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Calculus, SPEC2 2025 VCAA 3

A tank initially contains 5 kg of salt dissolved in 3000 litres of water. Salty water that contains 0.1 kg of salt per litre of water enters the tank at a rate of 20 litres per minute. The solution is kept thoroughly mixed and drains from the tank via a tap at the same rate of 20 litres per minute.

  1. By considering concentration, explain whether the quantity of salt in the tank increases with time.   (1 mark)

  2. Let \(Q\) denote the quantity of salt, in kilograms, in the tank at time \(t\) minutes.
  3. Show that \(Q\) satisfies the differential equation  \(\dfrac{d Q}{d t}=\dfrac{300-Q}{150}\).   (1 mark)

  4. Using Euler's method with a step size of 15 minutes, find \(Q(30)\), the approximate quantity of salt in the tank after 30 minutes.
  5. Give your answer in kilograms, correct to two decimal places.   (2 marks)

  6. Use calculus to solve the differential equation  \(\dfrac{d Q}{d t}=\dfrac{300-Q}{150}\), expressing \(Q\) in terms of \(t\).   (3 marks)

  7. What value does the quantity of salt in the tank approach as time approaches infinity?
  8. Give your answer in kilograms.   (1 mark)

  9. Find the time taken for the quantity of salt in the tank to reach 100 kg.   (1 mark)

  10. When the quantity of salt in the tank reaches 100 kg , the tap draining the tank is turned off. Assume that the tank does not overflow and there is no change to the inflow rate.
  11. After the tap is turned off, how many minutes does it take for the concentration of salt in the tank to reach  \(\dfrac{1}{20} \ \text{kg L}^{-1}\)?   (1 mark)

Show Answers Only

a.    \(\text{Initial concentration}=\dfrac{5}{3000}=0.001\dot{6} \ \text{kg/L}\)

\(\text{Concentration entering tank}=0.1 \ \text{kg/L}\)

\(\text{Since} \ \ 0.1>0.001\dot{6}, \ \text{salt concentration increases with time.}\)
 

b.    \(Q=\text{salt in tank at time}\  t\)

\(\dfrac{d Q}{d t}=\dfrac{d Q}{d t_{\text {in }}}-\dfrac{d Q}{d t_{\text {out }}}\)

\(\dfrac{d Q}{d t}=0.1 \times 20-\dfrac{Q}{3000} \times 20=2-\dfrac{Q}{150}=\dfrac{300-Q}{150}\)
 

c.    \(61.05 \ \text{kg}\)

d.    \(Q=300-295 e^{-\tfrac{t}{150}}\)

e.    \(\text{As} \ \ t \rightarrow \infty, Q \rightarrow 300\)

f.    \(t=150\, \log _e\left(\dfrac{59}{40}\right) \ \text {minutes }\)

g.    \(t=50 \ \text{minutes}\)

Show Worked Solution

a.    \(\text{Initial concentration}=\dfrac{5}{3000}=0.001\dot{6} \ \text{kg/L}\)

\(\text{Concentration entering tank}=0.1 \ \text{kg/L}\)

\(\text{Since} \ \ 0.1>0.001\dot{6}, \ \text{salt concentration increases with time.}\)

♦♦♦ Mean mark (a) 20%.

b.    \(Q=\text{salt in tank at time}\  t\)

\(\dfrac{d Q}{d t}=\dfrac{d Q}{d t_{\text {in }}}-\dfrac{d Q}{d t_{\text {out }}}\)

\(\dfrac{d Q}{d t}=0.1 \times 20-\dfrac{Q}{3000} \times 20=2-\dfrac{Q}{150}=\dfrac{300-Q}{150}\)
 

c.    \(Q_1=5+15 \times \dfrac{300-5}{150}=34.5 \ \text{kg}\)

\(Q_2=34.5+15 \times \dfrac{300-34.5}{150}=61.05 \ \text{kg}\)

♦ Mean mark (c) 46%.
d.     \(\dfrac{d Q}{d t}\) \(=\dfrac{300-Q}{150}\)
  \(\dfrac{d t}{d Q}\) \(=\dfrac{150}{300-Q}\)
  \(\displaystyle \int d t\) \(=\displaystyle \int \frac{150}{300-Q} d Q\)
  \( t\) \(=-150\, \log _e(300-Q)+c\)

\(\text{When} \ \ t=0, Q=5:\)

\(0=-150\, \log _e 295+c \ \ \Rightarrow \ \ c=150\, \log _e 295\)

\( t\) \(=150\, \log _e 295-150\, \log _e(300-Q)\)
\( t\) \(=150\, \log _e\left(\dfrac{295}{300-Q}\right)\)

\(\text{Solve for} \ Q \ \text{ by (CAS):}\)

\(Q=300-295 e^{-\tfrac{t}{150}}\)
 

e.    \(\text{As} \ \ t \rightarrow \infty, Q \rightarrow 300\)
 

f.    \(\text{Find \(t\) when \(Q=100\) (using part d):}\)

\(t=150\, \log _e\left(\dfrac{295}{300-100}\right)=150\, \log _e\left(\dfrac{59}{40}\right) \ \text {minutes }\)
 

g.    \(\text{After the tap is turned off:}\)

\(Q=100+0.1 \times 20 t=100+2 t\)

\(\text{Volume in tank}=3000+20 t\)

\(\text{Solve for \(t\):}\)

\(\dfrac{1}{20}=\dfrac{100+2 t}{3000+20 t}\)

\(t=50 \ \text{minutes}\)

♦♦♦ Mean mark (g) 18%.

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. \(\dfrac{d Q}{d t}=\dfrac{4 Q}{1600-t}\)
Show Answers Only

\(A\)

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)`
Show Answers Only

`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`

Show Answers Only

`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)`

Show Answers Only

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

  3. Solve the differential equation given in part a. to find `Q` as a function of `t`.
  4. Express your answer in the form  `Q = a/(16 + 2t)^(b/c)`, where `a, b` and `c` are positive integers.  (3 marks)

Show Answers Only
  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: Taking the common factor of `2`  from `16+2t` 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`
`Q` `= (16 + 2t)^(- 3/2)/(2^(-5))`
`Q` `= 32/((16 + 2t)^(3/2))`