I figured out that part "A" is -3/8, but i can't figure out part 2, a or b. please explain and help. thanks.
Sand is falling from a rectangular box container whose base measures 40 inches by 20 inches at a constant rate of 300 cubic inches per minute.
a) how is the depth of the sand in the box changing?
b) the sand is form is forming a conical pile. At a particular moment, the pile is 23 inches high and the diameter of the base is 16 inches. The diameter of the base at this moment is increasing at 1.5 inches per minute, at this moment ,
1. how fast is the area of the circular base of the cone increasing?
2. How fast is the height of the pile increasing?
(a) is correct. Not sure how the sand maintains a constant plane cross-section, but we'll posit that.
The volume of the conical pile is
v = 1/3 π r^2 h
If we assume that the aspect ratio of the cone does not change, then
r = 8/23 h, so dr/dt = 8/23 dh/dt
dr/dt = 3/4 in/min, so
dh/dt = 23/8 * 3/4 = 69/32 in/min
the area is
a = πr^2, so
da/dt = 2πr dr/dt
= 2π(8)(3/4) = 12π in^2/min
v = π/3 r^2h, so
dv/dt = π/3 (2rh dr/dt + r^2 dh/dt)
So, since dv/dt = 300 in^3/min
you can now solve for dh/dt
For part 2 i got .17955, would this be considered correct?
To solve both parts of the problem, we can use related rates, which involves using derivatives to find how two related variables are changing with respect to time.
a) To find how the depth of the sand in the box is changing, we need to relate the volume of the sand to the dimensions of the box.
Let's denote the depth as h (in inches). The volume V of the sand can be calculated using the formula for the volume of a rectangular prism: V = lwh, where l is the length, w is the width, and h is the depth.
Given that the base measures 40 inches by 20 inches, we have l = 40 and w = 20. We also know that the sand is falling into the box at a constant rate of 300 cubic inches per minute.
We need to find how dh/dt (the rate at which the depth of the sand is changing) relates to dV/dt (the rate at which the volume of the sand is changing). Since V = lwh, we can use the chain rule to differentiate both sides with respect to time:
dV/dt = dl/dt * w * h + l * dw/dt * h + l * w * dh/dt
Since dl/dt and dw/dt are both zero (as the length and width of the base are constant), we can simplify the equation to:
dV/dt = l * w * dh/dt
We can now solve for dh/dt by substituting the known values:
300 = 40 * 20 * dh/dt
Simplifying further, we find:
dh/dt = 300 / (40 * 20) = -3/8 inches per minute
Therefore, the depth of the sand in the box is decreasing at a rate of 3/8 inches per minute.
b) Moving on to the conical pile, we can use similar related rates concept. We need to find how fast the area of the circular base of the cone is increasing and how fast the height of the pile is increasing.
1. To find how fast the area of the circular base of the cone is increasing, we need to relate the variables involved. Let's denote the radius of the base as r (in inches) and the area of the circular base as A.
We know that the diameter of the base is increasing at a rate of 1.5 inches per minute, and the radius is half the diameter. So, dr/dt = 1.5/2 = 0.75 inches per minute.
To find dA/dt (the rate at which the area of the base is changing), we can differentiate the formula for the area of a circle A = πr^2 with respect to time:
dA/dt = d(πr^2)/dt = 2πr * dr/dt
At a particular moment, the pile is 23 inches high, and the diameter of the base is 16 inches. Thus, the radius is r = 16/2 = 8 inches.
Substituting the known values into the equation, we have:
dA/dt = 2π * 8 * 0.75 = 12π square inches per minute
Therefore, the area of the circular base of the cone is increasing at a rate of 12π square inches per minute.
2. To find how fast the height of the pile is increasing, we can use similar related rates concept.
We know that dh/dt represents the rate of change of the height of the cone. This rate is also given as -1.5 inches per minute since the height of the pile is decreasing due to sand being added to the base.
Therefore, the height of the pile is decreasing at a rate of 1.5 inches per minute.