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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?
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Solution
\[\text { Let x be the side andVbe the volume of the cube at any time t. Then },\]
\[V = x^3 \]
\[\Rightarrow\frac{dV}{dt}=3 x^2 \frac{dx}{dt}\]
\[\Rightarrow9=3 \left( 10 \right)^2 \frac{dx}{dt}\left[ \because x = 10 \text { cm and } \frac{dV}{dt} = {cm}^3 /\sec \right]\]
\[\Rightarrow\frac{dx}{dt}=0.03 cm/sec\]
\[\text { Let S be the surface area of the cube at any time t. Then },\]
\[S = 6 x^2 \]
\[\Rightarrow\frac{dS}{dt}=12x\frac{dx}{dt}\]
\[\Rightarrow\frac{dS}{dt}=12\times10\times0.03 \left[ \because x = 10 \text{ cm and } \frac{dx}{dt}= 0.03 cm/sec \right]\]
\[\Rightarrow\frac{dS}{dt} =\text{3.6 cm}^2 /sec\]
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