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an industrial manufacturing company uses an inverted conical (cone-shaped) tank to dispense liquid into containers. the tank measure 24 inches with a base radius of 48 inches. if the liquid flows out of the tank at a rate of 40 cubic inches per minute, at what rate is the height of the liquid falling when the height of the liquid is 10 inches deep?



Answer :

ayune

If the liquid flows out of the tank at a rate of 40 cubic inches per minute, the height of the liquid will decrease at rate 0.0318 in./minute

Referring to the attached picture, the two triangles in a cone are similar. Hence,

r/h = 48/24

or

r = 2h.

The volume of the liquid is given by:

V = 1/3 . πr²h

Substitute r = 2h,

V = 1/3 . π(2h)²h = 4/3 . πh³

Take the derivative with respect to t

dV/dt = 4/3 .  3πh² . dh/dt

dV/dt = 4 .  πh² . dh/dt

Substitute dV/dt = -40 and h = 10

-40 = 4 .  π(10)² . dh/dt

dh/dt = - 0.0318 in./minute

Learn more about derivative here:

https://brainly.com/question/22716418

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