DE

[ashish]

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

[Deadly_king]

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

[ashish]

i am getting a bit confused
dh/dt proportional to -h
or
dh/dt proportional to h ?

[Deadly_king]

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 

[ashish]

+ rep  yeah thanks  my doubt is clear !

hey do you have  Imran Amiran A/AS level pure mathematics book?

[Deadly_king]

+ 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

[ashish]

 
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

[ashish]

hey deadly king i managed to work it  

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....