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
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
Thank you so much for your time and help!
To solve these problems, we can use the given information and the formula for the volume of a cone.
A. After 5 minutes, what is the height of the pile?
First, we need to modify the formula for the volume of a cone to fit the given information. Since the height is always twice the length of the base diameter, we have h = 2d or 4r, where r is the radius. Substituting this into the volume formula, we get:
V = (π/3) * r^2 * h
V = (π/3) * (h/4)^2 * h
V = (π/48) * h^3
We know that after 5 minutes, 100 m^3 of sand has been dumped. So, we can set up the equation:
(π/48) * h^3 = 100
Solving for h, we have:
h^3 = (100 * 48) / π
h = ∛(4800/π)
h ≈ 11.51 m
Therefore, after 5 minutes, the height of the pile is approximately 11.51 meters.
B. After 5 minutes, how fast is the height increasing?
To find the rate at which the height is increasing after 5 minutes, we need to use derivatives. We know that the volume of the cone is increasing at a rate of 20 m^3/min. The derivative of the volume with respect to time, dv/dt, is given by:
dv/dt = (π/4) * h^2 * dh/dt
We can plug in the value of h (11.51) and the known volume rate (20):
20 = (π/4) * (11.51)^2 * dh/dt
Now, solving for dh/dt, we have:
dh/dt = (20 * 4/π) / (11.51)^2
Therefore, after 5 minutes, the height is increasing at a rate of approximately (20 * 4/π) / (11.51)^2 units per minute.
C. After 5 minutes, how fast is the base radius increasing?
Since the height is always twice the length of the base diameter, we can relate the rate of change of the height (dh/dt) to the rate of change of the base radius (dr/dt). We have:
dh/dt = 4 * dr/dt
Using the equation we found in part B, we can substitute for dh/dt:
4 * dr/dt = (20 * 4/π) / (11.51)^2
Simplifying, we find:
dr/dt = (20 * 4/π) / (11.51)^2 * (1/4)
Therefore, after 5 minutes, the base radius is increasing at a rate of approximately (20 * 4/π) / (11.51)^2 units per minute.
D. After 5 minutes, how fast is the area of the base increasing?
The area of the base of a cone is given by A = πr^2. We can relate the rate of change of the area (dA/dt) to the rate of change of the base radius (dr/dt) using the derivative. We have:
dA/dt = 2πr * dr/dt
Substituting the value of r from part A into the equation, we get:
dA/dt = 2π * (∛(4800/π)) * dr/dt
Using the equation we found in part C, we can substitute for dr/dt:
dA/dt = 2π * (∛(4800/π)) * ((20 * 4/π) / (11.51)^2)
Simplifying, we find:
dA/dt = 2π * (∛(4800/π)) * ((20 * 4/π) / (11.51)^2)
Therefore, after 5 minutes, the area of the base is increasing at a rate of approximately 2π * (∛(4800/π)) * ((20 * 4/π) / (11.51)^2) units per minute.