Rate of Water Rise in Cone-Shaped Vessel

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SUMMARY

The discussion focuses on calculating the rate of water rise in a cone-shaped vessel with a height of 2 meters and a diameter of 2 meters, where water flows in at a steady rate of 1 cubic meter per minute. The relationship between the radius and height of the water cone is established as \( r = \frac{1}{2}h \). By expressing the volume of water as a function of height and differentiating with respect to time, participants conclude that the height of the water when the vessel is 1/8 full is 0.25 meters, allowing for the calculation of the rate at which the water rises.

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markosheehan
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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 can't differentiate this
 
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At any point in time, how are the radius and height of the "cone of water" within the conical vessel related?
 
I don't know but I know they both get bigger with time.
 
markosheehan said:
I don't know but I know they both get bigger with time.

The vessel into which the water is flowing has a radius that is one half the height. Any "cone of water" partially or completely filling the vessel will be similar in shape to the vessel itself...so we may state (for the conical volume of water):

$$r=\frac{1}{2}h$$

And now, you can express the volume of water as a function of $h$ alone...and then implicitly differentiate with respect to time $t$ and given that you know $$\d{V}{t}$$, you may solve for $$\d{h}{t}$$

You will also need to know the height of the volume of water when the vessel is 1/8 full. By similarity, we see this is when the height of the water is 1/2 that of the vessel. ;)
 

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