If surface area of a cube is changing at a rate of `5 m^(2)//s`, find the rate of change of body diagonal at the moment when side length is `1 m`.

Options

A. `5 m//s`
B. `5sqrt(3) m//s`
C. `5/2sqrt(3) m//s`
D. `5/(4sqrt(3)) m//s`

Explanation

To solve the problem step by step, we need to find the rate of change of the body diagonal of a cube when the surface area is changing at a given rate.

### Step 1: Understand the given information
We know that:
- The rate of change of surface area \( \frac{dS}{dt} = 5 \, \text{m}^2/\text{s} \)
- The side length of the cube \( a = 1 \, \text{m} \)

### Step 2: Write the formulas for surface area and body diagonal
The surface area \( S \) of a cube is given by:
\[ S = 6a^2 \]
The body diagonal \( L \) of a cube is given by:
\[ L = a\sqrt{3} \]

### Step 3: Differentiate the surface area with respect to time
To find the rate of change of the body diagonal, we first differentiate the surface area with respect to time:
\[ \frac{dS}{dt} = 12a \frac{da}{dt} \]
This equation relates the rate of change of surface area to the rate of change of the side length \( a \).

### Step 4: Solve for \( \frac{da}{dt} \)
Rearranging the equation gives:
\[ \frac{da}{dt} = \frac{1}{12a} \frac{dS}{dt} \]
Substituting \( \frac{dS}{dt} = 5 \, \text{m}^2/\text{s} \) and \( a = 1 \, \text{m} \):
\[ \frac{da}{dt} = \frac{1}{12 \cdot 1} \cdot 5 = \frac{5}{12} \, \text{m/s} \]

### Step 5: Differentiate the body diagonal with respect to time
Now, we differentiate the body diagonal \( L \):
\[ L = a\sqrt{3} \]
Differentiating with respect to time gives:
\[ \frac{dL}{dt} = \sqrt{3} \frac{da}{dt} \]

### Step 6: Substitute \( \frac{da}{dt} \) into the equation for \( \frac{dL}{dt} \)
Substituting \( \frac{da}{dt} = \frac{5}{12} \, \text{m/s} \):
\[ \frac{dL}{dt} = \sqrt{3} \cdot \frac{5}{12} \]

### Step 7: Calculate the final result
Thus, the rate of change of the body diagonal is:
\[ \frac{dL}{dt} = \frac{5\sqrt{3}}{12} \, \text{m/s} \]

### Final Answer
The rate of change of the body diagonal at the moment when the side length is \( 1 \, \text{m} \) is:
\[ \frac{dL}{dt} = \frac{5\sqrt{3}}{12} \, \text{m/s} \]