# Classic Related Rates: Sand Conical

1. Oct 2, 2012

### jellofaceman

I've been trying to figure out where my mistake lies in the first solution. Some help would be appreciated. I did notice I got the same solution twice, so I assume I just calculated dr/dt twice and I need to use a different equation for dh/dt? Is dh/dt=(3/4)*dr/dt?

1. Sand falls from a conveyor belt at a rate of 12 m^3/min onto the top of a conical
pile. The height of the pile is always three-eights of the base diameter. How fast
(in cm/min) are the height and the radius changing when the pile is 2m high.

2. V=1/3πr2h
r= 4/3h = 8/3
h=3/4r = 2
dv/dt= 12
dh/dt= ?
dr/dt = ?
Meters*100=Cm

3. Solving for dh/dt /* Answer is incorrect*/
v=(1/3)π(4/3h)2h
v=(4/9)π(h)3
dv/dt=(4/3)πh2dh/dt
12=(16/3)π(dh/dt) /*Plug in values and rearrange*/
36/16π=dh/dt
dh/dt=.7162 /* Proposed correction: Multiply by 3/4?*/

Same for dr/dt /*Solution is Correct*/
v=(1/3)π(r)2(3/4)r
v=(3/12)πr^3
dv/dt=(9/12)πr2dr/dt
12=(9/12)π(8/3)2dr/dt
12/((9/12)π(8/3)2)=dr/dt
dr/dt=.7162

Last edited by a moderator: Oct 2, 2012
2. Oct 2, 2012

### Staff: Mentor

It's important to keep in mind which things are constant, and which things are variable. V, r, and h are all changing with time, and the only thing that remains constant is dV/dt. Think of r = r(t), h = h(t), and V = V(t) as functions of t such that at a particular moment t0, r(t0) = 8/3, h(t0) = 2, and V(t0) = whatever it happens to be at that time.

At any other time, r ≠ 8/3 and h ≠ 2.
Use one or the other of these equations -- r = (4/3)h or h = (3/4)r to write V as a function of one variable alone.

3. Oct 3, 2012