Help calc prob.

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1

<p>You have a square pryamid.
Length = 6 ft on square side of the base
Vertex from the base (i assume center) is 6 ft
The tank is filled to 4 ft. with the volume changing 1 c. feet / sec</p>

<p>What is dh / dt for this?</p>

<p>Note volume formula = 1/3bh for square pyramid.</p>

<p>I don't get it at all.... i have an idea but think its wrong (deals with similar triangles to get rid of one variable and then easy from there. It seems wrong though).</p>

2

<p>Does anyone know?</p>

3

<p>well the height=length in this case. V=1/3bh (b= l^2, by the way), which means V=1/3 h^3, therefore dv/Vt=h^2 dh/dt so, dh/dt when h=4 is 1/16 ft/sec (I can't tell if you mean the tank is draining or being filled once it's at 4, so this could be negative.</p>