Respuesta :
The level in the pot rising when the coffee in the cone is 5 in deep is 0.35 in/min.
Here the variables are the height of the coffee in the filter hf , the radius of the top of the coffee in the filter rf , and the height of the coffee in the pot hp:
We allowed Vp to represent the volume of coffee in the pot. We are interested in the instant when
hf = 5 in and dVp/dt = 10 in³/min.
The question is dhp/dt =? when dVp/dt = 10 in³/min and hf = 5 in
We need to relate hp to the other variables. In the pot (idealized as a cylinder), we have the volume Vp as Vp = π(3)²hp = 9πhp in³(since the volume of a cylinder of radius r and height h is V = πr²h).
Differentiation of Vp with respect to time gives dVp/dt = 9πhp/dt or
dhp/dt = (1/9π).(dVp/dt)
When dVp/dt = 10 in³/min and hf = 5 in, we have
dhp/ dt = 1/9π(10)
= 10/9π in/min
dhp/ dt ≈ 0.35 in/min (notice that the answer here is independent of hf ).
Hence, the level in the pot rising when the coffee in the cone is 5 in deep is 0.35 in/min.
To know more about volume check the below link:
https://brainly.com/question/463363
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