# differential equations questions again!!!Watch

#1
water is leaking slowly out of a tank. The depth of the water after t hours is h metres, and these variables are related by a differential equation of the form dh/dt= -ae^-0.1t
Initially the depth of water is 6 metres, after 2hours it has fallen to 5 metres. At what depth will the level eventually settle down?
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12 years ago
#2
hi, is the question meant to be at what time will the tank be empty?
Edit: ignore me, Im sure you've typed it out right, Im just having a go now...
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12 years ago
#3
Ok, I reckon what you have to do is work out an equation for h=, finding out a by substituting the values they give you in the question, then you can sketch the graph and find out the asymptote. Do you have the answer? I'll try and design some working now....
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12 years ago
#4
^ Thats what I've tried to do, but that doesnt give an actual depth, just something in relation to time....
oh its gone
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#5
um... its not wen the tank is empty its wen the level settles as in wen no more liquid flows out i think i'm sposed to equate dh/dt to zero but i dont get it rite
thanks for the help btw!!!
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12 years ago
#6
You sure the question is right? as dh/dt will never equal zero unless a=0, so that means the level will never settle but continue to drop.
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#7
yeah i copied the q off of the book its right... but i'm not sure if i'm sposed to equate dh/dt to zero
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12 years ago
#8
dh/dt can never be zero, it approaches zero as t approaches infinity.
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12 years ago
#9
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12 years ago
#10
dh = -ae^(-0.1t)dt
h(t) = K + 10ae^{-0.1t}
h(2) = 5 = K +10ae^(-0.2)
h(0) = 6 = K + 10a

Work out K and that's your answer because at t-> infinity the 10ae^(-0.2t) -> 0 and so h(t) -> K.
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#11
yup .483 is the answer how'd u get it?
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#12
um..alphanumeric i dont get how u did it!!! um... ok maybe i'm not trying hard enuff...lemme try once again!!!
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12 years ago
#13
(Original post by AlphaNumeric)
dh = -ae^(-0.1t)dt
h(t) = K + 10ae^{-0.1t}
h(2) = 5 = K +10ae^(-0.2)
h(0) = 6 = K + 10a

Work out K and that's your answer because at t-> infinity the 10ae^(-0.2t) -> 0 and so h(t) -> K.
K = 6-10a
K = 5-10ae^(-0.2)

=> 6-10a = 5-10ae^(-0.2)
=> a(10-10e^(-0.2)) = 6-5
=> a = 1/(10-10e^(-0.2))

As t -> infinity, h(t) -> K

K
= 6-10a
= 6-10(1/(10-10e^(-0.2))
= 6-10/(10-10e^(-0.2))
= 0.483m
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#14
thanks!!!
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