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Calculus, SPEC2 2021 VCAA 3

A thin-walled vessel is produced by rotating the graph of  `y = x^3-8`  about the `y`-axis for  `0 <= y <= H`.

All lengths are measured in centimetres.

    1. Write down a definite integral in terms of `y` and `H` for the volume of the vessel in cubic centimetres.   (1 mark)

    2. Hence, find an expression for the volume of the vessel in terms of `H`.   (1 mark)

Water is poured into the vessel. However, due to a crack in the base, water leaks out at a rate proportional to the square root of the depth `h` of water in the vessel, that is  `(dV)/(dt) = -4sqrth`, where `V` is the volume of water remaining in the vessel, in cubic centimetres, after `t` minutes.

    1. Show that  `(dh)/(dt) = (-4sqrth)/(pi(h + 8)^(2/3))`.   (2 marks)

    2. Find the maximum rate, in centimetres per minute, at which the depth of water in the vessel decreases, correct to two decimal places, and find the corresponding depth in centimetres.   (2 marks)

    3. Let  `H = 50`  for a particular vessel. The vessel is initially full and water continues to leak out at a rate of  `4 sqrth`  cm³ min`\ ^(-1)`.
    4. Find the maximum rate at which water can be added, in cubic centimetres per minute, without the vessel overflowing.   (1 mark)

  2. The vessel is initially full where  `H = 50`  and water leaks out at a rate of  `4sqrth`  cm³ min`\ ^(-1)`. When the depth of the water drops to 25 cm, extra water is poured in at a rate of  `40sqrt2`  cm³ min`\ ^(-1)`.
  3. Find how long it takes for the vessel to refill completely from a depht of 25 cm. Give your answer in minutes, correct to one decimal place.   (3 marks)

Show Answers Only
    1. `V = pi int_0^H (y + 8)^(2/3)\ dy`
    2. `V = (3pi)/5 [(H + 8)^(5/3)-32]`
    1. `text(See Worked Solutions)`
    2. `0.62\ text(cm/min when)\ \ h = 24.`
    3. `4sqrt50\ \ text(cm³/min)`
  3. `31.4\ text(mins)`
Show Worked Solution

a.i.   `y = x^3-8 \ => \ x = root3(y + 8) \ => \ x^2 = (y + 8)^(2/3)`

`:. V = pi int_0^H (y + 8)^(2/3)\ dy`

 

a.ii.   `V = (3pi)/5 [(H + 8)^(5/3)-32]`

 

b.i.   `(dV)/(dt) = -4sqrth`

`(dV)/(dh) = pi(h + 8)^(2/3) \ => \ (dh)/(dV) =1/(pi(h + 8)^(2/3))`

`(dh)/(dt)` `= (dV)/(dt) * (dh)/(dV)`
  `= (-4sqrth)/(pi(h + 8)^(2/3))`

 

b.ii.   `text(Solve)\ (d^2h)/(dt^2) = 0\ \ text(for)\ \ h\ \ text{(by CAS):}`

♦ Mean mark part (b)(ii) 42%.

`(d^2h)/(dt^2) = (2(h-24))/(3sqrth pi (h + 8)^(5/3))=0`

`=>h = 24`

`text(At)\ \ h=24, \ (dh)/(dt) = -0.62\ text(cm/min)`

`:.\ text(Max rate at which depth decreases is)`

`0.62\ text(cm/min when)\ \ h = 24.`

♦♦ Mean mark part (b)(iii) 24%.

 

b.iii.   `text(At)\ \ H = 50, text(vessel is full and losing water at)\ \ 4sqrt50\ \ text(cm³/min)`

`:. text(Water can be added at a max-rate of)\ \ 4sqrt50\ \ text(cm³/min and)`

`text(vessel will not overflow.)`

 

c.   `(dV)/(dt) = 40sqrt2-4sqrth`

♦♦♦ Mean mark part (c) 16%.

