Question

Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 13 feet high? Recall that the volume of a right circular cone with height h and radius of the base r is given by V=1/3πr^2h.

Answer 1

Argh!@!@ Fudge, I couldn't get out of the equations editor so I had to refresh the page, gotta retype it all...

Answer 2

oh ok

Answer 3

\[V = \frac {1}{12} d^2 h \rightarrow V = \frac {1}{12} \pi h^3\]\[\frac {dV}{dt} = \frac {1}{4} \pi h^2 \frac {dh}{dt}\]
\[V = \frac {1}{12} \pi h^3 = \frac {1}{12} \pi 13^3\]\[\frac {dV}{dt} = 10 ft^3 / \min\]\[h = 13 ft\]
Can you solve for dh/dt given that?

Answer 4

can you just demonstrate please

Answer 5

if you dont mind

Answer 6

\[\frac {dV}{dt} = \frac {1}{4} \pi h^2 \frac {dh}{dt}\]\[10 = \frac {1}{4} \pi \times 169 \times \frac {dh}{dt}\]

Answer 7

\[\frac {dh}{dt} = \frac {40 \pi}{169} \approx 0.744 ft/\min\]

Answer 8

webwork rejected the .744

Answer 9

-_____- all my hard work, for nothing T_T

Answer 10

i appreciate it regardless

Answer 11

Well, does my work make sense or do we have to do something else?

Answer 12

ohhhh, I just realized that I don't know how to divide :3

Answer 13

It should be \[\frac {dh}{dt} = \frac {40 }{169 \pi} \approx 0.075339618 ft/\min\]

Answer 14

that works. thank you very much

Answer 15

YAY!!!! :D:D