The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?
The surface area is increasing at the rate of 8/3 cm²/s when the length of an edge is 12 cm.
Explanation
This question is similar to Example 2 from the textbook. The key is to use the chain rule to relate the rate of change of volume to the rate of change of surface area through the edge length. First, find dx/dt from dV/dt, then use it to find dS/dt.
Solution Steps
Step 1: Let x be the length of an edge, V be the volume and S be the surface area of the cube. Then, V = x³ and S = 6x², where x is a function of time t.
Step 2: Given dV/dt = 8 cm³/s. Using chain rule: dV/dt = d(x³)/dt = 3x² · dx/dt. Therefore, 8 = 3x² · dx/dt, which gives dx/dt = 8/(3x²).
Step 3: Now, dS/dt = d(6x²)/dt = 12x · dx/dt (by chain rule).
Step 4: Substituting dx/dt = 8/(3x²): dS/dt = 12x · 8/(3x²) = 32/x.
Step 5: When x = 12 cm: dS/dt = 32/12 = 8/3 cm²/s.