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

Answer 2

The rate at which the water level is rising when the water level is 5 cm is;

dh/dt = 1.6 cm/s

We are told that;

length of square side; s = 4 cm

height of pyramid; h = 8 cm

volumetric rate; dV/dt = 30 cm³/s

Since the base is square shaped and we know that area of a square = s², then we can say that;

Volume of pyramid = ¹/₃s²h

The pyramid has a height of 8 cm and length of its' square base side is 8 cm.

Now, regardless of the water height, the surface of the water will always be square shaped. This implies that the ratio of the height to the top side of square will always be the same. Thus, we can say that; s = 4/8 h = h/2

Thus, Volume is;

V = ¹/₃ × (h/2)² × h

V = ¹/₃ × h³/4

V = h³/12

dV/dt = (3h²/4) dh/dt

Plugging in the relevant values gives;

30  = (3 × 5²/4) dh/dt

120 = 75 dh/dt

dh/dt = 120/75

dh/dt = 1.6 cm/s

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