Asked by Anonymous
sand is being dragged at the rate of 10m^3/min into a conical pile. if the height of the pile is always twice the base radius, at what rate is the height increasing when the pile is 2m high?
Answers
Answered by
Damon
The volume of anything pointy with straight sides is (1/3) base area * height
so
V = (1/3) pi r^2 h
but h = 2 r
so
V = (2/3) pi r^3
find dr/dV, worry about dh/dV later (note h = 2 r)
dV = (2/3) pi (3 r^2) dr = 2 pi r^2 dr
dV/dt = 2 pi r^2 dr/dt = 10 m^3/min
but r = h/2 and dr/dt = (1/2)dh/dt
so
10 = 2 pi h^2/4 (1/2) dh/dt
40 = pi h^2 dh/dt
when h = 2
40 = 4 pi dh/dt
so
dh/dt = 10/pi meters/minute
so
V = (1/3) pi r^2 h
but h = 2 r
so
V = (2/3) pi r^3
find dr/dV, worry about dh/dV later (note h = 2 r)
dV = (2/3) pi (3 r^2) dr = 2 pi r^2 dr
dV/dt = 2 pi r^2 dr/dt = 10 m^3/min
but r = h/2 and dr/dt = (1/2)dh/dt
so
10 = 2 pi h^2/4 (1/2) dh/dt
40 = pi h^2 dh/dt
when h = 2
40 = 4 pi dh/dt
so
dh/dt = 10/pi meters/minute
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