Question

An inverted pyramid is being filled with water at a constant rate of 30 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 4 cm, and the height is 8 cm. Find the rate at which the water level is rising when the water level is 5 cm.

Answer 1

Answer:

4.8cm/s

Step-by-step explanation:

We are given that an inverted pyramid.

$\frac{dV}{dt}=30cm^3/s$

Base of pyramid=4cm

Height of pyramid=8 cm

We have to find the rate at which the water level is rising when the water level is 5 cm.

Volume of pyramid=$\frac{1}{3}\times base\;area\times h$

As the height changes from 8 to h.

Therefore, the volume of pyramid for given height is given by

$V=\frac{1}{3}\times (4\times \frac{h}{8})^2\times h=\frac{1}{12}h^3$

Differentiate w.r.t time

$\frac{dV}{dt}=\frac{1}{4}h^2\frac{dh}{dt}$

Substitute h=5

$30=\frac{1}{4}(5)^2\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{30\times 4}{ 25}$

$\frac{dh}{dt}=\frac{24}{5}=4.8 cm/s$

Hence, the rate at which the water level is rising when the water level is 5cm=4.8 cm/s