b^2) |`
  `= | (3(–2)\-4(24) + 2)/sqrt(3^2 + (–4)^2 )|`
  `= | (-6\ -96 +2)/sqrt25 |`
  `= | -100/5 |`
  `= 20\ text(units)`

 

`:.\ _|_ text(dist of)\ B\ text(from)\ AD = 20\ text(units   … as required)`

 

(iii)   `E(18,39)\ \ \ C(22,42)`
`EC` `=sqrt ( (x_2\-x_1)^2 + (y_2\-y_1)^2 )`
  `= sqrt ( (22\-18)^2 + (42\-39)^2 )`
  `= sqrt (4^2 + 3^2)`
  `= sqrt25`
  `= 5\ text(units)`

 

`:.\ text(The length of)\ EC\ text(is 5 units.)`

 

(iv)    `text(Area of trapezium)` `=1/2 h (a+b)`
    `=1/2 h (EC + FD)`

`EC = 5\ text(units)`

 MARKER’S COMMENT: One of the most common cause of errors in this part occurred when students failed to use the result found in part (ii).

`text(Need to find)\ FD`

`text(S)text(ince)\ FD = 1/2 xx AD\ \ \ ( F\ text(is midpoint) )`

`A (–2,–1)\ \ D(22,17)`

`AD` `=sqrt( (22 + 2) + (17 + 1)^2 )`
  `= sqrt (24^2 + 18^2)`
  `= sqrt (900)`
  `= 30`

`=> FD = 1/2 xx 30 = 15`

`text(S)text(ince)\ ABCD\ text(is a parallelogram)`

`h` `= _|_ text(distance in part)\ text{(i)}`
  `= 20`
`:.\ text(Area of trapezium)` `=1/2 xx 20 (5 + 15)`
  `=200\  text(u²)`

Probability, STD2 S2 2011 HSC 25c

At another school, students who use mobile phones were surveyed. The set of data is shown in the table.

2UG 2011 25c

  1. How many students were surveyed at this school?   (1 mark)

  2. Of the female students surveyed, one is chosen at random. What is the probability that she uses pre-paid?   (1 mark)

Ten new male students are surveyed and all ten are on a plan. The set of data is updated to include this information.

  1. What percentage of the male students surveyed are now on a plan? Give your answer to the nearest per cent.   (1 mark)

Show Answers Only

a.    `580`

b.    `text(54%)`

c.    `text(42%)`

Show Worked Solution

a.    `text(# Students surveyed)=319+261=580`
 

b.    `Ptext{(Female uses prepaid)}`

`=text(# Females on prepaid)/text(Total females)=172/319=0.53918… ~~ 54\text{% (nearest %)}`
 

c.    `text(% Males on plan)` `=text(# Males on plan + 10)/text(Total males + 10)=(103+10)/(261+10)`
  `=113/271=0.4169…=42\text{%  (nearest %)}`

Probability, STD2 S2 2011 HSC 24b

A die was rolled 72 times. The results for this experiment are shown in the table.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Number obtained} \rule[-1ex]{0pt}{0pt} & \textit{Frequency} \\
\hline
\rule{0pt}{2.5ex} \ 1 \rule[-1ex]{0pt}{0pt} & 16 \\
\hline
\rule{0pt}{2.5ex} \ 2 \rule[-1ex]{0pt}{0pt} & 11 \\
\hline
\rule{0pt}{2.5ex} \ 3 \rule[-1ex]{0pt}{0pt} & \textbf{A} \\
\hline
\rule{0pt}{2.5ex} \ 4 \rule[-1ex]{0pt}{0pt} & 8 \\
\hline
\rule{0pt}{2.5ex} \ 5 \rule[-1ex]{0pt}{0pt} & 12 \\
\hline
\rule{0pt}{2.5ex} \ 6 \rule[-1ex]{0pt}{0pt} & 15 \\
\hline
\end{array}

  1. Find the value of `A`.   (1 mark)

  2. What was the relative frequency of obtaining a 4.   (1 mark)

  3. If the die was unbiased, which number was obtained the expected number of times?   (1 mark)

Show Answers Only

a.    \(10\)

b.    \(\dfrac{1}{9}\)

c.    \(5\)

Show Worked Solution

a.    \(\text{Since die rolled 72 times:}\)

\(A=72-(16+11+8+12+15)=10\)

