Question

question a tank is shaped like an upside - down square pyramid, with a base with sides that are 4 meters in length and a height of 12 meters. if water is being pumped into the pyramid at a rate of $\frac{2}{3}$ $\frac{m^{3}}{min}$, at what rate is the height of the water increasing when the water is 2 meters deep? (the volume of a pyramid with height $h$ and base area $b$ is given by $v=\frac{1}{3}bh$.) enter a fraction and do not include units. provide your answer below:

Explanation:

Step1: Find base - area formula

$B = s^{2}$, with $s = 4$, so $B=16$.

Step2: Differentiate volume formula

$V=\frac{1}{3}Bh=\frac{16}{3}h$, $\frac{dV}{dt}=\frac{16}{3}\frac{dh}{dt}$.

Step3: Solve for $\frac{dh}{dt}$

Given $\frac{dV}{dt}=\frac{2}{3}$, then $\frac{2}{3}=\frac{16}{3}\frac{dh}{dt}$, so $\frac{dh}{dt}=\frac{1}{8}$.

Answer:

$\frac{1}{8}$