(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
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?
2 answers
For part 2 i got .17955, would this be considered correct?
3 answers