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#1 2017-01-12 08:07:55
- markosheehan
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- Registered: 2016-06-15
- Posts: 51
cone
a vessel in the shape of a cone is standing on its apex. water flows in at a steady rate, of 1m^3 per minute. the vessel has a height of 2m and a diameter of 2m when the vessel is 1/8 full find the rate at which water is rising
v=1/3 pi r^2 h however i cant differentiate this
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#2 2017-01-12 15:14:35
- thickhead
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- Registered: 2016-04-16
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Re: cone
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#3 2017-01-13 22:10:04
- markosheehan
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- Registered: 2016-06-15
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Re: cone
i actually worked it out using dv/dt=dv/dh*dh/dt the answer i got was pi/4.im stuck on the next part though. what is the rate at which the free surface area of the water is increasing when the cone is 1/8th full
0
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#4 2017-01-13 23:02:19
- Bob
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Re: cone
hi markosheehan
For part 2:
As you want dA/dt I would work with dV/dt = dV/dA . dA /dt
As D = 2 when H = 2. => r = 0.5h.
Express both V and A in terms of h and eliminate h to get V in terms of A.
Differentiate that.
Bob
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