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Calculus, MET2 2024 VCAA 3

The points shown on the chart below represent monthly online sales in Australia.

The variable \(y\) represents sales in millions of dollars.

The variable \(t\) represents the month when the sales were made, where \(t=1\) corresponds to January 2021, \(t=2\) corresponds to February 2021 and so on.

  1. A cubic polynomial \(p ;(0,12] \rightarrow R, p(t)=a t^3+b t^2+c t+d\) can be used to model monthly online sales in 2021.

    The graph of \(y=p(f)\) is shown as a dashed curve on the set of axes above.

    It has a local minimum at (2,2500) and a local maximum at (11,4400).

    1. Find, correct to two decimal places, the values of \(a, b, c\) and \(d\).  (3 marks)
    1. Let \(q:(12,24] \rightarrow R, q(t)=p(t-h)+k\) be a cubic function obtained by translating \(p\), which can be used to model monthly online sales in 2022.

      Find the values of \(h\) and \(k\) such that the graph of \(y=q(t)\) has a local maximum at \((23,4750)\).  (2 marks)

  1. Another function \(f\) can be used to model monthly online sales, where

    \(f:(0,36] \rightarrow R, f(t)=3000+30 t+700 \cos \left(\dfrac{\pi t}{6}\right)+400 \cos \left(\dfrac{\pi t}{3}\right)\)

    Part of the graph of \(f\) is shown on the axes below.

    1. Complete the graph of \(f\) on the set of axes above until December 2023, that is, for \(t \in(24,36]\).Label the endpoint at \(t=36\) with its coordinates.  (2 marks)
    2. The function \(f\) predicts that every 12 months, monthly online sales increase by \(n\) million dollars.

      Find the value of \(n\).  (1 mark)

    1. Find the derivative \(f^{\prime}(t)\)  (1 mark)
    1. Hence, find the maximum instantaneous rate of change for the function \(f\), correct to the nearest million dollars per month, and the values of \(t\) in the interval \((0,36]\) when this maximum rate occurs, correct to one decimal place.  (2 marks)
Show Answers Only

ai.   \(a\approx -5.21, b\approx 101.65, c\approx -344.03, d\approx 2823.18\)

aii.  \(h=12, k=350\)

bi.  

bii.  \( n=360\)

biii. \(f^{\prime}(t)=30-\dfrac{350\pi}{3}\sin\left(\dfrac{\pi t}{6}\right)-\dfrac{400\pi}{3}\sin\left(\dfrac{\pi t}{3}\right)\)

biv. \(\text{Maximum rate occurs at }t=10.2, 22.2, 34.2\)

\(\text{Maximum rate}\ \approx 725\ \text{million/month}\)

Show Worked Solution
ai.   \(\text{Given}\ p(2)=2500, p(11)=4400, p^{\prime}(2)=0\ \text{and}\ p^{\prime}(11)=0\) 
  
\(p(t)=at^3+bt^2+ct+d\ \Rightarrow\ p^{\prime}(t)=3at^2+2bt+c\)
  
\(\text{Using CAS:}\)
    
\(\text{Solve}
\begin{cases}
8a+4b+2c+d=2500 \\
1331a+121b+11c+d=4400 \\
12a+4b+c=0  \\
363a+22b+c=0
\end{cases}\)
  

\(a\approx -5.21, b\approx 101.65, c\approx -344.03, d\approx 2823.18\)

aii. \(\text{Local maximim }p(t)\ \text{is}\ (11, 4400)\)

\(\therefore\ h\) \(=23-11=12\)
\(k\) \(=4750-4400=350\)

 

bi.   \(\text{Plotting points from CAS:}\)

\((24,4820), (26, 3930), (28, 3290), (30, 3600), (32, 3410), (34, 4170), (36, 5180)\)

bii.  \(\text{Using CAS: }\)

\(f(12)-f(0)\) \(=4460-4100=360\)
\(f(24)-f(12)\) \(=4820-4460=360\)
\(f(36)-f(24)\) \(=5180-4820=360\) 

\(\therefore\ n=360\)

biii.  \(f(t)\) \(=3000+30t+700\cos\left(\dfrac{\pi t}{6}\right)+400\cos\left(\dfrac{\pi t}{3}\right)\)
  \(f^{\prime}(t)\) \(=30-\dfrac{700\pi}{6}\sin\left(\dfrac{\pi t}{6}\right)-\dfrac{400\pi}{3}\sin\left(\dfrac{\pi t}{3}\right)\)
    \(=30-\dfrac{350\pi}{3}\sin\left(\dfrac{\pi t}{6}\right)-\dfrac{400\pi}{3}\sin\left(\dfrac{\pi t}{3}\right)\)

  
biv.  \(\text{Max instantaneous rate of change occurs when }f^{\prime\prime}(t)=0\)

