A large container has the shape of a frustum of a cone with top radius 5 m, bott

Question

A large container has the shape of a frustum of a cone with top radius 5 m, bottom radius 3 m, and height 12 m. The container is being filled with water at the constant rate of 2.8 m3/min. At what rate is the level of water rising at the instant the water is 1 m deep? Note: A frustum of a cone is the portion of a cone between two parallel planes. For example, the popular coffee cup has the shape of a frustum of a cone The water is rising at the rate of cm/min (Round to one decimal place as needed.)

Explanation / Answer

Solution:

When the water has depth y, the radius of the surface is

3 + (y/12)(5 - 3) = 3 + y/6

Its volume at that point is

v = 1/3 (R2 - r2) h

= (1/3)((3 + y/6)2 - 32)y

= y3/108 + y2/3

Now we are ready to begin. When y = 1,

v = 1/108 + 1/3 = 37/108                                 

dv/dt = (y2/108 + 2y/3) dy/dt

2.8 = (1/108 + 2/3) dy/dt

dy/dt = 8.17 m/min

The actual formula for the volume is

(/3)((18 + y)(3 + y/6)2 - 18*32)

= (y3/108 + y2/2 + 9y)

because the water is in the bottom of the cone

dv/dt = /36(y + 18)2 dy/dt

2.8 = /36*192 dy/dt

dy/dt = 0.08 m/min or 8 cm/min.

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