Question

The volume of a cube with sides of length $s$ is given by $V = {s^3}$. What is the rate of change of the volume with respect to $s$ when $s$ is $6\,cm$?

Answer 1

Here we have to find the rate of change of the volume with respect to $s$ when $s$ is $6\,cm$. So, we will use the concept of differentiation. Differentiation can be defined as the process in which we find the instantaneous rate of change in function based on one of its variables. In order to solve this question we first differentiate the volume with respect to $s$ and then put the value $s = 6\,$ in the function to find the rate of change of the volume at $s = 6\,cm$.

Complete step by step answer:
Here we have to find the rate of change of the volume with respect to $s$ when $s$ is $6\,cm$. By using the concept of differentiation. Differentiation can be defined as the process in which we find the instantaneous rate of change in function based on one of its variables. Let $x$ and $y$ be the two variables then the rate of change of $x$ with respect to $y$ is given by $\dfrac{{dy}}{{dx}}$. Here, we have the volume function i.e., $V = {s^3}$

Differentiate the volume function with respect to $s$.
$\Rightarrow \dfrac{{d(V)}}{{ds}} = \dfrac{{d({s^3})}}{{ds}}$
We know that $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$. Use the above formula to differentiate the volume function. So,
$\Rightarrow \dfrac{{d({s^3})}}{{ds}} = 3{s^2}$

Putting the value $s = 6$ in $3{s^2}$ to find the rate of change of the volume at $s = 6\,cm$.
We get,
$\Rightarrow \dfrac{{d({s^3})}}{{ds}} = 3{s^2} = 3{(6)^2}$
Squaring $6$ and multiplying it by $3$. We get,
$\therefore \dfrac{{d({s^3})}}{{ds}} = 3 \times 36 = 108\,c{m^3}$

Hence, the rate of change of the volume with respect to $s$ when $s$ is $6\,cm$ is equal to $108\,c{m^3}$.

Note: In order to solve this type of questions we must know the concept of differentiation. Also one can get confused with the rate asked in the question so we must check if the answer comes negative then it will be decreasing rate otherwise positive answer shows increasing rate. One must take care of that in respect to which variable we have to differentiate the function. With the help of differentiation we can also calculate the rate of change in surface area of the cube.