Question

The volume of a cube is increasing at the rate of 9 cm³/sec. How fast is the surface area increasing when the length of an edge is 10 cm?

Answer 1

\[\text { Let x be the side andVbe the volume of the cube at any time t. Then },\]

\[V = x^3 \]

\[\Rightarrow\frac{dV}{dt}=3 x^2 \frac{dx}{dt}\]

\[\Rightarrow9=3 \left( 10 \right)^2 \frac{dx}{dt}\left[ \because x = 10 \text { cm and } \frac{dV}{dt} = {cm}^3 /\sec \right]\]

\[\Rightarrow\frac{dx}{dt}=0.03 cm/sec\]

\[\text { Let S be the surface area of the cube at any time t. Then },\]

\[S = 6 x^2 \]

\[\Rightarrow\frac{dS}{dt}=12x\frac{dx}{dt}\]

\[\Rightarrow\frac{dS}{dt}=12\times10\times0.03 \left[ \because x = 10 \text{ cm and } \frac{dx}{dt}= 0.03 cm/sec \right]\]

\[\Rightarrow\frac{dS}{dt} =\text{3.6 cm}^2 /sec\]