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Calculus, 2ADV C3 2021 HSC 26

A particle is shot vertically upwards from a point 100 metres above ground level.

The position of the particle, `y` metres above the ground after `t` seconds, is given by

`y(t) = −5t^2 + 70t + 100`.

  1. Find the maximum height above ground level reached by the particle.  (2 marks)

  2. Find the velocity of the particle, in metres per second, immediately before it hits the ground, leaving your answer in the form  `asqrtb`,  where  `a`  and  `b`  are integers.  (3 marks)

Show Answers Only
  1. `345\ text(m)`
  2. `-20sqrt69\ text(m/s)`
Show Worked Solution

a.    `y(t) = -5t^2 + 70t + 100`

`y^(′)(t) = -10t + 70`

`y^(″)(t) = -10`

`text(Max height occurs when)\ \ y^(′)(t) = 0:`

`-10t + 70` `= 0`
`10t` `= 70`
`t` `= 7`

 

`:.\ text(Max height)\ ` `= -5(7)^2 + 70 xx 7 + 100`
  `= 345\ text(m)`

 

b.   `text(Particle hits ground when)\ \ y = 0:`

♦ Mean mark 38%.
`0` `= -5t^2 + 70t + 100`
`0` `= t^2 – 14t – 20`

 
`text(Using quadratic formula:)`

`t` `= (14 ± sqrt(14^2 + 4*20))/2`
  `= (14 + sqrt(276))/2\ \ \ (t > 0)`
  `= 7 + sqrt69`

 

`:. text(Velocity)\ (y^(′)(t))` `= -10(7 + sqrt69) + 70`
  `= -10sqrt69\ text(m/s)`

 
`:.a=-10 and b=69`

Calculus, 2ADV C4 2017 HSC 13d

The rate at which water flows into a tank is given by

`(dV)/(dt) = (2t)/(1 + t^2)`,

where `V` is the volume of water in the tank in litres and `t` is the time in seconds.

Initially the tank is empty.

Find the exact amount of water in the tank after 10 seconds.  (3 marks)

Show Answers Only

`text(ln)\ 101`

Show Worked Solution
`(dV)/(dt)` `= (2t)/(1 + t^2)`
`V` `= int (2t)/(1 + t^2)\ dt`
  `= text(ln)\ (1 + t^2) + c`

 
`text(When)\ \ t = 0,\ \ V = 0`

`0` `= text(ln)\ 1 + c`
`:. c` `= 0`

 
`text(Find)\ V\ text(when)\ t = 10:`

`V` `= text(ln)\ (1 + 10^2)`
  `= text(ln)\ 101`

Calculus, 2ADV C4 2006 HSC 9b

During a storm, water flows into a 7000-litre tank at a rate of `(dV)/(dt)` litres per minute, where `(dV)/(dt) = 120 + 26t-t^2` and `t` is the time in minutes since the storm began.

  1. At what times is the tank filling at twice the initial rate?  (2 marks)

  2. Find the volume of water that has flowed into the tank since the start of the storm as a function of `t`.  (1 mark)

  3. Initially, the tank contains 1500 litres of water. When the storm finishes, 30 minutes after it began, the tank is overflowing.

     

    How many litres of water have been lost?  (2 marks)

Show Answers Only
  1. `t = 6 or 20\ text(minutes)`
  2. `V = 120t + 13t^2 – 1/3t^3`
  3. `800\ text(litres)`
Show Worked Solution

i.  `(dV)/(dt) = 120 + 26t-t^2`

`text(When)\ t = 0, (dV)/(dt) = 120`

`text(Find)\ t\ text(when)\ (dV)/(dt) = 240`

`240 = 120 + 26t-t^2`

`t^2-26t + 120 = 0`

`(t-6)(t-20) = 0`

`t = 6 or 20`

`:.\ text(The tank is filling at twice the initial rate)`

`text(when)\ t = 6 and t = 20\ text(minutes)`

 

ii.  `V` `= int (dV)/(dt)\ dt`
  `= int 120 + 26t-t^2\ dt`
  `= 120t + 13t^2-1/3t^3 + c`

 
`text(When)\ t = 0, V = 0`

`=>  c = 0`

`:. V= 120t + 13t^2-1/3t^3`

 

iii.  `text(Storm water volume into the tank when)\ t = 30`

`= 120(30) + 13(30^2)-1/3 xx 30^3`

`= 3600 + 11\ 700-9000`

`= 6300\ text(litres)`
 

`text(Total volume)` `= 6300 + 1500`
  `= 7800\ text(litres)`

 

`:.\ text(Overflow)` `= 7800-7000`
  `= 800\ text(litres)`

Calculus, 2ADV C3 2005 HSC 6b

A tank initially holds 3600 litres of water. The water drains from the bottom of the tank. The tank takes 60 minutes to empty.

