3.1 Q-4

Question Statement

Find the approximate increase in the volume of a cube if the length of each edge changes from 5 cm to 5.02 cm.

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Background and Explanation

The formula for the volume of a cube is:
$V = x^3$
where $x$ represents the length of one edge of the cube.
To approximate the change in volume ($\mathrm{d}V$), we use the concept of differentials:
$\mathrm{d}V = \frac{\mathrm{d}V}{\mathrm{d}x} \cdot \mathrm{d}x$
This allows us to estimate the increase in volume based on a small change ($\mathrm{d}x$) in the edge length.

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Solution

Step 1: Identify the Initial Edge Length and Change

• The initial edge length of the cube is $x = 5 , \text{cm}$.
• After the change, the edge length becomes $5.02 , \text{cm}$.
• The change in edge length is:
  $\mathrm{d}x = 5.02 - 5 = 0.02 , \text{cm}$

Step 2: Differentiate the Volume Formula

The formula for the volume of a cube is $V = x^3$. Differentiating with respect to $x$:
$\frac{\mathrm{d}V}{\mathrm{d}x} = 3x^2$

Step 3: Apply the Differential Formula

Using the differential approximation formula $\mathrm{d}V = \frac{\mathrm{d}V}{\mathrm{d}x} \cdot \mathrm{d}x$, substitute:
$\mathrm{d}V = 3x^2 \cdot \mathrm{d}x$

Step 4: Substitute the Known Values

Substitute $x = 5 , \text{cm}$, $\mathrm{d}x = 0.02 , \text{cm}$:

1. Compute $x^2 = 5^2 = 25$.
2. Multiply $3x^2 = 3 \cdot 25 = 75$.
3. Multiply $\mathrm{d}V = 75 \cdot 0.02 = 1.5$.

Thus, the approximate increase in volume is:
$\mathrm{d}V \approx 1.5 , \text{cm}^3$.

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Key Formulas or Methods Used

1. Volume of a Cube:
   $V = x^3$
2. Differential Formula for Volume:
   $\mathrm{d}V = \frac{\mathrm{d}V}{\mathrm{d}x} \cdot \mathrm{d}x$
3. Derivative of Volume:
   $\frac{\mathrm{d}V}{\mathrm{d}x} = 3x^2$

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Summary of Steps

1. Calculate the initial edge length $x = 5$ and the change in edge length $\mathrm{d}x = 0.02$.
2. Differentiate the volume formula $V = x^3$ to find $\frac{\mathrm{d}V}{\mathrm{d}x} = 3x^2$.
3. Use $\mathrm{d}V = 3x^2 \cdot \mathrm{d}x$ to approximate the change in volume.
4. Substitute $x = 5$ and $\mathrm{d}x = 0.02$, and simplify to find $\mathrm{d}V$.
5. The final result is $\mathrm{d}V \approx 1.5 , \text{cm}^3$.