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#1 2017-01-12 08:07:55

markosheehan
Member
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
Posts: 1,086

Re: cone


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#3 2017-01-13 22:10:04

markosheehan
Member
Registered: 2016-06-15
Posts: 51

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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Registered: 2010-06-20
Posts: 10,053

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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