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

Answer 1

$$\begin{gathered} Volumeofacubeisgivenby \\ V=a^3 \\ \frac{dV}{dt}=\frac{d\left(a^3\right)}{dt} \\ \frac{dV}{dt}=3a^2\frac{da}{dt} \\ \frac{dV}{dt}=9c{m}^3/s \\ 3a^2\frac{da}{dt}=9 \\ \frac{da}{dt}=\frac{3}{a^2} \\ {\left.\frac{da}{dt}\right\}}_{ata=10cm}=\frac{3}{100} \\ SurfaceareaA=6a^2 \\ \frac{dA}{dt}=12a\frac{da}{dt} \\ {\left.\frac{dA}{dt}\right\}}_{ata=10cm}=12\times 10\times {\left.\frac{da}{dt}\right\}}_{ata=10cm} \\ {\left.\frac{dA}{dt}\right\}}_{ata=10cm}=12\times 10\times \frac{3}{100} \\ {\left.\frac{dA}{dt}\right\}}_{ata=10cm}=3.6c{m}^2/s \end{gathered}$$