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Model a water tank by a cone 4040 ft high with a circular base of radius 2020 feet at the top. Water is flowing into the tank at a constant rate of 8080 cubic ft/min . How fast is the water level rising when the water is 1212 ft deep

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Answer:

dh / dt = .13 ft / s

Step-by-step explanation:

Let at height h , radius be r within the cone.

20 / 40= r / ( 40-h )

2r = 40 -h

r = (40 - h ) / 2

Volume of upper cone

V = 1/3 πr² ( 40 - h )

= 1/3 π (40 - h )² x ( 40 - h ) / 4

= 1/12 x π (40 - h )³

Differentiating on both sides

dV /dt = 1/12 x3 x π (40 - h )² dh / dt

= π (40 - h )²/ 4 dh / dt

Given

dV /dt = 80 , h = 12

80 = π / 4 x ( 40 - 12 )² dh / dt

dh / dt = .13 ft / s

User Yoshua Nahar
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