Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
junior college 2 | H2 Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
part i and ii, thanks!!
= 5g/L x 2L/min
= 10g/min
Since 2L of salt water is entering the tank per minute and 1L of mixture flows out of the tank per minute,
the net change in the volume of the tank per minute = 2L - 1L = 1L
After t minutes, the volume of liquid in the tank
= 10L + 1L x t
= (10 + t)L
Question says that amount of dissolved salt in the tank is S g at time t.
So at time t, the amount of salt leaving the tank
= Sg/(10 + t)L x 1L/min
= S/(t + 10) g/min
Therefore rate of change of amount of salt in the tank at time t
= 10g/min - S/(t + 10) g/min
dS/dt = 10 - S/(t + 10)
(shown)
Assumption :
① The tank is well stirred and the contents are evenly mixed, such that the concentration of salt anywhere in the mixture is the same. i.e homogeneous mixture
② There is no evaporation or spillage or water
dS/dt = 10 - S/(t + 10)
dS/dt + S/(t + 10) = 10
Integrating factor is e^(∫ 1/(t+10) dt)
= e^(ln(t + 10))
= t + 10
So multiply both sides by (t + 10).
(Sometimes this is quite intuitive and there's no need to find the integrating factor. If you can see it directly just bring the term with S over and multiply by (t + 10) )
(t + 10) dS/dt + S = 10t + 100
The left hand side is now directly integrable. Reverse of product rule gives (t+10)S
Integrate both sides with respect to t,
∫ (t + 10) dS/dt + S) dt = ∫ (10t + 100) dt
(t + 10)S = 5t² + 100t + C , C is a constant.
When t = 0, S = 300 (from question which says initially the tank has 300 grams of salt mixed with 10 litres of water)
So (0 + 10)(300) = 5(0²) + 100(0) + C
C = 3000
Therefore,
(t + 10)S = 5t² + 100t + 3000
S = (5t² + 100t + 3000)/(t + 10)
S = ( (5t + 50)(t + 10) + 2500 )/(t + 10)
S = 5t + 50 + 2500/(t + 10)
Q = (t + 10)S
So dQ/dt = (t + 10)dS/dt + S ---①
(product rule)
Since dS/dt = 10 - S/(t + 10),
(t + 10) dS/dt = 10(t + 10) - S
(t + 10) dS/dt + S = 10(t + 10)
Substitute this into ①,
dQ/dt = 10(t + 10)
dQ/dt = 10t + 100
Q = ∫ (10t + 100) dt
Q = 5t² + 100t + C
So (t + 10)S = 5t² + 100t + C
When t = 0, S = 300,
(0 + 10)(300)= C
C = 3000
So (t + 10)S = 5t² + 100t + 3000
S = (5t² + 100t + 3000)/(t + 10)
S = 5t + 50 + 2500/(t+10)
See 1 Answer