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?
Hi, I am sorry to comment on such an old question, but I don't understand how you know what dr/dt is.
It would be wonder if someone could help explain!
sand being dumped from a funnel forms a conical file whose height is always one third the diameter of a base, of the sand is dumped at the rate of 2 cubic meters per minute, how fast is the pile rising when it is 1 meter deep?
The first thing you need to do is think of a formula in which you can "relate" the given. In this case, we have the rate of volume and we are finding the rate of height.
Given the formula for the volume of a conical pile,
V= (1/3)πr^2h
however, in this case, we don't have an exact value for r, but it says that "the height is always twice the base"
we can relate r to h
h= 2r ---> deriving > r= h/2
now, substitute the r in the formula above
V= (1/3)π(h/2)^2 h
V= (1/3)π (h^3/4) simplifying
V= (πh^3/12) *differentiate
dV/dt= (πh^3/12)
dV/dt= (π/12)(3h^2) dh/dt simplify..
dV/dt= (πh^2/4) substitute the value for h=8 and dV/dt which is 10
10= 64π/4 dh/dt
10= 16π dh/dt
dh/dt= 10/16π (just directly plug in to your calculator to get the answer.
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??