posted by :) on .
Sand is being dropped at the rate of 10 cubic meter per minute onto 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 8 meters high?
Volume = V = (1/3)π r^2 h
but h = 2r ---> r = h/2
V = (1/3)π (h^2/4)(h)
= (1/12)π h^3
dV/dt = (1/12)π h^2 dh/dt
10 = 2π (64) dr/dt
dr/dt = 10/(128π) m/min
= 5π/64 m/min
check my arithmetic
Thank you for answering my question. I just can't understand how it come up to the equation 10=2Π(64)dr/dt. I will be very thankful if you elaborate. :)
I can see why you are puzzled, since I have two typo errors.
from V = (1/12) π h^3, I should have had ...
dV/dt = (1/4)π h^2 dh/dt
now subbing in our given...
10 = (1/4)π(64) dh/dt
10 =16π dh/dt
dh/dt =10/(16π) = 5/(8π) m/min
I apologize for those blatant errors.
I'm very thankful for being kind to me. Last one thing, 8 meters is the height of the conical pile. Does it also mean that it is the same as the height of the sand?
Since we assume that the pile of sand forms a cone,
the height of the cone is equal to height of the sand pile
thank you so much. :) you really help me a lot. Godbless!
hii... am in the solution with the V = (1/3)π (h^2/4)(h) still have h in 2/4??
isn't the derivative of r = h/2 is just 2/4??