IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: ashish on October 08, 2010, 04:52:09 am
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two cylinders A and B are connected with a tube at their base.
cylinder A has a base area of 60cm^2, cylinder B has a base of 40cm^2!
initially the height of water in A was 120cm and in B was 20cm !
water is allowed to flow from A to B, the rate of flow is directly proportional to h, the difference in the two level of water!
can anyone please show me to get the differential equation in terms of t and h
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two cylinders A and B are connected with a tube at their base.
cylinder A has a base area of 60cm^2, cylinder B has a base of 40cm^2!
initially the height of water in A was 120cm and in B was 20cm !
water is allowed to flow from A to B, the rate of flow is directly proportional to h, the difference in the two level of water!
can anyone please show me to get the differential equation in terms of t and h
Volume of tubes is given by = Area of cross-section x height -----> V = (40h - 60h) = 20h
From this, we can find dv/dh = 20
Given dv/dt is proportional to -h ---> dv/dt = -kh
Therefore dh/dt = dh/dv x dv/dt
Hence dh/dt = 1/20 x -kh
Now you need to integrate ---> ln h = -(k/20)t + c
Given when t = 0, h = (120 - 20) = 100
ln100 = 0 + c ----> c = ln 100
Therefore the equation relating t and h will be
ln h = ln 100 - (k/20)t
Now you can rearrange it depending on the question :)
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i am getting a bit confused
dh/dt proportional to -h
or
dh/dt proportional to h ?
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i am getting a bit confused
dh/dt proportional to -h
or
dh/dt proportional to h ?
Yeah......you are right dude!
It is actually dh/dt proportional to -h since h is decreasing ;)
I'll modify my previous post :D
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+ rep yeah thanks my doubt is clear !
hey do you have Imran Amiran A/AS level pure mathematics book?
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+ rep yeah thanks my doubt is clear !
hey do you have Imran Amiran A/AS level pure mathematics book?
You're welcome :)
Hmm.....nopes. Never used that book though I heard a lot about its author who seems to be an extremely strict teacher :P
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here is another question
there are 2 cylinders A and B , A has base area of 60cm² and cylinder B has a base area of 40cm²
both of them are on the same level, and are connected by a tube at their bases. initially the water in A has a height of 120 cm while B has a height of 20cm. the rate of flow of water is proportional to the root of the height difference, h between the two cylinders.... show that dh/dt= -(k(h)^0.5)/24.....
can anyone show me how can i get this
i know that we should start by
dv/dt is proportional to (h)^0.5
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hey deadly king i managed to work it ;D
let say that cylinder A lost a volume V
new volume of water would be 7200(initial volume)- V
new height would be 120-(V/60)
on the other hand cylinder B has gained a volume V
new volume would be 800(initial volume)+ V
the new height would be 20+(V/40)
height difference h would be (20+(V/40))-(120-(V/60))
h=(V/24)-100
by making V subject of formula we obtain
V=24h+2400
now
as i said
dV/dt is proportional to root of h
since V= 24h+2400
d(24h+2400)/dt = -k(h)^0.5
since differential of a number is zero
24dh/dt=-k(h)^0.5
dh/dt=(-k/24)(h)^0.5
phew it was quite lengthy....