Question

It would be so helpful if someone could check my answers!

A machine starts dumping sand at the rate of 20 m3/min, forming a pile in the shape of a cone. The height of the pile is always twice the length of the base diameter. The volume formula for a right circular cone is V= 1/3pi*r2.*h

A. After 5 minutes, what is the height of the pile?
h = 2d or 4r
v = π/3 r^2 h
v = π/3 (h/4)^2 h
v = π/48 h^3

After 5 minutes, 100m^3 of sand would have been dumped. Therefore:
π/48 h^3 = 100
So 100 = π/48 h^3
48 (100) = (π/48 h^3) 48
4800 = π*h^3
4800/π = h^3
∛(4800/π) = h
approximately 11.51m

B. After 5 minutes, how fast is the height increasing?
dv/dt = π/4 h^2 dh/dt
20 = π/4 ∛(4800/π)^2 dh/dt
dh/dt = 20 * 4/π * ∛(π/4800)^2
(I was having some trouble simplifying this one and the two below it)

C. After 5 minutes, how fast is the base radius increasing?
since h = 4r, dh/dt = 4 dr/dt
4(dr/dt) = 20 * 4/π * ∛(π/4800)^2
dr/dt = (20 * 4/π * ∛(π/4800)^2)/4

D. After 5 minutes, how fast is the area of the base increasing?
base area: A=πr^2,
dA/dt = 2πr dr/dt
2π * (∛(4800/π))/4 * dr/dt
2π ∛(4800/4π) * dr/dt
2π ∛(4800/4π) * (20 * 4/π * ∛(π/4800)^2)/4

Answer 1

I see you fixed my error where I said h=2r.
(A) looks good
(B) 20 * 4/π * ∛(π/4800)^2 = 1/∛(45π) ≈ 0.19196
(C) as you say, dr/dt = 1/4 of dh/dt ≈ 0.04799
(D) Oops . why did you drag the 4 under the radical?
2π * (∛(4800/π))/4 * dr/dt
= π/2 * (∛(4800/π)) * 1/4 * 1/∛(45π) = ∛(5π/24) = 0.8682

Answer 2

Thank you so much for your time and help!