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.

Answer This Question
S
S

junior college 2 chevron_right H2 Maths chevron_right Singapore

part i and ii, thanks!!

Date Posted: 4 years ago
Views: 278
J
J
4 years ago
Rate of the amount of salt entering the tank
= 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
J
J
4 years ago
Method without using substitution :

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)
J
J
4 years ago
Method using substitution :

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

done {{ upvoteCount }} Upvotes
clear {{ downvoteCount * -1 }} Downvotes
Jiayang
Jiayang's answer
1883 answers (Tutor Details)
1st