Question

The volume V of a cube of side x meters is$V=V\left(x\right)=x^3$. Find the instantaneous rate of change of the volume with respect to the side x

at$x=3.$

Short answer

If the function for the volume of a cube is$V=V\left(x\right)=x^3$then the

instantaneous rate of change of the volume with respect to the side x

at $x=3$will be$27\frac{m^3}{m}.$

Step-by-step solution

Step 1. Given information   

Function for the volume of a cube is

$V=V\left(x\right)=x^3$

Side of cube$x=3$

Step 2. The Volume of the cube

Substitute$x=3$in the function for the volume of the cube

role="math" localid="1647033519318" $$\begin{gathered} V\left(x\right)=x^3 \\ V\left(3\right)=3^3 \\ V\left(3\right)=27 \end{gathered}$$

So the volume of the cube at the siderole="math" localid="1647033175289" $x=3$isrole="math" localid="1647033531007" $27m^3$

Step 3. Instantaneous rate of change of volume 

Use instantaneous rate of change formula with respect to side x at $x=3$

role="math" localid="1647033651982" $$\begin{gathered} V\text{'}\left(c\right)=\underset{x\to c}{\lim }\frac{V\left(x\right)-V\left(c\right)}{x-c} \\ V\text{'}\left(3\right)=\underset{x\to 3}{\lim }\frac{V\left(x\right)-V\left(3\right)}{x-3} \\ V\text{'}\left(3\right)=\underset{x\to 3}{\lim }\frac{x^3-27}{x-3} \\ V\text{'}\left(3\right)=\underset{x\to 3}{\lim }\frac{(x-3)(x^2+3x+9)}{x-3} \\ V\text{'}\left(3\right)=\underset{x\to 3}{\lim }{x}^2+3x+9 \\ V\text{'}\left(3\right)=3^2+3\left(3\right)+9 \\ V\text{'}\left(3\right)=27 \end{gathered}$$

So Instantaneous rate of change of volume of the cube is$27\frac{m^3}{m}$