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

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

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

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

Calculus, MET1 2023 VCAA SM-Bank 5

Let  \(f: R \rightarrow R\),  where  \(f(x)=2-x^2\).

  1. Calculate the average rate of change of \(f\) between \(x=-1\) and \(x=1\).  (1 mark)

  2. Calculate the average value of \(f\) between \(x=-1\) and \(x=1\).  (2 marks)

  3. Four trapeziums of equal width are used to approximate the area between the functions  \(f(x)=2-x^2\)  and the \(x\)-axis from \(x=-1\) to \(x=1\).
  4. The heights of the left and right edges of each trapezium are the values of \(y=f(x)\), as shown in the graph below.

  1. Find the total area of the four trapeziums.  (2 marks)

Show Answers Only

a.    \(0\)

b.    \(\dfrac{5}{3}\)

c.    \(\dfrac{13}{4}\)

Show Worked Solution
a.     \(\text{Average rate of change}\) \(=\dfrac{f(1)-f(-1)}{1-(-1)}\)
    \(\dfrac{1-1}{2}\)
    \(=0\)

 

b.    \(\text{Avg value}\) \(=\dfrac{1}{1-(-1)}\displaystyle\int_{-1}^{1} \Big(2-x^2\Big)\,dx\)
    \(=\dfrac{1}{2}\displaystyle\Bigg[2x-\dfrac{x^3}{3}\displaystyle\Bigg]_{-1}^{1}\)
    \(=\dfrac{1}{2}\displaystyle\Bigg[\Bigg(2.(1)-\dfrac{(1)^3}{3}\Bigg)-\Bigg(2.(-1)-\dfrac{(-1)^3}{3}\Bigg)\Bigg]\)
    \(=\dfrac{1}{2}\times\dfrac{10}{3}=\dfrac{5}{3}\)

 

c.     \(\text{Total Area}\) \(=2\times\ \text{Area from 0 to 1}\)
    \(=2\times \dfrac{h}{2}\Big(f(0)+2.f(0.5)+f(1)\Big)\quad \text{where }h=\dfrac{1}{2}\)
    \(=\dfrac{1}{2}\Big(2+2\times\dfrac{7}{4}+1\Big)=\dfrac{13}{4}\)

Calculus, MET1 2013 VCAA 6

Let  `g: R -> R,\ \ g(x) = (a-x)^2`,  where `a` is a real constant.

The average value of `g` on the interval  `[-1, 1]`  is  `31/12.`

Find all possible values of `a.`   (3 marks)

Show Answers Only

`+- 3/2`

Show Worked Solution

`text(Solution 1)`

♦ Mean mark 39%.
MARKER’S COMMENT: “Far too many” students misunderstood “average value” in this context.
`1/(1-(-1)) int_-1^1 (a-x)^2 dx` `= 31/12`
`[ax^2-ax^2 + x^3/3]_-1^1` `= 31/6`
`[(a^2-a+1/3)-(-a^2-a-1/3)]` `=31/6`
`2a^2+2/3` `=31/6`
`a^2` `=27/4`
`:. a` `=+- 3/2`

 

`text(Solution 2)`

`1/(1-(-1)) int_-1^1 (a-x)^2 dx` `= 31/12`
`1/2 [(a-x)^3/-3]_-1^1` `= 31/12`
`[(a-x)^3]_-1^1` `=-31/2`
`(a-1)^3-(a + 1)^3` `=-31/2`

`(a^3-3a^2 + 3a-1)-(a^3 + 3a^2 + 3a + 1)=-31/2`

`-6a^2-2` `=-31/2`
`6a^2` `= 27/2`
`a^2` `= 9/4`
`:. a` `= +- 3/2`

Calculus, MET1 2015 VCAA 4

Consider the function  `f:[-3,2] -> R, \ \ f(x) = 1/2(x^3 + 3x^2-4)`.

  1. Find the coordinates of the stationary points of the function.   (2 marks)

The rule for  `f` can also be expressed as  `f(x) = 1/2(x-1)(x + 2)^2`.

  1. On the axes below, sketch the graph of  `f`, clearly indicating axis intercepts and turning points.

     

    Label the end points with their coordinates.   (2 marks)


      

    met1-2015-vcaa-q4
     

  2. Find the average value of  `f` over the interval  `0<=x<=2.`   (2 marks)

Show Answers Only
  1. `(0,-2), (-2,0)`
  2.  

     

    met1-2015-vcaa-q4-answer

  3. `1`
Show Worked Solution

a.   `text(Stationary points when)\ \ f^{prime}(x)=0,`

`1/2(3x^2 + 6x)` `= 0`
`3x(x + 2)` `= 0`

 

`:. x = 0, -2`

 

`:.\ text(Coordinates of stationary points:)`

`(0, -2), (-2,0)`

 

b.    met1-2015-vcaa-q4-answer
♦ Part (c) mean mark 50%.
MARKER’S COMMENT: Most students recalled the average value definition but then did not integrate correctly.

 

c.    `text(Avg value)` `= 1/(2-0) int_0^2 f(x) dx`
    `= 1/2 int_0^2 1/2(x^3 + 3x^2-4)dx`
    `= 1/4[1/4x^4 + x^3-4x]_0^2`
    `= 1/4[(16/4 + 2^3-4(2))-0]`
    `= 1`