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Calculus, SPEC2 2023 VCAA 8 MC

Initially a spa pool is filled with 8000 litres of water that contains a quantity of dissolved chemical. It is discovered that too much chemical is contained in the spa pool water. To correct this situation, 20 litres of well-mixed spa pool water is pumped out every minute while 15 litres of fresh water is pumped in each minute.

Let \(Q\) be the number of kilograms of chemical that remains dissolved in the spa pool after \(t\) minutes. The differential equation relating \(Q\) to t is

  1. \(\dfrac{d Q}{d t}=\dfrac{4 Q}{t-1600}\)
  2. \(\dfrac{d Q}{d t}=\dfrac{-Q}{400}\)
  3. \(\dfrac{d Q}{d t}=\dfrac{3 Q}{t-1600}\)
  4. \(\dfrac{d Q}{d t}=\dfrac{3 Q}{1600-t}\)
  5. \(\dfrac{d Q}{d t}=\dfrac{4 Q}{1600-t}\)
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\(A\)

Show Worked Solution

\(\text{Volume}\ = 8000-5t \)

\(Q(t) = \dfrac{Q}{8000-5t} \)

\(\dfrac{dQ}{dt}=-\dfrac{20Q}{8000-5t} = \dfrac{4Q}{t-1600} \)

\(\Rightarrow A\)

Calculus, SPEC2 2020 VCAA 10 MC

A tank initially contains 300 grams of salt that is dissolved in 50 L of water. A solution containing 15 grams of salt per litre of water is poured into the tank at a rate of 2 L per minute and the mixture in the tank is kept well stirred. At the same time, 5 L of the mixture flows out of the tank per minute.

A differential equation representing the mass, `m` grams, of salt in the tank at time `t` minutes, for a non-zero volume of mixture is

  1. `(dm)/(dt) = 0`
  2. `(dm)/(dt) = −(5m)/(50 - 5t)`
  3. `(dm)/(dt) = 30 - m/10`
  4. `(dm)/(dt) = 30 - (5m)/(50 - 3t)`
  5. `(dm)/(dt) = 30 - (5m)/(50 - 5t)`
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`D`

Show Worked Solution

`V(t) = 50 + 2t – 5t = 50 – 3t`

`(dm)/(dt)\ text(in) = 15 xx 2 = 30\ text(g/min)`

`(dm)/(dt)\ text(out) = 5 xx m/(50 – 3t) = (5m)/(50 – 3t)`

`:. (dm)/(dt) = 30 – (5m)/(50 – 3t)`
 

`=> D`

Calculus, SPEC2 2013 VCAA 13 MC

Water containing 2 grams of salt per litre flows at the rate of 10 litres per minute into a tank that initially contained 50 litres of pure water. The concentration of salt in the tank is kept uniform by stirring and the mixture flows out of the tank at the rate of 6 litres per minute.

If `Q` grams is the amount of salt in the tank `t` minutes after the water begins to flow, the differential equation relating `Q` to `t` is

A.   `(dQ)/(dt) = 20 - (3Q)/(25 + 2t)`

B.   `(dQ)/(dt) = 10 - (3Q)/(25 + 2t)`

C.   `(dQ)/(dt) = 20 - (3Q)/(25 - 2t)`

D.   `(dQ)/(dt) = 10 - (3Q)/(25 - 2t)`

E.   `(dQ)/(dt) = 20 - (3Q)/25`

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`A`

Show Worked Solution
`text(Volume)` `= 50 + (10 – 6)t`
  `= 50 + 4t`

 
`text(Salt in tank at time)\ \ t=Q\ text(grams)`

`:.\ text(Concentration)\ = Q/(50 + 4t)\ text(grams per litre)`
 

`(dQ)/(dt)text(in) = 2 xx 10 = 20\ \ text(g/min)`

`(dQ)/(dt)text(out)` `= 6 xx Q/(50 + 4t)`
  `= (3Q)/(25 + 2t)`

 
`:. (dQ)/(dt) = 20 – (3Q)/(25 + 2t)`

`=> A`

Calculus, SPEC2 2014 VCAA 10 MC

A large tank initially holds 1500 L of water in which 100 kg of salt is dissolved. A solution containing 2 kg of salt per litre flows into the tank at a rate of 8 L per minute. The mixture is stirred continuously and flows out of the tank through a hole at a rate of 10 L per minute.

