Question

The volume of a cube is increasing at the rate of \( 6 \, \text{cm}^3/\text{s} \). How fast is the surface area of the cube increasing, when the length of an edge is \( 8 \, \text{cm} \)?

Hint:

For related rates problems, identify the variables, write their relationships, and differentiate with respect to time.

Answer 1

Step 1: {Relate volume and surface area of the cube}
The volume \( V \) of a cube with edge length \( x \) is \( V = x^3 \). Differentiating with respect to time \( t \), we obtain \( \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \). Given \( \frac{dV}{dt} = 6 \, \text{cm}^3/\text{s} \), we have \( 6 = 3(8)^2 \frac{dx}{dt} \). Solving for \( \frac{dx}{dt} \) yields \( \frac{dx}{dt} = \frac{6}{192} = \frac{1}{32} \, \text{cm}/\text{s} \).
Step 2: {Find the rate of change of surface area}
The surface area \( S \) of a cube with edge length \( x \) is \( S = 6x^2 \). Differentiating with respect to time \( t \), we get \( \frac{dS}{dt} = 12x \frac{dx}{dt} \). Substituting \( x = 8 \, \text{cm} \) and \( \frac{dx}{dt} = \frac{1}{32} \), we find \( \frac{dS}{dt} = 12(8)\left(\frac{1}{32}\right) = 3 \, \text{cm}^2/\text{s} \).
Conclusion: The surface area of the cube is increasing at \( 3 \, \text{cm}^2/\text{s} \).