\(\text{Maximum rate occurs at }t=10.2, 22.2, 34.2\)

\(\text{Maximum rate using CAS:}\)

\(f{\prime}(10.2)=f{\prime}(22.2)=f{\prime}(34.2)=725.396\approx 725\ \text{million/month}\)

Calculus, MET2 2022 VCAA 5

Consider the composite function `g(x)=f(\sin (2 x))`, where the function `f(x)` is an unknown but differentiable function for all values of `x`.

Use the following table of values for `f` and `f^{\prime}`.
 

  1. Find the value of `g\left(\frac{\pi}{6}\right)`.   (1 mark)

The derivative of `g` with respect to `x` is given by `g^{\prime}(x)=2 \cdot \cos (2 x) \cdot f^{\prime}(\sin (2 x))`.

  1. Show that `g^{\prime}\left(\frac{\pi}{6}\right)=\frac{1}{9}`.   (1 mark)

  2. Find the equation of the tangent to `g` at `x=\frac{\pi}{6}`.   (2 marks)

  3. Find the average value of the derivative function `g^{\prime}(x)` between `x=\frac{\pi}{8}` and `x=\frac{\pi}{6}`.   (2 marks)

  4. Find four solutions to the equation `g^{\prime}(x)=0` for the interval `x \in[0, \pi]`.   (3 marks)

Show Answers Only

a.    `3`

b.    `1/9`

c.    `y=1/9x+3-pi/54`

d.    `-48/pi`

e.    ` x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`

Show Worked Solution
 

a.  `g(pi/6)`
 `= f(sin(pi/3))`  
  `= f(sqrt3/2)`  
  `= 3`  

 

b.  `g\ ^{prime}(x)` `= 2\cdot\ cos(pi/3)\cdot\ f\ ^{prime}(sin(pi/3))`  
`g\ ^{prime}(pi/6)` `= 2 xx 1/2 xx f\ ^{prime}(sqrt3/2)`  
  `= 1/9`  

 

c.   `m = 1/9`  and  `g(pi/6) = 3`

`y  –  y_1` `= m(x  –  x_1)`  
`y  –  3` `= 1/9(x  –  pi/6)`  
`y` `= 1/9x + 3  –  pi/54`  

♦♦ Mean mark (c) 45%.
MARKER’S COMMENT: Some students did not produce an equation as required.

  
d.   The average value of `g^{\prime}(x)` between `x=\frac{\pi}{8}` and `x=\frac{\pi}{6}`

Average `= \frac{1}{\frac{\pi}{6}-\frac{\pi}{8}}\cdot\int_{\frac{\pi}{8}}^{\frac{\pi}{6}} g^{\prime}(x) d x`  
  `=24/pi \cdot[g(x)]_{\frac{\pi}{8}}^{\frac{\pi}{\6}}`  
  `= 24/pi \cdot(f(sqrt3/2) -f(sqrt2/2))`  
  `= 24/pi (3  –  5) = -48/pi`  

♦♦ Mean mark (d) 30%.
MARKER’S COMMENT: Those who used the Average Value formula were generally successful.
Some students substituted `g^{\prime}(x)`, not `g(x)`.

e.   `2 \cos (2 x) f^{\prime}(\sin (2 x)) = 0`

`:.\   2 \cos (2 x) = 0\ ….(1)`  or  ` f^{\prime}(\sin (2 x)) = 0\ ….(2)`

(1):   ` 2 \cos (2 x)`  `= 0`      `x \in[0, \pi]`
`\cos (2 x)` `= 0`      `2 x \in[0,2 \pi]`
`2x` `= pi/2 , (3pi)/2`  
`x` `= pi/4 , (3pi)/4`  
     
(2): ` f^{\prime}(\sin (2 x)) ` `= sqrt2/2`  
`2x` `= pi/4 , (3pi)/4`  
`x` `= pi/8 , (3pi)/8`  

  
`:. \  x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`


♦♦ Mean mark (e) 30%.
MARKER’S COMMENT: Some students were able to find `pi/4, (3pi)/4`. Some solved `2 cos(2x)=0` or `f^{\prime}(sin(2x))=0` but not both.

