Question

question a tank is shaped like an upside - down square pyramid, with base of 4 m by 4 m and a height of 6 m, as shown in the figure below. how fast does the height increase when the water is 1 m deep if water is being pumped in at a rate of 0.4 m³/sec? note that the volume of a square pyramid with length of the square bases side s and height h is given by v = 1/3 s²h.

Explanation:

Step1: Relate side - length and height

For a square - based pyramid, the ratio of the side - length of the base $s$ to the height $h$ is constant. Given the large pyramid has a base side - length of $4$m and height of $6$m, so $s=\frac{4}{6}h=\frac{2}{3}h$. The volume formula of a square - based pyramid is $V = \frac{1}{3}s^{2}h$. Substituting $s=\frac{2}{3}h$ into the volume formula, we get $V=\frac{1}{3}(\frac{2}{3}h)^{2}h=\frac{4}{27}h^{3}$.

Step2: Differentiate the volume formula with respect to time

Differentiate $V=\frac{4}{27}h^{3}$ with respect to time $t$ using the chain - rule. $\frac{dV}{dt}=\frac{4}{27}\times3h^{2}\frac{dh}{dt}=\frac{4}{9}h^{2}\frac{dh}{dt}$.

Step3: Substitute the known values

We know that $\frac{dV}{dt}=0.4$ m³/sec and $h = 1$m. Substitute these values into the equation $\frac{dV}{dt}=\frac{4}{9}h^{2}\frac{dh}{dt}$. So, $0.4=\frac{4}{9}\times(1)^{2}\times\frac{dh}{dt}$.

Step4: Solve for $\frac{dh}{dt}$

First, rewrite the equation as $\frac{dh}{dt}=\frac{0.4\times9}{4}$. Then, calculate $\frac{0.4\times9}{4}=\frac{3.6}{4}=0.9$ m/sec.

Answer:

$0.9$ m/sec