Qualification > Math

DE

(1/2) > >>

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:

--- Quote from: ashish on October 08, 2010, 04:52:09 am ---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

--- End quote ---

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:

--- Quote from: ashish on October 08, 2010, 05:15:35 am ---i am getting a bit confused
dh/dt proportional to -h
or
dh/dt proportional to h ?

--- End quote ---

Yeah......you are right dude!

It is actually dh/dt proportional to -h since h is decreasing  ;)

I'll modify my previous post  :D

ashish:
+ rep  yeah thanks  my doubt is clear !

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

Navigation

[0] Message Index

[#] Next page