Question

• An ice cube is slowly melting, losing 3cm^3 of water each hour. If it is always a perfect cube, (V=s^3), what is the rate of change of its side length when it has 8 cm^3 of ice left?

Answer 1

Given:

The volume is decreasing at the rate of 3 cm^3 per hour.

The volume of the left ice is 8 cm^3.

Aim:

We need to find the rate of change of the side of the cube.

Explanation:

Let the length of the cube is denoted as s.

Consider the volume of the cube.

$V=s^3$

Since the volume is decreasing at the rate of 3 cm^3 per hour. we can write,

$\frac{dV}{dt}=-3cm^3\/h$

where t represents time and the negative sign represents decreasing.

Differentiate the volume with respect to s.

$\frac{dV}{ds}=\frac{d}{ds}(s^3)=3s^2$

To find the rate of change of the side length, we use the chain rule.

$\frac{dV}{dt}=\frac{dV}{ds}\frac{ds}{dt}$

$\text{ Substitute }\frac{dV}{dt}=-3\text{ and }\frac{dV}{ds}=3s^2\text{ in the equation.}$

$-3=\frac{ds}{dt}(3s^2)$

$-\frac{3}{3s^2}=\frac{ds}{dt}$

$-\frac{1}{s^2}=\frac{ds}{dt}$

Since the left ice is 8 cm ^3.

$V=(s)^3=8$

$s^3=2^3$$s=2cm$

$Substitute\text{ s =2 in the equation}-\frac{1}{s^2}=\frac{ds}{dt}.$

$-\frac{1}{2^2}=\frac{ds}{dt}.$

$\frac{ds}{dt}=-\frac{1}{4}$

$\frac{ds}{dt}=-0.25cm\text{ per hour}$

Verification:

Let s =2 cm, then the volume is 8cm^3.

Let s =1.75cm, the volume is