Draining Cup Problem

[Jelfes]

The Problem

I have labelled a picture of this as such:


Now I have Volume of Cone: Cone = 1/3 * pi * 16(r^2) * 20(h) = (320 * pi)/3
Volume of Cup: Cup = (320 * pi)/3 - (135 * pi)/3 = (185 * pi)/3

Now that I have the volume of the of the cup containing the 3cm depth of water, how do I extract the rate of change of the water level from what I have?

 

[HallsofIvy]

The radius goes from 4cm to 3 cm over a height of 5 cm. At that same slope, it will got to 0 over an additional 3(5)= 15 cm. So you can think of this as a cone with base radius 4 cm and height 5+ 15= 20 cm. with the lower part, a cone with base radius 3 and height 15 cm, removed.
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[Jelfes]

The radius goes from 4cm to 3 cm over a height of 5 cm. At that same slope, it will got to 0 over an additional 3(5)= 15 cm. So you can think of this as a cone with base radius 4 cm and height 5+ 15= 20 cm. with the lower part, a cone with base radius 3 and height 15 cm, removed.
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Thank you, I understand the validity in using the cone volume formula now.

 

[Jelfes]

I still need help finding the rate of change of the water level at 3cm full. Thanks!

 

[Deleted member 4993]

I still need help finding the rate of change of the water level at 3cm full. Thanks!

Next, when the height of water is 'h' - calculate the volume of water contained in the cup (as a function of 'h'.

Next differentiate with respect to time..