Calculus, MET2 2022 VCAA 2

On a remote island, there are only two species of animals: foxes and rabbits. The foxes are the predators and the rabbits are their prey.

The populations of foxes and rabbits increase and decrease in a periodic pattern, with the period of both populations being the same, as shown in the graph below, for all `t \geq 0`, where time `t` is measured in weeks.

One point of minimum fox population, (20, 700), and one point of maximum fox population, (100, 2500), are also shown on the graph.

The graph has been drawn to scale.
 

The population of rabbits can be modelled by the rule `r(t)=1700 \sin \left(\frac{\pi t}{80}\right)+2500`.

  1.   i. State the initial population of rabbits.   (1 mark)

  2.  ii. State the minimum and maximum population of rabbits.   (1 mark)

  3. iii. State the number of weeks between maximum populations of rabbits.   (1 mark)

The population of foxes can be modelled by the rule `f(t)=a \sin (b(t-60))+1600`.

  1. Show that `a=900` and `b=\frac{\pi}{80}`.   (2 marks)

  2. Find the maximum combined population of foxes and rabbits. Give your answer correct to the nearest whole number.   (1 mark)

  3. What is the number of weeks between the periods when the combined population of foxes and rabbits is a maximum?   (1 mark)

The population of foxes is better modelled by the transformation of `y=\sin (t)` under `Q` given by
 

  1. Find the average population during the first 300 weeks for the combined population of foxes and rabbits, where the population of foxes is modelled by the transformation of `y=\sin(t)` under the transformation `Q`. Give your answer correct to the nearest whole number.   (4 marks)

Over a longer period of time, it is found that the increase and decrease in the population of rabbits gets smaller and smaller.

The population of rabbits over a longer period of time can be modelled by the rule

  1. Find the average rate of change between the first two times when the population of rabbits is at a maximum. Give your answer correct to one decimal place.   (2 marks)

  2. Find the time, where `t>40`, in weeks, when the rate of change of the rabbit population is at its greatest positive value. Give your answer correct to the nearest whole number.   (2 marks)

  3. Over time, the rabbit population approaches a particular value.
  4. State this value.   (1 mark)


Show Answers Only

ai.   `r(0)=2500`

aii.  Minimum population of rabbits `= 800`

Maximum population of rabbits `= 4200`

aiii. `160`  weeks

b.    See worked solution.

c.    `~~ 5339` (nearest whole number)

d.    Weeks between the periods is 160

e.    `~~ 4142` (nearest whole number)

f.    Average rate of change `= – 3.6` rabbits/week (1 d.p.)

g.    `t = 156` weeks (nearest whole number)  

h.    ` s → 2500`

Show Worked Solution

ai.  Initial population of rabbits

From graph when `t=0, \ r(0) = 2500`

Using formula when `t=0`

`r(t)` `= 1700\ sin \left(\frac{\pi t}{80}\right)+2500`  
`r(0)` `= 1700\ sin \left(\frac{\pi xx 0}{80}\right)+2500 = 2500`  rabbits  


aii. From graph,

Minimum population of rabbits `= 800`

Maximum population of rabbits `= 4200`

OR

Using formula

Minimum is when `t = 120`

`r(120) = 1700\ sin \left(\frac{\pi xx 120}{80}\right)+2500 = 1700 xx (-1) + 2500 = 800`

Maximum is when `t = 40`

`r(40) = 1700\ sin \left(\frac{\pi xx 40}{80}\right)+2500 = 1700 xx (1) + 2500 = 4200`

 
aiii. Number of weeks between maximum populations of rabbits `= 200  –  40 = 160`  weeks

 
b.  Period of foxes = period of rabbits = 160:

`frac{\2pi}{b} = 160`

`:.\  b = frac{\2pi}{160} = frac{\pi}{80}` which is the same period as the rabbit population.

Using the point `(100 , 2500)`

Amplitude when `b = frac{\pi}{80}`: 

`f(t)` `=a \ sin (pi/80(t-60))+1600`  
`f(100)` `= 2500`  
`2500` `= a \ sin (pi/80(100-60))+1600`  
`2500` `= a \ sin (pi/2)+1600`  
`a` `= 2500  –  1600 = 900`  

 
`:.\  f(t)= 900 \ sin (pi/80)(t  –  60) + 1600`

  

c.    Using CAS find `h(t) = f(t) + r(t)`:

`h(t):=900 \cdot \sin \left(\frac{\pi}{80} \cdot(t-60)\right)+1600+1700\cdot \sin \left(\frac{\pit}{80}\right) +2500`

  
`text{fMax}(h(t),t)|0 <= t <= 160`      `t = 53.7306….`
  

`h(53.7306…)=5339.46`

  
Maximum combined population `~~ 5339` (nearest whole number)


♦♦ Mean mark (c) 40%.
MARKER’S COMMENT: Many rounding errors with a common error being 5340. Many students incorrectly added the max value of rabbits to the max value of foxes, however, these points occurred at different times.

d.    Using CAS, check by changing domain to 0 to 320.

