If the surface area of a cube is increasing at a rate of 3.6 cm²/sec, retaining its shape; then the rate of change of its volume (in cm³/sec), when the length of a side of the cube is 10 cm, is :

Solution

Maths sir · 4 steps

Both S and V depend on the side a, so find da/dt from the area rate, then use it for the volume.

1. Link through the side. For a cube of side a, the surface area is S = 6a² and the volume is V = a³. Differentiate both with respect to time t.

   dS/dt = 12a da/dt, dV/dt = 3a² da/dt

2. Find da/dt. At a = 10 cm: 12 × 10 × da/dt = 3.6, so da/dt = 3.6/120 = 0.03 cm/s.
3. Rate of volume. dV/dt = 3 × 10² × 0.03 = 300 × 0.03 = 9 cm³/s.
4. So. The volume is increasing at 9 cm³/s, which is option (a).

The idea to keep. When two quantities share one variable, differentiate both with respect to time and connect them through that variable's rate. Here dV/dt = a/4dS/dt.

Answer from the NTA final answer key. Solution by PrepWiser.