1. A conical reservoir has a depth of 24 feet and a circular top of radius 12 feet. It is being filled so that the depth of water is increasing at a constant rate of 4 feet per hour. Determine the rate in cubic feet per hour at which water is entering the reservoir when the depth is 5 feet.

To determine the rate at which water is entering the reservoir when the depth is 5 feet, we need to find the rate of change of the volume of the water with respect to time.

We can start by finding an expression for the volume of a cone. The volume V of a cone is given by the formula:

V = (1/3) * π * r^2 * h

Where r is the radius of the circular top and h is the height of the cone.

Given that the radius r is 12 feet and the height h is 24 feet, we can substitute these values into the formula:

V = (1/3) * π * (12^2) * 24
V = (1/3) * π * 144 * 24
V = 1152π

Now, let's denote the height of the water in the reservoir as x. We are given that the depth of water is increasing at a constant rate of 4 feet per hour, so we can express the height of the water as a function of time x(t):

x(t) = 4t

Where t is the time in hours.

We can substitute this expression for x into the formula for the volume of the cone, with the radius and height being fixed values:

V(t) = (1/3) * π * r^2 * x(t)
V(t) = (1/3) * π * 144 * (4t)
V(t) = 192πt

To find the rate at which water is entering the reservoir, we need to differentiate the volume V(t) with respect to time t:

dV/dt = d/dt (192πt)
dV/dt = 192π

Therefore, the rate at which water is entering the reservoir is 192π cubic feet per hour when the depth is 5 feet.

To determine the rate at which water is entering the reservoir, we need to find the rate of change of the volume of water with respect to time, which is given by the derivative.

Let's start by finding an equation to represent the volume of water in the conical reservoir as a function of its depth.

We know that the volume of a cone is given by the formula V = (1/3) * π * r^2 * h, where V represents the volume, π is a constant approximately equal to 3.14, r represents the radius of the circular top, and h represents the depth.

Given that the radius is 12 feet, we can substitute this into the equation to get V = (1/3) * π * (12^2) * h = (1/3) * 144π * h = 48πh.

Now, let's differentiate this equation with respect to time to find the rate of change of volume with respect to time.

dV/dt = 48π * dh/dt.

We are given that the depth of water is increasing at a constant rate of 4 feet per hour, so dh/dt = 4.

Substituting this into the equation, we have dV/dt = 48π * 4 = 192π.

Therefore, the rate at which water is entering the reservoir when the depth is 5 feet is 192π cubic feet per hour.

dv = pi r^2 dh

where
r = (12/24)h = .5 h
so
dv = pi (.25)h^2 dh
dv/dt = .25 pi h^2 dh/dt
when h = 5
dv/dt = (25 pi/4 )dh/dt
dv/dt = (25 pi/4)(4)