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Calculus, 2ADV C3 2015 HSC 16c

The diagram shows a cylinder of radius `x` and height `y` inscribed in a cone of radius `R` and height `H`, where `R` and `H` are constants.
  

The volume of a cone of radius `r` and height `h` is  `1/3 pi r^2 h.`

The volume of a cylinder of radius `r` and height `h` is  `pi r^2 h.`

  1. Show that the volume, `V`, of the cylinder can be written as
     
         `V = H/R pi x^2 (R-x).`   (3 marks)

  2. By considering the inscribed cylinder of maximum volume, show that the volume of any inscribed cylinder does not exceed `4/9` of the volume of the cone.   (4 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.   `text(Show)\ \ V = H/R pi x^2 (R-x)`
 

♦♦ Mean mark (i) 16%.

 

`text(Consider)\ \ Delta ABC and Delta DEC`

`/_ ABC = /_DEC = 90°`

`/_ BCA\ \ text(is common)`

`:. Delta ABC\ \ text(|||)\ \ Delta DEC\ \ text{(equiangular)}`

`:.\ (DE)/(EC)`

`= (AB)/(BC)`

`\ text{(corresponding sides of}`

`\ text{similar triangles)}`
`y/(R-x)` `= H/R`  
`y` `= H/R (R-x)`  

 

`text(Volume of cylinder)`

`= pi r^2 h`

`= pi x^2 xx H/R (R-x)`

`= H/R pi x^2 (R-x)\ \ text(…  as required.)`
 

♦♦ Mean mark (ii) 21%.

ii.   `V= H/R pi x^2 (R-x)`

`(dV)/(dx)` `= H/R pi [(x^2 xx -1) + 2x (R-x)]`
  `= H/R pi [-x^2 + 2xR-2x^2]`
  `= H/R pi [2xR-3x^2]`
`(dV^2)/(dx^2)` `= H/R pi [2R-6x]`

 

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

`H/R pi [2xR-3x^2]` `= 0`
`2xR-3x^2` `= 0`
`x (2R-3x)` `= 0`
`x = 0\ \ or\ \ 3x` `= 2R`
`x` `= (2R)/3`

 

`text(When)\ \ x = 0:`

`(d^2V)/(dx^2) = H/R pi [2R-0] > 0\ \ =>\  text(MIN)`
 

`text(When)\ \ x = (2R)/3:`

`(d^2V)/(dx^2)` `= H/R pi [2R-6 xx (2R)/3]`
  `= H/R pi [-2R] < 0\ \ =>\ \text{MAX}`

 
`text(Maximum Volume of cylinder)`

`= H/R pi ((2R)/3)^2 (R-(2R)/3)`

`= H/R pi xx (4R^2)/9 xx R/3`

`= (4 pi H R^2)/27`

 

`text(Volume of cone)\ = 1/3 pi R^2 H`

`:.\ text(Max Volume of Cylinder)/text(Volume of cone)`

`= ((4 pi H R^2)/27)/(1/3 pi R^2 H)`

`= (4 pi H R^2)/27 xx 3/(pi R^2 H)`

`= 4/9`

 

`:.\ text(The volume of the inscribed cylinder does)`

`text(not exceed)\ 4/9\ text(of the cone volume.)`

Calculus, 2ADV C3 2005 HSC 8a

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
     
         `V = 2pih(R^2 − h^2).` (1 mark)

  2. Hence, or otherwise, show that the cylinder has a maximum volume when  `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.)`

Calculus, 2ADV C3 2010 HSC 5a

A rainwater tank is to be designed in the shape of a cylinder with radius  `r`  metres and height  `h`  metres.

2010 5a

The volume of the tank is to be 10 cubic metres. Let  `A`  be the surface area of the tank, including its top and base, in square metres.

  1. Given that  `A=2pir^2+2pi r h`,   show that  `A=2 pi r^2+20/r`.   (2 marks)

  2. Show that  `A`  has a minimum value and find the value of  `r`  for which the minimum occurs.  (3 marks)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `r~~1.17\ text(metres)`
Show Worked Solution

i.    `text(Show)\ A=2 pi r^2+20/r`

MARKER’S COMMENT: Students MUST know the volume formula for a cylinder. Those that did and stated `10=pi r^2 h` most often completed this question efficiently.

`text(S)text(ince)\ V=pi r^2h=10\ \ \ \ =>\ h=10/(pi r^2)`

`text(Substituting into)\ A`

`A` `= pi r^2+2 pi r (10/(pi r^2))`
  `=2 pi r^2+20/r\ \ \ text(…  as required)`

 

♦ Mean mark 44%
MARKER’S COMMENT: The “table method” or 1st derivative test for proving a minimum (i.e. showing how `(dA)/(dr)` changes sign) was also quite successful.
ii. `A`  `=2 pi r^2+20/r`
  `(dA)/(dr)` `=4 pi r-20/r^2`
  `(d^2A)/(dr^2)` `=4 pi+40/r^3>0\ \ \ \ \ (r>0)`

 
`=>\ text(MIN occurs when)\ \ (dA)/(dr)=0`

`4 pi r-20/r^2` `=0`
`4 pi r^3-20` `=0`
`4 pi r^3` `=20`
`r` `=root3 (5/pi)`
  `=1.16754…`
  `=1.17\ text{metres  (2 d.p.)}`