Posted by **:)** on Monday, May 14, 2012 at 11:49pm.

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?

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**Reiny**, Monday, May 14, 2012 at 11:57pm
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

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**:)**, Tuesday, May 15, 2012 at 1:24am
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. :)

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**Reiny**, Tuesday, May 15, 2012 at 8:54am
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.

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**:)**, Tuesday, May 15, 2012 at 9:30am
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?

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**Reiny**, Tuesday, May 15, 2012 at 10:09am
Since we assume that the pile of sand forms a cone,

the height of the cone is equal to height of the sand pile

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**:)**, Tuesday, May 15, 2012 at 7:49pm
thank you so much. :) you really help me a lot. Godbless!

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**z**, Tuesday, July 24, 2012 at 6:31am
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??

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