hmm.. maths help again..

CoolMyth SG

What module is this? Shocked

Paiseh, I very bad at making the starting equation. Laughing

105090 SG

nostalgic question Laughing

must use add. maths calculus

Embarassed

snop_coolz SG

i believe this is an add.maths statistic qns..u need to form a formula first then with the formula solve the qns..

papercut87 SG

MA1505.. past year paper

popikachu SG

i try and help... but may be wrong as i haven learn this in my sch...

First of all, find the time need for the tank to be full...

"A tap is turn on and salt solution of 2g/L at 4L/min"
i guess you understand this sentence...

120L-90L=30L (amount needed of solution to be full)
4L/min - 3L/min = 1L/min (the tank will gain 1L/min of water...)
so to have 30L of water, we need 30min...

__________________________________________________

"the WELL-STIRRED mixture" let just assume that there is magic here, and when the solution flow down, the whole mixture is mix...
"flow out of the tank at a rate of 3L/min" i guess you can understand this sentence...
every time, the amount of salt is different, now to find the speed of salt added...

since one min the will be 4L and 1L have 2 gram of salt, the speed of salt add is 4g/min
2g/L so 8g/4L
4L/min so 8g/min...
__________________________________________________

next we need to find the speed of salt lost... this is the biggest problem as the amount of salt in water in different in every second, 1/2 second, 1/4 second~~...

up til here is dunno liao...

popikachu SG

hmm so anyone know what lvl maths is this? JC? someone said past year paper right? then which lvl? got the answer for this?

papercut87 SG

here's how i came out with my ans:

M: amt of salt in g in the mixture at time t.

M(0)=90
dM/dt = 2(4) - (M/120)(3) --(*)
Solve (*) to get:
-40ln|8-M/40| = t + c
When t = 0, M = 90
c = -40ln|8-9/4|
At t = 30, (since it takes 30 mins to fill up the tank)
-40ln|8-M/40| 30 - 40ln|8-9/4|
M = 211g

seniors, hegu, did i do it correctly? somehow i think im wrong..

missallsunday SG

I think the part on dM/dt = 2(4) - (M/120)(3) is not correct as the concentration is not always M/120...

I am not sure about this too but this is how i do it...

the equation shld be dM/dt = 8-3M/V where V varies with time too..
Thus we need to take derivative wrt t again so as to account for the change in volume.
Thus M'' = -3M'/v+3m/(v^2)(dv/dt) (by chain rule) where M'' is 2nd derivative of M wrt t
dv/dt = 1 as 4-3 =1
solving this 2nd order DE,
we get M=Ce^(0.79129t/V)+De^(-3.7913t/V)
to solve for C and D,
we know initial M = 90
we also know initial dM/dt = 8-3(1) = 5 (1 is the initial concentration)
with these 2 conditions we can solve for C and D
.
.
.
By magic we get C to be 172.66 and D to be -82.658
Subbing t = 30 as 30 min is needed to fill the tank and V =120 to the equation, we get ans to be 178.39g

Hope i am right

Hamiltonian1125 SG

dM/dt = 2(4) - (M/120)(3) --(*) This part is doubtful....How can u divide M by 120 when u do not even know the vol of solution at time t.

Let,
N(t)=amt salt at time t
V(t)=90+4t-3t=90+t ( this is the vol at time t, initially at 90 but need to add in the vol pump in and minus the yol pump out.

dN(t)/dt=(2g/L)(4L/min)-(3L/min)(N(t)/V(t)) this last part is rate of flow out times conc at time t so we get amt of salt out per unit time...always include ur units in this ode as they help u to check if the eqn is making physical sense.

dN(t)/dt=(2g/L)(4L/min)-(3L/min)(N(t)/V(t))
=8-3[N/(90+t)]
Now we get a linear ode in terms of N and t. Pls note that in mi eqn N(t)=N...
For above ode use integrating factor to solve u should be able to get the ans.

Hamiltonian1125 SG

Yeah i tink ur method is correct.

missallsunday SG

I tink both our methods' answers are the same right? =p
Anyway i think your answer is much more elegant haha.

nightzip SG

wow....you're chim.... Laughing

and I am old... Sad Mr. Green

Hamiltonian1125 SG

You are flattering mi la Embarassed From the way u solve ur 2nd order D.E, are u an extinct dinosaurs( F math student)???
Anyway ur method is shorter Smile

missallsunday SG

Actually my method is flawed that 2nd ode cannot be solved easily i tink your answer is right shld be 202.03 using the integrating factor method

HyuugaNeji SG

my equation is also like this which is the same as hamiltonian.

x'(t) = 2(4) -3[x(t)/(90+t)]

Solve it and you get the answer

x(t) = in terms of just t + a constant k.

You solve for k by putting t=0 and x = 90 (initial amount of salt)

Classyfied SG

wow...ppl are using equations to reply.
interesting Laughing

papercut87 SG

Haha.. thks peeps! Feel so inferior Razz

papercut87 SG

missallsunday's method straightforward and easy to comprehend.
hami's method of solving more elegant.

me? me = kuku Mr. Green

hisoka SG

must use differential eqn isit?

ds/dt = 4* 2 - 3*s/l

dl/dt = 4 - 3

t = 0, l = 90, s = 90* 1

so l = t +90

erm i not sure how to solve the diff eqn lazy to try out but you check if your method is somethign liek that.

s = amoutn of salt, and l = amount of water

HyuugaNeji SG

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