♦ Mean mark 38%
IMPORTANT: Many students confused ‘relative frequency’ with ‘frequency’ and incorrectly answered 8.

 
b.    \(\text{Relative frequency of 4}=\dfrac{8}{72}=\dfrac{1}{9}\)
 

c.    \(\text{Expected frequency of any number}=\dfrac{1}{6}\times 72=12\)

\(\therefore\ \text{5 was obtained the expected number of times.}\)

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

a.    `34`

b.    `15/34`

Show Worked Solution

a.    `text(# People)=5+15+10+3+1=34`
  

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

Algebra, STD2 A4 2012 HSC 30b

A golf ball is hit from point `A` to point `B`, which is on the ground as shown. Point `A` is 30 metres above the ground and the horizontal distance from point `A` to point `B` is  300 m.
 

The path of the golf ball is modelled using the equation 

`h = 30 + 0.2d-0.001d^2` 

where 

`h` is the height of the golf ball above the ground in metres, and 

`d` is the horizontal distance of the golf ball from point `A` in metres.

The graph of this equation is drawn below.

  

  1. What is the maximum height the ball reaches above the ground?   (1 mark)

  2. There are two occasions when the golf ball is at a height of 35 metres.

     

    What horizontal distance does the ball travel in the period between these two occasions?   (1 mark)

  3. What is the height of the ball above the ground when it still has to travel a horizontal distance of 50 metres to hit the ground at point `B`?   (1 mark)

  4. Only part of the graph applies to this model.
  5. Find all values of `d` that are not suitable to use with this model, and explain why these values are not suitable.   (2 marks)

Show Answers Only

a.    `40 text(m)`

b.    `140 text(m)`

c.    `text(17.5 m)`

d.    `d < 0\ text(and)\ d>300`

Show Worked Solution

a.    `text(Max height) = 40 text(m)`
 

b.    `text(From graph:)`

`h = 35\ text(when)\ \ x = 30\ \ text(and)\ \ x = 170`

`:.\ text(Horizontal distance)= 170-30= 140\ text(m)`
 

c.    `text(Ball hits ground at)\ \ x = 300`

MARKER’S COMMENT: Responses for (c) in the range  `17<=\ h\ <=18`  were deemed acceptable.

`text(Find)\ y\ text(when)\ \ x = 250:`

`text(From graph,)\ y = 17.5\ text(m)`

`:.\ text(Height of ball is 17.5 m at a horizontal distance of 50 m before)\ B.`
 

d.    `text(Values of)\ d\ text(not suitable:)`

♦♦♦ Mean mark (d) 12%
MARKER’S COMMENT: Many students did not refer to the domain `d>300` as unsuitable to the model.

`text(If)\ d < 0 text(, it assumes the ball is hit away from point)\ B.`

`\text{This is not the case in our example.}`
 

`text(If)\ d > 300 text(,)\ h\ text(becomes negative which is not possible)`

`\text(– i.e. the ball cannot go below ground level.)`

Algebra, STD2 A2 2012 HSC 13 MC

Conversion graphs can be used to convert from one currency to another.  
  

 
  

Sarah converted  60  Australian dollars into Euros. She then converted all of these Euros
into New Zealand dollars.

How much money, in New Zealand dollars, should Sarah have? 

  1. $26  
  2. $45
  3. $78
  4. $135
Show Answers Only

`C`

Show Worked Solution

`text(Using the graphs)`

`$60\ text(Australian)=46\ text(Euro)`

`46 \ text(Euro)=$78\  text(New Zealand)`

`=>  C`

Statistics, STD2 S1 2013 HSC 8 MC

A high school has 100 students in each year group, Year 7 to Year 12. A survey is to be conducted to determine the average number of text messages sent per month by students at the school.

Which of the following would provide the most representative sample for this survey?

  1. All Year 7 students
  2. All physics students in Year 11 and 12
  3. 20 students chosen at random from each year group
  4. 120 students chosen at random from the school roll
Show Answers Only

`C`

Show Worked Solution

`text(The best sample would have an equal amount)`

`text(of people in each year randomly selected.)`

`=>C`

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, 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 2011 HSC 5b

To test some forensic science students, an object has been left in the park. At 10am the temperature of the object is measured to be 30°C. The temperature in the park is a constant 22°C. The object is moved immediately to a room where the temperature is a constant 5°C.

