Expert: Paul Klarreich - 3/30/2008

Question
I just want to make sure that I am on the right track with this problem.

Question:
As sand leaks out of a hole in a container, it forms a conical pile whose altitude is always the same as its radius.  If the height of the pile is increasing at a rate of 6 inches per minute, find the rate at which the sand is leaking out when the altitude is 10 inches.

This is what I have come up with:

Variables -
V= volume of cone
r= radius of cone
h= height of cone

Rates:
dh/dt = 6 in/min
dr/dt = (is this also 6 in/min since the radius is always the same as the height?)
dV/dt = unknown

V= (1/3)pi*r^2*h
r=h, so V= (1/3)pi*(h^2)*h = (1/3)pi*(h^3)

dV/dt = pi(h^2)(dh/dt)
h=10, dh/dt= 6

dV/dt = pi(100)(6)
     = 600 pi or 1884.954 in/min

Am I close, or just totally off??

Answer
Questioner:   Lauren
Category:  Calculus
Private:  No
 
Subject:  Related Rates - Cone
Question:  I just want to make sure that I am on the right track with this problem.

Question:
As sand leaks out of a hole in a container, it forms a conical pile whose altitude is always the same as its radius.  If the height of the pile is increasing at a rate of 6 inches per minute, find the rate at which the sand is leaking out when the altitude is 10 inches.

This is what I have come up with:

Variables -
V= volume of cone
r= radius of cone
h= height of cone

Rates:
dh/dt = 6 in/min
dr/dt = (is this also 6 in/min since the radius is always the same as the height?)  <<  YOU GOT IT.  BUT YOU COULD JUST WRITE  r = h
dV/dt = unknown

V= (1/3)pi*r^2*h  << AND MAKE THIS  (1/3) pi r^3,
which would simplify things.

r=h, so V= (1/3)pi*(h^2)*h = (1/3)pi*(h^3)

>> AHA! I spoke too soon.  YOU did get it.

dV/dt = pi(h^2)(dh/dt)
h=10, dh/dt= 6

dV/dt = pi(100)(6)
    = 600 pi or 1884.954 in/min

Am I close, or just totally off??

>> LOOKS GOOD TO ME.