A pyramid-shaped vat has square cross-section and stands on its tip. The dimensions at the top are $2\text{ m}\times2\text{ m}$,and the depth is $5\text{ m}$. If water is flowing into the vat at $3\text{ m}^3/\text{min}$, how fast is the water level rising when the depth of water (at the deepest point) is $4\text{ m}$? Note: the volume of any “conical” shape (including pyramids) is $(1/3)(\text{height})(\text{area of base})$.
2026-04-01 20:30:45.1775075445
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How fast is the water level rising?
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When it's $4$ meters deep, the surface area is $\left(\frac 4 5 \right)^2\times 4\text{ m}^2$. So $$ \left(\left(\frac 4 5 \right)^2\times 4\text{ m}^2\right) \times (\text{ ? } \text{ m per minute}) = 3\text{ m}^3\text{ per minute}. $$ Divide both sides by something to get $$ \text{ ? } \text{ m per minute} = \cdots\cdots $$
what the in coming water sees at the moment when the depth is $4 m$ has a square surface area of $\left(\frac 85\right)^2.$ looking at the volume added in time $dt$ and the change in depth $dh$ are related by $$\left(\frac 85\right) ^2\, dh = 3 \, dt $$ giving you $$ \frac{dh}{dt} = \frac{75}{64}\text{ m/min}.$$