The differential equation for `Q`, the number of kilograms of salt in the tank after `t` minutes, is given by

A.   `(dQ)/(dt) = 16 - (5Q)/(750 - t)`

B.   `(dQ)/(dt) = 16 - (5Q)/(750 + t)`

C.   `(dQ)/(dt) = 16 + (5Q)/(750 - t)`

D.   `(dQ)/(dt) = (100Q)/(750 - t)`

E.   `(dQ)/(dt) = 8 - Q/(1500 - 2t)`

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`A`

Show Worked Solution

`(dQ_text(in))/(dV)= 2\ text(kg/L),\ (dV_text(in))/(dt) = 8\ text(L/min)`

`V_0 = 1500,\ Q_0 = 100`

`(dV_text(out))/(dt) = 10\ text(L/min)`
  

`V(t)` `= 1500 + (8 – 10)t`
  `= 1500 – 2t`
  `= 2(750 – t)`

 

`(dQ_text(in))/(dt)` `= 2 xx 8 = 16 text(kg/min)`
`(dQ_text(out))/(dt)` `= Q/(v(t)) xx 10`
  `= (10Q)/(2(750 – t))`
  `= (5Q)/(750 – t)`

 
`:.(dQ)/(dt)= 16 – (5Q)/(150 – t)`

`=> A`

Calculus, SPEC1 2018 VCAA 8

A tank initially holds 16 L of water in which 0.5 kg of salt has been dissolved. Pure water then flows into the tank at a rate of 5 L per minute. The mixture is stirred continuously and flows out of the tank at a rate of 3 L per minute.

  1.  Show that the differential equation for `Q`, the number of kilograms of salt in the tank after `t` minutes, is given by
    `qquad (dQ)/(dt) = -(3Q)/(16 + 2t)`  (1 mark)
      
  2.  Solve the differential equation given in part a. to find `Q` as a function of `t`.
      
    Express your answer in the form  `Q = a/(16 + 2t)^(b/c)`, where `a, b` and `c` are positive integers.  (3 marks)
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  1.  `text(Proof)\ \ text{(See Worked Solutions)}`
  2.  `Q = 32/(16 + 2t)^(3/2)`
Show Worked Solution

a. `Q_0 = 0.5, \ V_0 = 16`

♦ Net mean mark of both parts 44%.

`V(t)` `= 16 + (5 – 3) t`
  `= 16 + 2t`

 

`text(Concentration)\ (C)` `= Q/V= Q/(16 + 2t)\ text(kg/L)`

 

`(dQ)/(dt)` `= 0 xx 5 – 3C`
  `= -(3Q)/(16 + 2t)`

MARKER’S COMMENT: Many students took the common factor of 2 from  `16+2t`. This wasn’t necessary and complicated the arithmetic in part b.

 

b.   `-1/(3Q) * (dQ)/(dt) = 1/(16 + 2t)`

`int -1/(3Q)\ dQ` `= int 1/(16 + 2t) dt`
`-1/3 int 1/Q\ dQ` `= 1/2 int 2/(16 + 2t)\ dt`
`-1/3 ln Q` ` = [1/2 ln(16 + 2t)] + c`

 
`text(When)\ \ t=0,\ \ Q=0.5,`

`-1/3 ln (1/2)` `= 1/2 ln (16) +c`
`c` `= -1/2 ln (16) -1/3 ln (1/2)`

 

`-1/3 ln Q` `= 1/2 ln(16 + 2t) -1/2 ln(16) – 1/3 ln (1/2)`
`-1/3 ln Q` `= 1/2 ln ((16 + 2t)/16) – 1/3 ln (1/2)`
`-1/3 ln Q` `= ln (((16 + 2t)^(1/2))/4) – ln (2^(-1/3))`
`ln (Q^(-1/3))` `= ln (((16 + 2t)^(1/2))/(2^2 ⋅ 2^(-1/3)))`
`Q^(-1/3)` `= ((16 + 2t)^(1/2))/(2^(5/3))`
`Q` `= (((16 + 2t)^(1/2))/(2^(5/3)))^-3`
  `= (16 + 2t)^(- 3/2)/(2^(-5))`
`:. Q` `= 32/((16 + 2t)^(3/2))`