Question

Volume of a cube is increasing at the rate of $0.003{\text{ }}{{\text{m}}^2}/\sec$. At the instant when the edge is 20 cm. Find the rate at which the edge is changing.

Answer 1

Hint: Related rate means we are usually going to have to take the derivative of our formula implicitly (in terms of another variable that is not part of the formula). In this case we want to take the derivative in terms of time (t). So, use this concept to reach the solution of the problem.

Complete step-by-step answer:
Volume of a cube in increasing at the rate of $0.003{\text{ }}{{\text{m}}^2}/\sec$ i.e., $\dfrac{{dv}}{{dt}} = 0.003$
We know that the volume of the cube of edge $a$ is given by $v = {a^3}$
As we have $\dfrac{{dv}}{{dt}} = 0.003$and substituting $v = {a^3}$, we have

$$\Rightarrow \dfrac{{d{{\left( a \right)}^3}}}{{dt}} = 0.003 \\\ \Rightarrow 3{a^2}\dfrac{{da}}{{dt}} = \dfrac{3}{{1000}} \\\ \Rightarrow {a^2}\dfrac{{da}}{{dt}} = \dfrac{3}{{1000}} \times \dfrac{1}{3} \\\ \Rightarrow {a^2}\dfrac{{da}}{{dt}} = \dfrac{1}{{1000}} \\\$$

At the instant the length of the edge is 20 cm i.e., $a = 20{\text{ cm }} = \dfrac{{20}}{{100}}{\text{ m}}$

$$\Rightarrow {\left( {\dfrac{{20}}{{100}}} \right)^2}\dfrac{{da}}{{dt}} = \dfrac{1}{{1000}} \\\ \Rightarrow \dfrac{{da}}{{dt}} = \dfrac{1}{{1000}} \times {\left( {\dfrac{{100}}{{20}}} \right)^2} \\\ \Rightarrow \dfrac{{da}}{{dt}} = \dfrac{1}{{1000}} \times \dfrac{{100}}{{20}} \times \dfrac{{100}}{{20}} \\\ \Rightarrow \dfrac{{da}}{{dt}} = \dfrac{1}{{40}} \\\ \therefore \dfrac{{da}}{{dt}} = 0.025{\text{ m /sec}} \\\$$

Thus, the edge is changing at the rate of $0.025{\text{ m /sec}}$.

Note: Here we have converted the length of edge which is given in centimetres (cm) to meters (m) as the increasing rate of volume of the cube is given in ${{\text{m}}^2}/\sec$. This we have done by using the conversion 100 cm = 1 m. Volume of the cube of edge $a$ is given by $v = {a^3}$.