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!!!

describing distributions is one of the key skills you’ll need to earn a high score on the ap statistics exam. if you need proof of this, just flip through some past exam questions, which can be found at the collegeboard website. you’ll notice that the first free response question is almost always a question requiring you to look at a graph and describe it.

step-by-step explanation: just trying to get points sorry : (