sand is falling into a conical pile so that the radius of the base of the pile is always equal to one half its altitude. of the sand is falling at the rate of 10 cubic feet per minute, how fast is the altitude of the pile increasing when the pile is 5 feet deep?
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2 answers

given: r = h/2
V = (1/3)π r^2 h
= (1/3)π (h/2)^2 (h)
= (1/12) π h^3
dV/dt = (1/4)π h^2 dh/dt
for the given data ...
10 = (1/4)π(25)dh/dt
dh/dt = 10(4)/((25π) = 1.6/π feet/min
check my arithmetic 👍
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answered by Reiny 
Just to appreciate the good work. I wanted to hammer it in a congruent way.
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answered by Wiseman
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