⦁ Sand is being poured on the ground from the office of an elevated pipe. It forms a conical pile whose altitude is always equal to 4/3 the radius of the base. Approximate all the answers to two decimal places.
⦁ How fast is the volume increasing when the radius of the base is 4 feet and is increasing at the rate of 0.01 ft. /s?
⦁ If sand is falling at the rate of 5 cubic ft. / s., how fast is the radius of the pile increasing when the radius is 5 feet?
V = 1/3 π r^2 h = 1/3 * π * r^2 * (4 r / 3) = (4 π / 9) * r^3
implicit differentiation ... dV = (4 π / 9) * 3 r^2 dr = (4 π / 3) * r^2 dr
use the equation by plugging in known (given) values and solving
To find the answers to these questions, we will need to use the concepts of related rates and the volume of a cone. Let's break down the problem and go step by step:
1. Given information:
- The altitude (height) of the cone is always equal to 4/3 times the radius of the base.
- The radius of the base is increasing at a rate of 0.01 ft/s.
- Sand is falling at a rate of 5 cubic ft/s.
2. Question 1: How fast is the volume increasing when the radius of the base is 4 feet and is increasing at a rate of 0.01 ft/s?
To find the rate of change of the volume, we can use the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
Given that the altitude (h) is always equal to (4/3) * r, we can substitute this value into the volume formula:
V = (1/3) * π * r^2 * (4/3) * r
V = (4/9) * π * r^3
Now, we can differentiate both sides of the equation with respect to time (t):
dV/dt = (4/9) * π * 3r^2 * (dr/dt)
dV/dt = (4/3) * π * r^2 * (dr/dt)
We are given that the radius (r) is 4 ft and is increasing at a rate of 0.01 ft/s:
dV/dt = (4/3) * π * (4^2) * (0.01)
dV/dt = (4/3) * π * 16 * 0.01
dV/dt = (2/3) * π * 0.64
dV/dt ≈ 1.34 ft³/s
So, the volume is increasing at a rate of approximately 1.34 ft³/s.
3. Question 2: How fast is the radius of the pile increasing when the radius is 5 feet?
To find the rate at which the radius is changing, we can differentiate the equation for the altitude (h = (4/3) * r) with respect to time:
dh/dt = (4/3) * dr/dt
We are given that sand is falling at a rate of 5 ft³/s. The volume formula for a cone is V = (1/3) * π * r^2 * h. We can differentiate this volume equation with respect to time to obtain another relationship:
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Again, substitute h = (4/3) * r into the equation and the given values:
5 = (1/3) * π * (2 * 5 * dr/dt * (4/3) * 5 + 5^2 * (4/3) * dr/dt)
5 = (1/3) * π * (40/3 * dr/dt + 100/3 * dr/dt)
5 = (1/3) * π * (140/3 * dr/dt)
15 = 140 * π * dr/dt
dr/dt ≈ 15 / (140 * π)
dr/dt ≈ 0.034 ft/s
So, when the radius is 5 feet, the radius of the pile is increasing at a rate of approximately 0.034 ft/s.