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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).

     i. Find, correct to two decimal places, the values of \(a, b, c\) and \(d\).   (3 mark)

    ii. 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)

  2. 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)
    1. 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.  \(\text{Max instantaneous rate of change occurs when }f^{\prime\prime}(t)=0\)

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

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

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

Graphs, MET2 2007 VCAA 15 MC

The graph of the function  `f: [0, oo) -> R`  where  `f(x) = 3x^(5/2)`  is reflected in the `x`-axis and then translated 3 units to the right and 4 units down.

The equation of the new graph is

`y = 3(x - 3)^(5/2) + 4`

`y = -3 (x - 3)^(5/2) - 4`

`y = -3 (x + 3)^(5/2) - 1`

`y = -3 (x - 4)^(5/2) + 3`

`y = 3(x - 4)^(5/2) + 3`

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ y= 3x^(5/2)`

`text(Reflect in the)\ x text(-axis:)`

`y= – 3x^(5/2)`

 

`text(Translate 3 units to the right:)`

`y=- 3(x-3)^(5/2)`

 

`text(Translate 4 units down:)`

`y=- 3(x-3)^(5/2) – 4`

`=>   B`

Graphs, MET2 2008 VCAA 9 MC

The transformation  `T: R^2 -> R^2` with rule

`T([(x), (y)]) = [(4, 0), (0, -2)] [(x), (y)] + [(1), (3)]`

maps the curve with equation  `y = x^3`  to the curve with equation

  1. `y = (-(x - 1)^3)/32 + 3`
  2. `y = (-(4x + 1)^3 + 3)/2`
  3. `y = (-(x + 1)^3)/32 - 3`
  4. `y = ((1 - x)^3)/64 - 3`
  5. `y = ((4x - 1)^3 + 3)/2`
Show Answers Only

`A`

Show Worked Solution

`text(Let)\ \ (x′,y′)\ \ text(be the transformation of)\ \ (x,y).`

♦ Mean mark 38%.
`x′` `= 4x + 1`
`x` `=(x′ – 1)/4`
`y′` `= – 2y + 3`
`y` `=(y′ – 3)/(- 2)`

 

`text(Substitute)\ \ x, y\ \ text(into)\ \ y = x^3:`

`(y′ – 3)/(- 2)` `= ((x′ – 1)/4)^3`
`y′ – 3` `= (- 2)/4^3 (x′ – 1)^3`
`y` `= (- (x – 1)^3)/32 + 3`

`=>   A`

Functions, MET1 2010 VCAA 6

The transformation  `T: R^2 -> R^2` is defined by
 

`T([(x), (y)]) = [(3, 0), (0, 2)] [(x), (y)] + [(– 1), (4)].`
 

The image of the curve  `y = 2x^2 + 1`  under the transformation `T` has equation  `y = ax^2 + bx + c.`

Find the values of `a, b,` and `c.`   (3 marks)

Show Answers Only

`a = 4/9,\ \ \ b = 8/9,\ \ \ c = 58/9`

Show Worked Solution

`text(Expanding the matrix equation:)`

`x^{′}` `= 3x-1` `qquad qquad y^{′}` `= 2y + 4`
`x` `= (x^{′}+1)/3` `qquad qquad y` `= (y^{′} -4)/2`

 

`text(Substitute)\ \ x, y\ \ text(into)\ \ y = 2x^2 + 1`

♦ Mean mark 40%.
`(y^{′}-4)/2` `= 2 ((x^{′} + 1)/3)^2 + 1`
`y^{′}-4` `= 4/9 ((x^{′})^2 + 2x^{′} + 1) + 2`
`y^{′}` `= 4/9 (x^{′})^2 + 8/9 x^{′} + (4/9 + 6)`
`:. y^{′}` `= 4/9 (x^{′})^2 + 8/9 x^{′} + 58/9`

 

`:. a = 4/9,\ \ \ b = 8/9,\ \ \ c = 58/9`

Graphs, MET2 2013 VCAA 20 MC

A transformation `T: R^2 -> R^2, T ([(x), (y)]) = [(1, 0), (0, -1)] [(x), (y)] + [(5), (0)]` maps the graph of a function `f` to the graph of `y = x^2, x in R.`

The rule of `f` is

  1. `f(x) = -(x + 5)^2`
  2. `f(x) = (5 - x)^2`
  3. `f(x) = -(x - 5)^2`
  4. `f(x) = -x^2 + 5`
  5. `f(x) = x^2 - 5`
Show Answers Only

`A`

Show Worked Solution

`text(Transformations are:)`

`x′` `=x+5`
`x` `=x′ -5`
`y′` `=-y`
`y` `=-y′`

 

`-y` `=(x+5)^2`
`f(x)` `= – (x+5)^2`
`-y′` `=-((x′-5)+5)^2`
`y′` `=(x′)^2`

 

`=>   A`