`(dV)/(dh) · (dh)/(dt) = 40sqrt2-4sqrth`

`pi(h + 8)^(2/3) · (dh)/(dt)` `= 40sqrt2-4sqrth`
`(dh)/(dt)` `= (40sqrt2-4sqrth)/(pi(h + 8)^(2/3)`
`(dt)/(dh)` `= (pi(h + 8)^(2/3))/(40sqrt2-4sqrth)`
`t` `= int (pi(h + 8)^(2/3))/(40sqrt2-4sqrth)\ dh`

 
`text(Time of vessel to refill from)\ \ h = 25\ \ text(to)\ \ h = 50:`

`t` `= int_25^50 (pi(h + 8)^(2/3))/(40sqrt2-4sqrth)\ dh`
  `~~ 31.4\ text(mins)`

Calculus, SPEC2 2013 VCAA 10 MC

The region bounded by the lines  `x = 0`, `y = 3`  and the graph of  `y = x^(4/3)`  where  `x ≥ 0`  is rotated about the `y`-axis to form a solid of revolution.

The volume of this solid is

A.   `(81pi3^(2/3))/11`

B.   `(12pi3^(3/4))/7`

C.   `(27pi3^(1/3))/7`

D.   `(18pi3^(1/2))/5`

E.   `(6pi3^(1/2))/5`

Show Answers Only

`D`

Show Worked Solution
`y` `= (x^(2/3))^2`
`y` `= (x^2)^(2/3)`
`y^(3/2)` `= x^2`

 

`:. V` `= pi int_0^3 x^2\ dy`
  `= pi int_0^3 y^(3/2)\ dy\ \ \ text{(by CAS)}`
  `= (18pisqrt3)/5`

 
`=> D`

Calculus, SPEC1 2015 VCAA 5

Find the volume generated when the region bounded by the graph of  `y = 2x^2 - 3`, the line  `y = 5`  and the `y`-axis is rotated about the `y`-axis.  (3 marks)

Show Answers Only

`16 pi`

Show Worked Solution

`V = pi int_(−3)^5 x^2\ dy`

`y` `= 2x^2 – 3`
`2x^2` `= y+3`
`x^2` `= 1/2(y+3)`

 

`:. V` `= pi/2 int_(−3)^5 (y + 3)\ dy`
  `= pi/2[(y^2)/2 + 3y]_(−3)^5`
  `= pi/2(25/2 + 15 – (9/2 – 9))`
  `= pi/2(16/2 + 24)`
  `= 16pi`

Calculus, SPEC1 2018 VCAA 9

A curve is specified parametrically by  `underset ~r(t) = sec(t) underset ~i + sqrt 2/2 tan(t) underset ~j, \ t in R`.

  1.  Show that the cartesian equation of the curve is  `x^2-2y^2 = 1`.   (2 marks)

  2.  Find the `x`-coordinates of the points of intersection of the curve  `x^2-2y^2 = 1`  and the line  `y = x-1`.   (1 mark)

  3.  Find the volume of the solid of revolution formed when the region bounded by the curve and the line is rotated about the `x`-axis.   (2 marks)

Show Answers Only

  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `x = 1 or x = 3`
  3. `(2 pi)/3`

Show Worked Solution

a.     `x = sec(t), qquad y = sqrt 2/2 tan(t)`

`x^2 = sec^2(t), qquad y^2 = 1/2 tan^2(t)`

`x^2 = sec^2(t), qquad 2y^2 = tan^2(t)`

`1 + tan^2(t)` `= sec^2(t)`
`1 + 2y^2` `= x^2`
`:.x^2-2y^2` `=1\ \ text(.. as required)`

 

b.    `x^2-2(x-1)^2` `= 1`
  `x^2-2(x^2-2x + 1)` `=1`
  `x^2-2x^2 + 4x-2` `=1`
  `-x^2 + 4x-2-1` `=0`
  `x^2-4x + 3` `=0`
  `(x-3) (x-1)` `=0`

 
`:. x = 1 or x = 3`

♦♦ Mean mark 30%.

 

c.   `x^2-{:2y_1:}^2` `=1`
  `{:2y_1:}^2` `=x^2-1`
  `{:y_1:}^2` `= (x^2-1)/2`
  `{:y_2:}^2` `= (x-1)^2`

 

`V` `=pi int_1^3 {:y_1:}^2-{:y_2:}^2 \ dx`
  `= pi int_1^3 (x^2-1)/2-(x-1)^2\ dx`
  `= pi [x^3/6-x/2-(x-1)^3/3]_1^3`
  `= pi [(3^3/6-3/2-2^3/3)-(1^3/6-1/2-0)]`
  `= pi (9/2-3/2-8/3-1/6 + 1/2)`
  `= pi (7/2-8/3-1/6)`
  `= pi ((21-16-1)/6)`
  `= (2 pi)/3`