Wheat runs out of a silo at a constant rate, forming a conical heap. After #1# minute the height of the heap is #20"cm"#. At what rate is its height increasing at that time?

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Answer 1 · 2015-08-18T23:14:44

The height is proportional to the cube root of the time, that is #H(t) = k t^(1/3)# for some constant #k#.

Hence #H'(t) = k/3 t^(-2/3)# giving rate #20/3 "cm"# per minute.

If #V(t)# gives the volume of the heap at time #t#, then #V(t) = k_1t# for some constant #k_1#, since the wheat runs from the silo at constant rate.

The height of the heap at time #t# is proportional to the cube root of the volume.

That is:

#H(t) = k_2 root(3)(V(t)) = k_2 root(3)(k_1 t) = k_2 root(3)(k_1) root(3)(t)#

for some constant #k_2#.

Let #k = k_2 root(3)(k_1)#. Then #H(t) = k root(3)(t)#

If #t# is measured in minutes, then #H(1) = k root(3)(1) = k = 20cm#

Then #H'(t) = d/(dt) k root(3)(t) = d/(dt) k t^(1/3) = k/3 t^(-2/3)#

So #H'(1) = 20/3"cm" = 6 2/3"cm"# per minute