IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: loggerhead on October 02, 2009, 10:28:16 pm
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Im stuck on this question can someone please walk me through it.
An underground storage tank is being filled with liquid as shown in the diagram. Initially the tank is
empty. At time t hours after filling begins, the volume of liquid is V m3 and the depth of liquid is h m.
It is given that V = 4/3h3
The liquid is poured in at a rate of 20m3 per hour, but owing to leakage, liquid is lost at a rate
proportional to h2.
When h = 1, dh/dt = 4.95.
(i) Show that h satisfies the differential equation
dh/dt = 5/h2 - 1/20
(ii) Verify that 20h2/(100 - h2) = 20 + 2000/((10 - h)(10 + h))
(iii) Hence solve the differential equation in part (i), obtaining an expression for t in terms of h. ???
This is CIE, 9709/03 (mathematics paper 3), October November 2008, question 8. I've attached the paper
Any help would be much appreciated.
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This question has been answered. I will look for the link tomorrow.
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I'm still not good with calculus...will try doing it though
And maybe I've seen the question before...so I'll look for the link
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Here is the link
https://studentforums.biz/index.php/topic,740.0.html
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I cant seem to follow the link, i just get this
"The topic or board you are looking for appears to be either missing or off limits to you."
what do i do??
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8iii) dh/dt= Separate variables to get dh=dt then use previous part and integrate
-20h-ln(100-h)+ln(100+h)+C=t so t=-20h+ln((100+h)/(100-h))+C
when h=0, t=0
then C=0 and t=t=-20h+ln((100+h)/(100-h))
This is what astar answered in that thread
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Thanks nid
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anytime :)
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thanks guys much appreciated. I'm sure there will be more questions to come.