A mathematical model predicts that the volume, `V`  litres, of water that will remain in the tank after  `t`  minutes is given by
  

`V = 3600(1 − t/60)^2,\ \ text(where)\ \ 0 ≤ t ≤ 60`.
 

  1. What volume does the model predict will remain after ten minutes?  (1 mark)

  2. At what rate does the model predict that the water will drain from the tank after twenty minutes?  (2 marks)

  3. At what time does the model predict that the water will drain from the tank at its fastest rate?  (2 marks)

Show Answers Only
  1. `text(2500 L)`
  2. `80\ text(liters per minute)`
  3. `0`
Show Worked Solution

i.    `V = 3600(1 − t/60)^2`

`text(When)\ t = 10,`

`V` `= 3600(1 − 10/60)^2`
  `= 3600 xx (5/6)^2`
  `= 2500\ text(L)`

 

ii.   `V = 3600(1 -t/60)^2`

`text(Using chain rule:)`

`(dV)/dt` `= 3600 xx 2 xx (1 – t/60) xx d/dt(1 – t/60)`
  `= 7200(1 – t/60) xx -1/60`
  `= −120(1 – t/60)`

 

`text(When)\ \ t =20`

`(dV)/dt` `= −120(1 – 20/60)`
  `= −80`

 

`:.\ text(After 20 minutes, the water will drain)`

`text(at 80 litres per minute.)`

 

iii. `(dV)/dt` `= −120(1 − t/60)`
    `= −120 + 2t`
  `(d^2V)/dt^2` `= 2`

 
`text(S)text(ince)\ (d^2V)/dt^2\ text(is a constant, no S.P.’s)`

 

`text(Checking limits of)\ \ 0 ≤ t ≤ 60`

`text(At)\ t = 0,`

`(dV)/dt = −120(1-0) = −120\ text(L/min)`

`text(At)\ t = 60,`

`(dV)/dt = −120(1 − 60/60) = 0\ text(L/min)`

 

`:.\ text(The model predicts water will drain)`

`text(out the fastest when)\ \ t = 0.`

Calculus, 2ADV C4 2011 HSC 9b

A tap releases liquid  `A`  into a tank at the rate of  `(2 + t^2/(t + 1))`  litres per minute, where  `t`  is time in minutes. A second tap releases liquid  `B`  into the same tank at the rate of  `(1 + 1/(t+1))`  litres per minute. The taps are opened at the same time and release the liquids into an empty tank. 

  1. Show that the rate of flow of liquid  `A`  is greater than the rate of flow of liquid  `B`  by  `t`  litres per minute.    (1 mark)

  2. The taps are closed after 4 minutes. By how many litres is the volume of liquid  `A`  greater than the volume of liquid  `B`  in the tank when the taps are closed?    (2 marks)

Show Answers Only
  1. `text(Proof)  text{(See Worked Solutions)}`
  2. `text(There is 8 litres more of liquid)\ A\ text(than)\ B.`
Show Worked Solution
♦♦♦ Mean mark 16%
MARKER’S COMMENT: Many students incorrectly differentiated in part (i). The 1 mark allocation indicates the answer will not require an involved multi-step process.

i.   `text(Show difference in flow rate)\ (D) = t`

`D` `= (2 + t^2/(t+1))-(1+ 1/(t+1))`
  `= (2(t+1) + t^2)/(t+1)-((t+1) + 1)/(t + 1)`
  `= (2t + 2 + t^2-t-2)/(t + 1)`
  `= (t^2 + t)/(t + 1)`
  `= (t (t + 1))/(t + 1)`
  `= t\ \ \ … text(as required)`

 

ii.   `text(Difference in Volume)`

♦♦♦ Mean mark 15%
MARKER’S COMMENT: Few students were able to answer this part. Previous parts of any question should be front and centre of your thinking when working out strategies.

`= int_0^4 (2 + t^2/(1+ t))\ dt\-int_0^4 (1 + t/(1+t))\ dt`

`= int_0^4 t\ dt\ \ \ \ \ text{(using part(i))}`

`= [t^2/2]_0^4`

`= 16/2\ – 0`

` = 8`
 

`:.\ text(There is 8 litres more of liquid)\ A\ text(than)\ B.`