VGravel is being dumped from a conveyor belt at a rate of 50 ft3/min. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 14 ft high

0.325 ft/min

Step-by-step explanation:

To solve we must follow the following steps:

First, the volume of a cone is given by the equation:

Vc = (1/3) * Pi * (r ^ 2) * h

where r is the radius and h is the height.

they tell us that the height is equal to twice the radius, therefore

h = 2r; r = h / 2

replacing r in the volume of the cone:

Vc = (1/3) * Pi * ((h / 2) ^ 2) * h

solving we have:

Vc = (1/3) * Pi * (1/4) * (h ^ 2) * h

Vc = (1/12) * Pi * h ^ 3

They give us the change in volume with respect to time, that is, dV / dt = 50

It can also be expressed in the following way:

dV / dt = (dV / dh) * (dh / dt)

We know V, so we can derive with respect to h, we are like this:

dV / dh = 3 * (1/12) * Pi * h ^ 2

dV / dh = (1/4) * Pi * h ^ 2

we know that h is equal to 14, therefore:

dV / dh = (1/4) * Pi * 14 ^ 2

dV / dh = (1/4) * 3.14 * 196 = 153.86

Now if we replace

dV / dt = (dV / dh) * (dh / dt)

50 = 153.86 * (dh / dt)

(dh / dt) =  50/158.36

(dh / dt) =  0.3249

This means that the change in height with respect to time is 0.325 feet per minutes.

The height of the pile is increasing at a rate of 0.325 ft/min

Step-by-step explanation:

Given

1) (dV/dt) = 50 ft³/min

2) The shape formed is a cone with diameter = height.

That is, d = h

2r = h

r = (h/2)

Find (dh/dt) when h = 14 ft

Volume of a cone = (πr²h/3)

with r = (h/2)

Volume of the cone = [π(h/2)²h/3]

V = (πh³/12)

But

(dh/dt) = (dh/dV) × (dV/dt)

V = (πh³/12)

(dV/dh) = (πh²/4)

(dh/dV) = (4/πh²)

(dh/dt) = (dh/dV) × (dV/dt)

(dh/dt) = (4/πh²) × 50 = (200/πh²)

when h = 14 ft

(dh/dt) = (200/π×14²)

(dh/dt) = 0.325 ft/min

Hope this Helps!!!

many reason,

step-by-step explanation: