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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)

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a.    ` (sqrt(3)pi)/6`

b.    `(3sqrt(3) +pi)/6`

c.    `k = pi/2`

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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 2023 VCAA 1b

Let  \(f(x)=\sin(x)e^{2x}\).

Find  \(f^{'}\Big(\dfrac{\pi}{4}\Big)\).   (2 marks)

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\(\dfrac{3\sqrt{2}}{2}e^{\frac{\pi}{2}}\ \text{or}\ \dfrac{3e^{\frac{\pi}{2}}}{\sqrt{2}}\)

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\(\text{Using the product rule}\)

\(f'(x)\) \(=e^{2x}\cos(x)+2e^{2x}\sin(x)\)
  \(=e^{2x}\Big(\cos(x)+2\ \sin(x)\Big)\)
\(\therefore\ f’\Big(\dfrac{\pi}{4}\Big)\) \(=e^{2(\frac{\pi}{4})}\Bigg(\cos(\dfrac{\pi}{4})+2\ \sin(\dfrac{\pi}{4})\Bigg)\)
  \(=e^{\frac{\pi}{2}}\Bigg(\dfrac{1}{\sqrt{2}}+\sqrt{2}\Bigg)\)
  \(=e^{\frac{\pi}{2}}\Bigg(\dfrac{1+\sqrt2 \times \sqrt2}{\sqrt2} \Bigg) \)
  \(=\dfrac{3\sqrt{2}}{2}e^{\frac{\pi}{2}}\ \ \text{or}\ \ \dfrac{3e^{\frac{\pi}{2}}}{\sqrt{2}}\)

Calculus, MET1 2007 ADV 2aii

Let  `y=xsinx.`  Evaluate  `dy/dx`  for  `x=pi`.   (3 marks)

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`-pi`

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`y = x sin x`

`(dy)/(dx)` `= x xx d/(dx) (sin x) + d/(dx) (x) xx sin x`
  `= x  cos x + sin x`

 
`text(When)\ \ x = pi,`

`(dy)/(dx)` `= pi xx cos pi + sin pi`
  `= pi (-1) + 0`
  `=-pi`

Calculus, MET1-NHT 2019 VCAA 1b

Let  `f(x) = x^2 cos(3x)`.
 
Find  `f ^{\prime} (pi/3)`.   (2 marks)

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`-(2pi)/3`

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  `f(x)` `= x^2 cos 3x`
  `f^{\prime}(x)` `= x^2 ⋅ 3(-sin 3x) + 2x cos 3x`
  `f^{\prime}(pi/3)` `= (pi/3)^2 ⋅ 3 (-sin pi) + 2 (pi/3) cos pi`
    `= -(2pi)/3`

Calculus, MET1 2011 VCAA 1b

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

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

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`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)^{prime}` `= f^{prime} h + fh^{prime}`
`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 2006 VCAA 3a

Let  `y = x tan(x)`. Evaluate  `(dy)/(dx)`  when `x = pi/6`.   (3 mark)

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`(3sqrt3 + 2pi)/9`

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`text(Using the product rule:)`

MARKER’S COMMENT: A common error was not knowing the exact trig function values.

`d/(dx)(uv)` `= u^{prime}v + uv^{prime}`
`(dy)/(dx)` `= tan(x) + x/(cos^2(x))`

 

`text(When)\ x = pi/6,`

`(dy)/(dx)` `= tan(pi/6) + (pi/6)/((cos(pi/6))^2)`
  `= 1/sqrt3 + pi/6 xx (2/sqrt3)^2`
  `= sqrt3/3 + (4pi)/18`
  `= (3sqrt3 + 2pi)/9`

Calculus, MET1 2006 ADV 2ai

Differentiate with respect to `x`:

Let  `f(x)=x tan x`.  Find  `f^{prime}(x)`.   (2 marks)

 

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`f^{prime}(x) = x  sec^2 x + tan x `

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`y = x tan x`

`text(Using product rule)`

`f^{prime} (uv)` `= u^{prime}v + uv ^{prime}`
`:.f^{prime}(x)` `= tan x + x xx sec^2 x`
  `= x sec^2 x + tan x`

Calculus, MET1 2014 VCAA 1a

If  `y = x^2sin(x)`, find  `(dy)/(dx)`.   (2 marks)

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`2xsin(x) + x^2cos(x)`

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`text(Using product rule:)`

MARKER’S COMMENT: Factorising your answer is not necessary.
`(fg)^{prime}` `= f^{prime}g + fg^{prime}`
`:. (dy)/(dx)` `= 2xsin(x) + x^2cos(x)`