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Calculus, MET2 2019 VCAA 15 MC

Let  `f: [2, oo) -> R, \ f(x) = x^2 - 4x + 2`  and  `f(5) = 7`. The function  `g`  is the inverse function of  `f`.

`g prime(7)`  is equal to

A.   `1/6`

B.   `5`

C.   `(sqrt7)/14`

D.   `6`

E.   `1/7`

Show Answers Only

`A`

Show Worked Solution

`f(x) = x^2 – 4x + 2`

`f(5) = 7\ \ =>\ \ g(7) = 5\ \ text{(inverse function)}`

`f prime(x) = 2x – 4`

`f prime(5) = 6`

`g prime(7) = 1/(f prime(5)) = 1/6`

 
`=>   A`

Calculus, MET2 2011 VCAA 1

Two ships, the Elsa and the Violet, have collided. Fuel immediately starts leaking from the Elsa into the sea.

The captain of the Elsa estimates that at the time of the collision his ship has 6075 litres of fuel on board and he also forecasts that it will leak into the sea at a rate of `(t^2)/5` liters per minute, where `t` is the number of minutes that have elapsed since the collision.

  1. At this rate how long, in minutes, will it take for all the fuel from the Elsa to leak into the sea?  (3 marks)

*Parts (b) - (d) are no longer in the syllabus. 

Show Answers Only
  1. `45\ text(min)`
Show Worked Solution

a.   `text(Let)\ \ f = text(fuel in Elsa at)\ t\ text(min)`

♦♦♦ Mean mark 23%.

`(df)/(dt) = (−t^2)/5\ \ text{(given)}`

 

`text(Find)\ f\ text(in terms of)\ t:`

`f` `= int −1/5t^2 dt`
`f` `= (−t^3)/15 + c`

 

`text(Substitute)\ \ (0,6075):`

`6075` `= 0/15 + c`
`:. c` `= 6075`

 

`text(Solve:)\ \ 0 = −(t^2)/15 + 6075\ \ text(for)\ t`

`:. t = 45\ text(min)`

 

*Parts (b) – (d) are no longer in the syllabus.

Calculus, MET2 2012 VCAA 4

Tasmania Jones is in the jungle, searching for the Quetzalotl tribe’s valuable emerald that has been stolen and hidden by a neighbouring tribe. Tasmania has heard that the emerald has been hidden in a tank shaped like an inverted cone, with a height of 10 metres and a diameter of 4 metres (as shown below).

The emerald is on a shelf. The tank has a poisonous liquid in it.

VCAA 2012 4a

  1. If the depth of the liquid in the tank is `h` metres.
    1. find the radius, `r` metres, of the surface of the liquid in terms of `h`.  (1 mark)
    2. show that the volume of the liquid in the tank is  `(pi h^3)/75\ text(m³)`.  (1 mark)

The tank has a tap at its base that allows the liquid to run out of it. The tank is initially full. When the tap is turned on, the liquid flows out of the tank at such a rate that the depth, `h` metres, of the liquid in the tank is given by

`h = 10 + 1/1600 (t^3 - 1200t)`,

where `t` minutes is the length of time after the tap is turned on until the tank is empty.

  1. Show that the tank is empty when  `t = 20`.  (1 mark)
  2. When  `t = 5` minutes, find.
    1. the depth of the liquid in the tank  (1 mark)
    2. No longer in syllabus
  3. The shelf on which the emerald is placed is 2 metres above the vertex of the cone.

    From the moment the liquid starts to flow from the tank, find how long, in minutes, it takes until  `h = 2`.

    (Give your answer correct to one decimal place.)  (2 marks)

  4. As soon as the tank is empty, the tap turns itself off and poisonous liquid starts to flow into the tank at a rate of 0.2 m³/minute.

    How long, in minutes, after the tank is first empty will the liquid once again reach a depth of 2 metres?  (2 marks)

  5. In order to obtain the emerald, Tasmania Jones enters the tank using a vine to climb down the wall of the tank as soon as the depth of the liquid is first 2 metres. He must leave the tank before the depth is again greater than 2 metres.

    Find the length of time, in minutes, correct to one decimal place, that Tasmania Jones has from the time he enters the tank to the time he leaves the tank.  (1 mark)

Show Answers Only
    1. `h/5`
    2. `(pih^3)/75`
  2. `text(See Worked Solutions)`
    1. `405/64\ text(m)`
    2. `text(No longer in course)`
  4. `12.2\ text(min)`
  5. `(8pi)/15\ text(min)`
  6. `9.5\ text(min)`
Show Worked Solution

a.i.   `text(Using similar triangles:)`

`r/h` `= 2/10`
`:. r` `= h/5`

 

a.ii.    `V` `= 1/3 pir^2h`
    `= 1/3 pi(h/5)^2h`
  `:. V` `= (pih^3)/75\ \ \ text(… as required)`

 

b.    `h(20)` `= 10 + 1/1600(20^3 – 1200 xx 20)`
    `= 10 + 1/1600(−16\ 000)`
    `= 10 – 10`
    `= 0`

 

c.i.   `text(When)\ \ t=5,`

`h(5)` `=10 + 1/1600 (5^3 – 1200xx5)`
  `= 405/64\ text(m)`

 

c.ii.   `text(No longer in syllabus.)`

 

d.    `text(Solve:)\ \ 10 + 1/1600 (t^3 – 1200t)` `= 2`

 

`t= 12.2\ text(min)quadtext(for)quadt ∈ (0,20)`

 

e.   `text(When)\ \ h=2,`

♦♦♦ Mean mark part (e) 17%.
`text(Volume)` `=(pi 2^3)/75=(8pi)/75\ text(m³)`
   

`text(Time to fill back up to)\ \ h=2`

`=(8pi)/75 -: 0.2`

`=(8pi)/15\ text(minutes)`

 

f.   `text(Length of time)`

♦♦♦ Mean mark part (f) 11%.

`=\ text(Time for 2m to empty + Time to Fill to 2m)`

`= (20 – 12.16799…) + (8pi)/15`

`= 9.5\ text(min)\ \ text{(1 d.p.)}`