Question

A brewpub is brewing beer in a large cylindrical container with radius 6 meters. Beer is being removed from the container at a rate of 15 cubic meters per hour. Find the rate at which the height of beer in the container is changing. The volume of a cylinder is given by Upper V equals pi r squared h​, where r is the radius and h is the height of the cylinder.

Answer 1

Answer:

$\frac{dh}{dt} = \frac{F_{in} }{36\pi } - \frac{5}{12}$

Explanation:

Thinking process:

Let the radius = 6 m

The rate of removal of the beer = 15 m³/hr

This can be written as: $F_{out} = 15 m^{3}/hr$

the rate or change $\frac{dV}{dt} = F_{in} - F_{out}$

this changes to $\frac{d(Ah)}{dt} = F_{in} - F_{out} \\A\frac{dh}{dt}= F_{in}-F_{out} \\\frac{dh}{dt } = \frac{1}{A}(F_{in}-F_{out})\\ =\frac{F_{in} }{36}-\frac{5}{12}$