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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}`.

`\quad x \quad` `\quad\quad 1/2\quad\quad` `\quad\quad(sqrt{2})/2\quad\quad` `\quad\quad(sqrt{3})/2\quad\quad`
`f(x)` `-2` `5` `3`
`\quad\quad f^{prime}(x)\quad\quad` `7` `0` `1/9`

 

  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 11 MC

If `\frac{d}{d x}(x \cdot \sin(x))=\sin (x)+x \cdot \cos(x)`, then `\frac{1}{k} \int x \ cos(x)dx` is equal to

  1. `k\left(x \cdot \sin (x)-\int \sin (x) d x\right)+c`
  2. `\frac{1}{k} x \cdot \sin (x)-\int \sin (x) d x+c`
  3. `\frac{1}{k}\left(x \cdot \sin (x)-\int \sin (x) d x\right)+c`
  4. `\frac{1}{k}(x \cdot \sin (x)-\sin (x))+c`
  5. `\frac{1}{k}\left(\int x \cdot \sin (x) d x-\int \sin (x) d x\right)+c`
Show Answers Only

`C`

Show Worked Solution

Given `\frac{d}{d x}(x \cdot \sin(x))=\sin (x)+x \cdot \cos(x)`, then

`x\cdot \cos (x)` `=\frac{d}{d x}(x\cdot \sin (x))-\sin (x)`  
`\frac{1}{k} \int x \cos (x) d x` `= \frac{1}{k}\left(\int \frac{d}{d x}(x \cdot \sin (x)) d x-\int \sin (x) d x\right)`  
  `= \frac{1}{k}\left(x \cdot \sin (x)-\int \sin (x)d x\right)+c`  

 
`=> C`

Functions, MET2 2023 VCAA 2

The following diagram represents an observation wheel, with its centre at point \(P\). Passengers are seated in pods, which are carried around as the wheel turns. The wheel moves anticlockwise with constant speed and completes one full rotation every 30 minutes.When a pod is at the lowest point of the wheel (point \(A\)), it is 15 metres above the ground. The wheel has a radius of 60 metres.
 

Consider the function \(h(t)=-60\ \cos(bt)+c\) for some \(b, c \in R\), which models the height above the ground of a pod originally situated at point \(A\), after time \(t\) minutes.

  1. Show that \(b=\dfrac{\pi}{15}\) and \(c=75\).   (2 marks)
  2. Find the average height of a pod on the wheel as it travels from point \(A\) to point \(B\).
  3. Give your answer in metres, correct to two decimal places.   (2 marks)
  4. Find the average rate of change, in metres per minute, of the height of a pod on the wheel as it travels from point \(A\) to point \(B\).   (1 mark)

After 15 minutes, the wheel stops moving and remains stationary for 5 minutes. After this, it continues moving at double its previous speed for another 7.5 minutes.

The height above the ground of a pod that was initially at point \(A\), after \(t\) minutes, can be modelled by the piecewise function \(w\):
 

\(w(t) = \begin {cases}
h(t)         &\ \ 0 \leq t < 15 \\
k         &\ \ 15 \leq t < 20 \\
h(mt+n) &\ \ 20\leq t\leq 27.5
\end{cases}\)

 
where \(k\geq 0, m\geq 0\) and \(n \in R\).

  1.   i.State the values of \(k\) and \(m\).   (1 mark)
       
     ii. Find all possible values of \(n\).   (2 marks)

      

    iii. Sketch the graph of the piecewise function \(w\) on the axes below, showing the coordinates of the endpoints.   (3 marks)


     
  1.  
Show Answers Only

a.    \(\text{See worked solution}\)

b.    \(\approx 36.80\ \text{m}\)

c.    \(8\)

d.i.  \(k=135, m=2\)

d.ii. \(n=30p+5,\ p\in Z\)

d.iii. 

Show Worked Solution
a.    \(\text{Period:}\) \(\dfrac{2\pi}{b}\) \(=30\)
    \(\therefore\ b\) \(=\dfrac{\pi}{15}\)

  
\(\text{Given }h(t)=-60\cos(bt)+c,\ \text{evaluate when }t=0, h=15\ \text{to find }c.\)

\(-60\ \cos(0)+c\)  \(=15\)
\(c\) \(=75\)

  \(\therefore\ h(t)=-60\cos(\dfrac{\pi t}{15})+75\)

b.     \(\text{Average height}\) \(=\dfrac{1}{\frac{30}{4}-0}\displaystyle\int_0^{\frac{30}{4}}\Bigg(-60\cos\Bigg(\dfrac{\pi}{15}t\Bigg)+75\Bigg)\,dt\)
    \(=\dfrac{2}{15}\left[75t-\dfrac{900\ \sin(\frac{\pi}{15}t)}{\pi}\right]_{0}^{7.5}\)
    \(=\dfrac{2}{15}\Bigg[75\times 7.5-\dfrac{900\ \sin(\frac{\pi}{15}\times 7.5)}{\pi}\Bigg]-\Bigg[0\Bigg]\)
    \(=\dfrac{75\pi-120}{\pi}=\dfrac{15(5\pi-8)}{\pi}\)
    \(=36.802\dots\approx 36.80\ \text{m}\)
 
♦♦ Mean mark (b) 45%.
MARKER’S COMMENT: \(\frac{1}{60}\int_0^{60} h(t)dt\) was a common error.
Others incorrectly found average rate of change instead of average value.

c.   \(\text{Av rate of change of height}\)

  \(=\dfrac{h(7.5)-h(0)}{7.5}\)
  \(=\dfrac{\Bigg(75-60\cos(\dfrac{\pi \times 7.5}{15})\Bigg)-\Bigg(75-60\cos(\dfrac{\pi\times 0}{15})\Bigg)}{7.5}\)
  \(=\dfrac{75-15}{7.5}=8\)
 
