Conveyor belt at a rate

Question

Gravel is being dumped from a conveyor belt at a rate of 30 ft³/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast (in ft/min) is the height of the pile increasing when the pile is 15 ft high? (Round your answer to two decimal places.)

Please I need help

Answer 1

Let V = volume, h = height, and r = radius

V = (1/3)πr²h h = height = diameter = 2r

V = (1/3)π(h/2)²h

V = (1/12)πh³

dV/dt = (1/4)πh²(dh/dt)

30 = (1/4)π(15)²(dh/dt)

Solve for dh/dt.

Answer 2

This is a related rates calculus problem. The rate given in this problem is a volumetric flow rate, meaning that we need the Volume of a Cone to solve this problem.

Let's first write our given values:

1. dV/dt = 30 ft³ / min
2. h = 15 ft
3. diameter (2r) = height (h)

1. Start with the volume of a cone

V = (1/3) pi * r² * h

Since height is always equal to the diameter at any given point in time:

d = 2r = h

We can re-write r in terms of h

r = h / 2

Squaring both sides, we get:

r² = (h/2)² = h² / 4

Substitute this into the volume equation.

V = (1/3) pi * r² * h

= (1/3) pi * (h² / 4) * h

= (1/3) pi * h³ / 4

V = (1/12) pi * h³

2. Take the derivative of both sides with respect to time

dV/dt = d/dt [(1/12) pi * h³]

dV/dt = (1/12) pi * 3 h² dh/dt

dV/dt = (1/4) pi * h² dh/dt

3. Now, plug in all the values we were given.

30 = (1/4) pi * (15)² dh/dt

30 = (1/4) pi (225) dh/dt

4. Solve for dh/dt