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Calculus, MET2 2023 VCAA 17 MC

A cylinder of height \(h\) and radius \(r\) is formed from a thin rectangular sheet of metal of length \(x\) and \(y\), by cutting along the dashed lines shown below.

The volume of the cylinder, in terms of \(x\) and \(y\), is given by

  1. \(\pi x^2y\)
  2. \(\dfrac{\pi xy^2-2y^3}{4\pi^2}\)
  3. \(\dfrac{2y^3-\pi xy^2}{4\pi^2}\)
  4. \(\dfrac{\pi xy-2y^2}{2\pi}\)
  5. \(\dfrac{2y^2-\pi xy}{2\pi}\)
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\(B\)

Show Worked Solution

\(y=2\pi r \to r=\dfrac{y}{2\pi}\ \ \ (1)\)

\(h\) \(=x-4r\)
  \(=x-4\times\dfrac{y}{2\pi}\)
  \(=x-\dfrac{2y}{\pi}\ \ \ (2)\)

 
\(\text{Substitute equations (1) and (2) into volume formula.}\)

\(V\) \(=\pi r^2h\)
  \(=\pi\times\Bigg(\dfrac{y}{2\pi}\Bigg)^2\times \Bigg(x-\dfrac{2y}{\pi}\Bigg)\)
  \(=\Bigg(\dfrac{y^2}{4\pi}\Bigg)\times \Bigg(\dfrac{\pi x-2y}{\pi}\Bigg)\)
  \(=\dfrac{\pi xy^2-2y^3}{4\pi^2}\)

♦♦ Mean mark 28%.
MARKER’S COMMENT: 26% incorrectly chose both C and D.

 
\(\Rightarrow B\)

Calculus, MET1 SM-Bank 27

A cone is inscribed in a sphere of radius `a`, centred at `O`. The height of the cone is `x` and the radius of the base is `r`, as shown in the diagram.

  1. Show that the volume, `V`, of the cone is given by
  2.    `V = 1/3 pi(2ax^2 - x^3)`.  (2 marks)
  3. Find the value of `x` for which the volume of the cone is a maximum. You must give reasons why your value of `x` gives the maximum volume.  (3 marks)
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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `x = 4/3 a`
Show Worked Solution

i.  `text(Show)\ V = 1/3 pi (2ax^2 – x^3)`
 

`V = 1/3 pi r^2 h`

`text(Using Pythagoras)`

`(x – a)^2 + r^2` `= a^2`
`r^2` `= a^2 – (x – a)^2`
  `= a^2 – x^2 + 2ax – a^2`
  `= 2ax – x^2`
`:. V` `= 1/3 xx pi xx (2ax – x^2) xx x`
  `= 1/3 pi (2ax^2 – x^3)\ …\ text(as required)`

 

ii.  `(dV)/(dx) = 1/3 pi (4ax – 3x^2)`

`(d^2V)/(dx^2) = 1/3 pi (4a – 6x)`

`text(Max or min when)\ (dV)/(dx) = 0`

`1/3 pi (4ax – 3x^2)` `= 0`
`4ax – 3x^2` `= 0`
`x(4a – 3x)` `= 0`
`3x` `= 4a,` ` \ \ \ \ x ≠ 0`
`x` ` =4/3 a`  

 

`text(When)\ \ x = 4/3 a`

`(d^2V)/(dx^2)` `= 1/3 pi (4a – 6 xx 4/3 a)`
  `= 1/3 pi (-4a) < 0`
`=>\ text(MAX)`

 

`:.\ text(Cone volume is a maximum when)\ \ x = 4/3 a.`

Calculus, MET1 SM-Bank 34

2005 8a

A cylinder of radius  `x`  and height  `2h`  is to be inscribed in a sphere of radius  `R`  centred at  `O`  as shown.

  1. Show that the volume of the cylinder is given by
  2.    `V = 2pih(R^2 − h^2).` (1 mark)
  3. Hence, or otherwise, show that the cylinder has a maximum volume when
  4.    `h = R/sqrt3.`  (3 marks)
Show Answers Only
  1. `text{Proof (See Worked Solutions).}`
  2. `text{Proof (See Worked Solutions).}`
Show Worked Solution
i.    `V` `= pir^2h\ \ \ \ text{(general form)}`
    `= pix^2 xx 2h`

 

`text(Using Pythagoras:)`

`R^2` `= h^2 + x^2`
`x^2` `= R^2 − h^2`
`:.V` `= 2pih(R^2 − h^2)\ \ …text(as required.)` 

 

ii. `V` `= 2pih(R^2 − h^2)`
    `= 2piR^2h − 2pih^3`
  `(dV)/(dh)` `= 2piR^2 − 3 xx 2pih^2`
    `= 2piR^2 − 6pih^2`
  `(d^2V)/(dh^2)` `= −12pih`

 
`text(Max or min when)\ (dV)/(dh) = 0`

`2piR^2 − 6pih^2` `= 0`
`6pih^2` `= 2piR^2`
`h^2` `= R^2/3`
`h` `= R/sqrt3,\ \ h > 0`

 
`text(When)\ h = R/sqrt3`

`(d^2V)/(dh^2) = −12pi xx R/sqrt3 < 0`

`:.\ text(Volume is a maximum when)\ h = R/sqrt3\ text(units.)`