Question

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

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 of the cube is: \[ V = x^3, \] where \( x \) is the length of an edge. Differentiating with respect to \( t \), we get: \[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt}. \] Substitute \( \frac{dV}{dt} = 6 \, \text{cm}^3/\text{s} \): \[ 6 = 3(8)^2 \frac{dx}{dt}. \] Solve for \( \frac{dx}{dt} \): \[ \frac{dx}{dt} = \frac{6}{192} = \frac{1}{32} \, \text{cm}/\text{s}. \] Step 2: Find the rate of change of surface area
The surface area of the cube is: \[ S = 6x^2. \] Differentiating with respect to \( t \), we get: \[ \frac{dS}{dt} = 12x \frac{dx}{dt}. \] Substitute \( x = 8 \, \text{cm} \) and \( \frac{dx}{dt} = \frac{1}{32} \): \[ \frac{dS}{dt} = 12(8)\left(\frac{1}{32}\right) = 3 \, \text{cm}^2/\text{s}. \] Conclusion: The surface area of the cube is increasing at \( 3 \, \text{cm}^2/\text{s} \).