Chapter 3.8, Problem 12E

a)

To determine

To find: The value of $\frac{dV}{dx}$ at value of $x=3\text{mm}$, if V is the volume of a cube with side length $x$, and explain its meaning.

Answer to Problem 12E

The value of $\frac{dV}{dx}$ is $\underset{\_}{27{\text{mm}}^{\text{3}}\text{/mm}}$.

Explanation of Solution

Calculate the value of $\frac{dV}{dx}$ when $x$ is $3\text{mm}$ and explain its meaning.

Use the below expression for volume.

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

Differentiate the volume equation.

$\frac{dV}{dx}=3x^2$ (1)

Substitute the value $3$ for $x$ in the above equation.

$\begin{array}{c}{\frac{dV}{dx}|}_{x=3}=3{\left(3\right)}^2\\ =27{\text{mm}}^{\text{3}}\text{/mm}\end{array}$

The above value of $27{\text{mm}}^{\text{3}}\text{/mm}$ is the rate at which the volume is increasing with the variable $x$.

Therefore, $\frac{dV}{dx}$ is $27{\text{mm}}^{\text{3}}\text{/mm}$.

b)

To determine

To show: The rate of change of the volume of a cube with respect to its edge length is equal to half the surface area of the cube and explain geometrically why this result is true.

Explanation of Solution

Each cube will have 6 sides, and area of one side of the cube is $x^2$.

Therefore, the surface area of the cube is as below.

$s\left(x\right)=6x^2$

Refer the equation (1).

$V\text{'}\left(x\right)=3x^2$

Simplify the equation as below.

$V\text{'}\left(x\right)=\frac{1}{2}6x^2$

Substitute $s\left(x\right)$ for $6x^2$.

$V\text{'}\left(x\right)=\frac{1}{2}s\left(x\right)$

Thus, the rate of change of the volume of the cube is approximately half of its surface area.

Assume the condition in which a cube’s side length is x and the change in length of side is $\Delta x$ on all sides.

Express the volume change, $\Delta V$ in terms of increased amount of length of side $\Delta x$.

$\Delta V=3x^2\left(\Delta x\right)+3x{\left(\Delta x\right)}^2+{\left(\Delta x\right)}^3$

The value of $\Delta x$ is smaller and negligible.

Rewrite the above equation as below.

$\begin{array}{l}\Delta V\simeq 3x^2\Delta x\\ \frac{\Delta V}{\Delta x}\simeq 3x^2\end{array}$

Hence, the result is geometrically true.

Therefore, the rate of change of the volume of the cube is approximately half of its surface area and the result is geometrically true.