The temperature of the object in the room can be modelled by the equation

`T=5+25e^(-kt)`,

where  `T`  is the temperature of the object in degrees Celcius, `t` is the time in hours since the object was placed in the room and `k` is a constant.

After one hour in the room the temperature of the object is 20°C.

  1. Show that  `k=ln(5/3)`    (2 marks)

  2. In a similar manner, the temperature of the object in the park before it was discovered can be modelled by an equation in the form  `T=A+Be^(-kt)`,  with the same constant  `k=ln(5/3)`.

     

    Find the time of day when the object had a temperature of 37°C.    (3 marks)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `8:46\ text(am)`
Show Worked Solution

i.   `T=5+25e^(-kt)`

`text(At)\ t=1,  T=20`

`20` `=5+25e^(-k)`
`25e^(-k)` `=15`
`e^(-k)` `=15/25=3/5`
`lne^(-k)` `=ln(3/5)`
`-k` `=ln(3/5)`
`k` `=-ln(3/5)`
  `=ln(3/5)^-1`
  `=ln(5/3)\ \ \ text(… as required)`

 

ii.   `T=A+Be^(-kt)\ \ text(where)\ \ k=ln(5/3)`

`text(S)text(ince park temp is a constant 22°`

`=>A=22`

`:.\ T=22+Be^(-kt)`
 

`text(At)\ t=0\ \ text{(10 am),}\  \ T=30`

`text(i.e.)\ \ 30=22+Be^0`

`=>B=8`

`:.\ T=22+8e^(-kt)`
 

`text(Find)\ \ t\ \ text(when)\ \ T=37`

MARKER’S COMMENT: Many students failed to recognise that a negative value for `t` was not invalid but represented the time before 10am.
`37` `=22+8e^(-kt)`
`8e^(-kt)` `=15`
`lne^(-kt)` `=ln(15/8)`
`-kt` `=ln(15/8)`
`:. t` `=-1/k ln(15/8),\  text(where)\ k=ln(5/3)`
  `=ln(15/8)/ln(5/3)`
  `=-1.23057` ….
  `=- text{1h 14m  (nearest minute)}`

 
`:.\ text{The object had a temp of  37°C  at  8:46 am}`

`text{(1h 14m  before 10am).}`

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.)`

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

a.    `$15\ text(million)`

b.    `$3327.5\  text(million)`

Show Worked Solutions
a.    `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.

 

b.    `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 12c

Jay is making a pattern using triangular tiles. The pattern has 3 tiles in the first row, 5 tiles in the second row, and each successive row has 2 more tiles than the previous row.

2012 12c

  1. How many tiles would Jay use in row 20?   (2 marks)

  2. How many tiles would Jay use altogether to make the first 20 rows?   (1 mark)

  3. Jay has only 200 tiles. How many complete rows of the pattern can Jay make?   (2 marks)

Show Answers Only

a.    `41`

b.    `440`

c.    `13\ text(rows)`

Show Worked Solutions
a.     `T_1` `=a=3`
  `T_2` `=a+d=5`
  `T_3` `=a+2d=7`

 
`=>\ text(AP where)\ \ a=3,\ \ d=2`

`\ \ \ \ \ vdots`

`T_20` `=a+19d`
  `=3+19(2)`
  `=41`

  
`:.\ text(Row 20 has 41 tiles.)`

 

MARKER’S COMMENT: Better responses stated the formula BEFORE any calculations were performed. This enabled students to get some marks if they made an error in their working.
b.    `S_20` `=\ text(the total number of tiles in first 20 rows)`
`S_20` `=n/2(a+l)`
  `=20/2(3+41)`
  `=440`

  
`:.\ text(There are 440 tiles in the first 20 rows.)`
  

c.   `text(If Jay only has 200 tiles, then)\ \ S_n<=200`

NOTE: Examiners often ask questions requiring `n` to be found using the formula `S_n=n/2[2a+(n-1)d]` as this requires the solving of a quadratic, and interpretation of the answer.
`n/2(2a+(n-1)d)` `<=200`
`n/2(6+2n-2)` `<=200`
`n(n+2)` `<=200`
`n^2+2n-200` `<=0`
`n` `=(-2+-sqrt(4+4*1*200))/(2*1)`
  `=(-2+-sqrt804)/2`
  `=-1+-sqrt201`
  `=13.16\ \ text{(answer must be positive)}`

  
`:.\ text(Jay can complete 13 rows.)`