# Related rates

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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?

• Related rates -

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

• Related rates -

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. :)

• Related rates -

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.

• Related rates -

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?

• Related rates -

Since we assume that the pile of sand forms a cone,
the height of the cone is equal to height of the sand pile

• Related rates -

thank you so much. :) you really help me a lot. Godbless!

• Related rates -

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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