Question

If the surface area of a cube is increasing at a rate of \(3.6\ \mathrm{cm^2/sec}\), retaining its shape; then the rate of change of its volume (in \(\mathrm{cm^3/sec}\)), when the length of a side of the cube is \(10\ \mathrm{cm}\), is :

• A. 9
• B. 10
• C. 18
• D. 20

Hint:

Write \(S=6a^2\) and \(V=a^3\); first find \(\frac{da}{dt}\) from \(\frac{dS}{dt}\), then use it in \(\frac{dV}{dt}\).

Answer 1

The Correct Option is A

Let the side be \(a\). \(S=6a^2\Rightarrow \frac{dS}{dt}=12a\frac{da}{dt}\).
Given \(\frac{dS}{dt}=3.6\) and \(a=10\): \(3.6=120\frac{da}{dt}\Rightarrow \frac{da}{dt}=0.03\ \mathrm{cm/s}\).
\(V=a^3\Rightarrow \frac{dV}{dt}=3a^2\frac{da}{dt}=3(100)(0.03)=9\).
Answer: (1) 9