differential equations questions again!!! Watch

frixis
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#1
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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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Twiglet
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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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Twiglet
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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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Twiglet
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^ 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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frixis
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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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silent ninja
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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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frixis
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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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silent ninja
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dh/dt can never be zero, it approaches zero as t approaches infinity.
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silent ninja
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Is the answer 0.483m?
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AlphaNumeric
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#10
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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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frixis
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yup .483 is the answer how'd u get it?
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frixis
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#12
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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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Christophicus
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(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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frixis
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#14
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thanks!!!
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