Sand falls from a conveyer belt at the rate of 10 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 are the a) height and b) radius changing when the pile is 4 m high?

Question

anonymous · Answer

I know that h = 3/4*r and dV/dt = 10 m^3/min, and V = pi r^2h

anonymous · Answer

@zepdrix

anonymous · Answer

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zepdrix · Answer

So they gave us a relationship between the height and the radius. That's very helpful, because we'll be able to Write the volume formula with ONLY H and V (to solve for h'), and then separately with ONLY R and V (to solve for r').

zepdrix · Answer

$$\large V=\frac{ 1 }{ 3 }\pi r^2h=\frac{ 1 }{ 3 }\pi r^2(\frac{ 3 }{ 4 }r)$$
See how we were able to write the RIGHT side in terms of only R?

anonymous · Answer

lol yea, so far I have dV/dt  = $$(\frac{ 4 }{ 3 } h) h ^{2} * \frac{ \pi }{ 3 }$$

anonymous · Answer

well i solved in terms of h to find dh/dt

zepdrix · Answer

Oh you're doing the other one first :) k cool

zepdrix · Answer

$$V=\frac{ 1 }{ 3 }\pi r^2h=\frac{ 1 }{ 3 }\pi (\frac{ 4 }{ 3 }h)^2h$$
$$=\frac{ 1 }{ 3 }\pi (\frac{ 16 }{ 9 }h^3)$$
Hmm I'm wondering if you plugged in your h correctly. Take a look at my setup.

anonymous · Answer

lol i haven't plugged in h yet

anonymous · Answer

oh wait 4/3 pi is squared?

anonymous · Answer

oops i see where i made my mistake lol