hmm.. maths help again..
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hmm so anyone know what lvl maths is this? JC? someone said past year paper right? then which lvl? got the answer for this?
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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.. -
I think the part on dM/dt = 2(4) - (M/120)(3) is not correct as the concentration is not always M/120...Originally posted by papercut87: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..
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 -
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.Originally posted by papercut87: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..
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. -
Yeah i tink ur method is correct.Originally posted by missallsunday: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 -
I tink both our methods' answers are the same right? =pOriginally posted by Hamiltonian1125:Yeah i tink ur method is correct.
Anyway i think your answer is much more elegant haha. -
wow....you're chim....
and I am old...
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You are flattering mi laOriginally posted by missallsunday:I tink both our methods' answers are the same right? =p
Anyway i think your answer is much more elegant haha.
From the way u solve ur 2nd order D.E, are u an extinct dinosaurs( F math student)???
Anyway ur method is shorter
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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
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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) -
wow...ppl are using equations to reply.
interesting -
Haha.. thks peeps! Feel so inferior
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missallsunday's method straightforward and easy to comprehend.
hami's method of solving more elegant.
me? me = kuku -
must use differential eqn isit?Originally posted by papercut87:am here back to bug u maths xperts with with another maths question again..
thanks in advance,
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 -
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) -
Use laplacian transformation for that