Asked by Fred
Can someone explain how to solve this problem?
Sand is being dumped on a conveyor belt onto a pile in such a way that the pile forms in the shape of a cone whose radius is always equal to its height. Assuming that the sand is being dumped at a rate of 10 cubic feet per minute, how fast is the height of the pile changing when there are 1000 cubic feet on the pile?
I got dh/dt = 1/(100000*pi) when I tried to solve it, but I don't think its the correct answer.
Thanks!
Sand is being dumped on a conveyor belt onto a pile in such a way that the pile forms in the shape of a cone whose radius is always equal to its height. Assuming that the sand is being dumped at a rate of 10 cubic feet per minute, how fast is the height of the pile changing when there are 1000 cubic feet on the pile?
I got dh/dt = 1/(100000*pi) when I tried to solve it, but I don't think its the correct answer.
Thanks!
Answers
Answered by
Steve
the cone has volume
v = 1/3 π r^2 h = 1/3 πr^3
dv/dt = πr^2 dr/dt
when v = 1000, r = ∛(3000/π)
so,
10 = π(3000/π)^(2/3) dr/dt
now you can find dr/dt, and since r=h, that's also dh/dt
v = 1/3 π r^2 h = 1/3 πr^3
dv/dt = πr^2 dr/dt
when v = 1000, r = ∛(3000/π)
so,
10 = π(3000/π)^(2/3) dr/dt
now you can find dr/dt, and since r=h, that's also dh/dt
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