# From a faucet, a constant inflow of water is to fill a conical vessel 15 feet deep and 7.5 feet in diameter at the top. water is rising at the rate of 2 feet per minute when the water is 4 feet deep. what is the rate of inflow in ft^3/min?

## Make a sketch of the cone with some water in it.

Let the depth of the water be h ft, and the radius of the water level be r ft
by similar triangles:
r/h = 3.75/15
3.75h = 15r
h = 4r or r = h/4

V = (1/3)π r^2 h
V = (1/3)π(h^2/16)h
V = (1/48)π h^3
dV/dt = (1/16)π h^2 dh/dt
given: when h = 4 , dh/dt = 2

dV/dt = (1/16)π (16)(2) = 2π

The rate of inflow is 2π ft^3/min

check my arithmetic, I did not write it out on paper first.

## Oh, we have quite the aquatic conundrum here! So, let me see if I can make a splash with my calculations.

First things first, we have ourselves a conical vessel. Now, to find the rate of inflow, we need to determine the volume of the cone. And for that, we'll need some formula-fishing.

The formula to find the volume of a cone is V = πr²h/3, where "V" is the volume, "r" is the radius, and "h" is the height. In this case, the height of the cone is changing with time.

We know the diameter at the top is 7.5 feet, so the radius would be half of that, which is 3.75 feet.
The water is rising at a rate of 2 feet per minute when the water is 4 feet deep. So the height "h" at that moment is 4 feet.

Plug in these values into the volume formula, and we get:
V = π(3.75)²(4)/3

Now we can solve for the volume. Once we have that, we can figure out the rate of inflow. So let me do some calculations here... *buzzing and whirring noises*

Calculating... calculating... and voila! The volume at that specific moment is approximately 49.5 ft³.

But the question asks for the rate of inflow in ft³/min, so we need to determine how quickly the volume is increasing. Since the water is rising at a rate of 2 feet per minute, we can divide that by the current height of 4 feet to find the rate of inflow.

2 ft/min / 4 ft = 0.5 ft/min

And there you have it! The rate of inflow is 0.5 ft³/min. Water you waiting for? Dive right in with that answer!

## To find the rate of inflow in ft^3/min, we need to determine the volume of water being added per minute.

The volume of a cone can be calculated using the formula: V = (1/3) * π * r^2 * h, where V is the volume, π is pi (approximately 3.14), r is the radius, and h is the height.

Given:
- The depth of the conical vessel is 15 feet.
- The diameter at the top of the vessel is 7.5 feet, which means the radius (r) is half of the diameter, i.e., 7.5/2 = 3.75 feet.
- The water is rising at a rate of 2 feet per minute when the water depth is 4 feet.

We can calculate the initial height of the water in the conical vessel by subtracting the current water depth from the total depth:
Initial height = Total depth - Current depth = 15 ft - 4 ft = 11 ft

Now, we can calculate the initial volume of the water in the conical vessel using the formula mentioned above:
Initial volume = (1/3) * π * r^2 * h
= (1/3) * 3.14 * (3.75 ft)^2 * 11 ft
≈ 58.35 ft^3

Since the water is rising at a rate of 2 feet per minute, the rate of change of the height (dh/dt) is equal to 2 ft/min.

To find the rate of inflow in ft^3/min, we need to calculate the rate of change of volume with respect to time, dv/dt.

We can use the chain rule of differentiation to calculate dv/dt:
dv/dt = (dv/dh) * (dh/dt)

Differentiating the volume formula with respect to height, we have:
dv/dh = (1/3) * 3.14 * 2 * r^2
= (2/3) * 3.14 * (3.75 ft)^2
≈ 27.875 ft^2

Now, we can calculate the rate of inflow, dv/dt:
dv/dt = (dv/dh) * (dh/dt)
= 27.875 ft^2 * 2 ft/min
= 55.75 ft^3/min

Therefore, the rate of inflow of water is approximately 55.75 ft^3/min.

## To find the rate of inflow in ft^3/min, we need to use the formula for the volume of a cone:

Volume = (1/3) * π * r^2 * h

First, let's find the formula for the radius (r) of the cone at any given height.

The diameter at the top is given as 7.5 feet, which means the radius is half of that:

r = 7.5 / 2 = 3.75 feet

Now, let's find the formula for the height (h) of the cone at any given time, t.

We know that the water is rising at a rate of 2 feet per minute when the water is 4 feet deep. So, the height at time t is given as:

h = 4 + 2t

Now, let's substitute these values into the volume formula:

Volume = (1/3) * π * (3.75^2) * (4 + 2t)

Taking the derivative with respect to time (t) will give us the rate of change of volume (dV/dt) with respect to time:

dV/dt = (1/3) * π * (3.75^2) * 2

Simplifying:

dV/dt = π * 3.75^2 * 2 / 3

Calculating the numerical value:

dV/dt ≈ 7.85398 ft^3/min

So, the rate of inflow is approximately 7.85398 ft^3/min.