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Calculus, 2ADV C4 2024 HSC 15

Initially there are 350 litres of water in a tank. Water starts flowing into the tank.

The rate of increase of the volume `V` of water in litres is given by  `\frac{dV}{dt}=300-7.5t`, where `t` is the time in hours.

Find the volume of water in the tank when  `\frac{dV}{dt}=0`.   (3 marks)

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`\text{6350 litres}`

Show Worked Solution

`\frac{dV}{dt}=300-7.5t`

`\text{Find}\ t\ \text{when}\ \frac{dV}{dt}=0:`

`300-7.5t=0\ \ =>\ \ t=40\ \text{hours}`

`V` `=\int 300-\frac{15}{2}t\ dt`  
  `=300t-\frac{15}{4}t^2+C`  

 
`V=350\ \ \text{when}\ \ t=0\ \ =>\ \ C=350`

`V=300t-\frac{15}{4}t^2+350`

`\text{Find}\ V\ \text{when}\ \ t=40:`

`V` `=300 xx 40-\frac{15}{4} xx 40^2+350`  
  `=12\ 000-6000+350`  
  `=6350\ \text{litres}`  

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)

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`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 2015 HSC 15c

Water is flowing in and out of a rock pool. The volume of water in the pool at time `t` hours is `V` litres. The rate of change of the volume is given by

`(dV)/(dt) = 80 sin(0.5t)`

At time  `t = 0`, the volume of water in the pool is 1200 litres and is increasing.

  1. After what time does the volume of water first start to decrease?  (2 marks)

  2. Find the volume of water in the pool when  `t = 3`.  (2 marks)

  3. What is the greatest volume of water in the pool?  (1 mark)

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  1. `2 pi\ text(hours)`
  2. `1349\ text(litres)`
  3. `1520\ text(litres)`
Show Worked Solution

i.   `(dV)/(dt) = 80 sin (0.5t)`

♦ Mean mark (i) 37%.

`text(Volume decreases when)\ \ (dV)/(dt) < 0`

`(dV)/(dt) < 0\ \ text(when)\ \ t > 2 pi\ text(hours)`

`:.\ text(After 2)\pi\ \text(hours, volume starts to decrease.)`

 

ii.   `V` `= int (dV)/(dt)\ dt`
  `= int 80 sin (0.5t)\ dt`
  `= -160 cos (0.5t) + c`

 

`text(When)\ \ t = 0,\ V = 1200`

`1200` `= -160 cos 0 + c`
`c` `= 1360`
`:.\ V` `= -160 cos (0.5t) + 1360`

 

`text(When)\ \ t = 3`

`V` `= -160 cos (0.5 xx 3) + 1360`
  `= 1348.68…\ \ text(litres)`
  `= 1349\ text{litres  (nearest litre)}`

 

iii.  `V = -160 cos (0.5t) + 1360`

♦♦ Mean mark (iii) 27%.

`=>\ text(Greatest volume occurs when)`

`cos (0.5t) = -1`

`:.\ text(Maximum volume)`

`= -160 (-1) + 1360`

`= 1520\ text(litres)`

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)

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  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 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)

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  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.`