Asked by Anonymous
Sands falls off a conveyor belt onto the ground at a steady rate of 200cm^3 per second forming a conical pile whose height always equal to half the diameter of the base, at what rate is the height increasing after 10 second
Answers
Answered by
Reiny
height of cone is 1/2 the diameter
---> height of cone equals its radius
V= (1/3)π r^2 h, but r = h
V = (1/3) π h^3
dV/dt = π h^2 dh/dt ***
dV/dt = 200 cm^3/s
so when t = 10
V = 2000 cm^3
2000 = (1/3)π h^3
h^3 = 6000/π
h = (6000/π)^(1/3) = appr 12.407
in ***
200 = π(12.407)^2 dh/dt
dh/dt = 200/(π(12.407)^2) = appr .414 cm/sec
check my arithmetic
---> height of cone equals its radius
V= (1/3)π r^2 h, but r = h
V = (1/3) π h^3
dV/dt = π h^2 dh/dt ***
dV/dt = 200 cm^3/s
so when t = 10
V = 2000 cm^3
2000 = (1/3)π h^3
h^3 = 6000/π
h = (6000/π)^(1/3) = appr 12.407
in ***
200 = π(12.407)^2 dh/dt
dh/dt = 200/(π(12.407)^2) = appr .414 cm/sec
check my arithmetic
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