`text{fMax}(h(t),t)|0 <= t <= 320`     `t = 213.7305…`
 
`h(213.7305…)=5339.4568….`
 
Therefore, the number of weeks between the periods is 160.
  

e.    Fox population:

`t^{\prime} = frac{90}{pi}t + 60`   →   `t = frac{pi}{90}(t^{\prime} – 60)`

`y^{\prime} = 900y+1600`   →   `y = frac{1}{900}(y^{\prime} – 1600)`

`frac{y^{\prime} – 1600}{900} = sin(frac{pi(t^{\prime} – 60)}{90})`

`:.\  f(t) = 900\ sin\frac{pi}{90}(t  –  60) + 1600`

  
Average combined population  [Using CAS]   
  
`=\frac{1}{300} \int_0^{300} left(\900 \sin \left(\frac{\pi(t-60)}{90}\right)+1600+1700\ sin\ left(\frac{\pi t}{80}\right)+2500\right) d t`
  

`= 4142.2646…..  ~~ 4142` (nearest whole number)


♦ Mean mark (e) 40%.
MARKER’S COMMENT: Common incorrect answer 1600. Some incorrectly subtracted `r(t)`. Others used average rate of change instead of average value.

f.   Using CAS

`s(t):= 1700e^(- 0.003t) dot\sin\frac{pit}{80} + 2500`
 

`text{fMax}(s(t),t)|0<=t<=320`        `x = 38.0584….`
 

`s(38.0584….)=4012.1666….`
 

`text{fMax}(s(t),t)|160<=t<=320`     `x = 198.0584….`
 

`s(198.0584….)=3435.7035….`
 

Av rate of change between the points

`(38.058 , 4012.167)`  and  `(198.058 , 3435.704)`

`= frac{4012.1666…. – 3435.7035….}{38.0584…. – 198.0584….} = – 3.60289….`
 

`:.` Average rate of change `= – 3.6` rabbits/week (1 d.p.)


♦ Mean mark (f ) 45%.
MARKER’S COMMENT: Some students rounded too early.
`frac{s(200) -s(40)}{200 – 40}` was commonly seen.
Some found average rate of change between max and min populations.

g.   Using CAS

`s^(”)(t) = 0` , `t = 80(n – 0.049) \ \forall n \in Z`

After testing `n = 1, 2, 3, 4` greatest positive value occurs for `n = 2` 

`t` `= 80(n  –  0.049)`  
  `= 80(2  –  0.049)`  
  `= 156.08`  

 
`:. \ t = 156` weeks (nearest whole number)


♦♦♦ Mean mark (g) 25%.
MARKER’S COMMENT: Many students solved `frac{ds}{dt}=0`.
A common answer was 41.8, as `s(156.11…)=41.79`.
Another common incorrect answer was 76 weeks.

h.   As `t → ∞`, `e^(- 0.003t) → 0`

`:.\ s → 2500`

Calculus, MET1 2022 VCAA 8

Part of the graph of `y=f(x)` is shown below. The rule `A(k)=k \ sin(k)` gives the area bounded by the graph of `f`, the horizontal axis and the line `x=k`.
 

  1. State the value of `A\left(\frac{\pi}{3}\right)`.   (1 mark)

  2. Evaluate `f\left(\frac{\pi}{3}\right)`.   (2 marks)

  3. Consider the average value of the function `f` over the interval `x \in[0, k]`, where `k \in[0,2]`.
  4. Find the value of `k` that results in the maximum average value.   (2 marks)

