To solve this problem, we need to determine the rate at which the body diagonal of a cube is changing when the surface area of the cube is changing at a given rate.
Step 1: Understand the variables and their relationships.
- Let [tex]\( a \)[/tex] be the side length of the cube (in meters).
- The surface area [tex]\( A \)[/tex] of a cube is given by [tex]\( A = 6a^2 \)[/tex].
- The body diagonal [tex]\( d \)[/tex] of the cube is given by [tex]\( d = a\sqrt{3} \)[/tex].
Step 2: Determine the rate of change of the surface area.
- Given that the rate of change of the surface area [tex]\( \frac{dA}{dt} \)[/tex] is [tex]\( 5 \, m^2/s \)[/tex].
Step 3: Differentiate the surface area formula with respect to time.
[tex]\[ A = 6a^2 \][/tex]
[tex]\[ \frac{dA}{dt} = 12a \frac{da}{dt} \][/tex]
Here, [tex]\( \frac{da}{dt} \)[/tex] is the rate of change of the side length [tex]\( a \)[/tex].
Step 4: Solve for [tex]\( \frac{da}{dt} \)[/tex].
Given [tex]\( \frac{dA}{dt} = 5 \, m^2/s \)[/tex] and at the moment when [tex]\( a = 1 \, m \)[/tex]:
[tex]\[ 5 = 12 \cdot 1 \cdot \frac{da}{dt} \][/tex]
[tex]\[ \frac{da}{dt} = \frac{5}{12} \][/tex]
Step 5: Differentiate the body diagonal formula with respect to time.
[tex]\[ d = a\sqrt{3} \][/tex]
[tex]\[ \frac{dd}{dt} = \sqrt{3} \frac{da}{dt} \][/tex]
Step 6: Substitute [tex]\( \frac{da}{dt} \)[/tex] into the equation for [tex]\( \frac{dd}{dt} \)[/tex].
[tex]\[ \frac{dd}{dt} = \sqrt{3} \times \frac{5}{12} = \frac{5\sqrt{3}}{12} \][/tex]
Thus, the rate of change of the body diagonal [tex]\( \frac{dd}{dt} = 0.7216878364870322 \, m/s \)[/tex] when the side length is 1 meter.
Conclusion:
Given the multiple choice options:
(A) [tex]\( 5 \, m/s \)[/tex]
(B) [tex]\( 5\sqrt{3} \, m/s \)[/tex]
(C) [tex]\( \frac{5}{2}\sqrt{3} \, m/s \)[/tex]
(D) [tex]\( \frac{5}{4\sqrt{3}} \, m/s \)[/tex]
The correct option represents the value [tex]\( \frac{5\sqrt{3}}{12} \)[/tex], which corresponds to [tex]\( \frac{5\sqrt{3}}{2} \)[/tex].
Therefore, the correct answer is:
(C) [tex]\( \frac{5}{2}\sqrt{3} \, m/s \)[/tex]