♦ Mean mark (c) 50%.
MARKER’S COMMENT: Common incorrect answer was – 8.
di.    \(\text{Period is 30 minutes, so after 15 minutes pod}\)
\(\text{is at the top of the wheel.}\)
  \(\therefore\ k=75-60\ \cos(\frac{\pi}{15}\times 15)=135\)

  
\(\text{Pod is travelling at twice its previous speed}\)
\(\text{so one revolution takes 15 minutes}\)

\(\therefore\ \text{Period}\rightarrow\) \(\dfrac{2\pi}{bm}\) \(=15\)
  \(bm\) \(=\dfrac{2\pi}{15}\)
  \(\dfrac{\pi}{15}\cdot m\) \(=\dfrac{2\pi}{15}\)
  \(m\) \(=\dfrac{2\pi}{15}\times \dfrac{15}{\pi}\)
    \(=2\)
 
♦♦ Mean mark (d)(i) 40%.
MARKER’S COMMENT: Many students could find \(k\) but not \(m\) with \(m=\frac{1}{2}\) a common error.
dii.   \(\text{When }t=20\ \text{the pod is at the top of the wheel and height is 135.}\)
  
\(\text{When }t=27.5\ \text{the pod is back at the start and height is 15.}\)

   
\(\text{Using CAS solve for }t=20:\rightarrow\ h(2(20)+n)=135\ \rightarrow\ n=30p+5\)

\(\text{Using CAS solve for }t=27.5:\rightarrow\ h(2(27.5)+n)=15\ \rightarrow\ n=30p+5\)

\(\therefore\ n=30p+5,\ p\in Z\)

 
♦♦♦ Mean mark (d)(ii) 20%.
MARKER’S COMMENT: Many students set up the correct equations to solve but did not provide the general solution. Some incorrectly stated the variable as an element of R.

diii.   

 
♦♦♦ Mean mark (d)(iii) 40%.
MARKER’S COMMENT: Students are reminded to include all endpoints and be particular about the curvature of graph.

Calculus, MET1 2023 VCAA 5

  1. Evaluate  \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\).   (1 mark)
  2. Hence, or otherwise, find all values of \(k\) such that \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx=\displaystyle \int_{0}^{\frac{\pi}{2}} \cos(x)\,dx\), where \(-3\pi<k<2\pi\).   (3 marks)
Show Answers Only

a.    \(\dfrac{1}{2}\)

b.    \(k=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

Show Worked Solution
a.    \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\) \(=\left[-\cos x\right]_0^\frac{\pi}{3}\)
    \(=-\cos\dfrac{\pi}{3}+\cos 0\)
    \(=-\dfrac{1}{2}+1\)
    \(=\dfrac{1}{2}\)

 

b.    \(\displaystyle \int_{k}^{\frac{\pi}{2}} \cos(x)\,dx\) \(=\left[\sin x\right]_k^\frac{\pi}{2}\)
    \(=\sin\bigg(\dfrac{\pi}{2}\bigg)-\sin (k)\)
    \(=1-\sin (k)\)

 
\(\text{Using part (a):}\)

\(1-\sin (k)\) \(=\dfrac{1}{2}\)
\(\sin (k)\) \(=\dfrac{1}{2}\)
\(\therefore\ k\) \(=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

Calculus, MET2 2019 VCAA 4 MC

`int_0^(pi/6) (a sin (x) + b cos(x))\ dx`  is equal to

  1. `((2 - sqrt 3)a - b)/2`
  2. `(b - (2 - sqrt 3) a)/2`
  3. `((2 - sqrt 3)a + b)/2`
  4. `((2 - sqrt 3) b - a)/2`
  5. `((2 - sqrt 3) b + a)/2`
Show Answers Only

`C`

Show Worked Solution

`int_0^(pi/6) (a sin (x) + b cos(x))\ dx`

`= [-a cos(x) + b sin(x)]_0^(pi/6)`

`= [-a ⋅ sqrt 3/2 + b/2 – (-a + 0)]`

`= (2a – sqrt 3 a + b)/2`

`= ((2 – sqrt 3) a + b)/2`

`=>   C`

Calculus, MET1 SM-Bank 5

The function with rule  `f(x)`  has derivative  `f^{prime}(x) =  cos\ 3x`.

If  `f(pi/6) = 1,`  find  `f(x).`  (3 marks)

Show Answers Only

`f(x)= 1/3 sin\ 3x + 2/3`

Show Worked Solution
`int f(x)\ dx` `=int cos\ 3x\ dx`
  `= 1/3 sin\ 3x + c`
`f(pi/6)`  `= 1/3\ [sin\ (3 xx pi/6) ]+ c`
`1`  `= 1/3\ sin\ pi/2+c`
`c`  `= 2/3`

 
`:.f(x)= 1/3 sin\ 3x + 2/3`

Calculus, MET1 2014 VCAA 7

If  `f^{prime}(x) = 2cos(x)-sin(2x)`  and  `f(pi/2) = 1/2`,  find  `f(x)`.   (3 marks)

Show Answers Only

`2sinx + 1/2cos(2x)-1`

Show Worked Solution
`f(x)` `= int(2cosx-sin2x)dx`
  `= 2sinx + 1/2cos(2x) + c`

  
`text(Substitute)\ \ f(pi/2) = 1/2:`

`1/2` `= 2sin(pi/2) + 1/2cos(pi) + c`
`1/2` `= 2-1/2 + c`
`c` `=-1`
`:. f(x)` `= 2sinx + 1/2cos(2x)-1`