Show Answers Only

a.    ` (sqrt(3)pi)/6`

b.    `(3sqrt(3) +pi)/6`

c.    `k = pi/2`

Show Worked Solution
a.  `A(pi/3)` `= pi/3sin(pi/3)`  
  `= pi/3 xx sqrt(3)/2`  
  `= (sqrt(3)pi)/6`  
b.   `f(k)` `= A^{\prime}(k)`  
`f(k)` `=d/dx(k\ sin\ k)`  
  `= sin\ k + k\ cos\ k`  
`f(pi/3)` `= sin\ pi/3 + pi/3\ cos\ pi/3`  
  `= sqrt(3)/2 + pi/3 xx 1/2`  
  `= sqrt(3)/2 + pi/6`  
  `=(3sqrt(3) +pi)/6`  

♦♦♦ Mean mark (b) 25%.
MARKER’S COMMENT: Common error was giving derivative of `A(k)` as `k\ cos (k)`.
c.   Average value `= (A(k))/k`  
  `= 1/k(int_0^k f(x) d x)`  
  `= 1/k[x\ sin (x)]_0^k`  
  `= 1/k[k\ sin (k)] = sin\ k`  

 
`:.` Average value has a maximum value of 1 when `k = pi/2`


♦♦♦ Mean mark (c) 25%.
MARKER’S COMMENT: Students often chose unnecessarily complicated methods or inconsistently applied nomenclature when attempting this question.

Calculus, MET1 2006 ADV 2aii

Differentiate with respect to `x`:

Let  `y=sin x/(x + 1)`.  Find  `dy/dx `.   (2 marks)

Show Answers Only

`dy/dx = {cos x (x + 1) – sin x} / (x + 1)^2`

Show Worked Solution

`y = sinx/(x + 1)`

`d/dx (u/v) = (u^{\prime} v – uv^{\prime})/v^2`

`u` `= sin x` `v` `= x + 1`
`u^{\prime}` `= cos x` `\ \ \ v^{\prime}` `= 1`

 

`:.dy/dx = {cos x (x + 1) – sin x} / (x + 1)^2`

Calculus, MET1 2019 VCAA 1b

Let  `g: R text(\ {−1}) -> R,\ \ g(x) = (sin(pi x))/(x + 1)`.

Evaluate  `g prime(1)`.  (2 marks)

Show Answers Only

`-pi/2`

Show Worked Solution
  `u = sin(pi x)` `v = x + 1`
  `u prime = pi cos(pi x)` `v prime = 1`

 

`g prime(x)` `= (v u prime – u v prime)/v^2`
  `= ((x + 1) ⋅ pi cos(pi x) – sin (pi x))/(x + 1)^2`
`g prime(1)` `= (2 pi cos(pi) – sin(pi))/2^2`
  `= (2 pi(-1) – 0)/4`
  `= -pi/2`

Calculus, MET1 2011 VCAA 1b

If  `g(x) = x^2 sin (2x)`,  find  `g prime (pi/6).`  (2 marks)

Show Answers Only

`(sqrt 3 pi)/6 + pi^2/36`

Show Worked Solution

`g(x) = x^2 sin (2x)`

MARKER’S COMMENT: Incorrect determination of exact trig values lost many students marks here.

`text(Using Product Rule:)`

`(fh)′` `= f′ h + fh′`
`g prime (x)` `= 2 x sin (2x) + 2x^2 cos (2x)`
   
`:. g prime (pi/6)` `= 2 (pi/6) sin (pi/3) + 2 (pi/6)^2 cos (pi/3)`
  `= pi/3 xx sqrt 3/2 + pi^2/18 xx 1/2`
  `= (sqrt 3 pi)/6 + pi^2/36`

Calculus, MET1 2012 VCAA 1b

If  `f(x) = x/(sin(x))`,  find  `f prime (pi/2).`  (2 marks)

Show Answers Only

`1`

Show Worked Solution

`text(Using Quotient Rule:)`

`(h/g)′` `= (h′ g – h g′)/g^2`
`f prime (x)` `= (1 xx sin (x) – x cos (x))/(sin x)^2`
`:. f prime (pi/2)` `= (sin (pi/2) – pi/2 xx cos (pi/2))/(sin(pi/2))^2`
  `= (1 – 0)/1^2`
  `= 1`

Calculus, MET1 SM-Bank 20

If   `f(x)= 2 sin 3x - 3 tan x`, find  `f′(0)`.   (2 marks)

Show Answers Only

`3`

Show Worked Solution
`y` `= 2 sin 3x – 3 tan x`
`(dy)/(dx)` `= 6 cos 3x – 3 sec^2 x`

 

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

`(dy)/(dx)` `= 6 cos 0 – 3 sec^2 0`
  `= 6 (1) – 3/(cos^2 0)`
  `= 6 – 3`
  `= 3`