Sand falls from an overhead hopper to form a right circular cone?

Question

If the cone formed has an angle a find the rate of change of volume with respect to height? If the height is changing at 2cms per minute, what volume of sand is falling from the hopper when the height of the cone is 3 meters? (the volume of a cone is (1/3)pie^2h where h is the height and r is the radius)

Answer 1

yes

Answer 2

Since it's a right triangle, the radius and the height are related by the angle a:

tan a = h/r, so r = h/tan a

(I'm assuming that the angle a is the angle at the base of the cone, not at the top.)

So, the volume of the cone is:

V = 1/3 π r² h

Substituting for r:

V(h) = 1/3 π (h/tan a)² h = 1/3 π h³ cot²a

Then, since your first question asks for the rate of change of volume with respect to height:

dV/dH = π h² cot²a dh/dh = π h² cot²a

Or differentiating V with respect to t:

dV/dt = V'(h) = π h² cot²a dh/dt

So, when h = 3 m and dh/dt = 2cm/min = 0.02 m/min

V'(3) = π (3 m)² cot²a (0.02 m/min) = 0.18π cot²a m³/min

Since they don't give you a, I think that's as good as you can do.

If they intend a to be the angle at the top, the answer would instead be 0.18π tan²(a/2) m³/min.

Answer 3

Let

V = volume of cone
r = radius of base of cone
h = height of cone

Given:

The cone has an angle a. This is ambiguous. I am assuming you mean the vertex of the cone has angle a.

dh/dt = 2 cm/min

Find dV/dh
Find dV/dt when h = 3 m
______________

tan(a/2) = r/h
r = h tan(a/2)

V = ⅓πr²h = ⅓π[h tan(a/2)]² h
V = ⅓π[tan²(a/2)] h³

dV/dh = (3)(⅓π[tan²(a/2)] h²)
dV/dh = π[tan²(a/2)] h²

dV/dt = (dV/dh)(dh/dt)
dV/dt = (π[tan²(a/2)] h²)(2) = 2π[tan(a/2)]² h²
dV/dt = 2π[tan²(a/2)] (300²)
dV/dt = 2π[tan²(a/2)] (90000)
dV/dt = 180,000π[tan²(a/2)] cm³/sec
dV/dt = 0.18π[tan²(a/2)] m³/sec

Answer 4

v = (1/3)(pi)r^2 h
r/h = tan(a/2)

Therefore v = (1/3)(pi)[tan(a/2)]^2 h^3

dv/dh = (pi)[tan(a/2)]^2 h^2

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