Question
Asked Oct 19, 2019
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Water flows from a tank of constant cross-sectional area 40ft^2 through an orifice of constant cross-sectional area 1/4ft^2 located at the bottom of the tank. Initially, the height of the water in the tank was 20ft, and t sec later it was given by the equation

2(h^1/2) + (1/20)t - 2(20^1/2) = 0         0<= t <= 40(20^1/2)

How fast was the height of the water decreasing when its height was 16ft?

 

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Expert Answer

Step 1

Compute the der...

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1 -t - 2/20 = 0. 20 Given that 2h Differentiate the above equation with respect to t. 1 dh 1 0 20 dh dt 20

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