- #1
jellofaceman
- 12
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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
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
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