Question

The volume of a cube is increasing at a rate of 7 cubic cm per second. The rate of change of its surface area when the length of an edge is 12 cm is

• A. $\frac{7}{3}{\mathrm{cm}}^2/\sec .$
• B. $\frac{3}{7}{\mathrm{cm}}^2/\sec .$
• C. $\frac{7}{5}{\mathrm{cm}}^2/\sec .$
• D. $\frac{5}{7}c{m}^2/sec.$

Answer 1

The correct option is A $\frac{7}{3}{\mathrm{cm}}^2/\sec .$

Let x be the length of an edge of the cube, V be the volume and S be the surface area at any time t. Then, $V=x^3\mathrm{and}S=6x^2.\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\frac{\mathrm{dV}}{\mathrm{dt}}=7{\mathrm{cm}}^3/\sec .\Rightarrow \frac{d}{\mathrm{dt}}\left(x^3\right)=7\Rightarrow 3x^2\frac{\mathrm{dx}}{\mathrm{dt}}=7\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{7}{3x^2}\mathrm{Now},S=6x^2\Rightarrow \frac{\mathrm{dS}}{\mathrm{dt}}=12x\frac{\mathrm{dx}}{\mathrm{dt}}\Rightarrow \frac{\mathrm{dS}}{\mathrm{dt}}=12x\times \frac{7}{3x^2}=\frac{28}{x}\Rightarrow {\left(\frac{\mathrm{dS}}{\mathrm{dt}}\right)}_{x=12}=\frac{28}{12}{\mathrm{cm}}^2/\sec .=\frac{7}{3}{\mathrm{cm}}^2/\sec .$